Chapter 2 STEADY STATE CONDUCTION

Transcription

1 Principles of Heat Transfer 8th Edition Kreith SOLUTIONS MANUAL Full clear download (no formatting errors) at: Chapter PROBLEM. STEADY STATE CONDUCTION GIVEN A plane wall, 7.5 cm thick, generates heat internally at the rate of 0 5 W/m 3. One side of the wall is insulated, and the other side is exposed to an environment at 90 C. The convection heat transfer coefficient between the wall and the environment is 500 W/(m K). If the thermal conductivity of the wall is W/(m K), calculate the maximum temperature in the wall. Plane wall with internal heat generation Thickness (L) 7.5 cm m Internal heat generation rate ( q G ) 0 5 W/m 3 One side is insulated Ambient temperature on the other side (T ) 90 C Convective heat transfer coefficient ( h ) 500 W/(m K) c Thermal conductivity (k) W/(m K) FIND The maximum temperature in the wall (T max) ASSUMPTIONS The heat loss through the insulation is negligible The system has reached steady state One dimensional conduction through the wall SKETCH SOLUTION The one dimensional conduction equation, given in Equation (.5), is

2 T k + c x q G T t T For steady state, 0 therefore t d T k d x + 0 q G

3 d T q G d x k This is subject to the following boundary conditions No heat loss through the insulation d T 0 at x 0 d x Convection at the other surface k d T d x h c (T T ) at x L Integrating the conduction equation once d T q G + C d x k C can be evaluated using the first boundary condition 0 q G (0) + C C 0 k Integrating again q G T k x + C The expression for T and its first derivative can be substituted into the second boundary condition to evaluate the constant C q G L q G L k h c C T C k k q G L h L + T k c Substituting this into the expression for T yields the temperature distribution in the wall T(x) x + q q G L G k h L + T c k q T(x) T + G L kl x k Examination of this expression reveals that the maximum temperature occurs at x 0 T max 90 C + h c q G L kl Tmax T + k 0 5 W / m 3 [W/(mK)] h c [ W/(mK)](0.075 m) (0.075 m) 500 W/(m K) 8 C

4 PROBLEM. GIVEN A small dam, which may be idealized by a large slab. m thick, is to be completely poured in a short period of time. The hydration of the concrete results in the equivalent of a distributed source of constant strength of 00 W/m 3. If both dam surfaces are at 6 C, determine the maximum temperature to which the concrete will be subjected, assuming steady-state conditions. The thermal conductivity of the wet concrete may be taken as 0.84 W/(m K). Large slab with internal heat generation Internal heat generation rate ( q G ) 00 W/m 3 Both surface temperatures (T s) 6 C Thermal conductivity (k) 0.84 W/(m K) FIND The maximum temperature (T max) ASSUMPTIONS Steady state conditions prevail SKETCH SOLUTION The dam is symmetric. Therefore, x will be measured from the centerline of the dam. The equation for one dimensional conduction is given by Equation (.5) T k + q G c t For steady state, T t T x 0 therefore d T k d x + 0 q G This is subject to the following boundary conditions. By symmetry, dt/dx 0 at x o. T T s at x L Also note that for this problem is a constant. q G Integrating the conduction equation

5 d T d x q G k x + C The constant C can be evaluated using the first boundary condition q 0 G (0) + C C 0 k Integrating once again q G T k x + C The constant C can be evaluated using the second boundary condition q T G s L + C C T s + q k G L k Therefore, the temperature distribution in the dam is q T T G s + (L x ) k The maximum temperature occurs at x 0 q G 00 W/m 3 T max T s + (L (0) ) 6 C + k [0.84 W/(m K)] (0.6 m) 37 C COMMENTS This problem is simplified significantly by choosing x 0 at the centerline and taking advantage of the problem s symmetry. For a more complete analysis, the change in thermal conductivity with temperature and moisture content should be measured. The system could then be analyzed by numerical methods discussed in chapter 4.

6 PROBLEM.3 GIVEN The shield of a nuclear reactor is idealized by a large 5 cm thick flat plate having a thermal conductivity of 3.5 W/(m K). Radiation from the interior of the reactor penetrates the shield and there produces heat generation that decreases exponentially from a value of 87.6 kw/ m 3. at the inner surface to a value of 8.76 kw/m 3 at a distance of.5 cm from the interior surface. If the exterior surface is kept at 38 C by forced convection, determine the temperature at the inner surface of the shield. Hint: First set up the differential equation for a system in which the heat generation rate varies according to q (x) q (0)e Cx. Large flat plate with non-uniform internal heat generation Thickness (L) 5 cm0.5 m Thermal conductivity (k) 3.5 W/(m K) Exterior surface temperature (T o) 38 C Heat generation is exponential with values of 87.6 kw/m 3 at the inner surface 8.76 kw/m 3 at.5 cm from the inner surface FIND The inner surface temperature (T i) ASSUMPTIONS One dimensional, steady state conduction The thermal conductivity is constant No heat transfer at the inner surface of the shield SKETCH SOLUTION From the hint, the internal heat generation is q (x) q (0) e cx where q (0) 87.6 kw/m 3 Solving for the constant c using the fact that q(x) 8.76 kw/m 3 at x.5 cm 0.5 m q ( x) c ln q (0) 8.76 kw /m ln 8.4 x 0.5 m 87.6 kw /m 3 m 3 The one dimensional conduction equation is given by Equation (.5)

7 The boundary conditions are dt T T k + q x G c 0 (steady state) t d T q G ( x) q (0) d x e k cx k dx 0 at x 0 Integrating the conduction equation T(L) T o 00 F at x L dt dx q (0) e cx + C ck The constant C can be evaluated by applying the first boundary condition q (0) 0 e c(0) + C C ck q ck(0) Integrating again T(x) q (0) q (0) e cx c k c k x + C The constant C can be evaluated by applying the second boundary condition q (0) T(L) T o q (0) e cl L + C c C T o + k c k q c (0) k L e c cl Therefore, the temperature distribution is q (0) T(x) T o + c k [e cl e cx + c(l x)] Solving for the temperature at the inside surface (x 0) q (0) T i T(0) T o + [e cl c + cl] k T i 38 C W / m 3 e (8.4/m ) 0.5m m 8.4 m 3.5 W /(m K ) m 38 C W / m W /(m K ) m * C

8 PROBLEM.4 GIVEN A plane wall 5 cm thick has a thermal conductivity given by the relation k T W/(m K) where T is in degrees Kelvin. If one surface of this wall is maintained at 50 C and the other at 50 C, determine the rate of heat transfer per square meter. Sketch the temperature distribution through the wall. A plane wall Thickness (L) 5 cm 0.5 m Thermal conductivity (k) T W/(m K) (with T in Kelvin) Surface temperatures: T h 50 C T c 50 C FIND (a) The rate of heat transfer per square meter (q/a) (b) The temperature distribution through the wall ASSUMPTIONS The wall has reached steady state Conduction occurs in one dimension SKETCH SOLUTION Simplifying Equation (.) for steady state conduction with no internal heat generation but allowing for the variation of thermal conductivity with temperature yields d k dt 0 dx dx with boundary conditions: T 43 K at x 0 T 33 K at x 0.5 m The rate of heat transfer does not vary with x dt q k constant dx A q ( T) dt dx A

9 Integrating.0T T q A x + C The constant can be evaluated using the first boundary condition.0 (43) (43) C q A (0) C (a) The rate of heat transfer can be evaluated using the second boundary condition:.0 (33) (33) q A (0.5 m) q 457 W/m k (b) Therefore, the temperature distribution is T +.0 T x COMMENTS Notice that although the temperature distribution is not linear due to the variation of the thermal conductivity with temperature, it is nearly linear because this variation is small compared to the value of the thermal conductivity. If the variation of thermal conductivity with temperature had been neglected, the rate of heat transfer would have been 333 W/m, an error of 8.5%.

10 PROBLEM.5 GIVEN Derive an expression for the temperature distribution in a plane wall in which there are uniformly distributed heat sources that vary according to the linear relation q G q w [ (T T w )] where q w is a constant equal to the heat generation per unit volume at the wall temperature T w. Both sides of the plate are maintained at T w and the plate thickness is L. A plane wall with uniformly distributed heat sources as in the above equation Both surface temperatures T w Thickness L FIND An expression for the temperature distribution ASSUMPTIONS Constant thermal conductivity (k) SKETCH SOLUTION The equation for one dimensional, steady state (dt/dt 0) conduction from Equation (.5) is d T dx With the boundary conditions dt dx q w q G q w q w q [ (T T w w )] (T T w ) k k k k 0 at x 0 T T w at x L Let T T w and m ( )/k then d m dx q w k This is a second order, linear, nonhomogeneous differential equation with constant coefficients. Its solution is the addition of the homogeneous solution and a particular solution. The solution to the homogeneous equation d m 0 dx

11 is determined by its characteristics equation. Substituting equation yields the characteristics equation e x m e x 0 m Therefore, the homogeneous solution has the form h C c mx + C e mx A particular solution for this problem is simply a constant: Substituting this into the differential equation q w q 0 m a w o a o k m k Therefore, the general solution is e x and its derivatives into the homogeneous a o q w C e mx + C e mx + m k With the boundary condition d dx 0 at x 0 0 at x L Applying the first boundary condition: d C me (0) C me (0) 0 C C C dx From the second boundary condition 0 C (e ml + e ml ) + The temperature distribution in the wall is q w m k C m k (e ml q w e ml ) q w T(x) T w (e mx + e mx ) + m ml k (e ml e ) m k q w q w T(x) T w + m k e mx e ml e e mx ml q w T(x) T w + m k cosh (mx) cosh (ml)

12 PROBLEM.6 A plane wall of thickness L has internal heat sources whose strength varies according to q 0 cos (ax) q G where q 0 is the heat generated per unit volume at the center of the wall (x 0) and a is a GIVEN constant. If both sides of the wall are maintained at a constant temperature of T w, derive an expression for the total heat loss from the wall per unit surface area. A plane wall with internal heat sources Heat source strength: q G q 0 cos (ax) Wall surface temperatures T w Wall thickness L FIND An expression for the total heat loss per unit area (q/a) ASSUMPTIONS Steady state conditions prevail The thermal conductivity of the wall (k) is constant One dimensional conduction within the wall SKETCH SOLUTION Equation (.5) gives the equation for one dimensional conduction. For steady state, dt/dt 0, therefore T k q G c 0 t T x With boundary conditions: d T q G q 0 cos (a x) d x k k dt 0 at x 0 (by symmetry) dx T T w at x L (given)

13 Integrating the conduction equation once dt - dx q o a k sin (ax) + C Applying the first boundary condition yields: C 0 The rate of heat transfer from one side of the wall is dt q k k A dx x L k A q G sin (al) a k q o A sin (al) a The total rate of heat transfer is twice the rate of heat transfer from one side of the wall q k A total q o sin (al) a An alternative method of solution for this problem involves recognizing that at steady state the rate of heat generation within the entire wall must equal the rate of heat transfer from the wall surfaces A L L q ( x)dx G q L q o L q o sin (al ) a COMMENTS q cos (ax) dx A q sin ( al) A q q o sin (al) A a The heat loss can be determined by solving for the temperature distribution and then the rate of heat transfer or via the conservation of energy which allows us to equate the heat generation rate with the rate of heat loss.

14 PROBLEM.7 GIVEN A very thin silicon chip is bonded to a 6-mm thick aluminum substrate by a 0.0-mm thick epoxy glue. Both surfaces of this chip-aluminum system are cooled by air at 5 0 C, where the convective heat transfer coefficient of air flow is 00 W/(m K). If the heat dissipation per unit area from the chip is 0 4 W/m under steady state condition, draw the thermal circuit for the system and determine the operating temperature of the chip. Silicon chip bonded to 6-mm thick aluminum substrate bye 0.0-mm thick epoxy glue Air temperature(t )5 0 C Convective heat transfer coefficient(h)00 W/(m K) Heat dissipation from chip(q/a) 04 W/m FIND Draw thermal circuit of system Operating temperature of the chip. ASSUMPTIONS -Dimensional Steady state conditions prevail Negligible heat loss from the sides Isothermal chip Negliglble radiation SKETCH SOLUTION From the figure Total heat transferred to the surrounding is sum of heat transferred from upper surface and lower surface. Thus... q q + q. q Tc T Tc T K L L h h ke ka m K W

15 . Tc 98K Tc T K q 6 * (m K / W ) 04 W/m 0 4 COMMENT *0 4 (Tc 98) 0 4 *0(Tc 98) 53.6 *0 4 * *T c T c30 K The heat transfer occurs on both sides through the chip to the surrounding. As there are both conductive and convective resistances on the lower side heat flow rate on the lower side will be less than that on the upper side which has only convective resistance.

16 PROBLEM.8 A thin, flat plate integrated circuit of 5 mm thickness is cooled on its upper surface by a dielectric liquid. The heat dissipation rate from the chip is 0,000 W/m and with the coolant flow at a free stream temperature of T,l 5 0 C, the convective heat transfer coefficient between the chip surface and the liquid is 000 W/(m K). On the lower surface, the chip is attached to a circuit board, where the thermal contact resistance between the chip and the board is 0-4 m.k/w. The thermal conductivity of board material is.0 W/m. K, and its other surface ( away from the chip) is exposed to ambient air at T,a 0 0 C where it is cooled by natural convection with the heat transfer coefficient of 30 W/(m K). (a) Determine the chip surface temperature under steady state condition for the described conditions. (b) If the maximum chip temperature is not to exceed 75 0 C, determine maximum allowable heat flux that is generated by the chip. (c) A colleague suggests that in order to improve the cooling, you use a high conductivity bonding base at chip-board interface that would reduce the thermal contact resistance at the interface to 0-5 m.k/w. Determine the consequent increase in the chip heat flux that can be sustained. GIVEN Heat dissipation rate ( q ) 0,000 W/m Coolant free stream temp (T,l) 5 0 C Ambient air temperature (T,a) 0 0 C Heat transfer coefficient (h) 000 W/(m K) Thermal contact resistance (R tc) 0-4 m.k/w Maximum chip temperature75 0 C FIND (a) Chip surface temperature under steady state condition (b) Maximum allowable heat flux generated by the chip (c) Consequent increase in chip heat flux if high conductivity bonding is used. ASSUMPTIONS Steady state conditions prevail The thermal conductivity of the wall (k) is constant One dimensional conduction Negligible radiation and thermal resistance between chip surface and the liquid. SKETCH The thermal circuit of problem is given by SOLUTION (a) A heat balance in the above problem gives... q q liquid + q air Substituting values from thermal circuits

17 0000 W/m Tc 98K Tc T K (m K / W ) W/m ( Tc 98 )*000 + ( Tc 93 )/ W/m 0000 W/m (000T c t c ) W/m Solving for T c, we get T c K or C (b) T c,max75 0 C 343 K Solving for q from above equation, we get Q50* *6.0 W/m W/m 5.4*0 4 W/m (c) Using the same equation as in (a), and changing only the value of thermal resistance, and using the value of T c as 343 K, we get q4.63*0 4 W/m, which is a decrease in allowable heat dissipation of around 56 W/m.

18 PROBLEM.9 In a large chemical factory, hot gases at 73 K are cooled by a liquid at 373 K with gas side and liquid side convection heat transfer coefficients of 50 and 000 W/(m K), respectively. The wall that separates the gas and liquid streams is composed of -cm thick slab of stainless steel on the liquid side. There is a contact resistance between the oxide layer and the steel of 0.05 m.k/w. Determine the rate of heat loss from hot gases through the composite wall to the liquid. GIVEN Hot gases at T g73 K cooled by liquid at T f373 K Convection heat transfer coefficients on gas side h g50 W/(m K) and h f00 W/(m K) Wall of stainless steel of thickness(l) cm 0.0 m Contact resistance (R cr ) 0.05 m.k/w FIND Rate of heat loss from hot gases through composite wall to liquid. ASSUMPTIONS Dimensional steady state heat transfer Thermal conductivity remain constant. Radiation is negligible. SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k) 4.4 W/(m K) SOLUTION Total resistance for the heat flow through the pipe is given by Rtotal L R " cr m.k/w hf k hg m.k/w m.k/w m.k/w Heat flux for the above resistance for given temperature difference is given by q Tg Tf Rtotal W/m W/m 334 W/m

19 PROBLEM.0 The conversion of solar energy into electric power by means of photovoltaic panels will be an important part of the transition from fossil fuels to sustainable energy resources. As described in detail in Principles of Sustainable Energy, a typical PV panel consists of a top layer of glass attached with a thin optically clear adhesive to a very thin layer of photoelectric material such as doped-silicon in which the incident solar irradiation is converted into electric energy. Experiments have shown that the solar to electric efficiency ƞ t silicon, where T silicon is the silicon temperature in K. In a typical installation where solar irradiation is G700 W/m, 7% is reflected from the top surface of the glass, 0% is absorbed by the glass, and 83% is transmitted to the photovoltaic active layer. A part of irradiation absorbed by photovoltaic material is converted into heat and the remainder is converted into electric energy. The silicon layer is attached by a 0.0-mm thick layer of solder to a 3-mm thick aluminum nitride substrate as shown in the schemetic. Determine the electric power produced by this PV panel, assuming the following properties for the pertinent materials: conductivity of the glass k g.4 W/(m K), conductivity of the adhesive k a45 W/(m K), the emmisivity of the glass is 0.90, heat transfer coefficient from the top of the panel to the surrounding is 35 W/(m K), and the surrounding air temperature is T air0 0 C. The solar PV panel is 5 m long and m wide and is situated on the roof where the bottom is considered insulated. (Hint: Start by applying first law of thermodynamics to the photovoltaic-active layer and note that some of the irradiation will be converted to electricity and some of it transmitted thermally). GIVEN Electric efficiency ƞ t silicon Solar irradiation is G700 W/m Thickness of solder(t s0.0 mm Al substrate thickness (t Al)3 mm0.003 m Conductivity of the glass k g.4 W/(m K) Conductivity of the adhesive k a45 W/(m K) Emissivity of the glass is 0.90 Heat transfer coefficient from the top of the panel to the surrounding(h c) Surrounding air temperature is T air0 0 C. Solar PV panel area 5 m*m FIND Electric power produced by the PV panel. ASSUMPTIONS Dimensional steady state heat transfer Thermal conductivity remains constant. SKETCH 35 W/(m K),

20 PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k) 4.4 W/(m K) SOLUTION The energy which is not converted to electrical energy is transferred to the ambience through the adhesive and glass layer. Tsi Tg, t Tsi Tg, t Q loss0.83g s(-ƞ) L a > 0.83*700*( T L si) g 0 5 3*0 3 ka kg 45.5 Tsi Tg, t T si > T si t g,t Also under steady state the heat transferred to the glass should be equal to total heat loss through glass to ambience. 0.83G s(-ƞ)+0.g s Tg, t T 0.9 T 4 4 g, t T \ hc 0.83*700*( T si)+0.*700 35(Tg, t 93) 0.9 * 5.67 *0 8 * T 4 g, t *0-8 T g, t T g,t T si 5.03*0-8 T g, t T g,t t si Solving the above two equation in mathematical computational software (eg. Mathematica) we get T si K & T g,t305.6 K Total power0.83g s (-ƞ)*a 0.83*700*( *306.6)**5 98 W.98 KW. COMMENTS

21 The capacity of PV panel also depends on its cross sectional area. Thus more power can be generated if larger cross section of photovoltaic panels are used.

22 48 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

23 PROBLEM. For the design of novel type of nuclear power plant, it is necessary to determine the temperature distribution in a large slab-type nuclear fuel element. Volumetric heat is generated uniformly in the fuel element at the rate of *0 7 W/m 3. This slab fuel is insulated on one side, while on other side it is covered by a stainless steel cladding of 0.3 cm thickness. Heat is removed from the fuel slab by a liquid at 00 0 C that flows on the other side of the steel cladding with the convective heat transfer coefficient of 0,000 W/(m K). Determine the maximum temperature in the fuel element and sketch the temperature distribution. GIVEN Volumetric heat generated ( q )*0 7 W/m Stainless steel cladding thickness (y) 0.3 cm m Coolant liquid temperature (T ) 00 0 C Heat transfer coefficient(h) 000 W/(m K) Fuel element thickness (x).5 cm 0.05 m FIND Maximum temperature in fuel element Sketch temperature distribution. ASSUMPTIONS Dimensional steady state heat transfer Constant properties. Uniform heat generation. Negligible contact resistance between surfaces. SKETCH Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

24 PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k) 4.4 W/(m K) SOLUTION Heat transfer equation in dimension with steady state and uniform heat generation is given by T q 0 where x varies from L to L x kf Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

25 Integrating twice we get dt dx T q x C and kf q x Cx C where C and C are constants. kf Applying boundary conditions. The wall is insulated at the end thus q (x-l)0 ( which implies dt 0) dx Substituting this in first order equation above we get dt q (L) C dx(x L) C q L kf q kf. Using energy conservation equation q conduction convection q conduction equation considering unit width q (L) ks Ts Ts b q convection equation is ks Ts Ts h Ts T b Equating these two above equations at L we get T xl q(lb) q(l) T(L)T s T ks h The differential equation at xl gives q q L L C kf kf Equating both and solving for C we get C T ql b 3L k s h kf Substituting C and C in the integrated differential equation we get T q x q Lx T ql b 3L kf kf k s h kf T max occurs at dt 0 dx 5 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

26 Applying this condition in differential equation above we get q x q L 0 kf kf which implies Tmax is at L. Thus q L T ql b 3L T max kf k s h kf Substituting values to get T max we get T *0 max (7.5 *0 3 ) T *0 7 * 7.5 *0 3 * 3*0 3* 7.5 *0 * * 60 T max K 65 K 5 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

27 PROBLEM. Nomads in the desert make ice by exposing a thin water layer to cold air during the night. The icing or freezing of thin layers of water is often also referred to as ice making by nocturnal ( or night time ) cooling, where the surface temperature of water is lowered considerably by radioactive and convective cooling, and it had been practiced extensively in ancient India. To model this process and evaluate the conduction process in water layer, consider a 5-mm thick horizontal layer gets cooled such that its top surface is at temperature of C. After a while the water begins to freeze at top surface and the ice front expands downwards through the water layer. Determine the location of solid liquid(ice-water) interface if the bottom surface temperature of the liquid water is at 3 0 C. State your assumptions for the model and calculations. GIVEN Thickness of the layer(t) 5 mm Surface temperature (T s) -5 0 C68 K Water temperature(t w) 3 0 C 76 K FIND Location of solid liquid interface. ASSUMPTIONS Dimensional steady state heat transfer Constant properties throughout time Negligible radiation Negligible convection by liquid. SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0, For water at 73 K, k w0.569 W/mK For ice at 73 K, k i.88 W/mK SOLUTION If x is the distance of solid liquid interface from liquid surface, the energy balance over the interface gives kw *(Tw 73) ki *(73 Ts ) * (76 73).88 * (73 68) > x Solving for x we get 5 x x 5 x.707(5-x) 9.4 x > x0.77 mm Thus the interface is at distance of 0.77 mm from liquid surface or 4.3 mm from ice surface Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

28 STEADY STATE CONDUCTION IN CYLINDERS PROBLEM.3 GIVEN The heat conduction equation in cylindrical coordinates is T T T T T c k t r r r r z +q G (a) Simplify this equation by eliminating terms equal to zero for the case of steady-state heat flow without sources or sinks around a right-angle corner such as the one in the accompanying sketch. It may be assumed that the corner extends to infinity in the direction perpendicular to the page. (b) Solve the resulting equation for the temperature distribution by substituting the boundary condition. (c) Determine the rate of heat flow from T to T. Assume k W/(m K) and unit depth. Steady state conditions Right-angle corner as shown below No sources or sinks Thermal conductivity (k) W/(m K) FIND (a) Simplified heat conduction equation (b) Solution for the temperature distribution (c) Rate of heat flow from T to T ASSUMPTIONS Corner extends to infinity perpendicular to the paper No heat transfer in the z direction Heat transfer through the insulation is negligible SKETCH SOLUTION The boundaries of the region are given by m r m Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

29 Assuming there is no heat transfer through the insulation, the boundary condition is T 0 at r m r T 0 at r m r T 00 C at 0 T 0 C at (a) The conduction equation is simplified by the following Steady state T t No sources or sinks 0 q k 0 No heat transfer in the z direction T 0 z Since T r 0 over both boundaries, T r 0 throughout the region (Maximum principle); therefore, Substituting these simplifications into the conduction equation r T 0 throughout the region also. 0 k 0 0 r T 0 (b) Integrating twice T 0 T c + c The boundary condition can be used to evaluate the constants At 0, T 00 C : 00 C c At, T 0 C : 0 C c ( /) + 00 C c 00 o C Therefore, the temperature distribution is Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

30 C T( ) C (c) Consider a slice of the corner as follow The heat transfer flux through the shaded element in the k T k ( T q thickness r T direction is ) In the limit as 0, q k dt r d Multiplying by the surface area drdz and integrating along the radius r o 00 C k r q drdz o dr 00 C k r q ln o r r r r q 00 C k [ W/(m K)] ln( m/ m) 44. W/m 44.W per meter in the z direction COMMENTS Due to the boundary conditions, the heat flux direction is normal to radial lines Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

31 PROBLEM.4 GIVEN Calculate the rate of heat loss per foot and the thermal resistance for a 5 cm schedule 40 steel pipe covered with a 7.5 cm thick layer of 85% magnesia. Superheated steam at 50 C flows inside the pipe [ h c 70 W/(m K)] and still air at 6 C is on the outside [ h c 30 W/(m K)]. A 6 in. standard steel pipe covered with 85% magnesia Magnesia thickness 5 cm0.5 m Superheated steam at T s 50 C flows inside the pipe Surrounding air temperature (T ) 7 C Heat transfer coefficients Inside ( h ) 70 W/(m K) ci Outside ( h ) 30 W/(m K) FIND co (a) The thermal resistance (R) (b) The rate of heat loss per foot (q/l) ASSUMPTIONS Constant thermal conductivity The pipe is made of % carbon steel SKETCH PROPERTIES AND CONSTANTS From Appendix, Tables 0,, and 4 For a 6 in. schedule 40 pipe Inside diameter (D i) in.0.54 m Outside diameter (D o) 6.65 in m Thermal Conductivities 85% Magnesia (k I) W/(m K) at 0 C % Carbon steel (k s) 43 W/(m K) at 0 C SOLUTION The thermal circuit for the insulated pipe is shown below Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

32 (a) The values of the individual resistances can be calculated using Equations (.4) and (.39) R co (m K)/W D i L h co A o h co [30W /(m K )] m * L L r ln i r o R ki L k i ( ) m ln m * L *0.059 W /(m K) L (m K)/W R ci h ci A i The total resistance is r ln o r ln i R ks (m K)/W L k s * 43W /( m K ) L h ci D i L [70W /(m 0.06 (m K)/W K )] 0.54 L L R total R co + R ki + R ks + R ci R total ( ) (m K)/W L R total.0536 (m K)/W L (b) The rate of heat transfer is given by T 50 C 7 C q R total.0536 (mk )/W L q COMMENTS L 6.7 W/m Note that almost all of the thermal resistance is due to the insulation and that the thermal resistance of the steel pipe is negligible Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

33 PROBLEM.5 GIVEN Suppose that a pipe carrying a hot fluid with an external temperature of T i and outer radius r i is to be insulated with an insulation material of thermal conductivity k and outer radius r o. Show that if the convective heat transfer coefficient on the outside of the insulation is h and the environmental temperature is T, the addition of insulation can actually increases the rate of heat loss if r o < k / h and that maximum heat loss occurs when r o k/ h. This radius, r c, is often called the critical radius. An insulated pipe External temperature of the pipe T i Outer radius of the pipe r i Outer radius of insulation r o Thermal conductivity k Ambient temperature T Convective heat transfer coefficient h FIND Show that (a) The insulation can increase the heat loss if r o < k/ h (b) Maximum heat loss occurs when r o k/ h ASSUMPTIONS The system has reached steady state The thermal conductivity does not vary appreciably with temperature Conduction occurs in the radial direction only SKETCH SOLUTION Radial conduction for a cylinder of length L is given by Equation (.37) qk T i T L k o ln r r 0 i Convection from the outer surface of the cylinder is given by Equation (.0) q c h c A T h r o L (T o T ) For steady state q k q c L k T i T o h ro L (T o T ) Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

34 ln r 0 r i 6 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

35 The outer wall temperature, T o, is an unknown and must be eliminated from the equation Solving for T i T o T i T o hr o k ln r o (T o T ) r i T i T (T i T o) + (T o T ) hr o k ln r o (T o T ) + (T o T ) r i h r o r T ln o i T (T o T ) k T r T i or T o T i h r o ln r o k Substituting this into the convection equation T i T T T q q h r o r i c h r o L ln o > q r k r i ln o r i r o L h Lk Examining the above equation, the heat transfer rate is a maximum when the term r ln r o i r o L h Lk d k r ln o k L dr r i o r h o r i is a minimum, which occurs when its differential with respect to r o is zero 0 > k d d + ln r o r 0 h dr o dr i o r o k k + 0 > r o h r o h r o The second derivative of the denominator is k h r 3 r o o which is greater than zero at r o k/h, therefore r o k/h is a true minimum and the maximum heat loss occurs when the diameter is r o k/h. Adding insulation to a pipe with a radius less than k/h will increase the heat loss until the radius of k/h is reached. COMMENTS A more detailed solution taking into account the dependence of h c on the temperature has been obtained by Sparrow and Kang, Int. J. Heat Mass Transf., 8: , Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

36 PROBLEM.6 GIVEN A solution with a boiling point of 8 C boils on the outside of a.5 cm tube with a No. 4 BWG gauge wall. On the inside of the tube flows saturated steam at 40 kpa(abs). The convection heat transfer coefficients are 8.5 kw/(m K) on the steam side and 6. kw/(m K) on the exterior surface. Calculate the increase in the rate of heat transfer if a copper tube is used instead of a steel tube. Tube with saturated steam on the inside and solution boiling at 8 C outside Tube specification: in. No. 4 BWG gauge wall Saturated steam in the pipe is at 40 kpa(abs) Convective heat transfer coefficients Steam side ( h ci ) : 8.5 kw/(m K) Exterior surface ( hco ) : 6. kw/(m K) FIND The increase in the rate of heat transfer for a copper over a steel tube ASSUMPTIONS The system is in steady state Constant thermal conductivities SKETCH PROPERTIES AND CONSTANTS: From Appendix, Tables 0,, 3 and 4 Temperature of saturated steam at 60 psia (T s) 44 C Thermal conductivities Copper (k c) 39 W/(m K) at 7 C % Carbon steel (k s) 43 W/(m K) at 0 C Tube inside diameter (D i) 0.0 m SOLUTION The thermal circuit for the tube is shown below 6 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

37 The individual resistances are R ci (m K)/W D i L h ci A i h ci [8500W /(m K )] 0.0m L L R co h co A o h co D i L (m K)/W [600W /(m K )] 0.05m L L ln r o ln 0.05m R kc r i 0.0m (m K)/W R ks ln Lk c r o r s Lk s The rate of heat transfer is L * 39W /(m K ) 0.05m ln 0.0m L * 43W /(m K ) L L (m K)/W T q Rtotal T s R ci R k T R co For the copper tube q c L 44 C 8 C ( (mk )/W 5967 W For the steel tube q s L 44 C 8 C ( (mK )/W 4,008 W The increase in the rate of heat transfer per unit length with the copper tube is Increase q c L q s L 958 Per cent increase 4, W 00 4% COMMENTS The choice of tubing material is significant in this case because the convective heat transfer resistances are small making the conductive resistant a significant portion of the overall thermal resistance Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

38 PROBLEM.7 GIVEN Steam having a quality of 98% at a pressure of N/m is flowing at a velocity of m/s through a steel pipe of.7 cm OD and. cm ID. The heat transfer coefficient at the inner surface, where condensation occurs, is 567 W/(m K). A dirt film at the inner surface adds a unit thermal resistance of 0.8 (m K)/W. Estimate the rate of heat loss per meter length of pipe if; (a) the pipe is bare, (b) the pipe is covered with a 5 cm layer of 85% magnesia insulation. For both cases assume that the convective heat transfer coefficient at the outer surface is W/(m K) and that the environmental temperature is C. Also estimate the quality of the steam after a 3-m length of pipe in both cases. A steel pipe with steam condensing on the inside Diameters Outside (D o).7 cm 0.07 m Inside (D i). cm 0.0 m Velocity of the steam (V) m/s Initial steam quality (X i) 98% Steam pressure N/m Heat transfer coefficients Inside (h ci) 567 W/(m K) Outside (h co) W/(m K) Thermal resistance of dirt film on inside surface (R f) 0.8 (m K)/W Ambient temperature (T ) C FIND The heat loss per meter (q/l) and the change in the quality of the steam per 3 m length for (a) A bare pipe (b) A pipe insulated with 85% Magnesia: thickness (L i) 0.05 m ASSUMPTIONS Steady state conditions exist Constant thermal conductivity Steel is % carbon steel Radiative heat transfer from the pipe is negligible Neglect the pressure drop of the steam SKETCH PROPERTIES AND CONSTANTS From Appendix, Tables 0,, and 3 The thermal conductivities are: % carbon steel (k s) 43 W/(m K) at 0 C 85% Magnesia (k i) W/(m K) at 0 C 5 N/m : Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

39 Temperature (T st) 07 C Heat of vaporization (h fg) 37 kj/kg Specific volume ( s).39 m 3 /kg SOLUTION (a) The thermal circuit for the uninsulated pipe is shown below Evaluating the individual resistances R co h co A o h co D o L [W/(m K)] (0.07 m ) L L.07 (mk)/w R ks ln r o ln r i Lk i [43 W/ ( mk)] L (mk)/w r f R f A r f D L i 0.8 m K/W L (0.0m) L.78 (mk)/w R ci h ci A i h ci D i L [567 W/(m K)] (0.0m)L L (mk)/w The rate of heat transfer through the pipe is T q R total R T st R ks T R i R ci q 07 C C.5 W/m L ( ) (m K)/W The total rate of transfer of a three meter section of the pipe is q.5 W/m (3 m) 67.4 W The mass flow rate of the steam in the pipe is D V i (0.0m m s A i V ) (m/s) 4 4 (.39 m g/s /kg)(kg/000g) s s The mass rate of steam condensed in a 3 meter section of the pipe is equal to the rate of heat transfer divided by the heat of vaporization of the steam m c q h fg 67.4 W g/s 37 J/g(Ws/J) The quality of the saturated steam is the fraction of the steam which is vapor. The quality of the steam after a 3 meter section, therefore, is (original vapor mass) (mass o f vapor condensed) m s m c Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

40 X i total mass of steam X i m s g/ X i 0.98 (0.49 g /s) s % 0.49 g /s Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

41 The quality of the steam changed by %. The thermal circuit for the pipe with insulation is shown below The convective resistance on the outside of the pipe is different than that in part (a) because it is based on the outer area of the insulation R co h co A o h co ( D o L i ) L [W/(m K)] (0.07 m+0.m)l L 0.8 (mk)/w The thermal resistance of the insulation is R ki ln D o L i ln r i 0.07 Lk i [0.059 W/(m K)] L 4.8 (mk)/w The rate of heat transfer is q T R total R R ki T si T R ks R f R ci q 07 C C L ( )(m K)/W.0 W/m Therefore, the rate of steam condensed in 3 meters is q.0* 3W m c 0.06 g/s h fg 37 J/g (Ws/J) The quality of the steam after 3 meters of pipe is 0.06 g/ X f 0.98 (0.49 g/s) s 0.9 9% 0.49 g/s The change in the quality of the steam is 6%. COMMENTS Notice that the resistance of the steel pipe and the convective resistance on the inside of the pipe are negligible compared to the other resistances. The resistance of the dirt film is the dominant resistance for the uninsulated pipe Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

42 PROBLEM.8 GIVEN Estimate the rate of heat loss per unit length from a 5 cm ID, 6 cm OD steel pipe covered with high temperature insulation having a thermal conductivity of 0. W/(m K) and a thickness of. cm. Steam flows in the pipe. It has a quality of 99% and is at 50 C. The unit thermal resistance at the inner wall is (m K)/W, the heat transfer coefficient at the outer surface is 7 W/(m K), and the ambient temperature is 6 C. Insulated, steam filled steel pipe Diameters ID of pipe (D i) 5 cm0.05 m OD of pipe (D o) 6 cm0.06 m Thickness of insulation (L i). cm0.0 m Steam quality 99% Steam temperature (T s) 50 C Unit thermal resistance at inner wall (A R i) 0.06 (m K)/W Heat transfer coefficient at outer wall (h o) 7 W /(m K) Ambient temperature (T ) 6 C Thermal conductivity of the insulation (k I) 0. W /(m K) FIND Rate of heat loss per unit length (q/l) ASSUMPTIONS % carbon steel Constant thermal conductivities Steady state conditions SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0 The thermal conductivity of % carbon steel (k s) 43 W/(m K) at 0 C SOLUTION The outer diameter of the insulation (D I) 6 cm + (. cm) 8.4 cm The thermal circuit of the insulated pipe is shown below The values of the individual resistances are AR R i i AR (m K)/W i (m K)/W D i L A i L 0.05m L Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

43 R ks ln D o D i Lk s ln 0.06m 0.05m * L * 43W /(m K ) L (m K)/W R ki ln D I D o Lk i ln 0.084m 0.06m L * 0.W /(m K ) L (m K)/W R co h co A o L h co D I L [7W / (mk )] 0.084m L 0.3 (m K)/W The rate of heat transfer is q L T q R total R i T s T R ks R ki R co 50 C 6C ( )(h ft F )/Btu 84 W/m Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

44 PROBLEM.9 GIVEN The rate of heat flow per unit length q/l through a hollow cylinder of inside radius r i and outside radius r o is q/l ( A k T)/(r o r i) where A (r o r i)/ln(r o/r i). Determine the per cent error in the rate of heat flow if the arithmetic mean area (r o + r i) is used instead of the logarithmic mean area A for ratios of outside to inside diameters (D o/d i) of.5,.0, and 3.0. Plot the results. A hollow cylinder Inside radius r i Outside radius r o Heat flow per unit length as given above FIND (a) Per cent error in the rate of heat flow if the arithmetic rather than the logarithmic mean area is used for ratios of outside to inside diameters of.5,.0, and 3.0. (b) Plot the results ASSUMPTIONS Radial conduction only Constant thermal conductivity Steady state prevails SKETCH SOLUTION The rate of heat transfer per unit length using the logarithmic mean area is q (r o r ) i k T k T L log r o r r ln o i r ln o r i r i The rate of heat transfer per unit length using the arithmetic mean area is q L arith The per cent error is T (r o + r i) k r o r i k T r o r i r o r i Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

45 q L log q L arith % error q 00 L log ln k T r o r i k T r o k T ln r o r i r o r i r i 00 % error r o r ln o r i r i r o r i 00 For a ratio of outside to inside diameters of.5.5 % error ln (.5) 00.37%.5 The percent errors for the other diameter ratios can be calculated in a similar manner with the following results (b) Diameter ratio % Error COMMENTS For diameter ratios less than, use of the arithmetic mean area will not introduce more than a 4% error Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

46 PROBLEM.0 GIVEN A.5-cm-OD, -cm-id copper pipe carries liquid oxygen to the storage site of a space shuttle at 83 C and 0.04 m 3 /min. The ambient air is at C and has a dew point of 0 C. How much insulation with a thermal conductivity of 0.0 W/(m K) is needed to prevent condensation on the exterior of the insulation if h c + h r 7 W/(m K) on the outside? Insulated copper pipe carrying liquid oxygen Inside diameter (D i) cm 0.0 m Outside diameter (D o).5 cm 0.05 m LOX temperature (T ox) 83 C LOX flow rate (m ox) 0.04 m 3 /min Thermal conductivity of insulation (k i) 0.0 W/(m K) Exterior heat transfer coefficients (h o h c + h r) 7 W/(m K) Ambient air temperature (T ) C Ambient air dew point (T dp) 0 C FIND Thickness of insulation (L) needed to prevent condensation ASSUMPTIONS Steady-state conditions have been reached The thermal conductivity of the insulation does not vary appreciably with temperature Radial conduction only The thermal resistance between the inner surface of the pipe and the liquid oxygen is negligible, therefore T wi T ox SKETCH PROPERTIES AND CONSTANTS From Appendix, Table, thermal conductivity of copper (k c) 40 W/(m K) at 0 C SOLUTION The thermal circuit for the pipe is shown below The rate of heat transfer from the pipe is Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

47 q TT ox T R total ln D I ln D o D o D i h o A I Lk I Lk c The rate of heat transfer by convection and radiation from the outer surface of the pipe is Equating these two expressions q T R o T h o A i T I h o A I T ln D I T ox ln D o D o D i T I Lk I Lk c T h o A I ln D I ln D o D o D i T T ox D I L Lk I Lk h o c T T I D I L T T ox + h o D h o ln D I D o ln D o D i T T I I k I k c D I DI ln D ln D ln D o I o D i k I k I k c h o T T T ox T I ln 0.05 ln D I ln (0.05) o 0.0 C) W/(m K) 0.0 W/(m K) 40W/(m K) 7 W/(m K) o C ( ln D D I I (m o C 0 o K)/W C 0.0 Solving this by trial and error D I m 5.4 cm Therefore, the thickness of the insulation is D o.5 cm L D I 5.4 cm.5 cm COMMENTS Note that the thermal resistance of the copper pipe is negligible compared to that of the insulation Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

48 PROBLEM. GIVEN A salesperson for insulation material claims that insulating exposed steam pipes in the basement of a large hotel will be cost effective. Suppose saturated steam at 5.7 bars flows through a 30-cm- OD steel pipe with a 3 cm wall thickness. The pipe is surrounded by air at 0 C. The convective heat transfer coefficient on the outer surface of the pipe is estimated to be 5 W/(m K). The cost of generating steam is estimated to be $5 per 0 9 J and the salesman offers to install a 5 cm thick layer of 85% magnesia insulation on the pipes for $00/m or a 0-cm-thick layer for $300/m. Estimate the payback time for these two alternatives assuming that the steam line operates all year long and make a recommendation to the hotel owner. Assume that the surface of the pipe as well as the insulation have a low emissivity and radiative heat transfer is negligible. Steam pipe in a hotel basement Pipe outside diameter (D o) 30 cm 0.3 m Pipe wall thickness (L s) 3 cm 0.03 m Surrounding air temperature (T ) 0 C Convective heat transfer coefficient (h c) 5 W/(m K) Cost of steam $5/0 9 J Insulation is 85% magnesia FIND Payback time for (a) Insulation thickness (L Ia) 5 cm 0.05 m; (b) Insulation thickness (L Ib) 0 cm 0.0 m; Make a recommendation to the hotel owner. ASSUMPTIONS Cost $00/m Cost $300/m The pipe and insulation are black (.0) The convective resistance on the inside of the pipe is negligible, therefore the inside pipe surface temperature is equal to the steam temperature The pipe is made of % carbon steel Constant thermal conductivities SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 5: The Stefan-Boltzmann constant ( 8 W/(m K 4 ) From Appendix, Table 0 and Thermal conductivities: % Carbon Steel (k s) 43 W/(m K) at 0 C 85% Magnesia (k I) W/(m K) at 0 C ** 7 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

49 From Appendix, Table 3 The temperature of saturated steam at 5.7 bars (T s) 56 C SOLUTION The rate of heat loss and cost of the uninsulated pipe will be calculated first. The thermal circuit for the uninsulated pipe is shown below Evaluating the individual resistances r ln o r i R ks Lk s T q Rtotal 0.5 ln (m K)/W [43 W/( m K)] L R co (m K)/W (0.5 m) h c A o h c r o L [5 W/(m K)] L L The rate of heat transfer for the uninsulated pipe is The cost to supply this heat loss is T s R ks T R co q 56 o C 0 o C 348 W/m L ( ) (K m) / W cost (348 w/m) (J/W s) (3600 s/h) (4 h/day) (365 days/yr) ($5/09J) $496/(yr m) For the insulated pipe the thermal circuit is The resistance of the insulation is given r by: ln Ia 0. r ln o 0.5 R kia Lk I [0.059 W/(mK)] L (m K)/W ln R kib r Io 0.5 r ln o (m K)/W Lk I [0.059 W/(mK)] L (a) The rate of heat transfer for the pipe with 5 cm of insulation is T q Rtotal T s T R ks R kia R co q 56 o C 0 o C 66 W/m L ( ) ( Km)/W Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

50 The cost of this heat loss is cost (66 w/m) (J/W s) (3600 s/h) (4 h/day) (365 days/yr) ($5/0 9 J) $6/yr m Comparing this cost to that of the uninsulated pipe we can calculate the payback period Payback period Cost of installation $00 / m uninsulated cost insulated cost $496 /(yr m) $6 /(yr m) Payback period 0.43 yr 5 months (b) The rate of heat loss for the pipe with 0 cm of insulation is T T T q s Rtotal 0 o C q 56 o C 95.7 W/m L ( ) ( Km)/W The cost of this heat loss R ks R kib R co cost (95.7 w/m) (J/W s) (3600 s/h) (4 h/day) (365 days/yr) ($5/0 9 J) $5/yr m Comparing this cost to that of the uninsulated pipe we can calculate the payback period $300 / m Payback period 0.6 yr 7.5 months $496 / yr m $5 / yr m COMMENTS The 5 cm insulation is a better economic investment. The 0 cm insulation still has a short payback period and is the superior environmental investment since it is a more energy efficient design. Moreover, energy costs are likely to increase in the future and justify the investment in thicker insulation Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

51 PROBLEM. GIVEN A cylindrical liquid oxygen (LOX) tank has a diameter of. m, a length of 6. m, and hemispherical ends. The boiling point of LOX is 79.4 C. An insulation is sought which will reduce the boil-off rate in the steady state to no more than.3 kg/h. The heat of vaporization of LOX is 4 kj/kg. If the thickness of this insulation is to be no more than 7.5 cm, what would the value of its thermal conductivity have to be? Insulated cylindrical tank with hemispherical ends filled with LOX Diameter of tank (D t). m Length of tank (L t) 6. m Boiling point of LOX (T bp) C Heat of vaporization of LOX (h fg) 4 kj/kg Steady state boil-off rate ( m ). kg/h Maximum thickness of insulation (L) 7.5 cm m FIND The thermal conductivity (k) of the insulation necessary to maintain the boil-off rate below. kg/h. ASSUMPTIONS The length given includes the hemispherical ends The thermal resistance of the tank is negligible compared to the insulation The thermal resistance at the interior surface of the tank is negligible SKETCH SOLUTION The tank can be thought of as a sphere (the ends) separated by a cylindrical section, therefore the total heat transfer is the sum of that through the spherical and cylindrical sections. The steady state conduction through a spherical shell with constant thermal conductivity, from Equation (.50), is K r o r i (T T i ) q s 4 o r o r i The rate of steady state conduction through a cylindrical shell, from Equation (.37), is q c L c k T o ln r o r i T i (L c L t. m 4.88 m) The total heat transfer through the tank is the sum of these Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

52 q q s + q c 4 k r o r i (T o T i ) + L c k r o r i (T o ln r o r i T i ) k (T o T i) r o r i L c r o r i r ln o r i The rate of heat transfer required to evaporate the liquid oxygen at m is m h fg, therefore COMMENTS m s h fg k (T o T i) k k (T mh fg r o r i L c r o r i r o ln r r L T ) o i c o i r r o i ln r o.3kg /h*(/3600 s) (54kJ /kg) *(000J / kj ) k (0.685) (0.6) 4.88 [0C (79.4C)] k 0.09 W/(m K) r i r i ln Based on data given in Appendix, Table, no common insulation has such low value of thermal conductivity. However, Marks Standard Handbook for Mechanical Engineers lists the thermal conductivity of expanded rubber board, Rubatex, at 80 C to be W/(m K) Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

53 PROBLEM.3 GIVEN The addition of insulation to a cylindrical surface, such as a wire, may increase the rate of heat dissipation to the surroundings (see Problem.5). (a) For a No. 0 wire (0.6 cm in diameter), what is the thickness of rubber insulation [k 0.6 W/(m K)] that maximizes the rate of heat loss if the heat transfer coefficient is 0 W/(m K)? (b) If the current-carrying capacity of this wire is considered to be limited by the insulation temperature, what per cent increase in capacity is realized by addition of the insulation? State your assumptions. An insulated cylindrical wire Diameter of wire (D w) 0.6 cm m Thermal conductivity of rubber (k) 0.6 W/(m K) FIND Heat transfer coefficient ( h c ) 0 W/(m K) (a) Thickness of insulation (L i) to maximize heat loss (b) Per cent increase in current carrying capacity ASSUMPTIONS The system is in steady state The thermal conductivity of the rubber does not vary with temperature SKETCH SOLUTION (a) From Problem.5, the radius that will maximize the rate of heat transfer (r c) is: k r c h 0.6 W/(mK ) 0 W/(m K ) 0.06 m The thickness of insulation needed to make this radius is m L i r c r w 0.06 m 0.05 m.5 cm (b) The thermal circuit for the insulated wire is shown below where R ki ln r o r i Lk and R c h c A h c r o L The rate of heat transfer from the wire is Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

54 T T q T Ii L (T Ii T ) R c R total R ki ln r o k h c r o If only a very thin coat of insulation is put on the wire to insulate it electrically then r o r i D w/ m. The rate of heat transfer from the wire is q L 0 (T Ii T ) 0 W/(m K)(0.003 m) For the wire with the critical insulation thickness r i 0.08 (T Ii T ) q L ln (T Ii T ) W/(m K) 0 W/(m K)(0.06 m) 0.86 (T Ii T ) The current carrying capacity of the wire is directly related to the rate of heat transfer from the wire. For a given maximum allowable insulation temperature, the increase in current carrying capacity of the wire with the critical thickness of insulation over that of the wire with a very thin coating of insulation is q L r L q thin coat a % increase % q 0.08 L thin coat COMMENTS This would be an enormous amount of insulation to add to the wire changing a thin wire into a rubber cable over an inch in diameter and would not be economically justifiable. Thinner coatings of rubber will achieve smaller increases in current carrying capacity Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

55 PROBLEM.4 GIVEN A standard 4 0 cm steel pipe (ID cm., OD.5 cm) carries superheated steam at 650 C in an enclosed space where a fire hazard exists, limiting the outer surface temperature to 38 C. To minimize the insulation cost, two materials are to be used; first a high temperature (relatively expensive) insulation is to be applied to the pipe and then magnesia (a less expensive material) on the outside. The maximum temperature of the magnesia is to be 35 C. The following constants are known. Steam-side coefficient h 500 W/(m K) High-temperature insulation conductivity k 0. W/(m K) Magnesia conductivity k W/(m K) Outside heat transfer coefficient h W/(m K) Steel conductivity k 43 W/(m K) Ambient temperature T a C (a) Specify the thickness for each insulating material. (b) Calculate the overall heat transfer coefficient based on the pipe OD. (c) What fraction of the total resistance is due to () steam-side resistance, () steel pipe resistance, (3) insulation (combination of the two), and (4) outside resistance? (d) How much heat is transferred per hour, per foot length of pipe? Steam filled steel pipe with two layers of insulation Pipe inside diameter (D i) cm m Pipe outside diameter (D o).5 cm0.5 m Superheated steam temperature (T s) 650 C Maximum outer surface temperature (T so) 38 C Maximum temperature of the Magnesia (T m) 35 C Thermal conductivities High-temperature insulation (k h) 0. W/(m K) Magnesia (k m) W/(m K) Steel (k s) 43 W/(m K) Heat transfer coefficients Steam side ( h ) 500 W/(m K) ci Outside ( hco ) W/(m K) Ambient temperature (T a) C FIND (a) Thickness for each insulation material (b) Overall heat transfer coefficient based on the pipe OD (c) Fraction of the total resistance due to Steam-side resistance Steel pipe resistance Insulation Outside resistance (d) The rate of heat transfer per unit length of pipe (q/l) ASSUMPTIONS The system is in steady state Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

56 Constant thermal conductivities Contact resistance is negligible SKETCH SOLUTION The thermal circuit for the insulated pipe is shown below The values of the individual resistances can be evaluated with Equations (.4) and (.39) R co h co A o h co r 4 L r 4 r 3 ln R km L k m ln R kh ln R ks r 3 r L k h r r L k s R ci h ci A i h ci The variables in the above equations are r L r m r m r 3? r 4? k m W/(m K) Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

57 k s 43 W/(m K) k h 0. W/(m K) h co W/(m K) h ci 500 W/(m K) The temperatures for this problem are T s 650 C T? T? T 3 35 C T 4 38 C T a C There are five unknowns in this problem: q/l, T, T, r 3, and r 4. These can be solved for by writing the equation for the heat transfer through each of the five resistances and solving them simultaneously.. Steam side convective heat transfer T q h ci r L (T s T l) L [500 W /(m K)] m (650 C T ) Rci q T W/m [] L. Conduction through the pipe wall T q k L s (T T ) L[43W /(m K )] (T T ) q R ks ln r r ln (T T ) W/m [] L 3. Conduction through the high temperature insulation T q Rkh k h L (T T 3) r ln 3 r L[0.W / m K )] ln r 3 ln (T 35 C) q 0.68 L ln r (T 35) W/m [3] 4. Conduction through the magnesia insulation k m L L[0.076W /(m K )] T q R km ln (T 3 T 4) ln (r 4 ) ln(r 3 ) r 4 (35 C 38 C) r 3 q 3.3 L ln (r 4 ) ln (r 3 ) W/m [4] 8 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

58 5. Air side convective heat transfer 8 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

59 T q Rco h co r 4 L (T 4 T a) L r 4 (W / (m K )) (38 C C) q L r 4 W/m [5] To maintain steady state, the heat transfer rate through each resistance must be equal. Equations [] through [5] are a set of five equations with five unknowns, they may be solved through numerical iterations using a simple program or may be combined algebraically as follows Substituting Equation [] into Equation [] yields T.065 T 4. Substituting this into Equation [3] and combining the result with Equation [] ln r T T.878 Substituting this into Equation [4] and combining the result with Equation [] r 4 exp 0.669T , T Finally, substituting this into Equation [5] and combining the result with Equation [] 6, T exp U q L D (T s T a ) 0.669T , T Solving this by trial and error: T C This result can be substituted into the equations above to find the unknown radii r m.64 cm r m 8. cm The thickness of the high temperature insulation r 3 r 6.05 cm The thickness of the magnesia insulation r 4 r cm (b) Substituting T 647 C into [] yields a heat transfer rate of 89 W/m. The overall heat transfer coefficient based on the pipe outside area must satisfy the following equation q U A (T s T a) U D L (T s T a) U.30 W/(m K) (c) The overall resistance for the insulated pipe is R total 89 W/m 0.5 (650 C C) UA [.30W /(m K) ] 0.5 L L.8 (K/W) () The thermal resistance of the steam side convection is R ci r L h ci A i h ci (500 W /(m K )) L L K/W The fraction of the resistance due to steam side convection / () The thermal resistance of the steel pipe is 8 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

60 r ln ln r R ks K/W Lk s L[43W /(m K )] L The fraction of the resistance due to the steel pipe / (3) The thermal resistance of the magnesia insulation is r 4 ln r ln R km K/W Lk m L[0.076W /(mk)] L The thermal resistance of the high temperature insulation is r 3 ln 0.64 ln r R kh.5 K/W Lk h L[0.W /( m K )] L The fraction of the resistance due to the insulation.09/ (4) The convective thermal resistance on the air side is Rco 0.08 K/W h co A o h co r 4 L (W /(m K )) (0.8)L L The fraction of the resistance due to air side convection 0.08/ (d) The rate of heat transfer is q U A (T s T a) U D L (T s T a) q.3 W/(m K) (650 C C) 88 W/m L COMMENTS Notice that the insulation accounts for 97% of the total thermal resistance and that the thermal resistance of the steel pipe and the steam side convection are negligible Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

61 PROBLEM.5 Show that the rate of heat conduction per unit length through a long hollow cylinder of inner radius r i and outer radius r o, made of a material whose thermal conductivity varies linearly with temperature, is given by q k L T o T i (ro r i ) / k m A GIVEN where T i temperature at the inner surface T o temperature at the outer surface A (r o r i)/ln r o r i k m k o [ + k (T i + T o)/] L length of cylinder A long hollow cylinder The thermal conductivity varies linearly with temperature Inner radius r i Outer radius r o FIND Show that the rate of heat conduction per unit length is given by the above equation ASSUMPTIONS Conduction occurs in the radial direction only Steady state prevails SKETCH SOLUTION The rate of radial heat transfer through a cylindrical element of radius r is q L dt k A dr k r dt dr a constant But the thermal conductivity varies linearly with the temperature k k o ( + T) q L dt r k o ( + T) dr q dr k o ( + T) dt Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

62 L r Integrating between the inner and outer radii Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

63 q L r o r i r dr k o T o ( T i T )dt q (ln r o ln r i ) k o L q r ln o k o L r i o i i T o T T T i o i (T o T ) (T T ) q (r o r i ) ln r k o (T o T i ) L o (r o r i ) q r A i k m (T o T i) L (r o r i ) (T o T i ) q L T o r o T i r i k A m where A (r o r i)/ln r o r i k m k o [ + k (T i + T o)/] Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

64 PROBLEM.6 GIVEN A long, hollow cylinder is constructed from a material whose thermal conductivity is a function of temperature according to k T, where T is in C and k is in W/(m K). The inner and outer radii of the cylinder are.5 and 5 cm, respectively. Under steady-state conditions, the temperature at the interior surface of the cylinder is 47 C and the temperature at the exterior surface is 93 C. (a) Calculate the rate of heat transfer per meter length, taking into account the variation in thermal conductivity with temperature. (b) If the heat transfer coefficient on the exterior surface of the cylinder is 7 W/(m K), calculate the temperature of the air on the outside of the cylinder. A long hollow cylinder Thermal conductivity (k) T [T in C, k in W/(m K)] Inner radius (r i).5 cm Outer radius (r o) 5 cm Interior surface temperature (T wi) 47 C Exterior surface temperature (T wo) 93 C Exterior heat transfer coefficient ( h ) 7 W/(m K) o Steady-state conditions FIND (a) The rate of heat transfer per meter length (q/l) (b) The temperature of the air on the outside (T ) ASSUMPTIONS Steady state heat transfer Conduction occurs in the radial direction only SKETCH SOLUTION (a) The rate of radial conduction is given by Equation (.37) dt q k A dr q ( T) r L dt dr Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

65 L dr - r q ( T) dt Integrating this from the inside radius to the outside radius r o dr L T wo ( T )dt r i r q T wi L In r o ln r i q [0.5 (T wo T wi) (T wo T wi )] ln r o r i L q [05 (Two Twi ) (T wo T wi )] q L q L [0.5 (T wo T wi) (T wo T wi )] r ln o r i ln 5 [0.5 (47 93) (47 93 )] W/m.5 q L 87. W/m (b) The conduction through the hollow cylinder must equal the convection from the outer surface in steady state q L Solving for the air temperature T T wo q h A o T h r o (T wo T ) o r o o 93 C 87. W/m * L h o 7 W/(m K ) * 0.5m 3 C Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

66 PROBLEM.7 GIVEN Derive an expression for the temperature distribution in an infinitely long rod of uniform cross section within which there is uniform heat generation at the rate of W/m. Assume that the rod is attached to a surface at T s and is exposed through a convective heat transfer coefficient h to a fluid at T f. An infinitely long rod with internal heat generation Temperature at one end T s Heat generation rate ( q G A) W/m Convective heat transfer coefficient h c Ambient fluid temperature T f FIND Expression for the temperature distribution ASSUMPTIONS The rod is in steady state The thermal conductivity (k) is constant SKETCH SOLUTION Let A the cross sectional area of the rod D /4 An element of the rod with heat flows is shown at the right Conservation of energy requires that Energy entering the element + Heat generation Energy leaving the element k A dt q A x k A dt G + h dx c D x [T(x) T f] x dx x x Dividing by x and letting x dt dt ka h q dx x x dx c D x (T T f) G A x x 0 yields d T ka dx h c D (T T f ) q A G Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

67 d T dx 4 h c (T T f ) D k 4 h Let T T f and m c (D k ) d dx m q G k q G k This is a second order, linear, nonhomogeneous differential equation with constant coefficients. Its solution is the addition of the homogeneous solution and a particular solution. The solution to the homogeneous equation d dx m 0 is determined by its characteristic equation. Substituting equation yields the characteristic equation e x m e x 0 ± m e x and its derivatives into the homogeneous Therefore, the homogeneous solution has the form h C c mx + C e mx A particular solution for this problem is simply a constant a o Substituting this into the differential equation q G q 0 m a G o a o k m k Therefore, the general solution is With the boundary conditions q C e mx + C e mx + q G m k a finite number as x T s T f at x 0 From the first boundary condition, as x e mx, therefore C 0 From the second boundary condition q T G s T f C + C T s T f m k m k q G The temperature distribution in the rod is q T(x) T f T s q T G f m k q G e mx + m k T(x) T f + T s q T G f m k q G e mx + m k Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

68 PROBLEM.8 GIVEN Heat is generated uniformly in the fuel rod of a nuclear reactor. The rod has a long, hollow cylindrical shape with its inner and outer surfaces at temperatures of T i and T o, respectively. Derive an expression for the temperature distribution. A long, hollow cylinder with uniform internal generation Inner surface temperature T i Outer surface temperature T o FIND The temperature distribution ASSUMPTIONS Conduction occurs only in the radial direction Steady state prevails SKETCH SOLUTION Let r i the inner radius r o the outer radius q G the rate of internal heat generation per unit volume r dt dr q G r dt dr k the thermal conductivity of the fuel rod The one dimensional, steady state conduction equation in cylindrical coordinates is given in Equation (.) d r dr d dr With boundary conditions + 0 k rq G k T T i at r r i T T o at r r o Integrating the conduction equation once r dt dr r q G k + C Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

69 Integrating again dt r q G C k r r q G Applying the first boundary condition r i q G dr T + C ln (r) + C 4 k T i + C ln (r i ) + C 4 k r C T i + i 4 k q G C i ln (r i) Applying the second boundary condition r o q T G o + C ln (r o) + C 4 k T o r o q G 4 k q G r i + C ln (r o ) + T i + 4 k C ln (r i) C T q G (r r ) 4 k i o i r ln o r i Substituting the constants into the temperature distribution T o q T G r q o G T i (r o r i ) 4 k r i T + ln (r) + T i + 4 k r ln o 4 k r i r (r o r i ) ln r i i r o r i q G (T o q T G (r r ) + 4 k ln T o T i (r o r i ) q G r o 4 k ln r r i T i ) ln r i + T i r ln o r i 9 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

70 PROBLEM.9 In a cylindrical fuel rod of a nuclear reactor, heat is generated internally according to the equation where q G q r r o q g local rate of heat generation per unit volume at r GIVEN r o outside radius q rate of heat generation per unit volume at the centerline Calculate the temperature drop from the center line to the surface for a.5 cm diameter rod having a thermal conductivity of 6 W/(m K) if the rate of heat removal from its surface is.6 MW/ m. A cylindrical rod with internal generation and heat removal from its surface Outside diameter (D o).5 cm0.05 m Rate of heat generation is as given above Thermal conductivity (k) 6 W/(m K) Heat removal rate (q/a).6 MW/m FIND The temperature drop from the center line to the surface ( T) ASSUMPTIONS The heat flow has reached steady state The thermal conductivity of the fuel rod is constant One dimensional conduction in the radial direction SKETCH SOLUTION The equation for one dimensional conduction in cylindrical coordinates is given in Equation (.) d r dt q + G 0 r dr dr k d dr r dt dr k r q r r o 9 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

71 With the boundary conditions dt 0 at r 0 dr Integrating once r dt dr T T s at r r o r q k 4 + r q + C 4 k r o From the first boundary condition: C 0, therefore Integrating again q A dt dr q k q r 3 r o r 4 r T + C k 8 r o r r o 0 r Evaluate this expression at the surface of the cylinder and at the centerline of the sphere and subtracting the results gives us the temperature drop in the cylinder T T 0 T (0) 4 (0) r 4 r o o 8 r 8 r A L q r dr r r o q k r 4 r o o o 3q r o 6 k The rate of heat generation at the centerline (q ) can be evaluated using the conservation of energy. The total rate of heat transfer from the cylinder must equal the total rate of heat generation within the cylinder q A r o L L q q A r o r o r r 4 r o 4 r o 0 r o r o q q q q ro A m [.6*06 W/ m ] W/m Therefore, the temperature drop within the cylinder is 3[5.0 8 W / m )] 0.05 T 6[6W / (m K )] 577 C Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

72 PROBLEM.30 GIVEN An electrical heater capable of generating 0,000 W is to be designed. The heating element is to be a stainless steel wire, having an electrical resistivity of ohm-centimeter. The operating temperature of the stainless steel is to be no more than 60 C. The heat transfer coefficient at the outer surface is expected to be no less than 70 W/(m K) in a medium whose maximum temperature is 93 C. A transformer capable of delivering current at 9 and V is available. Determine a suitable size for the wire, the current required, and discuss what effect a reduction in the heat transfer coefficient would have. (Hint: Demonstrate first that the temperature drop between the center and the surface of the wire is independent of the wire diameter, and determine its value.) A stainless steel wire with electrical heat generation Heat generation rate ( Q G ) 0,000 W Electrical resistivity ( ) ohms-cm Maximum temperature of stainless steel (T max) 60 C Heat transfer coefficient ( h c ) 700 W/(m K) Maximum temperature of medium (T ) 93 C Voltage (V) 9 or V FIND (a) A suitable wire size: diameter (d w) and length (L) (b) The current required (I) (c) Discuss the effect of reduction in the heat transfer coefficient ASSUMPTIONS Variation in the thermal conductivity of stainless steel is negligible The system is in steady-state Conduction occurs in the radial direction only SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k) 4.4 W/(m K) SOLUTION The rate of heat generation per unit volume is Q G Q G r w L q G volume The temperature distribution in a long cylinder with internal heat generation is given in Section Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

73 q G r T(r) C 4 k where C is a constant determined by boundary conditions. Therefore q G r w q G r T(0) T(r w w ) [C 0] C 4 k 4 k 4 The convective heat transfer from the outer surface must equal the internal heat generation q c h c A [T(r w) T ] Q T(r G w ) T r w L h c Adding the two temperature differences calculated above yields Q G Q G kl Q G [T(0) T(r w )] + [T(r w ) T ] + 4 kl Q G r w L h c Q G T(0) T kl r w Lh c The wire length and its radius are related through an expression for the electric power dissipation Q G V V V P e L L L R e Q G V k r w 3 r w h c T(0) T r w [T(0) T ] Q V G r w k h c 0 A r w V r w Q G For the volt case 3 r w (60 C 90 C) 6 (0, 000W) (80 0 ohm -cm) r w 0 (V ) (00 cm/m) 4.4 W/(mK) 700 (W/(m K)) After checking the units, they are dropped for clarity 67 r w ( r ) 0 Solving by trial and error r w m.5 mm For the volt case, the suitable wire diameter is d w (r w) 5 mm The length of the wire required is L ( V) ( m) ohm -cm (0, 000 W) The electrical resistance of this wire is L ohm-cm (0.353 m) R e r w (0.005 m) *(00 cm/m) Therefore, the current required for the volt case is (00 cm /m) m ohm Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

74 I V V 833 amps R e ohm This same procedure can be used for the 9 volt case yielding d w 6.3 mm L m R e ohm I amps COMMENTS The 5-mm-diameter wire would be a better choice since the amperage is less. However, 833 amps is still extremely high. The effect of a lower heat transfer coefficient would be an increase in the diameter and length of the wire as well as an increase in the surface temperature of the wire Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

75 STEADY STATE CONDUCTION IN SPHERE PROBLEM.3 GIVEN A hollow sphere with inner and outer radii of R and R, respectively, is covered with a layer of insulation having an outer radius of R 3. Derive an expression for the rate of heat transfer through the insulated sphere in terms of the radii, the thermal conductivities, the heat transfer coefficients, and the temperatures of the interior and the surrounding medium of the sphere. An insulated hollow sphere Radii Inner surface of the sphere R Outer surface of the sphere R Outer surface of the insulation R 3 FIND Expression for the rate of heat transfer ASSUMPTIONS Steady state heat transfer Conduction in the radial direction only Constant thermal conductivities SKETCH SOLUTION Let k the thermal conductivity of the sphere k 3 the thermal conductivity of the insulation h the interior heat transfer coefficient h 3 the exterior heat transfer coefficient T i the temperature of the interior medium T o the temperature of the exterior medium The thermal circuit for the sphere is shown below The individual resistances are From Equation (.5) R c h A h 4 R L Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

76 The rate of heat transfer is R k 4 R R k R R R R 3 R k3 4 k 3 R 3 R Rc3 h 3 A 3 h 3 4 R 3 L T q Rtotal Rc T R k R k 3 R c3 q 4 T R R R 3 R R h k R R k 3 R 3 R R 3 h 3 4 T q R R R 3 R R h k R R k 3 R 3 R R 3 h Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

77 PROBLEM.3 GIVEN The thermal conductivity of a material can be determined in the following manner. Saturated steam N/m is condensed at the rate of 0.68 kg/h inside a hollow iron sphere that is.3 cm thick and has an internal diameter of 5 cm. The sphere is coated with the material whose thermal conductivity is to be evaluated. The thickness of the material to be tested is 0 cm and there are two thermocouples embedded in it, one.3 cm from the surface of the iron sphere and one.3 cm from the exterior surface of the system. If the inner thermocouple indicates a temperature of 0 C and the outer themocouple a temperature of 57 C, calculate (a) the thermal conductivity of the material surrounding the metal sphere, (b) the temperatures at the interior and exterior surfaces of the test material, and (c) the overall heat transfer coefficient based on the interior surface of the iron sphere, assuming the thermal resistances at the surfaces, as well as the interface between the two spherical shells, are negligible. Hollow iron sphere with saturated steam inside and coated with material outside Steam pressure N/m Steam condensation rate ( m) s 0.68 kg/h Inside diameter (D i) 5 cm 0.5 m Thickness of the iron sphere (Ls).3 cm 0.03 m Thickness of material layer (L m) 0 cm 0. m Two thermocouples are located.3 cm from the inner and outer surface of the material layer Inner thermocouple temperature (T ) 0 C Outer thermocouple temperature (T ) 57 C FIND (a) Thermal conductivity of the material (k m) (b) Temperatures at the interior and exterior surfaces of the test material (T mi, T mo) (c) Overall heat transfer coefficient based on the inside area of the iron sphere (U) ASSUMPTIONS Thermal resistance at the surface is negligible Thermal resistance at the interface is negligible The system has reached steady-state The thermal conductivities are constant One dimensional conduction radially SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 3: For saturated steam at N/m, Saturation temperature (T s) 5 C Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

78 Heat of vaporization (h fg) 87 kj/kg SOLUTION (a) The rate of heat transfer through the sphere must equal the energy released by the condensing steam: k m h q m s h fg 0.68 kg/h 87 kj/kg 000 J/kJ (Ws)/J 43. W 3600s The thermal conductivity of the material can be calculated by examining the heat transfer between the thermocouple radii T q R k Solving for the thermal conductivity q (r r ) 4 T r 4 r r (T T ) D i r + L s m T r k m r r 0.5m m m 0.8 m D i 0.5m r + L s + L m 0.03 m m + 0. m 0.03 m k m 43.W(0.355 m 0.8 m ) 4 (0.355 m )(0.8m) (0 o C 57 o C ) 0.46 W/(m K) (b) The temperature at the inside of the material can be calculated from the equation for conduction through the material from the inner radius, the radius of the inside thermocouple T q R ki T mi T i r r i 4 k m r r i Solving for the temperature of the inside of the material q (r r i ) Tmi T + 4 k m r r i D i 0.5 m r i + L m m 0.68 m T mi 0 C + T q 4 [0.46 W /(mk)](0.8 m) (0.68 m) T r o 4 T mo r k m r o r 43.W (0.03 m) C The temperature at the outside radius of the material can be calculated from the equation for conduction through the material from the radius of the outer thermocouple to the outer radius R k o Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

79 Solving for the temperature of the outer surface of the material 0 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

80 T mo T q (r o r ) 4 k m r o r D i r o + L s + L m 0.5 m m m m T mo 57 C 43.W (0.03 m) 4 [0.46 W /(mk)](0.368 m) (0.355 m) 50 C (c) The heat transfer through the sphere can be expressed as q U A i T U D (T s T mo) q 43. W U Di 6.74 W/(m K) (T s T mo ) (0.5m) (5 C 50 C ) 0 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

81 PROBLEM.33 For the system outlined in Problem.3, determine an expression for the critical radius of the insulation in terms of the thermal conductivity of the insulation and the surface coefficient between the exterior surface of the insulation and the surrounding fluid. Assume that the temperature difference, R, R, the heat transfer coefficient on the interior, and the thermal conductivity of the material of the sphere between R and R are constant. GIVEN An insulated hollow sphere Radii Inner surface of the sphere R Outer surface of the sphere R Outer surface of the insulation R 3 FIND An expression for the critical radius of the insulation ASSUMPTIONS Constant temperature difference, radii, heat transfer coefficients, and thermal conductivities Steady state prevails SKETCH SOLUTION Let k the thermal conductivity of the sphere k 3 the thermal conductivity of the insulation h the interior heat transfer coefficient h 3 the exterior heat transfer coefficient T i the temperature of the interior medium T o the temperature of the exterior medium From Problem., the rate of heat transfer through the sphere is R R R 3 R R h k R R k 3 R 3 R R 3 h 3 q 4 T R R R 3 R R h k R R k 3 R 3 R R 3 h 3 The rate of heat transfer is a maximum when the denominator of the above equation is a minimum. This occurs when the derivative of the denominator with respect to R 3 is zero d dr 3 0 > h 3 R k 3 R 3 k 3 h 3 The maximum heat transfer will occur when the outer insulation radius is equal to k 3/h Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

82 PROBLEM.34 Show that the temperature distribution in a sphere of radius r o, made of a homogeneous material in which energy is released at a uniform rate per unit volume q G, is q G ro r T(r) T o + 6 k r o GIVEN A homogeneous sphere with energy generation Radius r o FIND Show that the temperature distribution is as shown above. ASSUMPTIONS Steady state conditions persist The thermal conductivity of the sphere material is constant Conduction occurs in the radial direction only SKETCH SOLUTION Let k the thermal conductivity of the material T o the surface temperature of the sphere Equation (.3) can be simplified to the following equation by the assumptions of steady state and radial conduction only d r dr r dt dr d dr q G r dt dr k 0 r q G k With the following boundary conditions dt 0 at r 0 dr T T o at r r o Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

83 Integrating the differential equation once r dt dr 3 r q G 3 k + C From the first boundary condition C 0 Integrating once again r q G T + C 6 k Applying the second boundary condition T o + C C T o + 6 k r o q G 6 k r o q G Therefore, the temperature distribution in the sphere is r q G r o q G T + T o + 6 k 6 k T(r) T o + q G r o r 6 k r o Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

84 PROBLEM.35 Liquid oxygen is to be stored on a service module of NASA s new Orion Spacecraft (NASA, Orion Quick Facts, FS JSC, Lyndon B. Johnson Space Center, Houston, TX) in a spherical stainless steel container. The service module is depicted in the figure below (note the various spherical containers shown), and the entire Orion Spacecraft is shown later in figure. 3.(Chapter 3). The container has an outside diameter of.0 m and a wall thickness of 0 mm. The boiling point of liquid oxygen is 90 K, and its latent heat is 3 kj/kg. The tank is to be installed in an environment where loss of oxygen from the container is not to exceed.0 kg/day. Tank is to be installed in service module where ambient temperature is 5 K and convection coefficient is 5 W/ (m K). To ensure this calculate the thermal conductivity of material used for insulation if its thickness is 0 cm. GIVEN Liquid oxygen container with.0 m diameter and 0 mm thickness. Boiling point of oxygen (T b) 90 K Ambient temperatue (T )5 K Convection coefficient (h) 5 W/ (m K) Latent heat of vaporization (h fg) 3 kj/kg Loss of oxygen (ṁ).0 kg/day/86400 kg/sec FIND Suitable layer of insulation and its thickness. ASSUMPTIONS Steady state conditions persist The insulation is uniform. Properties remain constant. SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k s) 4.4 W/(m K). The material with lowest conductivity available is selected pertaining to light weight requirement for spacecraft. From Appendix, Table, glass fiber is selected with conductivity of (k g)0.035 W/ (m K) Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

85 SOLUTION The inner and outer radius of the stainless steel sphere is r m r 0.5 m r m The rate of heat loss by vaporization is given by q /86400*3*000 J/sec.465 W This amount should be equal to heat loss by conduction through the sphere. Thus W 4 * *4.4* 0.50 * * * ks * 0.5 * * * 5 * 0.6 q T T r r r 3 * * ks * r * r * * kg * *4.4* 0.50 * * * ks * 0.5 * * * 5 * 0.6 W.465 * k k Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

86 k0.065/54.73 W/(m K) W/(m K) COMMENTS Currently no material is available whose conductivity is as low as W/(m K). Scientists are working on developing low conductivity material so that the thickness of insulation material required reduces which is critical for developing light weight spacecraft Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

87 PROBLEM.36 GIVEN The development of contact lenses has transformed the solutions that are available today for vision impairments. However, wearing them also poses several problems that includes the condition of dry eyes due to lack of cooling, oxygenation, and moisturizing or lubrication of cornea, among others. In a development phase of a new protective liquid for contact lenses, the optometrist would like to know rate of heat loss from the anterior chamber of the eye through the cornea and the contact lens. The system is modelled as a partial hollow sphere, as shown in schematic diagram. The inner and outer radius of the cornea are respectively r 0 mm and r.5 mm, and the outer radius of the fitted contact lens is r 34.5 mm. The anterior chamber, which contains the aqueous humor that provides nutrients to the cornea (as the later tissue has no blood vessels), exposes the inner surface of the cornea to a temperature of T i C with a convective heat transfer coefficient of h i0 W/(m K). The thermal conductivity of transparent tissue of cornea is 0.35 W/(m K), and that of the contact lens material is 0.8 W/(m K). The outer surface of the contact lens is exposed to room air at T 0 0 C and has a convection coefficient of h 05 W/(m K). Draw a thermal circuit for the system showing all of the temperature potentials and thermal resistances. Then estimate the rate of heat loss from the anterior eye assuming that steady state exists and eye aperture spherical angle is of Inner radius of cornea r 0 mm0.0 m Outer radius of cornea r.5 mm0.05 m Radius of contact lens r 34.5 mm0.045 m Inner cornea temperature T i C Inner convective heat transfer coefficient h i0 W/(m K) Conductivity of cornea tissue (k )0.35 W/(m K) Conductivity of lens material (k )0.8 W/(m K) Room air temperature T 0 0 C Outer convection coefficient of h 05 W/(m K) FIND Draw thermal circuit showing all temperature potentials and thermal resistances Estimate rate of heat loss from anterior eye. ASSUMPTIONS Steady state conditions persist Properties remain constant. Eye is treated as partial hollow sphere sweeping an angle of θ 00 0 SKETCH Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

88 09 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

89 PROPERTIES AND CONSTANTS From Appendix, Table 0, Thermal conductivity of stainless steel (k) 4. W/(m K) SOLUTION Since solid angle for complete sphere is 70 0 or 4π the surface area reduces by the factor of 70 resistances increase by the factor of. The thermal circuit diagram for the problem is given as follows 70 and thus Total heat loss with the lens is given by q Ti T 0 Rtotal R totalr + R + R 3+ R 4 R 70 7.* 4 hi * r 4 *0 * K/W R 70 r r 4 k * rr 7. * * 0.35 * 0.05 * K/W R 3 70 r3 r 4 k * r r3 7. * * 0.8 * * K/W R * 4 ho * r 3 4 * 5 * K/W R totalr + R + R 3+ R K/W Total heat loss with the lens is Ti T q W 0.09 W Rtotal Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

90 PROBLEM.37 GIVEN In cryogenic surgery, a small spherical probe is brought into contact with the diseased tissue which is frozen and thereby destroyed. One such probe can be modeled as a 3-mm diameter sphere whose surface is maintained at 40 K when the surrounding tissue is at 34 K. During the surgical procedure, a thin layer of tissue freezes around the probe at a temperature of 73 K. Assuming that the thermal conductivity of a frozen tissue is.5 W/(m K) and the heat transfer mechanism at the surface is described by the effective convective coefficient of 50 W/(m K), estimate the thickness of frozen tissue formed during a 30 min long operation. Diameter of sphere(d ) 3 mm, r m Probe surface temperature(t p) 40 K Surrounding tissue temperature(t t) 34 K Tissue freezing temperature(t f) 73 K Thermal conductivity of frozen tissue(k t).5 W/(m K) Effective convective coefficient (h)50 W/(m K) Operation time (t) 30 min FIND Thickness of a frozen tissue formed ASSUMPTIONS dimensional Steady state conditions persist Negligible contact resistance between probe and frozen tissue. Constant properties. SKETCH SOLUTION The thermal circuit diagram for the given as follows 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

91 The thickness of tissue in 30 mins is obtained by energy balance that is when rate of heat loss through conduction is equal to rate of heat loss through convection, which means q q conduction q convection If r is the radius of frozen tissue at 30 mins The convection heat loss convection h * 4 r T T 50*4π* r *(34-73) 576 r t f Conduction heat loss is given by Tf Tp q conduction r r r ktrr 4 *.5* * r 0.933r r Equating above two equations we get 576 r 0.933r r *r r Solving the above equation, we get r m or 7 mm. Thus the thickness of frozen tissuer -r 5.5 mm 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

92 EXTENDED SURFACES OR FINS PROBLEM.38 GIVEN The addition of aluminum fins has been suggested to increase the rate of heat dissipation from one side of an electronic device -m-wide and -m-tall. The fins are to be rectangular in cross section,.5-cm-long and 0.5-cm-thick. There are to be 00 fins per meter. The convective heat transfer coefficient, both for the wall and the fins, is estimated at 35 W/(m K). With this information, determine the per cent increase in the rate of heat transfer of the finned wall compared to the bare wall. Aluminum fins with a rectangular cross section Dimensions:.5-cm-long and 0.5-mm-thick Number of fins per meter 00 The convective heat transfer coefficient ( h c ) 35 W/(m K) FIND The per cent increase in the rate of heat transfer of the finned wall compared to the bare wall ASSUMPTIONS Steady state heat transfer SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of aluminum (k) 40 W/(m K) at 7 C SOLUTION Since the fins are of uniform cross section, Table. can be used to calculate the heat transfer rate from a single fin with convection at the tip where M sin h (ml) q f M cosh (ml) h c P k A s h c (m k ) h c mk cosh ( ml) h c (t si nh ( ml) w) k (tw ) s s T s T 3 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

93 For a m width (w m) M 35 W/(m K) (.005 m) 40W/(m K) (0.05 m ) s 6.49 s W/K m L h c P L ka h c (t k (tw) w) 0.05 m 35 W/(m K) (.005m) 40W/(m K) (0.005 m ) L m 0.05 m m h c 35 W/(m K) m K W/(m K) m Therefore, the rate of heat transfer from one fin, -meter-wide is: sin h (0.7) c os h (0.7) q f 6.49 s W/K cos h (0.7) sin h (0.7) q f.79 s W/K In m of wall area there are 00 fins covering 00 tw 00 (0.005 m) ( m) 0.5 m of wall area leaving 0.75 m of bare wall. The total rate of heat transfer from the wall with fins is the sum of the heat transfer from the bare wall and the heat transfer from 00 fins. q tot q bare + 00 q fin h A bare s + 00 q fin q tot 35 W/(m K) (0.75 m ) s + 00 (.79) s W/K 05.3 s W/K The rate of heat transfer from the wall without fins is q bare h c A s 35 W/(m K) ( m ) s 35.0 W/K The percent increase due to the addition of fins is COMMENTS % increase % This problem illustrates the dramatic increase in the rate of heat transfer that can be achieved with properly designed fins. The assumption that the convective heat transfer coefficient is the same for the fins and the wall is an oversimplification of the real situation, but does not affect the final results appreciably. In later chapters, we will learn how to evaluate the heat transfer coefficient from physical parameters and the geometry of the system Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

94 PROBLEM.39 GIVEN The tip of a soldering iron consists of a 0.6-cm diameter copper rod, 7.6 cm long. If the tip must be 04 C, what are the required minimum temperature of the base and the heat flow, in watts, into the base? Assume that h.7 W/(m K) and T air C. Tip of soldering iron consists of copper rod Outside diameter (D) 0.6 cm m Length (L) 7.6 cm m Temperature of the tip (T L) 04 C Heat transfer coefficient ( h ).7 W/(m K) Ambient temperature (T ) C FIND (a) Minimum temperature of the base (T s) (b) Heat flow into the base (q) in W ASSUMPTIONS The tip is in steady state The thermal conductivity of copper is uniform and constant, i.e., not a function of temperature The copper tip can be treated as a fin SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of copper (K) 388 W/(m K) at 7 C SOLUTION (a) From Table., the temperature distribution for a fin with a uniform cross section and convection from the tip is h cosh [m(l x)] sinh[m(l x)] mk s h cosh(ml) si nh(ml ) mk where T T and s (0) T s T 5 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

95 L m L h P L ka h D k D 4 4 h kd m 4.7W/(m K) 388W/(m K) (0.006 m) L m m m h.7 W/(m K) mk W/ ( mk) m Evaluating the temperature at x L L T L s T s T T Solving for the base temperature cosh (0) sinh (0) cosh(0.475) sinh(0.475) T o C T s T + T L C + 04o C C (b) To maintain steady state conditions, the rate of heat transfer into the base must be equal to the rate of heat loss from the rod. From Table., the rate of heat loss is q f M sin h (ml) cos h(ml) h mk h mk cosh ( ml) si nh( ml) 3 where M h PkA s h k D (T s T ) 4 M.7W/(m K) 388W/ (m K) (0.006 m) 3 4 (6 C C) W q f W sinh (0.475) c osh (0.475) cosh(0.475).00936s inh(0.475) 6.3 W COMMENTS A small soldering iron such as this will typically be rated at 30 W to allow for radiation heat losses and more rapid heat-up Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

96 PROBLEM.40 GIVEN One end of a 0.3-m-long steel rod is connected to a wall at 04 C. The other end is connected to a wall that is maintained at 93 C. Air is blown across the rod so that a heat transfer coefficient of 7 W/(m K) is maintained over the entire surface. If the diameter of the rod is 5 cm and the temperature of the air is 38 C, what is the net rate of heat loss to the air? A steel rod connected to walls at both ends Length of rod (L) 0.3 m Diameter of the rod (D) 5 cm 0.05 m Wall temperatures T s 04 C T L 93 C Heat transfer coefficient ( h c ) 7 W/(m K) Air temperature (T ) 38 C FIND The net rate of heat loss to the air (q f) ASSUMPTIONS The wall temperatures are constant The system is in steady state The rod is % carbon steel The thermal conductivity of the rod is uniform and not dependent on temperature One dimensional conduction along the rod SKETCH PROPERTIES AND CONSTANTS From Appendix, Table 0 The thermal conductivity of % carbon steel (k) 43 W/(m K) (at 0 C) SOLUTION The rod can be idealized as a fin of uniform cross section with fixed temperatures at both ends. From Table. the rate of heat loss is cos h (ml) L s q f M sin h (ml) 7 08 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

97 where L T L T 93 C 38 C 55 C and s T s T 04 C 38 C 66 C L m L h c P L ka h c D k D 4 L 4 h c kd 0.3 m 4 7 W/(m K) 4 7 W/(m K) (0.05 m).687 M h P k A s h D k s 4 3 7W/(m K) (0.05 m) W/(m K) (66 C) 78.8 W COMMENTS 55 cosh (.687) q f 78.8 W 66 sinh (.687) 74.4 W In a real situation the convective heat transfer coefficient will not be uniform over the circumference. It will be higher over the side facing the air stream. But because of the high thermal conductivity, the temperature at any given section will be nearly uniform Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

98 PROBLEM.4 GIVEN Both ends of a 0.6-cm copper U-shaped rod, are rigidly affixed to a vertical wall as shown in the accompanying sketch. The temperature of the wall is maintained at 93 C. The developed length of the rod is 0.6 m, and it is exposed to air at 38 C. The combined radiation and convection heat transfer coefficient for this system is 34 W/(m K). (a) Calculate the temperature of the midpoint of the rod. (b) What will the rate of heat transfer from the rod be? U-shaped copper rod rigidly affixed to a wall Diameter (D) 0.6 cm m Developed length (L) 0.6 m Wall temperature is constant at (T s) 93 C Air temperature (T ) 38 C Heat transfer coefficient ( h ) 34 W/(m K) FIND (a) Temperature of the midpoint (T Lf) (b) Rate of heat transfer from the rod (M) ASSUMPTIONS The system is in steady state Variation in the thermal conductivity of copper is negligible The U-shaped rod can be approximated by a straight rod of equal length Uniform temperature across any section of the rod SKETCH PROPERTIES AND CONSTANTS From Appendix, Table, thermal conductivity of copper (k) 396 W/(m K) at 64 C SOLUTION By symmetry, the conduction through the rod at the center must be zero. Therefore, the rod can be thought of as two pin fins with insulated ends as shown in the sketch above. (a) From Table., the temperature distribution for a fin of uniform cross section with an adiabatic tip is cosh [m(l f x)] s cosh (ml) where T T, s T s T and L f length of the fin m hp ka k h D D 4 4 h 4 34W/(m K) kd 396W/(m K) (0.006 m) 7.57 m

99 Evaluating the temperature of the tip of the pin fin ( L f ) s cosh [m(l f L f cosh (m L f ) )] cosh (m L f ) The length of the fin is half of the wire length (L f 0.3 m) ( L f ) T T (Lf ) s T T 7.57 s cos h (0.3m) m 0.05 T(L f) 0.05 (T s T ) + T 0.05 (93 C 38 C) + 38 C 49. C The temperature at the tip of the fin is the temperature at the midpoint of the curved rod (49. C). (b) From Table., the heat transfer from the fin is q fin M tanh (m L f) where M h P k A s h ( D)k D (T s T ) 4 M 34W/(m K) 396W/(m K) (0.006 m) 3 4 (93 C 38 C) W q fin W tanh 7.57 m (0.3 m) 4.56 W The rate of heat transfer from the curved rod is approximately twice the heat transfer of the pin fin q rod q fin (4.56 W) 9. W

100 PROBLEM.4 GIVEN A circumferential fin of rectangular cross section, 3.7-cm-OD and 0.3-cm-thick surrounds a.5- cm-diameter tube as shown below. The fin is constructed of mild steel. Air blowing over the fin produces a heat transfer coefficient of 8.4 W/(m K). If the temperatures of the base of the fin and the air are 60 C and 38 C, respectively, calculate the heat transfer rate from the fin. A mild steel circumferential fin of a rectangular cross section on a tube Tube diameter (D t).5 cm 0.05 m Fin outside diameter (D f) 3.7 cm m Fin thickness (t) 0.3 cm m Heat transfer coefficient ( h c ) 8.4 W/(m K) Fin base temperature (T s) 60 C Air temperature (T ) 38 C FIND The rate of heat transfer from the fin (q fin) ASSUMPTIONS The system has reached steady state The mild steel is % carbon steel The thermal conductivity of the steel is uniform Radial conduction only (temperature is uniform across the cross section of the fin) The heat transfer from the end of the fin can be accounted for by increasing the length by half the thickness and assuming the end is insulated SKETCH PROPERTIES AND CONSTANTS Thermal conductivity of % carbon steel (k) 43 W/(m K) at 0 C SOLUTION The rate of heat transfer for the fin can be calculated using the fin efficiency determined from the efficiency graph for this geometry, Figure.7. The length of a fin (L) (D f D t)/ m The parameters needed are D r i t D 0.5 m r o t + L 0.5 m m m

101 r o t r i 3 3 h c k t(r o r i ) m m 0.05 m r o r i 8.4w/(m K) 43W/(m K) (0.003 m)(0.085m 0.05 m ) t m m m 0.76 From Figure.7, the fin efficiency for these parameters is: f 98% The rate of heat transfer from the fin is q fin f h c A fin (T s T ) f h c r o (T s T ) t r i q fin (0.98) 8.4W/(m K) [(0.085 m m) (0.05 m) ] (60 C 38 C) 9.46 W

102 PROBLEM.43 GIVEN A turbine blade 6.3-cm-long, with cross-sectional area A m and perimeter P 0. m, is made of stainless steel (k 8 W/(m K). The temperature of the root, T s, is 48 C. The blade is exposed to a hot gas at 87 C, and the heat transfer coefficient h is 454 W/(m K). Determine the temperature of the blade tip and the rate of heat flow at the root of the blade. Assume that the tip is insulated. Stainless steel turbine blade Length of blade (L) 6.3 cm m Cross-sectional area (A) m Perimeter (P) 0. m Thermal conductivity (k) 8 W/(m K) Temperature of the root (T s) 48 C Temperature of the hot gas (T ) 87 C Heat transfer coefficient ( h c ) 454 W/(m K) FIND (a) The temperature of the blade tip (T L) (b) The rate of heat flow (q) at the roof of the blade ASSUMPTIONS Steady state conditions prevail The thermal conductivity is uniform The tip is insulated The cross-section of the blade is uniform One dimensional conduction SKETCH SOLUTION (a) The temperature distribution in a fin of uniform cross-section with an insulated tip, from Table., is cosh [m( L x)] s cosh (ml) h P 454W/(m K)(0. m) where m 8. k A 4 8W/(m K)(4.6 0 m ) m T T

103 At the blade tip, x L, therefore L T T L cosh [m(0)] s T cosh (ml) cosh (ml) T s T L T + T T s cosh (ml) 87 C + cosh 48 o C 87 o C 8. (0.063 m) m 866 C (b) The rate of heat transfer from the fin is given by Table. to be q M tanh (m L) where M h c P k A s COMMENTS M 454W/(m K)(0. m) 8W/(m K) ( m ) (48 C 87 C) 6 W q ( 6 W) tanh 8. (0.063 m) m 6 W (out of the blade) In a real situation, the heat transfer coefficient will vary over the surface with the highest value near the leading edge. But because of the high thermal conductivity of the blade, the temperature at any section will be esentially uniform.

104 PROBLEM.44 GIVEN To determine the thermal conductivity of a long, solid.5 cm diameter rod, one half of the rod was inserted into a furnace while the other half was projecting into air at 7 C. After steady state had been reached, the temperatures at two points 7.6 cm apart were measured and found to be 6 C and 9 C, respectively. The heat transfer coefficient over the surface of the rod exposed to the air was estimated to be.7 W/(m K). What is the thermal conductivity of the rod? A solid rod, one half inserted into a furnace Diameter of rod (D).5 cm 0.5 m Air temperature (T ) 7 C Steady state has been reached Temperatures at two points 7.6 cm apart T 6 C T 9 C The heat transfer coefficient ( h c ).7 W/(m K) FIND The thermal conductivity (k) of the rod ASSUMPTIONS Uniform thermal conductivity One dimensional conduction along the rod The rod approximates a fin of infinite length protruding out of the furnace SKETCH SOLUTION This problem can be visualized as the following pin fin problem shown below The fin is of uniform cross section, therefore Table. can be used. The temperature distribution for a fin of infinite length, from Table., is s e mx h D where m c P h c 4 h c ka kd k D Substituting this into the temperature distribution and solving for k

105 s 4 h c 4 h exp x k c kd ln s D x at x L L T L T 9 C 7 C 64 C s T W T 6 C 7 C 99 C L 64 s Therefore k 4.7w/(m K) 0.05 ln (0.6465) m 0 W/(m K) COMMENTS Note that this procedure can only be used if the assumption of an infinite length fin is valid. Otherwise, the location of the temperature measurements along the fin must be specified to determine the thermal conductivity.

106 PROBLEM.45 GIVEN Heat is transferred from water to air through a brass wall (k 54 W/(m K)). The addition of rectangular brass fins, 0.08-cm-thick and.5-cm-long, spaced.5 cm apart, is contemplated. Assuming a water-side heat transfer coefficient of 70 W/(m K) and an air-side heat transfer coefficient of 7 W/(m K), compare the gain in heat transfer rate achieved by adding fins to: (a) the water side, (b) the air side, and (c) both sides. (Neglect temperature drop through the wall.) A brass wall with brass fins between air and water Thermal conductivity of the brass (k) 54 W/(m K) Fin thickness (t) 0.08 cm m Fin length (L).5 cm 0.05 m Fin spacing (d).5 cm 0.5 m Water-side heat transfer coefficient ( h cw ) 70 W/(m K) FIND Air-side heat transfer coefficient ( h ca ) 7 W/(m K) Compare the heat transfer rate with fins added to (a) the water side, q (a) (b) the air side, q (b) (c) both sides, q (c) ASSUMPTIONS The thermal resistance of the wall is negligible Steady state conditions prevail Constant thermal conductivity One dimensional conduction Heat transfer from the tip of the fins is negligible SKETCH SOLUTION The fins are of uniform cross-section, therefore Table. may be used. To simplify the analysis, the heat transfer from the end of the fin will be neglected. For a fin with adiabatic tip, the rate of heat transfer is q f M tanh (m L) where M h c PkA s h c (w)k (wt) s w h c kt s m h c P ka h c (w) kwt h c kt

107 The number of fins per square meter of wall is number of fins m (0.033 m/fin)m width) 75. fins/m Fraction of the wall area not covered by fins is A bare A n m 75. (-m-width) (0.008 m) m The rate of heat transfer from the wall with fins is equal to the sum of the heat transfer from the bare wall and from the fins q h c A bare s + (number of fins) [M tanh (m L)] q h c A bare M 75. A s w tanh (ml) s s R c where A w is the total base area, i.e., with fins removed. Therefore, the thermal resistance of a wall with fins based on a unit of base area is For fins on the water side M w R c s m w A W A h bare c A w M 75. tanh (ml) s -m-width 70W/(m K)() 54W/(m K) ( m) 3.83 W/K 70W/(m K) 54W/(m K)(.0008 m) 88.7 m For fins on the air side tan h (m 88.7 a L) tanh (0.05 m) m M a s -m-width 7W(m K) () 54W/(m K) (0.0008). W/K m a 7W/(m K) 54W/(m K) ( m) 8.05 m tan h m 8.05 a L tan h (0.05 m) m The thermal circuit for the problem is The values of thermal resistances with and without fins are (R ca) nofins A w h ca w A 7W/(m K) A w (m K)/W

108 (R cw) nofins A w h cw w A 70 W/(m K) A w (m K)/W (R ca) fins A 7 W/(m w K)(0.94) 75. m.w/k (0.605) A w 0.04 (m K)/W (R cw) fins A w 70 W/(m K)(0.94) 75. m 3.83 W/K (0.977) A w (m K)/W (a) The rate of heat transfer with fins on the water side only is T q (a) (R ca ) no fins (R cw ) fins q (a ) A w T ( )(m K)/W 6.4 T W/(m K) (b) The rate of heat transfer with fins on the air side only is T q (b) (R ca ) fins (R cw ) no f ins q (b) A w T ( )(m K)/W 50. T W/(m K) (c) With fins on both sides, the rate of heat transfer is T q (c) (R ca ) fins (R cw ) no f ins q (c ) A w T ( )(m K)/W 6. T W/(m K) As a basis of comparison, the rate of heat transfer without fins on either side is: T q A w ( )(m W)/K 5.5 T W/(m K) The following percent increase over the no fins case occurs Case % Increase COMMENTS (a) fins on water side 5.8 (b) fins on air side 3 (c) fins on both sides 94 Placing the fins on the side with the larger thermal resistance, i.e., the air side, has a much greater effect on the rate of heat transfer.

109 The small gain in heat transfer rate achieved by placing fins on the water side only would most likely not be justified due to the high cost of attaching the fins.

110 PROBLEM.46 GIVEN The wall of a liquid-to-gas heat exchanger has a surface area on the liquid side of.8 m (0.6m 3m) with a heat transfer coefficient of 55 W/(m K). On the other side of the heat exchanger wall flows a gas, and the wall has 96 thin rectangular steel fins 0.5-cm-thick and.5-cmhigh [k 3 W/(m K)]. The fins are 3-m-long and the heat transfer coefficient on the gas side is 57 W/(m K). Assuming that the thermal resistance of the wall is negligible, determine the rate of heat transfer if the overall temperature difference is 38 C. The wall of a heat exchanger has 96 fins on the gas side Surface area on the liquid side (A L).8 m (0.6 m 3 m) Heat transfer coefficient on the liquid side (h cl) 55 W/(m K) The wall has 96 thin steel fins 0.5-cm-thick and.5-cm-high Thermal conductivity of the steel (k) 3 W/(m K) Fin length (w) 3 m, Fin height (L).5 cm 0.05 m Fin thickness (t) 0.5 cm m Heat transfer coefficient on the gas side (h cg) 57 W/(m K) The overall temperature difference ( T) 38 C FIND The rate of heat transfer (q) ASSUMPTIONS The thermal resistance of the wall is negligible The heat transfer through the wall is steady state The thermal conductivity of the steel is constant SKETCH SOLUTION The heat transfer from a single fin can be calculated from Table. for a fin with convection from the tip sinh (ml) h ) c q cosh (ml f M mk h cosh (ml) c sinh (ml) mk

111 where m h c P ka h w) 0.0 m) c (t 57 W/(m K)(6 m k (wt ) 3 W/(m K)(3 m)(0.005 m) 87.5 m ml 87.5 h c 57 W/(m K) (0.05 m).09 and 0.78 m mk W/(m K) m M h c Pk A s 57 W/(m K) (6.0m) 3W/(m K) (3 m)(0.005 m) (T s T g) 3.96 (T s T g)w/k q f 3.96 (T s T g )W/K sinh (.09) 0.78 c os h (.09) cosh (.09) 0.78s in h (.09) (T s T g ) W/K The rate of heat transfer on the gas side is the sum of the convection from the fins and the convection from the bare wall between the fins. The bare area is The total rate of heat transfer to the gas is A bare A wall (number of fins) (Area of one fin).8 m (96 fins) [(3 m) (0.005 m)/fin] 0.36 m q g q bare + (number of fins) q f h cg A bare (T s T g ) + 96(3.395) (T s T g ) W/K q 57 W/(m K) (0.36 m g ) 96 (3.395) (T s T g ) W/K (T s T g ) W/K T s T g R g The thermal resistance on the gas side is R g K/W The thermal resistance on the liquid side is K/W R L h cl A w 55 W/(m K) (.8 m ) K/W The rate of heat transfer is q T T 38 o C 7500 W R tot R g R L ( )K/W COMMENTS Note that despite the much lower heat transfer coefficient on the gas side, the thermal resistance is no larger than on the liquid side. This is the result of balancing the fin geometries which is a desirable situation from the thermal design perspective. Adding fins on the liquid side would not increase the rate of heat transfer appreciably.

112 PROBLEM.47 GIVEN The top of a 30 cm I-beam is maintained at a temperature of 60 C, while the bottom is at 93 C. The thickness of the web is.5 cm. Air at 60 C is blowing along the side of the beam so that h 40 W/(m K). The thermal conductivity of the steel may be assumed constant and equal to 43 W/(m K). Find the temperature distribution along the web from top to bottom and plot the results. A steel in. I-beam Temperature of the top (T L) 60 C Temperature of the bottom (T s) 93 C Thickness of the web (t).5 cm Air temperature (T ) 600 C Heat transfer coefficient ( h c ) 40 W/(m K) Thermal conductivity of the steel (k) 43 W/(m K) FIND The temperature distribution along the web and the plot the results ASSUMPTIONS The thermal conductivity of the steel is uniform The beam has reached steady state conditions One dimensional through the web The beam is very long compared to the web thickness SKETCH SOLUTION The web of the I beam can be thought of as a fin with a uniform rectangular cross section and a fixed tip temperature. From Table., the temperature distribution along the web is sinh [m (L x)] L sinh (m x ) s s sinh (ml) where T T

113 m h c P ka h c (w k wt t) h kt c 40 W/(m K) 43 W /( m K) 0.05m. m ml & sinh (m L) 9.54 s T s T 93 0 C 60 C 67 C L T L T 0 Substitute these into the temperature distribution T ( x) s T 0.05 sinh [3.666 ( x)] T (x) 60 0 C sinh [3.66 ( x)] This temperature distribution is plotted below COMMENTS In a real situation, the heat transfer coefficient is likely to vary with distance and this would require a numerical solution.

114 PROBLEM.48 GIVEN The handle of a ladle used for pouring molten lead is 30-cm-long. Originally the handle was made of.9.5 cm mild steel bar stock. To reduce the grip temperature, it is proposed to form the handle of tubing 0.5-cm-thick to the same rectangular shape. If the average heat transfer coefficient over the handle surface is 4 W/(m K), estimate the reduction of the temperature at the grip in air at C. A steel handle of a ladle used for pouring molten lead Handle length (L) 30 cm 0.3 m Original handle:.9 by.5 cm mild steel bar stock New handle: tubing 0.5 cm thick with the same shape The average heat transfer coefficient ( h c ) 4 W/(m K) Air temperature (T ) C FIND The reduction of the temperature at the grip ASSUMPTIONS The lead is at the melting temperature The handle is made of % carbon steel The ladle is normally in steady state during use The variation of the thermal conductivity is negligible One dimensional conduction Heat transfer from the end of the handle can be neglected SKETCH PROPERTIES From Appendix, Tables 0 and Thermal conductivity of % carbon steel 43 W/(m K) at 0 C Melting temperature of lead (T s) 60 K 38 C SOLUTION The ladle handle can be treated as a fin with an adiabatic end as shown below The temperature distribution in the handle, from Table. is s cosh [m( L x)] cosh (ml) where T(x) T s T s T 38 C C 307 C

115 where h m c P ka P w + t (0.09 m) + (0.05 m) m The only difference in the two handles is the cross-sectional area Solid handle A s wt (0.09 m) (0.05 m) m m L 0.3 m 4 W/(m K)(0.063 m) 43 W/(m K)( m ).788 L s cosh (0) cosh (.788) 0.66 L TL T 0.6 s Hollow handle T L T s C (307 C) 60 C A H wt [w (0.005 m)] [t (0.005 m)] (0.09 m) (0.05 m) (0.06) ( m) m m L 0.3 m 4 W/(m K)(0.063 m) 43 W/(m K)( m ) 4.65 L s cosh (0) cosh (4.647) 0.09 T L T s C (307 C) 7 C The temperature of the grip is reduced 33 C by using the hollow handle. COMMENTS The temperature of the hollow handle would be comfortable to the bare hand. Therefore, no insulation is required. This will reduce the cost of the item without reducing utility.

116 PROBLEM.49 A 0.3-cm-thick aluminum plate has rectangular fins cm, on one side, spaced 0.6 cm apart. The finned side is in contact with low pressure air at 38 C and the average heat transfer coefficient is 8.4 W/(m K). On the unfinned side water flows at 93 C and the heat transfer coefficient is 84 W/(m K). (a) Calculate the efficiency of the fins (b) calculate the rate of heat transfer per unit area of wall and (c) comment on the design if the water and air were interchanged. GIVEN Aluminum plate with rectangular fins on one side Plate thickness (D) 0.3 cm m Fin dimensions (t L) m m Fin spacing (s) m apart Finned side Air temperature (T a) 38 C Heat transfer coefficient ( h a ) 8.4 W/(m K) Unfinned side Water temperature (T w) 93 C Heat transfer coefficient ( h w ) 84 W/(m K) FIND (a) The fin efficiency ( f) (b) Rate of heat transfer per unit wall area (q/a w) (c) Comment on the design if the water and air were interchanged ASSUMPTIONS Width of fins is much longer than their thickness The system has reached steady state The thermal conductivity of the aluminum is constant SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of aluminum (k) 38 W/(m K) at 65 C SOLUTION (a) The fin efficiency is defined as the actual heat transfer rate divided by the rate of heat transfer if the entire fin were at the wall temperature. Since the fin is of uniform cross section, Table. can be used to find an expression for the heat transfer from a fin with a convection from the tip

117 q f M sinh (ml) cosh (ml) h a h (ml) cos mk (h a mk) sinh (ml) where m h a P h a w h ka k (wt) k t M h a PkA s w h a t k s where s T sa T a If the entire fin were at the wall temperature (T sa) the rate of heat transfer would be q f h a A f (T sa T a ) h a w(l + t) (T sa T a ) a The fin efficiency is f q q f f sinh (ml) M cosh (ml) h a w (L h a mk h a mk cosh (ml) sinh (ml) t ) (T sa T a ) h a m kt 8.4 W/(m K) 38 W/(m K)(0.006 m). m m L. m (0.006 m) M w (T sa T a) 8.4 W/(m K) (0.006 m) 38 W/(m K) 4.65 w (T sa T w )s W/(mK) h a 8.4 W/(m K) mk f. m 4.65W/(m K) 38W/(m K) sinh (0.0733) cosh (0.0733) cosh (0.0733) sinh (0.0733) 8.4 W/(m K) [()0.006 m m] (b) The heat transfer to the air is equal to the sum of heat transfer from the fins and the heat transfer from the wall area not covered by fins. The number of fins per meter height is m (0.076 m/fin) 3.6 fins The wall area not covered by fins per m of total wall area is A bare m (3. 6 fins) m/fin ( m width) m The surface area of the fins per m of wall area is The rate of heat transfer to the air is A fins 3.6 fins ((0.006 m) m) ( m width).79 m

118 q a h a A bare (T sa T a) + h a f A fins (T sa T a) q a h a (A bare + f A fins) (T sa T a) Therefore, the resistance to heat transfer on the air side (R a) is T sa T a R ca

119 R ca h a ( A bare f A fins ) The thermal circuit for the wall is shown below h a A total The individual resistance based on m of wall area are R cw K/W h w A w 38.7 W/(m K)(m ) R k D k A w m 38.7 W/m K(m ) K/W R ca m f A fins ) (0.998)( h a ( A bare 8.4 W/(m K) [0.789 m.79 m ) K/W The rate of heat transfer through the wall is T q Rtot T T a 38 o w 93 o C C R cw R k R ca 307 W (per m of wall) ( )K/ W (c) Note that the air side convective resistance is by far the dominant resistance in the problem. Therefore, the fins will enhance the overall heat transfer much less on the water side. For fins on the water side m 83.7 W/(m K) 38 W/(m K) (0.006 m) 38.6 m and m L 38.6 m (0.006 m) 0.36 M w (T sw T w) 83.7 W/(m K) (0.006 m) 38 W/(m K) 4.70 w (T sw T w )W/m K f h 83.7 W/(m K) w mk 38.6 m 4.70 W/(m K) 38 W/(m K) sinh (0.36) c osh (0.36) cosh (0.36) sinh (0.36) 83.7 W/(m K) [ (0.006 m) m 38 o C q T w T a D 93 o C ( ) (m K)/W h ca k h cw ( )

120 COMMENTS 50 W/m The fins are most effective in the medium with the lowest heat transfer coefficient. With no fins, the rate of heat transfer would be 49 W/m. Fins on the water side increase the rate of heat transfer 6%. Fins on the air side increase the rate of heat transfer 6%. Therefore, installing fins on the water side would be a poor design.

121 MULTIDIMENSIONAL STEADY STATE CONDUCTION PROBLEM.50 GIVEN Compare the rate of heat flow from the bottom to the top in the aluminum structure shown in the sketch below with the rate of heat flow through a solid slab. The top is at 0 C, the bottom at 0 C. The holes are filled with insulation which does not conduct heat appreciably. The aluminum structure shown in the sketch below Temperature of the top (T T) 0 C Temperature of the bottom (T B) 0 C The holes are filled with insulation which does not conduct heat appreciably FIND Compare the rate of heat flow from the bottom to the top with the rate of heat flow through a solid slab ASSUMPTIONS The structure is in steady state Heat transfer through the insulation is negligible The thermal conductivity of the aluminum is uniform The edges of the structure are insulated Two dimensional conduction through the structure SKETCH PROPERTIES AND CONSTANTS The thermal conductivity of aluminum (k) 36 W/(m K) at 0 C SOLUTION Because of the symmetry of the structure, we can draw the flux plot for just one of the twenty-four equivalent sections

122 (a) The total number of flow lanes in the structure, (M) () (4) (48). Each flow lane consists of curvilinear squares (6 on top as shown, and 6 on bottom. Therefore, the shape factor is 48 S M 4 N The heat flow per meter, from Equation (.85), is The total rate of heat flow is q ks T overall 36 W/m K (4) (0 C ( 0 C)) 9440 W/m q o q (length of structure) 9440 W/m (3 m) 8,30 W (b) For a solid aluminum plate, the total heat flow from Equation (.3), is q TOT Ak t T (3 m) (0.3 m) 36 W/(m K) 0.05 (0 C) 4,500 W Therefore, the insulation filled tubes reduce the heat transfer rate by 33%. COMMENTS The shape factor was determined graphically and can easily be in error by 0%. Also, the surface temperature will not be uniform in the insulated structure.

123 PROBLEM.5 GIVEN FIND Determine by means of a flux plot the temperatures and heat flow per unit depth in the ribbed insulation shown in the accompanying sketch. The sketch below (a) The temperatures (b) The heat flow per unit depth ASSUMPTIONS Steady state conditions Two dimensional heat flow The heat loss through the insulation is negligible The thermal conductivity of the material is uniform SKETCH SOLUTION The total number of heat flow lanes (M) The number of curvilinear squares per lane (N) 8 Therefore, the shape factor is S M.38 N 8 The rate of heat transfer for unit depth is given by Equation.85 q ks T (0.5 W/(m K)) (.38) (00 C 30 C) 48.3 W/m

124 PROBLEM.5 GIVEN Use a flux plot to estimate the rate of heat flow through the object shown in the sketch. The thermal conductivity of the material is 5 W/(m K). Assume no heat is lost from the sides. The shape of object shown in the sketch The thermal conductivity of the material (k) 5 W/(m K) The temperatures at the upper and lower surfaces (30 C & 0 C) FIND The rate of heat flow through the object (By means of a flux plot) ASSUMPTIONS No heat is lost from the sides and ends Uniform thermal conductivity Two dimensional conduction Steady state SKETCH SOLUTION The flux plot is shown below The number of heat flow lanes (M) 0 0 The number of curvilinear squares in each lane (N) Therefore, the shape factor for this object is S M N 0.67 The rate of heat transfer per unit length from Equation (.85) is q ks The total rate of heat transfer is T overall [5 W/(m K)] (.67) (0 C) 500 W/m q tot ql (500 W/m) (0 m) 0,000 W

125 PROBLEM.53 GIVEN Determine the rate of heat transfer per unit length from a 5-cm-OD pipe at 50 C placed eccentrically within a larger cylinder of 85% Magnesia wool as shown in the sketch. The outside diameter of the larger cylinder is 5 cm and the surface temperature is 50 C. A pipe placed eccentrically within a larger cylinder of 85% Magnesia wool as shown in the sketch Outside diameter of the pipe (D p) 5 cm 0.05 m Temperature of the pipe (T s) 50 C Outside diameter of the larger cylinder (D o) 5 cm 0.5 m Temperature of outer pipe (T o) 50 C FIND The rate of heat transfer per meter length (q) ASSUMPTIONS Two dimensional heat flow (no end effects) The system is in steady state Uniform thermal conductivity SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of 85% Magnesia wool (k) W/(m K) (at 0 C). SOLUTION The rate of heat transfer can be estimated from a flux plot The number of flow lanes (M) 5 30 The number of squares per lane (N) 5

126 Therefore, the shape factor is 30 S M 6 N 5 Equation (.85) can be used to find the rate of heat transfer per unit length q ks T ks(t s T o) [0.059 W/(m K)] (6) (50 C 50 C) 35.4 W/m COMMENTS This problem can also be solved analytically (see Table.) S cosh D d 4 z Dd 6.53 (z the distance between the centers of the circular cross sections) q ks T 38.5 W/m The answer from the graphical solution is 8% less than the analytical value.

127 PROBLEM.54 GIVEN Determine the rate of heat flow per foot length from the inner to the outer surface of the molded insulation in the accompanying sketch. Use k 0.7 W/( m K). The object with a cross section as shown in the sketch below The thermal conductivity (k) 0.7 W/(m K) FIND The rate of heat flow per foot length from the inner to the outer surface (q) ASSUMPTIONS The system has reached steady state The thermal conductivity does not vary with temperature Two dimensional conduction SKETCH SOLUTION A flux plot for the object is shown below The number of heat flow lanes (M) 8 6 The number of curvilinear squares per lane (N) 4 Therefore, the shape factor is S The heat flow per unit length, from Equation (.85) is COMMENTS q ks T overall [0.7 W/(m K)] (4) (94 C) 3 W/m The problem can also be solved analytically. From Table. S q ks π ln.08 W/D T 34.5 W/m π ln The analytical solution yields a rate of heat flow that is about % larger than the value obtained from the flux plot.

128 PROBLEM.55 GIVEN A long -cm-diameter electric copper cable is embedded in the center of a 5 cm square concrete block. If the outside temperature of the concrete is 5 C and the rate of electrical energy dissipation in the cable is 50 W per meter length, determine the temperatures at the outer surface and at the center of the cable. A long electric copper cable embedded in the center of a square concrete block Diameter of the pipe (D p) cm 0.0 m Length of a side of the block 5 cm 0.5 m The outside temperature of the concrete (T o) 5 C The rate of electrical energy dissipation ( Q G / L ) 50 W/m FIND The temperatures at the outer surface (T s) and at the center of the cable (T c) ASSUMPTIONS Two dimensional, steady state heat transfer Uniform thermal conductivities SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of concrete (k b) 0.8 W/(m K) at 0 C From Appendix, Table The thermal conductivity of copper (k c) 396 W/(m K) at 63 C SOLUTION For steady state, the rate of heat transfer through the concrete block must equal the rate of electrical energy dissipation. The heat transfer rate can be estimated with a flux plot of one quarter of the block: The number of flow lanes (M) The number of squares per lane (N) 0

129 Therefore, the shape factor is S M N The rate of heat flow per unit length is given by Equation (.85) q k b S T k b S(T s T o ) Solving for the surface temperature of the cable Q G L T s T o + QL G k b S 5 C + 50W/m [0.8W/(m K)](.4) 53 C From Equation (.56) the temperature in the center of the cable is q G r T 0 c T s + 4kC Where q G heat generation per unit volume q G r 0 L Q G L T C T s + 53 C + 4 k C 50W/m 4 396W/( m K) 53 C C 53 C COMMENT The thermal conductivity of the cable is quite large and therefore its temperature is essentially uniform. The analytical solution for this geometry, given in Table., is π π S 5 cm In 0.8 W.9 In.08 D cm This would lead to a cable temperature of 639 C, 0% higher than the flux plot estimate. The high error is probably due to the difficulty in drawing the flux plot close to the cable and may be improved by drawing a larger scale flux plot is geometries that involve tight curves.

130 PROBLEM.56 GIVEN A large number of 3.8 cm OD pipes carrying hot and cold liquids are embedded in concrete in an equilateral staggered arrangement with center line. cm apart as shown in the sketch. If the pipes in rows A and C are at 6 C while the pipes in rows B and D are at 66 C, determine the rate of heat transfer per foot length from pipe X in row B. A large number of pipes embedded in concrete as shown below Outside diameter of pipes (D) 3.8 cm The temperature of the pipes in rows A and C 6 C The temperature of the pipes in rows B and D 66 C FIND The rate of hat transfer per foot length from pipe X in row B ASSUMPTIONS Steady state, two dimensional heat transfer Uniform thermal conductivity in the concrete SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of concrete (k b) 0.8 W/(m K) at 0 C SOLUTION A flux diagram for this problem is shown below

131 By symmetry, the total heat transfer from the tube X is four times that shown in the flux diagram. The number of heat flow lanes (M) The number of curvilinear squares per lane (N) 7 Therefore, the shape factor is S M N The heat transfer per unit length from Table., from Equation (.85) is q KS T overall [0.8 W/(m K)] (4.6) (66 0 C 6 C) 9.44 W/m

132 PROBLEM.57 GIVEN A long -cm-diameter electric cable is imbedded in a concrete wall (k 0.3 W/(m K)) which is m* m, as shown in the sketch. If the lower surface is insulated, the surface of the cable is 00 C and the exposed surface of the concrete is 5 C, estimate the rate of energy dissipation per meter of cable. A long electric cable imbedded in a concrete wall with cable diameter (D) cm 0.0 m Thermal conductivity of the wall (k) 0.3 W/(m K) Wall dimensions are m by m, as shown in the sketch below The lower surface is insulated The surface temperature of the cable (T s) 00 C The temperature of the exposed concrete surfaces (T o) 5 C FIND The rate of energy dissipation per meter of cable (q/l) ASSUMPTIONS The system is in steady state The thermal conductivity of the wall is uniform Two dimensional heat transfer SKETCH SOLUTION By symmetry, only half of the flux plot needs to be drawn The number of heat flow lanes (M) 4 8 The number of curvilinear squares per lane (N) 6 Therefore, the shape factor is S M N For steady state, the rate of energy dissipation per unit length in the cable must equal the rate of heat transfer per unit length from the cable which, from Equation (.85), is q ks(t s T o) (0.3 W/(m K) (4.7)) (00 C 5 C) 46 W/m

133 PROBLEM.5

134 PROBLEM.53 GIVEN 5 Determine the temperature distribution and heat flow rate per meter length in a long concrete block having the shape shown below. The cross-sectional area of the block is square and the hole is centered. A long concrete block having the shape shown below The cross-sectional area of the block is square The hole is centered FIND (a) The temperature distribution in the block (b) The heat flow rate per meter length ASSUMPTIONS The heat flow is two dimensional and in steady state The thermal conductivity in the block is uniform SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of concrete (k b) 0.8 W/(m K) at 0 C SOLUTION The temperature distribution and heat flow rate may be estimated with a flux plot

135 (a) The temperature distribution is given by the isotherms in the flux plot. (b) The number of flow lanes (M) 4 The number of squares per lane (N) 4 Therefore, the shape factor is S M N From Equation (.85), the rate of heat flow per unit length is COMMENTS q ks T [0.8 W/(m K)] (0.5) (40 C) 54 W/m If the lower surface were not insulated, the shape factor from Table., would be S 4.8 q 75.6 W/m W In.08 D The rate of heat transfer with the insulation as calculated with the flux plot is about 9% less than the analytical result without insulation. We would expect a reduction of slightly less than 5%.

136 PROBLEM.59 GIVEN A 30-cm-OD pipe with a surface temperature of 90 C carries steam over a distance of 00 m. The pipe is buried with its center line at a depth of m, the ground surface is 6 C, and the mean thermal conductivity of the soil is 0.7 W/(m K). Calculate the heat loss per day, and the cost of this loss if steam heat is worth $3.00 per 0 6 kj. Also, estimate the thickness of 85% magnesia insulation necessary to achieve the same insulation as provided by the soil with a total heat transfer coefficient of 3 W/(m K) on the outside of the pipe. A buried steam pipe Outside diameter of the pipe (D) 30 cm 0.3 m Surface temperature (T s) 90 C Length of pipe (L) 00 m Depth of its center line (Z) m The ground surface temperature (T g) 6 C The mean thermal conductivity of the soil (k) 0.7 W/(m K) Steam heat is worth $3.00 per 0 6 kj The heat transfer coefficient (h c) 3 W/(m K) for the insulated pipe FIND (a) The heat loss per 4 hour day (b) The value of the lost heat (c) The thickness of 85% magnesia insulation necessary to achieve the same insulation ASSUMPTIONS Steady state conditions Uniform thermal conductivity Two dimensional heat transfer from the pipe SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of 85% magnesia (k i) W/(m K) (at 0 C) SOLUTION (a) The shape factor for this problem, from Table., is

137 S L z cosh D If z/l < Note that the condition Z/L << is satisfied in this problem. (00 m) S 43 m (m) cosh 0.3 m From Equation (.85), the rate of heat transfer is q ks T 0.7 W/(m K) (43 m) (90 C ( 6 C)) ki 3600 s 4 h q 6,300 W (J/Ws) kj/day 000 J h day (b) The cost of this heat loss is Cost kj/day $ kj $4.3/day (c) The thermal circuit for the pipe covered with insulation is The rate of heat loss from the pipe is T s T g T s T g q R ki R c ln r o Lk i r Lr o h c 6,300 W 6,300 W L ( T s T g ) r ln o L(00 m )[90 C ( 6 )] ln r o 0.5 m r 3 o k i r i r o h c 0.059W/(m K) W/(m K) ln r o r o By trial and error: r o 0.84 m Insulation thickness r o r i 0.84 m 0.5 m m 3.4 cm COMMENTS The value of the heat loss per year is 365 $4.3 $544. Hence insulation will pay for itself quite rapidly.

138 PROBLEM.60 GIVEN Two long pipes, one having a 0-cm-OD and a surface temperature of 300 C, the other having a 5-cm-OD and a surface temperature of 00 C, are buried deeply in dry sand with their centerlines 5 cm apart. Determine the rate of heat flow from the larger to the smaller pipe per meter length. Two long pipes buried deeply in dry sand Pipe Diameter (D ) 0 cm 0. m, Surface temperature (T ) 300 C Pipe Diameter (D ) 5 cm 0.05 m, Surface temperature (T ) 00 C Spacing between their centerlines (s) 5 cm 0.5 m FIND The rate of heat flow per meter length (q/l) ASSUMPTIONS The heat flow between the pipes is two dimensional The system has reached steady state The thermal conductivity of the sand is uniform SKETCH PROPERTIES AND CONSTANTS From Appendix, Table Thermal conductivity of dry sand (k) 0.58 W/(m K) at 0 C SOLUTION The shape factor for this geometry, from Table., is S L r cosh r where r r 5 cm and L 5 cm 6 r.5 cm r.5 cm S π cosh The rate of heat transfer per unit length is q Sk T (.96) 0.58 W/(m K) (300 C 00 C) 67 W/m

139 PROBLEM.6 GIVEN A radioactive sample is to be stored in a protective box with 4-cm-thick walls having interior dimensions 4 cm* 4 cm* cm. The radiation emitted by the sample is completely absorbed at the inner surface of the box, which is made of concrete. If the outside temperature of the box is 5 C, but the inside temperature is not to exceed 50 C, determine the maximum permissible radiation rate from the sample, in watts. A radioactive sample in a protective concrete box Wall thickness (t) 4 cm 0.4 m Box interior dimensions: 4 4 cm All radiation emitted is completely absorbed at the inner surface of the box The outside temperature of the box (T o) 5 C The maximum inside temperature (T i) 50 C FIND The maximum permissible radiation rate from the sample, q (in watts) ASSUMPTIONS The system is in steady state SKETCH PROPERTIES AND CONSTANTS From Appendix, Table The thermal conductivity of concrete (k b) 0.8 W/(m K) at 0 C SOLUTION The box consists of 4 wall sections: A 4 cm cm wall sections: A 4 cm 4 cm 4 edge sections: D cm long 8 edge sections: D 4 cm long 8 corner sections: L 4 cm thick The shape factors for this geometry (when all interior dimensions are greater than one-fifth of the wall thickness, as in this case) is given on Section.5. of the text For the wall sections S A L (4 cm) ( cm) 4 cm m and S A (4 cm) (4 cm) 4 cm L 4 cm For the edge sections

140 S D 0.54 ( cm) 6.48 cm For the corner sections S L 0.5 (4 cm) 0.6 cm and S D 0.54 (4 cm).6 cm Multiplying each shape factor by the number of elements having that shape factor and summing them S 4 S + S + 4 S S S 5 S 4 ( cm) + (4 cm) + 4(6.48 cm) + 8(.6 cm) + 8(0.6 cm) 04 cm The rate of heat transfer is COMMENTS q ks T 0.8 W/(m K) (04 cm) (m/00 cm) (50 C 5 C) 3.3 W The conductivity of the concrete was evaluated at 0 C while the actual temperature is between 50 C and 5 C. Therefore, the actual rate of heat flow may be slightly different than that calculated, but no better property value is available in the text.

141 PROBLEM.6 GIVEN A 5 cm-od pipe is buried with its centerline.5 m below the surface of the ground [k of soil is 0.35 W/(m K)]. An oil having a density of 800 kg/m 3 and a specific heat of. kj/(kg K) flows in the pipe at 5.6 L/s. Assuming a ground surface temperature of 5 C and a pipe wall temperature of 95 C, estimate the length of pipe in which the oil temperature decreases by 5.5 C. An oil filled pipe buried below the surface of the ground Pipe outside diameter (D) 5 cm 0.5 m Depth of centerline (z).5 m Thermal conductivity of the soil (k) 0.35 W/(m K) Specific gravity of oil (Sp. Gr.) 0.8 Specific heat of oil (c p). kj/(kg K) Flows rate of oil m 5.6 L/s m 3 /s The ground surface temperature (T s) 5 C The pipe wall temperature (T p) 95 C FIND The length of pipe (L) in which the oil temperature decreases by 5.5 C ASSUMPTIONS Steady state condition Two dimensional heat transfer SKETCH SOLUTION The rate of heat flow from the pipe can be calculated using the shape factor from Table. for an infinitely long cylinder S cosh Z D π cosh (.5) The rate of heat transfer per unit length is given by Equation (.85) q ks T overall (0.35 W/(m K)) (.79) (95 0 C 5 0 C) 56.4 W/m

142 The total heat loss required to decrease the oil by 5.5 C is q t m c p T [m 3 /s] (00 J/(kg K)) (5.5 C)*800 (kg/m 3 ) 5744 W We can estimate the length of pipe in which the oil temperature drops 0 F by assuming the rate of heat loss from the pipe per unit length is constant, then: COMMENTS q t ql L q t q m The heat loss from the pipe will actually be less because as the oil temperature and therefore also the pipe temperature decreases with distance from the inlet. This means the length will be slightly longer than the estimate above. If the calculation is based on an arithmetic mean pipe temperature of 50 C, the estimated length is 954 m about 4% more.

143 PROBLEM.63 A.5-cm-OD hot steam line at 00 C runs parallel to a 5.0-cm-OD cold water line at 5 C. The pipes are 5 cm center to center and deeply buried in concrete with a thermal conductivity of 0.87 W/(m K). What is the heat transfer per meter of pipe between the two pipes? GIVEN Hot pipe outside diameter (D h).5 cm 0.05 m Hot pipe temperature (T h) 00 C Cold pipe outside diameter (D c) 5.0 cm 0.05 m Cold pipe temperature (T c) 5 C Center to center distance between pipes (l) 5 cm 0.05 m Thermal conductivity of concrete (k) 0.87 W/(m K) FIND The heat transfer per meter of pipe (q/l) ASSUMPTIONS Two dimensional heat transfer between the pipes Steady state conditions Uniform thermal conductivity SKETCH PROPERTIES AND CONSTANTS Specific heat of water (c p) 487 J/(kg K) SOLUTION The shape factor for this geometry is in Table. Where L S L r cosh D h S 4 r 0.05 m cosh 0.05 m 4 and r r c D c 0.05 cosh r h D h 0.05 The rate of heat transfer per unit length, from Equation (.85), is COMMENTS q ks T overall 0.87 W/(m K) (3.763) (00 C 5 C) 78 W/m Normally, the temperature of both fluids will change as heat is transferred between them. Hence, for any appreciable length of pipe, an average temperature difference must be used.