ME 33 Homework Assignment #6 Problem Statement: ater at 30 o C flows through a long.85 cm diameter tube at a mass flow rate of 0.020 kg/s. Find: The mean velocity (u m ), maximum velocity (u MAX ), and the pressure difference (Δp) between two sections Δx = 50 m apart in the tube. Schematic: T = 303.5 K D =.85 cm m = 0.020 kg s Assumptions: L = 50 m Analysis:. Incompressible fully developed flow. 2. Constant properties for water at 30 o C = 303.5K. Interpolating from Table A.6, ρ = 995.8 kg/m 3, μ = 8.0 0 4 N s/m 2 From the mass flow rate, density, and cross sectional area, the mean velocity is found using eq. 8.5: u m = m ρa c = kg s 4 (0.085 m)2 995.8 kg m π = 0.020 4 D2 π ρ m 3 = 0.07472 m s u m = 7.5 0 2 m s Determine whether the flow is laminar or turbulent. Re D = ρu md μ = kg 995.8 m 3 0.07472m 0.085 m s =,78.8 8.0 0 4 N s/m 2 The Reynolds number is less than the critical number of 2300; thus the flow is laminar. The velocity profile is given by equation 8.5:
u(r) = 2 u m [ ( r 2 ) ] r o The maximum velocity at r = 0 is: u MAX = u(r = 0) = 2 u m = 0.494 m/s u MAX = 0.5 m s The pressure gradient is related to the mean velocity by equation 8.4: u m = 8μ (dp dx ) r o 2 For fully developed laminar flow, the pressure gradient is constant: dp dx = 8μu 8 8.0 0 4 s N m 2 = m 2 0.07472m s r o (0.00925 m) 2 = 5.5948 Pa/m The pressure drop over 50 m length of pipe is: Δp = 840 Pa Δp = dp Pa L = 5.5948 50 m = 839.294 Pa dx m
Problem 2 Statement: You want to heat water from 2 o C to 70 o C as it flows through a 2.0 cm internal diameter, 7.0 m long tube. The tube is equipped with an electric resistance heater, which provides uniform heating at the outer surface of the tube. The heater is well insulated, so that in steady operation all the thermal energy generated in the heater is transferred to the water in the tube. You want the system to provide hot water at a rate of 8.0 L/min. Find: The required power rating of the resistance heater, and the inner surface temperature at the outlet (T s,o ). Schematic: q" = constant T m,i = 2 C D = 2 cm T m,o = 70 C V = 8.0 L/min Assumptions: L = 7 m. Steady flow conditions. 2. Uniform surface heat flux. 3. Inner surface of the tube is smooth. 4. Properties evaluated at the mean fluid temperature of (70+2)/2 = 4 o C: ρ = 992. kg/m 3 k = 0.63 /mk ν = μ ρ = 0.658 0 6 m 2 /s c = 479 J/kgK Pr = 4.32 Analysis: The mass flow rate through the tube and the required power rating of the resistance heater are: m = ρv = (992. kg/m 3 )(0.008 m 3 /min) = 7.937 kg/min = 0.32 kg/s q = m c (T m,o T m,i ) = (0.32 kg s J ) (479 ) (70 2)K = 32.06 k kgk q = 32 k
To determine the inner surface temperature at the tube outlet we first determine whether the flow is laminar or turbulent. u m = V = ( 8 60 0 3 ) m 3 s A π(0.02 m) 2 c 4 = 0.4244 m s Re = u md ν = (0.4244m )(0.02 m) s = 2,890 0.658 0 6 m 2 /s This value is greater than the critical Reynolds number of 2300 for pipe flow. Therefore, the flow is turbulent and the entry length is estimated at 0 pipe diameters: x fd,t 0D = 0(0.02 m) = 0.2 m Since the entry length is much shorter than the total tube length, the thermal and hydrodynamic profiles of the turbulent flow are assumed to be fully developed at L = 7 m. The local Nusselt number is determined from: Nu = hd k = 0.023Re0.8 Pr 0.3 = 0.023(2890) 0.8 (4.32) 0.4 = 80.2 The local heat transfer coefficient is: h = k D 0.63 Nu = mk 0.02 m (80.2) = 2530 /m2 K Using q = ha(t s,o T m,o ), the inner surface temperature of the tube at the outlet is: T s,o = q ha + T m,o = 2530 32,060 π(0.02 m)(7 m) + 70 C = 98.8 C T s,o = 99 C
Problem 3 Statement: A typical automotive exhaust pipe carries exhaust gas at a mean velocity of 6.0 m/s. If the gas leaves the manifold at a temperature of 450 C and the muffler is located 2.6 m away, estimate the temperature of the gas as it enters the muffler. Assume an ambient air temperature of 40 C and a heat transfer coefficient of 2 / for the outside surface of the exhaust pipe. The exhaust pipe is thin walled and has a diameter of 0 cm. The exhaust pipe inner roughness Is ε = 0.020 mm. Find: The approximate temperature of the exhaust gas as it enters the muffler (T m,o ). Schematic: Assumptions:. Incompressible, fully developed flow. 2. Constant properties of air at 723 K are assumed for the exhaust gas. ρ = 0.482 kg/m 3 c p =.08 0 3 J/kgK Pr = 0.698 μ = 3.46 0 5 N s/m 2 ν = 7.93 0 5 m 2 /s k = 5.36 0 2 /mk Analysis: Re D = u md ν = 6.0 m 0. m s 7.93 0 5 m 2 /s = 8.34 03 > 2300 The flow is turbulent, but possibly transitional (not fully turbulent) since 2300 < Re D < 0,000. ε D = 0.020 0 3 m = 0.0002 0.m From Figure 8.3, the friction factor is: f = 0.033
For fully developed turbulent flow in a smooth tube, the local Nusselt number and local heat transfer coefficient are: ( f 8 ) (Re D 000)Pr Nu D = + 2.7 ( f 8 ) 2 (Pr 2/3 ) = 25.586 h i = Nu D k D = 3.74 m 2 K However, since Re D < 0,000 and since L = 26 < 60, the calculated h D i could be either an overestimate or an underestimate of h i. Lacking any better estimate, use D Nu Nu D, h i h i. R tot = R conv,i + R conv,o = h i PL + h o PL = π 0.m 2.6m ( 3.74 + 2 ) m2 K = 0.929 K Then, from equation 8.45b, T T m,o = exp ( ) T T m,i m c p R tot T m,o = T (T T m,i ) exp ( ) = 37.4 C m c p R tot T m,o = 370 C Comment: q = ΔT lm R tot = 900 The negative sign indicates that heat flows from gas in the exhaust pipe to the ambient air environment.
Problem 4 Statement: A blood purification system removes blood from a patient through a 3.0 mm diameter tube at a mass flow rate of 0.090 kg/min. The blood exits the patient at 37 C and cools to 33 C just prior to entering a filter and heat exchanger that raises the blood temperature back to 37 C. Approximate the properties of blood with those of water. Find: Estimate the rate of heat transfer from the blood and the heat transfer coefficient, assuming fully developed blood flow in the tube. Schematic: Assumptions:. Incompressible, fully developed flow. 2. Constant properties of water at 35 C = 308.5 K are assumed for blood. ρ = 993.8 kg/m 3 k = 0.625 /mk μ = 7.22 0 4 N s / m 2 c = 478 J/kgK Analysis: 3. Constant surface heat flux or constant surface temperature may be assumed. q = m c (T m,o T m,i ) = 0.090 kg 60 s 478 J (33 37)K = 25.07 kgk q = 25 The negative sign indicates that the blood is transfering heat to its surrounding environment. The Reynolds number is: Re D = u md ν = m D ρa c ν = 88.7 < Re cr
For fully developed laminar flow, constant surface heat flux, Nu D = hd k = 4.36 h = 4.36 k D = 908.3 m 2 K h = 90 m 2 K For fully developed laminar flow, constant surface temperature, Nu D = hd = 3.66 k h = 3.66 k D = 762.5 m 2 K h = 760 m 2 K Comments: This process of blood cooling is unlikely to be either constant heat flux or surface temperature. It is also unlikely for it to be fully developed. Therefore, these answers are first order estimates. If more information is given about the outer convection condition, this problem could be solved more accurately using a similar method as in problem 3.
Problem 5 Statement: The condenser of a steam power plant contains N = 000 brass tubes (k t = 0 /m-k), each of inner and outer diameter, D i = 25 mm and D o = 28 mm. Steam condensation on the outer surfaces of the tubes is characterized by a convection coefficient of h o = 0,000 /m 2 -K. Find: (a) If cooling water from a nearby lake is pumped through the condenser tubes at a mass flow rate of ṁ c = 400 kg/s, what is the overall heat transfer coefficient U o based on the tube outer surface area? ater properties may be approximated as constant: μ = 9.60 x 0-4 N-s/m 2, k = 0.60 /m-k, and Pr = 6.6. Schematic: (b) If after extended operation, fouling results in an additional resistance of R f,i = 0-4 m 2 -K/, at the inner surface of the tubes, what is the value of U o? (c) If water is extracted from the lake at 5 C, and 0 kg/s of steam at 0.0622 bar are to be condensed, what is the temperature of the cooling (lake) water leaving the condenser? The specific heat of the water is 480 J/kg-K. Assumptions:. ater is incompressible, and viscous dissipation is negligible. 2. Fully developed flow in tubes. 3. Negligible fouling on outer tube surfaces, D o. 4. ater properties inside tubes are given as follows: c = 480 J/kg K, μ = 9.6 0-4 N s/m 2, k = 0.60 /m K, Pr = 6.6. 5. From Table A-6, properties outside tubes are for saturated vapor / liquid water at p = 0.0622 bar: Tsat = 30 K, h fg = 2.44 0 6 J/kg.
Analysis: (a) ithout fouling resistance, Eq..5 yields: = ( D D o ln ( D o) o D ) + i U o h i D i 2k t + h o Re Di = 4m = 2,220; thus the flow in the tubes is turbulent. From eq 8.60: πd i μ U o = [ ( D o ) + D Do o ln( ) D i + ] h i D i 2k t h o h i = ( k 4/5 ) 0.023Re D Di Pr 0.4 = 3398 / i = [ 3398 / (28 25 28 0.028m ln( 25 ) + ) + ] 2 0 /mk 0,000 / = 2252 U o = 2250 (b) ith fouling, Eq..5 yields: = ( D D o ln ( D o) o D ) + i U o h i D i 2k t + + ( D o ) R h o D f,i i U o = [4.44 0 4 / + ( D o ) R D f,i ] = 798 i U o = 800 (c) The rate of heat transfer to water inside the tubes equals the rate at which thermal energy is extracted from steam outside of the tubes. T m,o = T m,i + m hh fg m cc = 5 C + 0kg J 2.44 06 s kg 400 kg s m hh fg = m c c (T m,o T m,i ) 480 J kgk = 29.4 C T m,o = 30 C Comments: () The largest contribution to the thermal resistance is from convection at the interior surface of the tubes. To increase U o, h i could be increased by increasing m c. (2) Note that the outlet temperature of the cooling water T m,o = 29.4 C = 302.5 K remains below T sat = 30 K, as required.
Problem 6 Statement: In a Rankine power system,.5 kg/s of steam leaves the turbine as saturated vapor at 0.5 bar. The steam is condensed to saturated liquid by passing it over the tubes of a shell-and-tube heat exchanger, while cooling liquid water, having an inlet temperature of T c,i = 280 K, is passed through the tubes. The condensing heat exchanger contains 00 thin-walled tubes, each 0 mm diameter, and the total cooling water flow rate through the tubes is 5 kg/s. The average convection coefficient associated with condensation on the outer surface of the tubes may be approximated as h o = 5000 /m 2 -K. Appropriate property values for the cooling liquid water are c = 478 J/kg-K, μ = 700 x 0-6 kg/s-m, k = 0.628 /m-k, and Pr = 4.6. Find: (a) hat is the liquid water outlet temperature from the tubes? Schematic: (b) hat is the required tube length (per tube)? (c) After extended use, deposits accumulating on the inner and outer tube surfaces result in a cumulative fouling factor of 0.0003 m 2 -K/. For the prescribed inlet conditions and the computed tube length, what mass fraction of the vapor is condensed? Assumptions:. Negligible heat loss to the surroundings. 2. Negligible wall conduction resistance and fouling (initially). ith fouling, R f = 0.0003 m 2 K/. 3. Cooling liquid water properties treated as constant: c = 478 J/kg K, μ = 700 0-6 kg/s m, k = 0.628 /m K, Pr = 4.6. 4. From Table A.6, for saturated steam at p = 0.5 bar, T sat = 355 K, h fg = 2.304 0 6 J/kg. Analysis: (a) From an overall energy balance, q h = m hh fg = q c = m cc c (T c,o T c,i )
T c,o = T c,i + m.5 hh kg fg s 2.304 J 06 kg = 280 K + m cc c 5 kg = 335. K J 478 s kg T c,o = 335 K (b) ith C r = C min /C max 0, NTU = ln ( ε), where ε = q = m cc c (T c,o T c,i ) 335. 280 = q MAX m cc c (T h,i T c,i ) 355 280 = 0.7347 NTU = ln( 0.7347) =.327 = UA/C min ith negligible tube wall thickness, the overall heat transfer coefficient U is: = +. U h i h o The Reynolds number for internal flow inside a singe tube is: Re D = 4m kg c πdμ = 4 5 s /00 π(0.0) 700 0 6 kg/s m = 27,284 Using the Dittus-Boelter correlation (equation 8.60) for fully developed turbulent pipe flow, Nu D = 0.023Re D 4/5 Pr n = 0.023(27,284) 4/5 (4.6) 0.4 = 49.8 Assuming the tube length is sufficiently long such that the internal flow is fully developed for the majority of the tube length (L > 60 D): h i h i = ( k D ) Nu D = 9,408 ith U = [( ) + ( 9408 5000 )] = 3,265, the heat transfer area is: The tube length is L = A = m cc c ( NTU U ) = 5 kg s A NπD = 25.5m2 00π(0.0m) J (478 kgk ) (.327 3,265 ) = 25.5m 2 = 8. m, satisfying the condition that L > 60 *0.0 m = 0.6 m. L = 8. m (c) ith fouling, the overall heat transfer coefficient is /U wf = /U wof + R f. Hence, U wf = (3.063 0 4 + 3 0 4 )m 2 K/
U wf = 649 /m 2 K NTU = U wf A/C min = (649 /m 2 K 25.5m 2 )/( 5 kg /s 478 J /kg K) = 0.670 From equation.35a, ε = exp( NTU) = exp( 0.670) = 0.488 q = εq max = 0.488 5 kg s 478 J kg K (355 280)K = 2.296 06 ithout fouling, the heat rate was: q = m hh fg =.5 kg s 2.304 J 06 kg = 3.456 06 Hence, m h,wf m h,wof = 2.296 06 3.456 0 6 = 0.664 The reduced condensation rate with fouling is then: m h,wf = 0.664.5 kg s = 0.996 kg s m h,wf =.0 kg s