2-21. for gage pressure, the high and low pressures are expressed as. Noting that 1 psi = kpa,

Transcription

1 - -58E The systolic and diastolic pressures of a healthy person are given in mmhg. These pressures are to be expressed in kpa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be 000 kg/m and,600 kg/m, respectively. Analysis Using the relation P P high low ρgh ρgh high low Noting that psi kpa, P ρgh for gage pressure, the high and low pressures are expressed as N (,600 kg/m (9.8 m/s (0. m kg m/s kpa 000 N/m N kpa (,600 kg/m (9.8 m/s (0.08 m kg m/s 000 N/m 6.0 kpa 0.7 kpa psi psi P high (6.0 Pa. psi kpa and P low (0.7 Pa.55 psi kpa For a given pressure, the relation P ρgh can be expressed for mercury and water as P ρ water ghwater and P ρ mercury gh mercury. Setting these two relations equal to each other and solving for water height gives ρ P ρ Therefore, h h mercury water ghwater ρ mercuryghmercury hwater hmercury ρ water water, high water, low ρ ρ ρ ρ mercury water mercury water h h mercury,high mercury,low,600 kg/m 000 kg/m,600 kg/m 000 kg/m (0. m.6 m (0.08 m.09 m Discussion Note that measuring blood pressure with a water monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices. h POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

2 - -6 Water is poured into the U-tube from one arm and oil from the other arm. The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given. The height of each fluid in that arm is to be determined. Assumptions Both water and oil are incompressible substances. Properties The density of oil is given to be ρ 790 kg/m. We take the density of water to be ρ 000 kg/m. Analysis The height of water column in the left arm of the monometer is given to be h w 0.70 m. We let the height of water and oil in the right arm to be h w and h a, respectively. Then, h a h w. Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as P ρ bottom Patm + w ghw and Pbottom Patm + ρ w ghw + a gha Setting them equal to each other and simplifying, ρ h w Water ρ ρ ρ ρ ρ ρ + ( ρ ρ h w ghw w ghw + a gha w hw w hw + a ha hw hw a / w Noting that h a h w, the water and oil column heights in the second arm are determined to be 0.7 m h w + (790/000 hw hw 0.68 m 0.7 m 0.68 m + (790/000h a ha 0.67 m Discussion Note that the fluid height in the arm that contains oil is higher. This is expected since oil is lighter than water. oil a h a h w POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

3 -7-8 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible. The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ.0 kg/m. We take the density of water to be 000 kg/m. Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is V A ( π D / [ π (0.5 m /](0 m 0.5 m Then the buoyancy force becomes F B ρ gv (000 kg/m (9.8 m/s kn (0.5 m 000 kg m/s 0 m.6 kn D 5 cm Discussion The upward force exerted by water on the duct is.6 kn, which is equivalent to the weight of a mass of 5 kg. Therefore, this force must be treated seriously. F B POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

4 -8-85 A helium balloon tied to the ground carries people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ.6 kg/m. The density of helium gas is /7 th of this. Analysis The buoyancy force acting on the balloon is V balloon F B The total mass is πr / π(5 m / 5.6 m ρ air gv balloon N (.6 kg/m (9.8m/s (5.6 m 5958 N kg m/s m m He total The total weight is.6 ρ HeV kg/m (5.6 m 86.8 kg 7 m + m kg He people N W m g (6.8 kg(9.8 m/s kg m/s total Thus the net force acting on the balloon is F B net F W Then the acceleration becomes 7 N 5 N F a m net total 7 N kg m/s 6.8 kg N 6.5 m/s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

5 - -9 A ski lift is operating steadily at 0 km/h. The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions Air drag and friction are negligible. The average mass of each loaded chair is 50 kg. The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor. Analysis The lift is 000 m long and the chairs are spaced 0 m apart. Thus at any given time there are 000/0 50 chairs being lifted. Considering that the mass of each chair is 50 kg, the load of the lift at any given time is oad (50 chairs(50 kg/chair,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 00 m is W g kj 000 kg m ( z (,500 kg(9.8 m/s (00 m,55 kj mg z At 0 km/h, it will take distance km Δt 0. h 60 s velocity 0 km / h to do this work. Thus the power needed is W& g Wg Δt,55 kj 68.kW 60 s The velocity of the lift during steady operation, and the acceleration during start up are m/s V ( 0 km/h.778 m/s.6 km/h ΔV a Δt.778 m/s m/s 5 s During acceleration, the power needed is W& a m( V V / Δt (,500 kg kj/kg ((.778 m/s 0 /(5 s 9.6 kw /s 000 m Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be and h at sinα at 00 m 000 m (0.556 m/s (5 s (0..9 m /s Thus, W& g W & total kj/kg 000 kg m ( z z / Δ (,500 kg(9.8 m/s (.9 m /(5 s. kw mg t W& a + W& kw g /s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

6 -6-75 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pumpmotor unit and the pressure difference between the inlet and the exit of the pump are to be determined. Assumptions The elevations of the tank and the lake remain constant. Frictional losses in the pipes are negligible. The changes in kinetic energy are negligible. The elevation difference across the pump is negligible. Properties We take the density of water to be ρ 000 kg/m. Analysis (a We take the free surface of the lake to be point and the free surfaces of the storage tank to be point. We also take the lake surface as the reference level (z 0, and thus the potential energy at points and are pe 0 and pe gz. The flow energy at both points is zero since both and are open to the atmosphere (P P P atm. Further, the kinetic energy at both points is zero (ke ke 0 since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point are m& ρv & (000 kg/m (0.070 m /s 70 kg/s pe kj/kg (9.8m/s (0 m 000 m /s gz 0 m 0.96 kj/kg Then the rate of increase of the mechanical energy of water becomes Pump ΔE & mech, fluid m& ( emech,out emech,in m& ( pe 0 m& pe (70 kg/s(0.96 kj/kg.7 kw The overall efficiency of the combined pump-motor unit is determined from its definition, η ΔE& W.7 kw 0. kw mech,fluid pump -motor & elect,in 0.67 or 67.% Storage tank (b Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is.7 kw: ΔE& mech, fluid m& ( emech,out emech,in Solving for ΔP and substituting, Δ &E mech,fluid ΔP V &.7 kj/s kpa m m /s kj P P m& V & ΔP ρ 96 kpa Therefore, the pump must boost the pressure of water by 96 kpa in order to raise its elevation by 0 m. Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor. POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

7 -8 - A rigid container that is filled with -a is heated. The temperature and total enthalpy are to be determined at the initial and final states. Analysis This is a constant volume process. The specific volume is v V 0 v 0.00 m m.0 m 0 kg The initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A- by interpolation T T 00 kpa 0.6 C /kg P -a 00 kpa 0 kg and x v v f v fg ( m /kg ( m /kg v h h f + x h fg (0.009( kj/kg The total enthalpy is then H (0 kg(5.5 kj/kg 55. kj mh The final state is also saturated mixture. epeating the calculations at this state, T T 600 kpa.55 C x v v f v fg ( m /kg 0.07 ( m /kg h h f + x h fg 8.5+ (0.07( kj/kg H (0 kg(8.6 kj/kg 86. kj mh POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

8 -9 - A piston-cylinder device that is filled with -a is cooled at constant pressure. The final temperature and the change of total internal energy are to be determined. Analysis The initial specific volume is V. m v 0. m /kg m 00 kg The initial state is superheated and the internal energy at this state is -a 00 kpa 00 kg. m P 00 kpa v 0. m /kg The final specific volume is u 6.08 kj/kg (Table A - P / kg v 0. m v m This is a constant pressure process. The final state is determined to be saturated mixture whose temperature is T T 00 kpa 0.09 C /kg (Table A - The internal energy at the final state is (Table A- v v f ( m /kg x 0.60 v ( m /kg fg v u u f + x u fg (0.60( kj/kg Hence, the change in the internal energy is Δu u u kj/kg POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

9 -9-79 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. The volume of the tire remains constant. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-. Analysis Initially, the absolute pressure in the tire is P P + Patm kPa g Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from PV P V T K P P (0 kpa T T T 98 K Thus the pressure rise is ΔP P P kpa 6 kpa The amount of air that needs to be bled off to restore pressure to its original value is P V (0 kpa(0.05 m m kg T (0.87 kpa m /kg K(98 K m P V (0 kpa(0.05 m kg T (0.87 kpa m /kg K( K Δm m m kg Tire 5 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

10 - -8 A piston-cylinder device containing argon undergoes an isothermal process. The final pressure is to be determined. Assumptions At specified conditions, argon behaves as an ideal gas. Properties The gas constant of argon is 0.08 kj/kg K (Table A-. Analysis Since the temperature remains constant, the ideal gas equation gives P V PV m P V PV T T which when solved for final pressure becomes P V V P P (00 kpa 00 kpa V V P Argon 0. kg 0.05 m 00 kpa POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

11 -6-9 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-, 0.65 kpa m /kg K, T cr 67. K, P cr.06 MPa Analysis (a From the ideal gas equation of state, v T P (0.65 kpa m /kg K(7 K m /kg 500 kpa (.7% error (b From the compressibility chart (Fig. A-5, Thus, P T P P cr T T cr.5 MPa.06 7 K 67. K 0.59 MPa. Z 0.96 H O.5 MPa 50 C v Zv ideal (0.96(0.095 m /kg m /kg (0.% error (c From the superheated steam table (Table A-6, T P 50.5 MPa C } v m /kg POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

12 -9-95 Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-, 0.65 kpa m /kg K, T cr 67. K, P cr.06 MPa Analysis (a From the ideal gas equation, T v T ( K( 6 K v (b The pressure of the steam is P P Psat@50 C 6,59 kpa From the compressibility chart at the initial state (Fig. A-5, T P T T cr P P At the final state, Thus, P v T cr P v 6 K K 6.59 MPa MPa 0.79 Pv Z ( P Z v (c From the superheated steam table, P T cr cr Z Z 6,59 kpa , v (.50(67. K,060 kpa K Water 50 C sat. vapor Q T 50 C v m /kg x (Table A- P 6,59 kpa v v m T /kg 77 C 750 K (from Table A-6 or EES POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

13 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are 0.87 kj/kg.k, k. (Table A-a. Analysis For the isothermal expansion process: V V mt (0.5 kg(0.87 kj/kg.k( K 0.0 m P (000 kpa mt (0.5 kg(0.87 kj/kg.k( K m P (500 kpa V PV ln (000 kpa(0.0 m V m ln 0.0 m W b, For the polytropic compression process: P 7.8 kj n n.. V PV (500 kpa(0.056 m (000 kpa V V m W b, PV PV n For the constant pressure compression process: W b (000 kpa( m (500 kpa(0.056 m., P ( V V (000 kpa( m kj The net work for the cycle is the sum of the works for each process W W + W + W (.86 + ( kj net b, b, b, Air MPa 50 C -.86 kj POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

14 5-8 Closed System Energy Analysis 5-9 Saturated water vapor is isothermally condensed to a saturated liquid in a piston-cylinder device. The heat transfer and the work done are to be determined. Assumptions The cylinder is stationary and thus the kinetic and potential energy changes are zero. There are no work interactions involved other than the boundary work. The thermal energy stored in the cylinder itself is negligible. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E E in out Net energy transfer by heat, work, and mass W b,in Q Q out out system Change in internal, kinetic, potential, etc.energies ΔU m( u u W m( u u b,in ΔE (since KE PE 0 Water 00 C sat. vapor Heat The properties at the initial and final states are (Table A- T 00 C v v g 0.7 m / kg x u u g 59. kj/kg P P 55.9 kpa T 00 C v v f x 0 u u f The work done during this process is m kj/kg / kg T v kj ( (55.9 kpa( m /kg PdV P v v kpa m w b, out That is, w b,in 96.0 kj/kg Substituting the energy balance equation, we get q out w u u w + u kj/kg b,in ( b, in fg 96.0 kj/kg POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

15 5-5-5 An insulated rigid tank is initially filled with a saturated liquid-vapor mixture of water. An electric heater in the tank is turned on, and the entire liquid in the tank is vaporized. The length of time the heater was kept on is to be determined, and the process is to be shown on a P-v diagram. Assumptions The tank is stationary and thus the kinetic and potential energy changes are zero. The device is well-insulated and thus heat transfer is negligible. The energy stored in the resistance wires, and the heat transferred to the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E E in out Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc.energies We,in ΔU m( u VIΔt m( u u ΔEsystem u The properties of water are (Tables A- through A-6 (since Q KE PE 0 W e H O V const. P 00kPa v f x 0.5 u 0.000, f 7.0, v.69 m /kg u g fg 088. kj/kg T v v f u u f + x v + x u fg fg [ 0.5 ( ] ( kj/kg 0. m /kg v v 0. m /kg u u 556. kj/kg g@0.m /kg sat.vapor Substituting, 000 VA (0 V(8 A Δt (5 kg( kJ/kg kj/s Δt 986 s 5. min v POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

16 5-5- Two tanks initially separated by a partition contain steam at different states. Now the partition is removed and they are allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of heat lost from the tanks are to be determined. Assumptions The tank is stationary and thus the kinetic and potential energy changes are zero. There are no work interactions. Analysis (a We take the contents of both tanks as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as E E in out Net energy transfer by heat, work, and mass Q out Change in internal, kinetic, potential, etc.energies ΔU A ΔEsystem + ΔU B [ m u u ] + [ m( u u ] (sincew KE PE 0 ( A The properties of steam in both tanks at the initial state are (Tables A- through A-6 P T u, A, A, B, B 000 kpa v, 00 C u, T, B 50 C v f x 0.50 u v v u f f + x v + x u fg fg f A A m 79.7 kj/kg , 6.66, g u fg /kg The total volume and total mass of the system are V V A m m A + V B + m B m Av, A + mbv, + 5 kg v 0.98 m /kg 97. kj/kg + [ 0.50 ( ] ( kj/kg Now, the specific volume at the final state may be determined v V 0.7 m m.06 m 5 kg which fixes the final state and we can determine other properties (b Substituting, T P 00 kpa x v 0.7 m /kg u Q out ΔU A + ΔU B B B m /kg ( kg( m /kg + ( kg( m /kg.06 m /kg 00 kpa.5 C v v f v g v f u + x u 56.+ f fg [ m u u ] + [ m( u u ] ( A ( kj/kg ( kg( kJ/kg + ( kg( kJ/kg 959 kj B TANK A kg MPa 00 C Q TANK B kg 50 C x0.5 or Q out 959 kj POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

17 A student living in a room turns her 50-W fan on in the morning. The temperature in the room when she comes back 0 h later is to be determined. Assumptions Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - C and.77 MPa. The kinetic and potential energy changes are negligible, Δke Δpe 0. Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. All the doors and windows are tightly closed, and heat transfer through the walls and the windows is disregarded. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-. Also, c v 0.78 kj/kg.k for air at room temperature (Table A-. Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be expressed as Ein E out Net energy transfer by heat, work, and mass The mass of air is W W e, in e, in Change in internal, kinetic, potential, etc. energies ΔU V 6 6 m ΔEsystem m( u u mcv ( T T P V (00 kpa( m m 7. kg T (0.87 kpa m /kg K(88 K The electrical work done by the fan is W W& Δ t (0.5 kj / s(0 600 s 500 kj e e OOM Substituting and using the c v value at room temperature, 500 kj (7. kg(0.78 kj/kg C(T - 5 C T 58. C Discussion Note that a fan actually causes the internal temperature of a confined space to rise. In fact, a 00-W fan supplies a room with as much energy as a 00-W resistance heater. Fan m 6 m 6 m POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

18 Air in a closed system undergoes an isothermal process. The initial volume, the work done, and the heat transfer are to be determined. Assumptions Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of.5 K and.77 MPa. The kinetic and potential energy changes are negligible, Δ ke Δpe 0. Constant specific heats can be used for air. Properties The gas constant of air is 0.87 kj/kg K (Table A-. Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as E in out Net energy transfer by heat, work, and mass Q Q E in in W W The initial volume is mt V P b,out b,out Q in Change in internal, kinetic, potential, etc.energies ΔU mc ( T 0 W (sincet b,out ΔEsystem v T T ( kg(0.87 kpa m /kg K(7 K 600 kpa 0.55 m Using the boundary work relation for the isothermal process of an ideal gas gives Air 600 kpa 00 C Q W b,out dv v m Pdv mt mt ln v v ( kg(0.87 kpa m P mt ln P 600 kpa /kg K(7 Kln 57.kJ 80 kpa From energy balance equation, Q in Wb, out 57.kJ POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

19 ong cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven is to be determined. Assumptions The thermal properties of the rods are constant. The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the steel rods are given to be ρ 78 kg/m and c p 0.65 kj/kg. C. Analysis Noting that the rods enter the oven at a velocity of m/min and exit at the same velocity, we can say that a -m long section of the rod is heated in the oven in min. Then the mass of the rod heated in minute is m ρv ρa ρ( πd / (78 kg/m ( m[ π (0. m We take the -m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as E E in out Net energy transfer by heat, work, and mass Substituting, Q in Change in internal, kinetic, potential, etc.energies ΔU rod ΔEsystem m( u u mc( T T / ] Q in mc( T T (8.6 kg(0.65 kj/kg. C(700 0 C 57,5 kj 8.6 kg Noting that this much heat is transferred in min, the rate of heat transfer to the rod becomes Q & Q / Δt (57,5 kj/(min 57,5 kj/min kw in in Oven, 900 C Steel rod, 0 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

20 6-6- Air flows steadily in a pipe at a specified state. The diameter of the pipe, the rate of flow energy, and the rate of energy transport by mass are to be determined. Also, the error involved in the determination of energy transport by mass is to be determined. Properties The properties of air are 0.87 kj/kg.k and c p.008 kj/kg.k (at 50 K from Table A-b Analysis (a The diameter is determined as follows T v P & mv A V D A π 0.87 kj/kg.k( K 0.9 m (00 kpa ( (8 / 60 kg/s(0.9 m 5 m/s ( m π /kg /kg m m (b The rate of flow energy is determined from W & flow mp & v (8/ 60 kg/s(00 kpa(0.9 m /kg 0. kw (c The rate of energy transport by mass is E& mass m& ( h + ke m& cpt + V 00 kpa 77 C kj/kg (8/60 kg/s (.008 kj/kg.k( K + (5 m/s 000 m /s 05.9 kw (d If we neglect kinetic energy in the calculation of energy transport by mass E& mass mh & mc & pt (8/60 kg/s(.005 kj/kg.k( K 05.8 kw Therefore, the error involved if neglect the kinetic energy is only 0.09%. Air 5 m/s 8 kg/min POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

21 6-6-7 Air is decelerated in a diffuser from 0 m/s to 0 m/s. The exit temperature of air and the exit area of the diffuser are to be determined. Assumptions This is a steady-flow process since there is no change with time. Air is an ideal gas with variable specific heats. Potential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-. The enthalpy of air at the inlet temperature of 00 K is h kj/kg (Table A-. Analysis (a There is only one inlet and one exit, and thus m& m& m&. We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as or, E & & 0 (steady in Eout ΔEsystem 0 ate of net energy transfer by heat, work, and mass h m& ( h From Table A-, E& in ate of change in internal, kinetic, potential, etc.energies E& out & + V / m& ( h + V / (since W& Δpe 0 V V, 0 h h + ( 0 m/s ( 0 m/s kj/kg 000 m /s V V h kj/kg T 5.6 K (b The specific volume of air at the diffuser exit is v T ( kpa m /kg K( 5.6 K ( 00 kpa m /kg P From conservation of mass, AI 6.98 kj/kg m& v A V A m& v V ( kg/s(. m 0 m/s /kg m POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

22 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined. Assumptions This is a steady-flow process since there is no change with time. Potential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A- through 6 and P 0 MPa v m /kg T 50 C h. kj/kg P 0 MPa T 50 C V 80 m/s P 0 kpa h x 0.9 h + x f h fg kj/kg Analysis (a The change in kinetic energy is determined from V Δke V ( 50 m/s (80 m/s kj/kg 000 m /s.95 kj/kg (b There is only one inlet and one exit, and thus m& m& m&. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as E & & 0 (steady in Eout ΔEsystem 0 ate of net energy transfer by heat, work, and mass m& ( h + V E& in / W& out ate of change in internal, kinetic, potential, etc.energies E& W& out out m& h & + m& ( h h + V V + / V (since Δpe 0 Then the power output of the turbine is determined by substitution to be W & out ( kg/s( kj/kg 0. MW (c The inlet area of the turbine is determined from the mass flow rate relation, STEAM m kg/s P 0 kpa x 0.9 V 50 m/s W m& A V v A m& v V ( kg/s( m 80 m/s /kg m POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

23 6-6-6 Air is compressed in an adiabatic compressor. The mass flow rate of the air and the power input are to be determined. Assumptions This is a steady-flow process since there is no change with time. The compressor is adiabatic. Air is an ideal gas with constant specific heats. Properties The constant pressure specific heat of air at the average temperature of (0+00/0 C8 K is c p.06 kj/kg K (Table A-b. The gas constant of air is 0.87 kpa m /kg K (Table A-. Analysis (a There is only one inlet and one exit, and thus m & m& m&. We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as E & & 0 (steady in Eout ΔEsystem 0 ate of net energy transfer by heat, work, and mass V m& h + E& in + W& W& in in ate of change in internal, kinetic, potential, etc. energies E& out m& h m& h & V + h V + V m& c p ( T V T + The specific volume of air at the inlet and the mass flow rate are v T ( 0.87 kpa m /kg K(0 + 7 K m P 00 kpa V /kg.8 MPa 00 C Compressor 00 kpa 0 C 0 m/s m& A V v (0.5 m (0 m/s 5.5kg/s m /kg Similarly at the outlet, v V T ( 0.87 kpa m /kg K( K 0.07 m P 800 kpa & v m (5.5 kg/s(0.07 m /kg A 0.08 m (b Substituting into the energy balance equation gives 7.77 m/s /kg W& in m& c p ( T V T + V (7.77 m/s (0 m/s (5.5 kg/s (.06 kj/kg K(00 0K + 08 kw kj/kg 000 m /s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

24 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions Steady operating conditions exist. The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. Changes in the kinetic and potential energies of fluid streams are negligible. Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < T 50 kpa 7. C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h h 80 C 5.0 kj/kg h h 0 C 8.95 kj/kg h h C kj/kg Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: Energy balance: 0 (steady in m& out Δm& system 0 m& + m& m& m & Ein E & & out ate of net energy transfer by heat, work, and mass E& in m& h + m& h ate of change in internal, kinetic, potential, etc. energies E& out m& h 0 (steady ΔEsystem & 0 (since W& Δke Δpe 0 Combining the two relations and solving for &m gives ( m& m& m & & + h + m h h m& h h & h h m Substituting, the mass flow rate of cold water stream is determined to be ( kj/kg m& ( 0.5 kg/s kg/s kj/kg ( T 80 C m 0.5 kg/s T 0 C m H O (P 50 kpa T C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

25 Oil is to be cooled by water in a thin-walled heat exchanger. The rate of heat transfer in the heat exchanger and the exit temperature of water is to be determined. Assumptions Steady operating conditions exist. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Changes in the kinetic and potential energies of fluid streams are negligible. Fluid properties are constant. Properties The specific heats of water and oil are given to be.8 and.0 kj/kg. C, respectively. Analysis We take the oil tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as E E & & in out ate of net energy transfer by heat, work, and mass E& in mh & out 0 (steady ΔEsystem & ate of change in internal, kinetic, potential, etc.energies 0 E& out out + mh & (since Δke Δpe 0 mc & ( T T p Then the rate of heat transfer from the oil becomes [ mc & p( Tin Tout ] oil ( kg/s(. kj/kg. C(50 C 0 C 8 kw Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is determined from 8 kj/s [ mc & p( Tout Tin ] water Tout Tin + C C m& c (.5 kg/s(.8 kj/kg. C water p Cold water C.5 kg/s 0 C Hot oil 50 C kg/s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

26 The mass flow rate of a compressed air line is divided into two equal streams by a T-fitting in the line. The velocity of the air at the outlets and the rate of change of flow energy (flow power across the T- fitting are to be determined. Assumptions Air is an ideal gas with constant specific heats. The flow is steady. Since the outlets are identical, it is presumed that the flow divides evenly between the two. Properties The gas constant of air is 0.87 kpa m /kg K (Table A-. Analysis The specific volumes of air at the inlet and outlets are v v T ( 0.87 kpa m /kg K(0 + 7 K m P 600 kpa T ( 0.87 kpa m /kg K(6 + 7 K v m P 00 kpa Assuming an even division of the inlet flow rate, the mass balance can be written as AV AV A v V V V 8.m/s v v A v The mass flow rate at the inlet is m& AV πd V π (0.05 m 50 m/s v v m /kg /kg /kg 0.7 kg/s.6 MPa 0 C 50 m/s while that at the outlets is m& 0.7 kg/s m& m& 0.86 kg/s Substituting the above results into the flow power expression produces W & flow m& P v m& P v (0.86 kg/s(00 kpa(0.065 m /kg (0.7 kg/s(600 kpa(0.056 m /kg 0.96 kw. MPa 6 C. MPa 6 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

27 A reversible heat pump with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satisfies the increase in entropy principle. Assumptions The heat pump operates steadily. Analysis Since the heat pump is completely reversible, the combination of the coefficient of performance expression, first aw, and thermodynamic temperature scale gives C 00 kw COP 6.7 HP W & net HP, rev T / T H (8 K /(9 K The power required to drive this heat pump, according to the coefficient of performance, is then 00 kw W& net, in H.7 kw COP 6.7 HP, rev According to the first law, the rate at which heat is removed from the low-temperature energy reservoir is W& 00 kw.7 kw 96.6 kw H net, in The rate at which the entropy of the high temperature reservoir changes, according to the definition of the entropy, is H 00 kw ΔS& H 0.0 kw/k T 9 K H and that of the low-temperature reservoir is 96.6 kw ΔS& 0.0 kw/k T 8 K The net rate of entropy change of everything in this system is ΔS & ΔS& H + ΔS& kw/k total as it must be since the heat pump is completely reversible. Q & 0 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

28 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically at constant pressure. The entropy change of the water during this process is to be determined. Assumptions The kinetic and potential energy changes are negligible. The cylinder is well-insulated and thus heat transfer is negligible. The thermal energy stored in the cylinder itself is negligible. The compression or expansion process is quasi-equilibrium. Analysis From the steam tables (Tables A- through A-6, Also, v v f P 50 kpa h h sat. liquid s kpa kpa kpa m /kg 67. kj/kg.7 kj/kg K 00 kj H O 50 kpa Sat. liquid V m m.75 kg v m /kg We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in out Net energy transfer by heat, work, and mass W E e,in W b,out W e,in Change in internal, kinetic, potential, etc.energies ΔU m( h h ΔEsystem since ΔU + W b ΔH during a constant pressure quasi-equilibrium process. Solving for h, Thus, h W 00 kj kg e,in h + m P 50 kpa x h 90. kj/kg s h s f h h fg + x Then the entropy change of the water becomes Δ S f 90. kj/kg s.7 + fg ( 0.08( kj/kg K ( (.75 kg(.68.7 kj/kg 5.7 kj/k m s s K POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

29 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions This is a steady-flow process since there is no change with time. The process is isentropic (i.e., reversible-adiabatic. Analysis There is only one inlet and one exit, and thus m & m& m&. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as 0 (steady Ein Eout Esystem 0 & & & Δ MPa 60 C ate of net energy transfer by heat, work, and mass E& in mh & W& out ate of change in internal, kinetic, potential, etc.energies E& out mh & m& ( h h The inlet state properties are P MPa T 60 C s + W& out h 59.9 kj/kg kj/kg K (Table A - 6 For this isentropic process, the final state properties are (Table A-5 P s 00 kpa s kj/kg K x h s s s fg h + x f f h 7.5+ (0.997( kj/kg fg T MPa 00 kpa Turbine 00 kpa s Substituting, w out h h ( kj/kg 6.0 kj/kg POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

30 Computer chips are cooled by placing them in saturated liquid -a. The entropy changes of the chips, -a, and the entire system are to be determined. Assumptions The system is stationary and thus the kinetic and potential energy changes are zero. There are no work interactions involved. There is no heat transfer between the system and the surroundings. Analysis (a The energy balance for this system can be expressed as E E in out Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc.energies 0 ΔU ΔEsystem [ m( u u ] chips [ m( u u ] -a The heat released by the chips is [ m( u u ] + [ m( u u ] chips -a [ 0 ( 0 ] K 0.8 kj Q chips mc( T T (0.00 kg(0. kj/kg K The mass of the refrigerant vaporized during this heat exchange process is Q a Q a 0.8 kj m g, kg u u u 07.0 kj/kg g f 0 C Only a small fraction of -a is vaporized during the process. Therefore, the temperature of -a remains constant during the process. The change in the entropy of the -a is (at -0 F from Table A- ΔS a mg,s g, + m f,s f, m f,s f, ( ( ( (0 (0.005( kJ/K (b The entropy change of the chips is T ( 0 + 7K ΔS chips mc ln (0.00 kg(0. kj/kg Kln kj/k T (0 + 7K (c The total entropy change is Δ S S ΔS + ΔS ( kj/k total gen -a chips The positive result for the total entropy change (i.e., entropy generation indicates that this process is possible. POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

31 Nitrogen is compressed in an adiabatic compressor. The minimum work input is to be determined. Assumptions This is a steady-flow process since there is no change with time. The process is adiabatic, and thus there is no heat transfer. Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at an anticipated average temperature of 00 K are c p.0 kj/kg K and k.97 (Table A-b. Analysis There is only one inlet and one exit, and thus m & m& m&. We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as E & & 0 (steady in Eout ΔEsystem 0 ate of net energy transfer by heat, work, and mass E& in mh & + W& W& in in ate of change in internal, kinetic, potential, etc.energies E& out mh & & m& ( h h For the minimum work input to the compressor, the process must be reversible as well as adiabatic (i.e., isentropic. This being the case, the exit temperature will be 600 kpa T Nitrogen compressor 0 kpa 0 C 600 kpa T ( k / k 0.97 /.97 P 600 kpa T (0 K P 0 kpa 79 K 0 kpa s Substituting into the energy balance equation gives win p h h c ( T T (.0 kj/kg K(79 0K 8 kj/kg POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

32 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the rate of entropy generation within the heat exchanger are to be determined. Assumptions Steady operating conditions exist. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Changes in the kinetic and potential energies of fluid streams are negligible. Fluid properties are constant. Properties The specific heats of cold and hot water are given to be.8 and.9 kj/kg. C, respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein E & & out ate of net energy transfer by heat, work, and mass in E& in + mh & in ate of change in internal, kinetic, potential, etc.energies E& out mh & mc & ( T 0 (steady ΔEsystem & p (since Δke Δpe 0 T 0 Then the rate of heat transfer to the cold water in this heat exchanger becomes in [ mc & ( Tout Tin ] cold water (0.5 kg/s(.8 kj/kg. C(5 C 5 C.5 kw p Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is determined to be.5 kw [ mc & p( Tin Tout ] hot water Tout Tin 00 C 97.5 C mc & ( kg/s(.9 kj/kg. C (b The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: 0 (steady Sin Sout + S& gen ΔS& & & system { m& cold ate of net entropy transfer by heat and mass s + m& hot s m& cold s m& ate of entropy generation m& s + m& s m& s m& s + S& hot s + S& S& gen gen gen 0 0 m& p ate of change of entropy (since Q 0 cold ( s s + m& hot ( s s Noting that both fluid streams are liquids (incompressible substances, the rate of entropy generation is determined to be S& gen m& cold T cp ln T + m& hot T cp ln T Hot water 00 C kg/s 5 C (0.5 kg/s(.8 kj/kg.kln + ( kg/s(.9 kj/kg.kln kw/k Cold water 5 C 0.5 kg/s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

33 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer, the outlet temperature of the air, and the rate of entropy generation are to be determined. Assumptions Steady operating conditions exist. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Changes in the kinetic and potential energies of fluid streams are negligible. Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be.005 and.0 kj/kg. C, respectively. The gas constant of air is 0.87 kj/kg.k (Table A-. Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as Ein E & & out ate of net energy transfer by heat, work, and mass E& in mh & out ate of change in internal, kinetic, potential, etc.energies E& out out + mh & mc & ( T 0 (steady ΔEsystem & p T 0 (since Δke Δpe 0 Then the rate of heat transfer from the exhaust gases becomes [ mc & p( T Tout] gas. The mass flow rate of air is (. kg/s(.kj/kg. C(80 C 95 C 05.7 kw in PV (95 kpa(.6 m /s m& &.808 kg/s T (0.87 kpa.m /kg.k (9 K Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes [ mc & T T ] T T p( out in air out in + mc & p 0 C kw (.808 kg/s(.005 kj/kg.. C C The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: m& exhaust Sin S & & out ate of net entropy transfer by heat and mass s + m& air s m& exhaust s m& air S& gen { ate of entropy generation m& s + m& s m& s m& s + S& + s + S& Then the rate of entropy generation is determined to be S& gen m& exhaust T cp ln T + m& air T cp ln T S& gen gen gen 0 (steady ΔS& system 0 0 m& ate of change of entropy (since Q 0 exhaust Air 95 kpa 0 C.6 m /s ( s s + m& air ( s s (. kg/s(.kj/kg.kln + (.808 kg/s(.005 kj/kg.kln kw/k Exhaust gases. kg/s, 95 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

34 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked and the amount of entropy generation associated with this heat transfer process are to be determined. Assumptions The egg is spherical in shape with a radius of r 0.75 cm. The thermal properties of the egg are constant. Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible. There are no changes in kinetic and potential energies. Properties The density and specific heat of the egg are given to be ρ 00 kg/m and c p. kj/kg. C. Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as E E in out Net energy transfer by heat, work, and mass Q in Change in internal, kinetic, potential, etc. energies ΔU egg ΔEsystem m( u u mc( T T Boiling Water Then the mass of the egg and the amount of heat transfer become πd π (0.055 m m ρv ρ (00 kg/m kg 6 6 Q mc ( T T ( kg(. kj/kg. C(70 8 C 8. in p kj Egg 8 C We again take a single egg as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the egg and its immediate surroundings so that the boundary temperature of the extended system is at 97 C at all times: S S in out Net entropy transfer by heat and mass + Q T in b Sgen { Entropy generation + S gen ΔSsystem Change in entropy ΔS system S gen Q T in b + ΔS system where Substituting, T ΔS system m( s s mcavg ln ( kg(. kj/kg.k ln kj/k T S gen Qin + ΔS T b system 8. kj kj/k kj/k (per egg 70 K POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

35 iquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of entropy generation are to be determined. Assumptions This is a steady-flow process since there is no change with time. Kinetic and potential energy changes are negligible. There are no work interactions. Properties Noting that T < T 00 kpa 0. C, the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From Tables A- through A-6, 00 kj/min P 00 kpa 8.9 kj/kg h h o C T 0 C T P 00 kpa 50 C s P 00 kpa h h T 60 C s s f s s h o C 769. kj/kg 7.80 kj/kg K C C kj/kg K o 5.8 kj/kg 0.8 kj/kg K Analysis (a We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as 0 (steady Mass balance: m & in m& out Δm& system 0 m& + m& m& Energy balance: E E & & out 0 (steady in ΔEsystem 0 ate of net energy transfer by heat, work, and mass E& in m& h + m& h ate of change in internal, kinetic, potential, etc.energies E& out out & + m& h Combining the two relations gives m& h + m& h ( m& + m& h m& ( h h + m& ( h h out Solving for &m and substituting, the mass flow rate of the superheated steam is determined to be Also, m& out m& h h ( h h (00/60kJ/s (.5 kg/s( m & m& + m& kg/s ( kJ/kg kj/kg 0.66 kg/s (b The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the extended system is 5 C at all times. It gives 0 Sin Sout + S& gen ΔS& system 0 & & { ate of net entropy transfer by heat and mass m& s + m& s m& s T ate of entropy generation out b,surr + S& gen ate of change of entropy 0 Substituting, the rate of entropy generation during this process is determined to be out S& gen m& s m& s m& s + Tb,surr (.666 kg/s( 0.8 kj/kg K ( 0.66 kg/s( 7.80 kj/kg K (00/ 60 kj/s (.5 kg/s( kj/kg K + 98 K 0. kw/k 0 C.5 kg/s 50 C MIXING CHAMBE 00 kpa 60 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

36 A horizontal cylinder is separated into two compartments by a piston, one side containing nitrogen and the other side containing helium. Heat is added to the nitrogen side. The final temperature of the helium, the final volume of the nitrogen, the heat transferred to the nitrogen, and the entropy generation during this process are to be determined. Assumptions Kinetic and potential energy changes are negligible. Nitrogen and helium are ideal gases with constant specific heats at room temperature. The piston is adiabatic and frictionless. Properties The properties of nitrogen at room temperature are kpa.m /kg.k, c p.09 kj/kg.k, c v 0.7 kj/kg.k, k.. The properties for helium are.0769 kpa.m /kg.k, c p 5.96 kj/kg.k, c v.56 kj/kg.k, k.667 (Table A-. Analysis (a Helium undergoes an isentropic compression process, and thus the final helium temperature is determined from T He, P T P.7 K ( k / k (.667 / kpa (0 + 7K 95 kpa (b The initial and final volumes of the helium are Q N 0. m He 0. kg V He, mt P (0. kg(.0769 kpa m /kg K(0 + 7 K m 95 kpa V He, mt P Then, the final volume of nitrogen becomes V (0. kg(.0769 kpa m /kg K(.7 K m 0 kpa V + V m N, N, He, V He, (c The mass and final temperature of nitrogen are m T P V (95 kpa(0. m N T (0.968 kpa m /kg K(0 + 7 K PV N, m (0 kpa(0.88 m (0.85 kg(0.968 kpa m /kg K 0.85 kg 55. K The heat transferred to the nitrogen is determined from an energy balance Q in ΔU + ΔU N He v ( N v He [ mc T T ] + [ mc ( T T ] (0.85 kg(0.7 kj/kg.k( (0. kg(.56 kj/kg.k( kj (d Noting that helium undergoes an isentropic process, the entropy generation is determined to be S gen T P Qin ΔS N + ΔSsurr mn c p ln ln + T P T 55. K 0 kpa 6.6 kj (0.85 kg (.09 kj/kg.kln (0.968 kj/kg.kln + 9 K 95 kpa ( K kj/k POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

37 An insulated rigid tank is connected to a piston-cylinder device with zero clearance that is maintained at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into the cylinder. The final temperatures in the tank and the cylinder are to be determined. Assumptions Both the tank and cylinder are well-insulated and thus heat transfer is negligible. The water that remains in the tank underwent a reversible adiabatic process. The thermal energy stored in the tank and cylinder themselves is negligible. The system is stationary and thus kinetic and potential energy changes are negligible. Analysis (a The steam in tank A undergoes a reversible, adiabatic process, and thus s s. From the steam tables (Tables A- through A-6, v v g P 500 kpa u u sat. vapor s s P 50 kpa s s v ( sat. mixture, A g@500 kpa v u, kpa g@500 kpa f + x u f 0.78 m /kg kj/kg kj/kg K, A x v + x, A fg, A The initial and the final masses in tank A are Thus, m m, A, B V A v m, A, A T s 0. m.067 kg 0.78 m /kg m, A, A, A T s kg (b The boundary work done during this process is sat@50 kpa.5 C (0.905( m u fg s fg ( V, B PB m, B B W b, out PdV PB 0 v, Taking the contents of both the tank and the cylinder to be the system, the energy balance for this closed system can be expressed as or, Thus, E E in out Net energy transfer by heat, work, and mass P m h B, B v W, B ΔEsystem Change in internal, kinetic, potential, etc.energies b,out ΔU ( ΔU A + ( ΔU B Wb,out + ( ΔU A + ( ΔU B 0 ( mu mu A + ( mu B 0 m h + ( m u m u 0 +, B, B A f (0.905(05. kj/kg 76.6 kj/kg and m, A V A v ( mu mu (.067( ( 0.7( 76.6, A A, B m, B m 0.7 kg.0789 m /kg kj/kg At 50 kpa, h f 67. and h g 69. kj/kg. Thus at the final state, the cylinder will contain a saturated liquid-vapor mixture since h f < h < h g. Therefore,.5 C T, B Tsat@50 kpa /kg Sat. vapor 500 kpa 0. m 50 kpa POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

38 A double-pane window consists of two layers of glass separated by a stagnant air space. For specified indoors and outdoors temperatures, the rate of heat loss through the window and the inner surface temperature of the window are to be determined. Assumptions Heat transfer through the window is steady since the indoor and outdoor temperatures remain constant at the specified values. Heat transfer is one-dimensional since any significant temperature gradients will exist in the direction from the indoors to the outdoors. Thermal conductivities of the glass and air are constant. Heat transfer by radiation is negligible. Properties The thermal conductivity of the glass and air are given to be k glass 0.78 W/m C and k air 0.06 W/m C. Analysis The area of the window and the individual resistances are A (. m ( m. m i conv, 0.07 C/W h A (0 W/m. C(. m 0.00 m glass C/W ka (0.78 W/m. C(. m 0.0 m air 0.9 C/W ka (0.06 W/m. C(. m o o conv, C/W o h A (5 W/m. C(. m total conv, conv, ( C/W The steady rate of heat transfer through window glass then becomes T T [ ( 5] C Q & W total 0.59 C/W The inner surface temperature of the window glass can be determined from T T Q & T o T Q & conv, C ( W(0.07 C/W 9. C conv, T i Air o T POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

39 Two cylindrical aluminum bars with ground surfaces are pressed against each other in an insulation sleeve. For specified top and bottom surface temperatures, the rate of heat transfer along the cylinders and the temperature drop at the interface are to be determined. Assumptions Steady operating conditions exist. Heat transfer is one-dimensional in the axial direction since the lateral surfaces of both cylinders are wellinsulated. Thermal conductivities are constant. Properties The thermal conductivity of aluminum bars is given to be k 76 W/m C. The contact conductance at the interface of aluminum-aluminum plates for the case of ground surfaces and of 0 atm MPa pressure is h c,00 W/m C (Table 0-. Analysis (a The thermal resistance network in this case consists of two conduction resistance and the contact resistance, and they are determined to be contact 0.07 C/W h c A (,00 W/m C[ π (0.05 m /] c 0.5 m plate 0. C/W ka (76 W/m C[ π (0.05 m /] Then the rate of heat transfer is determined to be ΔT ΔT (50 0 C Q &. W total contact + bar ( C/W Therefore, the rate of heat transfer through the bars is. W. (b The temperature drop at the interface is determined to be ΔT (. W(0.07 C/W 6. C interface contact Bar Interface Bar i glass o T T POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

40 A composite wall consists of several horizontal and vertical layers. The left and right surfaces of the wall are maintained at uniform temperatures. The rate of heat transfer through the wall, the interface temperatures, and the temperature drop across the section F are to be determined. Assumptions Heat transfer is steady since there is no indication of change with time. Heat transfer through the wall is one-dimensional. Thermal conductivities are constant. Thermal contact resistances at the interfaces are disregarded. Properties The thermal conductivities are given to be k A k F, k B 8, k C 0, k D 5, k E 5 W/m C. Analysis (a The representative surface area is the individual thermal resistances are A m. The thermal resistance network and 5 7 T T mid, mid, total A B D E F ka C ka ka ka ka A B D E F 0.0m 0.0 C/W ( W/m C(0. m ka + 5 C 0.05 m 0.06 C/W (0 W/m C(0.0 m 0.05 m 0.6 C/W (8 W/m C(0.0 m 0.m 0. C/W o (5 W/m C(0.06 m 0.m 0.05 (5 W/m C(0.06 m C/W 0.06 m 0.5 C/W ( W/m C(0. m T T total mid, mid, (00 00 C 57 W 0.9 C/W 7 + mid, o mid, 0.0 C/W 0.05 C/W C/W (for a 0. m m section Then steady rate of heat transfer through entire wall becomes (5 m(8 m 5 Q & total (57 W.9 0 W 0. m (b The total thermal resistance between left surface and the point where the sections B, D, and E meet is total + mid, C/W Then the temperature at the point where the sections B, D, and E meet becomes T T Q & T T Q & total 00 C (57 W(0.065 C/W 6 C total (c The temperature drop across the section F can be determined from ΔT Q & ΔT Q & F (57 W(0.5 C/W C F 6 POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

41 Steam flows in a steel pipe, which is insulated by gypsum plaster. The rate of heat transfer from the steam and the temperature on the outside surface of the insulation are be determined. Assumptions Heat transfer is steady since there is no indication of any change with time. Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. Thermal conductivities are constant. The thermal contact resistance at the interface is negligible. Properties (a The thermal conductivities of steel and gypsum plaster are given to be 50 and 0.5 W/m C, respectively. Analysis The thermal resistances are Insulation T i i steel ins o T o Steam i C/W h i Ai (800 W/m C π (0.06 m(0 m ln( D / D ln(8 / 6 steel C/W πk steel π (50 W/m C(0 m ln( D / D ln(6 / 8 ins 0.00 C/W πk ins π (0.5 W/m C(0 m o C/W h A (00 W/m C π (0.6 m(0 m o o The total thermal resistance and the rate of heat transfer are C/W total i steel ins o T T (00 0 C i o 5,957 W total m C/W (b The temperature at the outer surface of the insulation is determined from Ts To Ts ( 0 C 5,957 W Ts 7.9 C m C/W o POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

42 An electric hot water tank is made of two concentric cylindrical metal sheets with foam insulation in between. The fraction of the hot water cost that is due to the heat loss from the tank and the payback period of the do-it-yourself insulation kit are to be determined. Assumptions Heat transfer is steady since there is no indication of any change with time. Heat transfer is one-dimensional since there is thermal symmetry about the center line and no variation in the axial direction. Thermal conductivities are constant. The thermal resistances of the water tank and the outer thin sheet metal shell are negligible. 5 Heat loss from the top and bottom surfaces is negligible. Properties The thermal conductivities are given to be k 0.0 W/m C for foam insulation and k 0.05 W/m C for fiber glass insulation Analysis We consider only the side surfaces of the water heater for simplicity, and disregard the top and bottom surfaces (it will make difference of about 0 percent. The individual thermal resistances are A πd π ( 0.0 m( m.5m A i o o o i h A i i C/W (50 W/m. C(.5m πd π ( 0.6 m( m o h A foam total o o.89 m 0.09 C/W ( W/m. C(.89 m ln( r / r ln( / C/W π k π (0.0 W/m C( m i + o + foam C/W The rate of heat loss from the hot water tank is Tw T (55 7 C Q & 68.8 W total 0.07 C/W The amount and cost of heat loss per year are Q Q & Δt ( kw(65 h/yr 60.7 kwh/yr Cost of Energy (Amount of energy(unit cost (60.7 kwh($0.08 / kwh $8. $8. f % $80 If cm thick fiber glass insulation is used to wrap the entire tank, the individual resistances becomes Ao πdo π ( 0.5 m( m.67 m o i foam fiberglass o o 0.06 C/W h o o Ao ( W/m C(.67 m T w T ln( r / r ln( / 0 foam 0.7 C/W πk π (0.0 W/m C( m ln( r / r ln(6 / fiberglass 0.79 C/W πk π (0.05 W/m C( m total i + o + foam + fiberglass C/W The rate of heat loss from the hot water heater in this case is Tw T (55 7 C Q & 0.9 W total 0.68 C/W The energy saving is saving W The time necessary for this additional insulation to pay for its cost of $0 is then determined to be Cost ( kw(time period($0.08 / kwh $0 Then, Time period,90 hours 58 days.5 years T w i foam o T POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

43 0-9 An electric wire is tightly wrapped with a -mm thick plastic cover. The interface temperature and the effect of doubling the thickness of the plastic cover on the interface temperature are to be determined. Assumptions Heat transfer is steady since there is no indication of any change with time. Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. Thermal properties are constant. The thermal contact resistance at the interface is negligible. 5 Heat transfer coefficient accounts for the radiation effects, if any. Properties The thermal conductivity of plastic cover is given to be k 0.5 W/m C. Analysis In steady operation, the rate of heat transfer from the wire is equal to the heat generated within the wire, W& e VI ( 8 V( A 0 W plastic conv T The total thermal resistance is T conv 0.58 C/W h A ( W/m. C[ π (0.00 m(0 m] plastic total o ln( r / r ln(./ C/W πk π (0.5 W/m. C(0 m o conv + plastic Then the interface temperature becomes T T Q & T T + Q & total The critical radius of plastic insulation is C/W total 0 C + (0 W(0.8 C/W 70.0 C k 0.5 W/m. C r cr m 6.5 mm h W/m. C Doubling the thickness of the plastic cover will increase the outer radius of the wire to mm, which is less than the critical radius of insulation. Therefore, doubling the thickness of plastic cover will increase the rate of heat loss and decrease the interface temperature. POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

44 Circular aluminum fins are to be attached to the tubes of a heating system. The increase in heat transfer from the tubes per unit length as a result of adding fins is to be determined. Assumptions Steady operating conditions exist. The heat transfer coefficient is constant and uniform over the entire fin surfaces. Thermal conductivity is constant. Heat transfer by radiation is negligible. Properties The thermal conductivity of the fins is given to be k 86 W/m C. Analysis In case of no fins, heat transfer from the tube per meter of its length is A no fin no fin πd π (0.05 m( m 0.57 m ha no fin ( T b T (0 W/m. C(0.57m (80 5 C 97 W The efficiency of these circular fins is, from the efficiency curve, Fig. 0- / c ( D D / ( / m r + ( t / (0.00/. r 0.05 / h t h η + kap kt o W/m C o (86 W/m C(0.00m Heat transfer from a single fin is A fin fin π ( r η fin fin,max 0.97(0 W/m. C( m (80 5 C.5 W r + πr t π (0.0 fin η ha fin ( T b T fin π (0.0( m Heat transfer from a single unfinned portion of the tube is Aunfin πd s π (0.05 m(0.00 m m unfin haunfin ( Tb T (0 W/m. C( m (80 5 C.9 W There are 50 fins and thus 50 interfin spacings per meter length of the tube. The total heat transfer from the finned tube is then determined from Q & n + 50( W total, fin ( fin unfin Therefore the increase in heat transfer from the tube per meter of its length as a result of the addition of the fins is Q & W increase total,fin no fin 80 C 5 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

45 A hot plate is to be cooled by attaching aluminum pin fins on one side. The rate of heat transfer from the m by m section of the plate and the effectiveness of the fins are to be determined. Assumptions Steady operating conditions exist. The temperature along the fins varies in one direction only (normal to the plate. Heat transfer from the fin tips is negligible. The heat transfer coefficient is constant and uniform over the entire fin surface. 5 The thermal properties of the fins are constant. 6 The heat transfer coefficient accounts for the effect of radiation from the fins. Properties The thermal conductivity of the aluminum plate and fins is given to be k 7 W/m C. Analysis Noting that the cross-sectional areas of the fins are constant, the efficiency of the circular fins can be determined to be m η hp ka c hπd kπd / h kd - fin tanh m tanh(5.7 m 0.0 m m m 0.0 m (5 W/m. C 5.7 m (7 W/m. C(0.005 m 0.95 The number of fins, finned and unfinned surface areas, and heat transfer rates from those areas are n A m (0.006 m(0.006 m fin unfinned A finned unfinned πd 7777 πd m 5,00 W 07 W ( T 7,777 π ( π (0.005(0.0 + ( πd π 7777 η η ha ( T T fin fin,max 0.95(5 W/m. C(6.68 m (00 0 C ha unfinned b fin T fin b 0.86 m (5 W/m. C(0.86 m (00 0 C Then the total heat transfer from the finned plate becomes Q & total, fin + 5, W 7. kw finned unfinned The rate of heat transfer if there were no fin attached to the plate would be A no fin no fin (m(m m ha no fin ( T b T Then the fin effectiveness becomes fin 7,00 ε fin no fin (5 W/m. C(m (00 0 C 50 W - cm D0.5 cm 0.6 cm POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

46 An ice chest made of 0-cm thick styrofoam is initially filled with 5 kg of ice at 0 C. The length of time it will take for the ice in the chest to melt completely is to be determined. Assumptions Heat transfer is steady since the specified thermal conditions at the boundaries do not change with time. Heat transfer is one-dimensional. Thermal conductivity is constant. The inner surface temperature of the ice chest can be taken to be 0 C at all times. 5 Heat transfer from the base of the ice chest is negligible. Properties The thermal conductivity of styrofoam is given to be k 0.0 W/m C. The heat of fusion of water at atm is h.7 kj/kg. Analysis Disregarding any heat loss through the bottom of the ice chest, the total thermal resistance and the heat transfer rate are determined to be A chest conv total if A (0. 0.0( (0. 0.0( ( ( m i o (0.(0. + (0.(0.5 + (0.( m 0.0 m.597 C/W ka i (0.0 W/m. C( m C/W ha o (8 W/m. C(0.7 m chest + conv C/W T T (8 0 C s 6.79 W.6678 C/W total The total amount of heat necessary to melt the ice completely is Q mhif ( 50 kg(.7 kj/kg 6,685 kj Then the time period to transfer this much heat to the cooler to melt the ice completely becomes Q,685,000 J Δt s 76 h.5 days 6.79 J/s Ice chest T s chest conv T POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

47 Two persons are wearing different clothes made of different materials with different surface areas. The fractions of heat lost from each person s body by perspiration are to be determined. Assumptions Heat transfer is steady. Heat transfer is one-dimensional. Thermal conductivities are constant. Heat transfer by radiation is accounted for in the heat transfer coefficient. 5 The human body is assumed to be cylindrical in shape for heat transfer purposes. Properties The thermal conductivities of the leather and synthetic fabric are given to be k 0.59 W/m C and k 0. W/m C, respectively. Analysis The surface area of each body is first determined from A πd / π (0.5 m(.7 m/ m A A m The sensible heat lost from the first person s body is leather conv total ka 0.00 m C/W (0.59 W/m. C( m C/W ha (5 W/m. C( m leather + conv C/W The total sensible heat transfer is the sum of heat transferred through the clothes and the skin clothes skin sensible T T total T T conv clothes ( 0 C 8. W C/W ( 0 C 0.0 W C/W W skin Then the fraction of heat lost by respiration becomes f & respiration total total total sensible epeating similar calculations for the second person s body synthetic conv total ka 0.00 m C/W (0. W/m. C(.5 m C/W ha (5 W/m. C(.5 m Q & leather + conv T T C/W ( 0 C C/W sensible total f & respiration total total total sensible 5.9 W T leather conv T synthetic conv T T POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

48 A 6-m-diameter spherical tank filled with liquefied natural gas (NG at -60 C is exposed to ambient air. The time for the NG temperature to rise to -50 C is to be determined. Assumptions Heat transfer can be considered to be steady since the specified thermal conditions at the boundaries do not change with time significantly. Heat transfer is one-dimensional since there is thermal symmetry about the midpoint. adiation is accounted for in the combined heat transfer coefficient. The combined heat transfer coefficient is constant and uniform over the entire surface. The temperature of the thin-shelled spherical tank is said to be nearly equal to the temperature of the NG inside, and thus thermal resistance of the tank and the internal convection resistance are negligible. Properties The density and specific heat of NG are given to be 5 kg/m and.75 kj/kg C, respectively. The thermal conductivity of super insulation is given to be k W/m C. Analysis The inner and outer surface areas of the insulated tank and the volume of the NG are A πd π ( m 50.7 m A πd π (.0 m 5.8 m V πd / 6 π ( m / 6.5 m The rate of heat transfer to the NG is r r (.05.0 m insulation.07 C/W πkr r π ( W/m. C(.0 m(.05 m o total C/W h A ( W/m. C(5.8m o o + T T Q & insulation total NG C/W [ ( 55] C.75 W.57 C/W We used average NG temperature in heat transfer rate calculation. The amount of heat transfer to increase the NG temperature from -60 C to -50 C is m ρv (5 kg/m (.5m, kg Q p T mc ΔT (, kg(.75 kj/kg. C insulation 5 [( 50 ( 60 C].95 0 kj Assuming that heat will be lost from the NG at an average rate of 5.7 W, the time period for the NG temperature to rise to -50 C becomes 5 Q.95 0 kj 7 Δt.55 0 s 90 h 88 days kw o T NG tank -60 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

49 - - The temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the initial ΔT is to be determined. Assumptions The junction is spherical in shape with a diameter of D 0.00 m. The thermal properties of the junction are constant. The heat transfer coefficient is constant and uniform over the entire surface. adiation effects are negligible. 5 The Biot number is Bi < 0. so that the lumped system analysis is applicable (this assumption will be verified. Properties The properties of the junction are given to be k 5 W/m. C, ρ 8500 kg/m, and c 0 J/kg. C. p Analysis The characteristic length of the junction and the Biot number are V πd / 6 D 0.00 m c m A surface πd 6 6 hc (90 W/m. C(0.000 m Bi < 0. k (5 W/m. C Since Bi < 0., the lumped system analysis is applicable. Then the time period for the thermocouple to read 99% of the initial temperature difference is determined from T ( t T T T i 0.0 ha h b ρc V ρc p p c (8500 kg/m Gas 90 W/m. C 0.65 s (0 J/kg. C(0.000 m h, T Junction D T(t - T ( t T T T i e bt 0.0 e (0.65 s - t t 7.8 s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

50 -9-0 The heating times of a sphere, a cube, and a rectangular prism with similar dimensions are to be determined. Assumptions The thermal properties of the geometries are constant. The heat transfer coefficient is constant and uniform over the entire surface. Properties The properties of silver are given to be k 9 W/m ºC, ρ 0,500 kg/m, and c p 0.5 kj/kg ºC. Analysis For sphere, the characteristic length and the Biot number are V πd / 6 D 0.05 m c m 5 cm Air Asurface πd 6 6 h, T hc ( W/m. C(0.008 m Bi < 0. k (9 W/m. C Since Bi < 0., the lumped system analysis is applicable. Then the time period for the sphere temperature to reach to 5ºC is determined from b ha ρc V ρ p h c p c (0,500 kg/m W/m. C s (5 J/kg. C(0.008 m - Cube: c V A h Bi k b T ( t T T T i surface c ha ρc 6 e bt 5 e 0 - ( s t 0.05 m m 6 6 ( W/m. C(0.008 m < 0. (9 W/m. C V ρ p h c p c (0,500 kg/m t 8 s 0.5 min 5 cm 5 cm 5 cm W/m. C s (5 J/kg. C(0.008 m - Air h, T T ( t T T T i e bt ectangular prism: c V A h Bi k b surface (0,500 kg/m c ha ρc 5 e 0 - ( s t ( W/m. C( m < 0. (9 W/m. C V ρ p t 8 s 0.5 min (0.0 m(0.05 m(0.06 m (0.0 m(0.05 m + (0.0 m(0.06 m + (0.05 m(0.06 m h c p c W/m. C s (5 J/kg. C( m - cm 5 cm 6 cm m Air h, T T ( t T T T i e bt 5 e 0 - ( s t t 6s 9. min The heating times are same for the sphere and cube while it is smaller in rectangular prism. POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

51 -9-6 Tomatoes are placed into cold water to cool them. The heat transfer coefficient and the amount of heat transfer are to be determined. Assumptions The tomatoes are spherical in shape. Heat conduction in the tomatoes is one-dimensional because of symmetry about the midpoint. The thermal properties of the tomatoes are constant. The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0. so that the one-term approximate solutions (or the transient temperature charts are applicable (this assumption will be verified. Properties The properties of the tomatoes are given to be k 0.59 W/m. C, α m /s, ρ 999 kg/m and c p.99 kj/kg. C. Analysis The Fourier number is αt τ r o 6 (0. 0 m /s( 600 s 0.65 (0.0 m which is greater than 0.. Therefore one-term solution is applicable. The ratio of the dimensionless temperatures at the surface and center of the tomatoes are θ θ T T sin( λ s λ τ Ae s, sph Ti T Ts T λ sin( λ 0,sph T0 T T λ 0 T τ A λ e T T i Substituting, 7. 7 sin( λ λ λ From Table -, the corresponding Biot number and the heat transfer coefficient are Bi. hr Bi k o h kbi r o (0.59 W/m. C(. 59 W/m. C (0.0 m The maximum amount of heat transfer is m 8ρV 8ρπD / 6 8(999 kg/m [ π (0.08 m / 6]. kg Qmax mc p[ Ti T ] (. kg(.99 kj/kg. C(0 7 C 96.6 kj Then the actual amount of heat transfer becomes Q Q max cyl T0 T Ti T Q Q max Q (96.6 kj 88 kj sin λ λ cos λ 0 7 sin(.00 (.00 cos( λ (.00 Water 7 C Tomato T i 0 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

52 -5 - A long cylindrical shaft at 00 C is allowed to cool slowly. The center temperature and the heat transfer per unit length of the cylinder are to be determined. Assumptions Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the center line. The thermal properties of the shaft are constant. The heat transfer coefficient is constant and uniform over the entire surface. The Fourier number is τ > 0. so that the one-term approximate solutions (or the transient temperature charts are applicable (this assumption will be verified. Properties The properties of stainless steel 0 at room temperature are given to be k.9 W/m. C, ρ 7900 kg/m, c p 77 J/kg. C, α m /s Analysis First the Biot number is calculated to be hr Bi k o (60 W/m. C(0.75 m (.9 W/m. C The constants λ and A corresponding to this Biot number are, from Table -, λ.090 and A.58 The Fourier number is 6 αt (.95 0 m /s(0 60 s τ 0.58 (0.75 m which is very close to the value of 0.. Therefore, the one-term approximate solution (or the transient temperature charts can still be used, with the understanding that the error involved will be a little more than percent. Then the temperature at the center of the shaft becomes T0 T λ τ (.090 (0.58 θ 0, cyl Ae (.58 e Ti T T T0 90 C The maximum heat can be transferred from the cylinder per meter of its length is m ρv ρπro (7900 kg/m [ π (0.75 m (m] 760. kg Q mc [ T T ] (760. kg(0.77 kj/kg. C(00 50 C 90,60 kj max p i Once the constant J is determined from Table - corresponding to the constant λ.090, the actual heat transfer becomes Q Q max cyl To T Ti T J( λ λ Q 0.76(90,60 kj 5,960 kj Air T 50 C Steel shaft T i 00 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

53 -0-6 A rib is roasted in an oven. The heat transfer coefficient at the surface of the rib, the temperature of the outer surface of the rib and the amount of heat transfer when it is rare done are to be determined. The time it will take to roast this rib to medium level is also to be determined. Assumptions The rib is a homogeneous spherical object. Heat conduction in the rib is one-dimensional because of symmetry about the midpoint. The thermal properties of the rib are constant. The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0. so that the one-term approximate solutions (or the transient temperature charts are applicable (this assumption will be verified. Properties The properties of the rib are given to be k 0.5 W/m. C, ρ 00 kg/m, c p. kj/kg. C, and α m /s. Analysis (a The radius of the roast is determined to be m. kg m ρv V m ρ 00 kg/m V ( m V πr o r o m π π The Fourier number is 7 αt (0.9 0 m /s( s τ 0.7 ( m r o which is somewhat below the value of 0.. Therefore, the one-term approximate solution (or the transient temperature charts can still be used, with the understanding that the error involved will be a little more than percent. Then the one-term solution can be written in the form T0 T λ τ 60 6 λ (0.7 θ 0, sph Ae Ae Ti T.5 6 It is determined from Table - by trial and error that this equation is satisfied when Bi 0, which corresponds to λ.07 and A Then the heat transfer coefficient can be determined from hr Bi k o h kbi r o (0.5 W/m. C( W/m. C ( m This value seems to be larger than expected for problems of this kind. This is probably due to the Fourier number being less than 0.. (b The temperature at the surface of the rib is T ( r, sin( / o t T λ sin(.07 rad τ λro ro (.07 (0.7 θ ( ro, t sph Ae (.9898 e Ti T λro / ro.07 T ( ro, t T ( ro, t 59.5 C.5 6 (c The maximum possible heat transfer is Q max mc p ( T Ti (. kg(. kj/kg. C(6.5 C 080 kj Then the actual amount of heat transfer becomes Q sin( λ λ cos( λ sin(.07 (.07 cos(.07 θ o, sph ( Q λ (.07 max Q 0.78Q max (0.78(080 kj 69 kj (d The cooking time for medium-done rib is determined to be T0 T λ 7 6 τ θ 0, sph Ae (.9898 e T T.5 6 i τro (0.6( m t α 7 (0.9 0 m /s (.07 τ 0,866 s 8 min hr Oven 6 C τ 0.6 ib.5 C This result is close to the listed value of hours and 0 minutes. The difference between the two results is due to the Fourier number being less than 0. and thus the error in the one-term approximation. Discussion The temperature of the outer parts of the rib is greater than that of the inner parts of the rib after it is taken out of the oven. Therefore, there will be a heat transfer from outer parts of the rib to the inner parts as a result of this temperature difference. The recommendation is logical. POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

54 -7-5 A person puts apples into the freezer to cool them quickly. The center and surface temperatures of the apples, and the amount of heat transfer from each apple in h are to be determined. Assumptions The apples are spherical in shape with a diameter of 9 cm. Heat conduction in the apples is one-dimensional because of symmetry about the midpoint. The thermal properties of the apples are constant. The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0. so that the one-term approximate solutions (or the transient temperature charts are applicable (this assumption will be verified. Properties The properties of the apples are given to be k 0.8 W/m. C, ρ 80 kg/m, c p.8 kj/kg. C, and α. 0-7 m /s. Analysis The Biot number is Air hro (8 W/m. C(0.05 m Bi 0.86 T -5 C k (0.8 W/m. C The constants λ and A corresponding to this Apple Biot number are, from Table -, T i 0 C λ.76 and A.90 The Fourier number is 7 αt (. 0 m /s(h 600 s/h τ 0. > 0. r o (0.05 m Then the temperature at the center of the apples becomes T0 T 0 ( 5 λ τ T (.76 (0. θ 0, sph Ae (.9 e T0. C Ti T 0 ( 5 The temperature at the surface of the apples is θ ( r o, t sph T( ro, t T T T i A e T ( ro, t ( T ( r 0 ( 5 λ τ The maximum possible heat transfer is m ρv ρ πr Q max mc o ( T p i o sin( λ r λ r o o / r, t.7 C / r o o (.9 e (.76 (0.. (80 kg/m π (0.05 m 0.06 kg (0.06 kg(.8 kj/kg. C T Then the actual amount of heat transfer becomes Q Q max θ Q 0.0Q max o, sph (0.0(.75 kj 7. kj sin(.76 rad.76 [ 0 ( 5 ] C.75 kj sin( λ λ cos( λ sin(.76 rad (.76 cos(.76 rad ( λ (.76 POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

55 -90-0 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve temperature to drop to specified temperatures and the maximum heat transfer are to be determined. Assumptions The thermal properties of the valves are constant. The heat transfer coefficient is constant and uniform over the entire surface. Depending on the size of the oil bath, the oil bath temperature will increase during quenching. However, an average canstant temperature as specified in the problem will be used. The Biot number is Bi < 0. so that the lumped system analysis is applicable (this assumption will be verified. Properties The thermal conductivity, density, and specific heat of the balls are given to be k 8 W/m. C, ρ 780 kg/m, and c p 0 J/kg. C. Analysis (a The characteristic length of the balls and the Biot number are c V.8( πd /.8D.8(0.008 m m A πd 8 8 s hc (800 W/m. C(0.008 m Bi 0.0 < 0. k 8 W/m. C Therefore, we can use lumped system analysis. Then the time for a final valve temperature of 00 C becomes has 8h 8(800 W/m. C b 0.88 s ρc pv.8ρc D p.8(780 kg/m (0 J/kg. C(0.008 m T ( t T bt (0.88 s t e e t 5.9 s T T i (b The time for a final valve temperature of 00 C is T ( t T bt (0.88 s e e T T i (c The time for a final valve temperature of 5 C is T ( t T bt 5 50 (0.88 s e e T T i - t t t.5 s t 5. s (d The maximum amount of heat transfer from a single valve is determined from.8πd.8π (0.008 m (0.0 m m ρv ρ (780 kg/m kg Q mc [ T T ] ( kg(0 J/kg. C( C,00 J. kj (per valve p f i Oil T 50 C - Engine valve T i 800 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

56 - - Wind is blowing parallel to the wall of a house. The rate of heat loss from that wall is to be determined for two cases. Assumptions Steady operating conditions exist. The critical eynolds number is e cr adiation effects are negligible. Air is an ideal gas with constant properties. Properties The properties of air at atm and the film temperature of (T s + T / (+5/ Air 8.5 C are (Table A- V 55 km/h T k 0.08 W/m C 5 C -5 ν. 0 m /s T s C Pr 0.70 Analysis Air flows parallel to the 0 m side: The eynolds number in this case is V [( / 600m/s](0 m 7 e.08 0 ν 5. 0 m /s which is greater than the critical eynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient and then heat transfer rate are determined to be h 0.8 / Nu (0.07 e 87 Pr [0.07( ](0.70 k k 0.08 W/m. C h Nu (.6 0. W/m. C 0 m As w ( m(0 m 0 m ha ( T T (. W/m If the wind velocity is doubled: s s. C(0 m ( 5 C 9080 W V [(0 000 / 600m/s](0 m e.6 0 ν 5. 0 m /s 7 / 9.08 kw.6 0 which is greater than the critical eynolds number. Thus we have combined laminar and turbulent flow. Using the proper relation for Nusselt number, the average heat transfer coefficient and the heat transfer rate are determined to be h 0.8 / Nu (0.07 e 87 Pr [0.07( ](0.70 k k 0.08 W/m. C h Nu ( W/m. C 0 m Q & ha ( T (57.88 W/m. C(0 m ( 5 C 6,0 W 6. kw s T s /.8 0 POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

57 -9-8 Ambient air flows over parallel plates of a solar collector that is maintained at a specified temperature. The rates of convection heat transfer from the first and third plate are to be determined. Assumptions Steady operating conditions exist. The critical eynolds number is m e cr adiation effects are negligible. Atmospheric pressure is taken atm. V, T Properties The properties of air at the film temperature of (5+0/.5 C are m (Table A- k W/m. C ν.8 0 Pr 0.70 Analysis (a The critical length of the plate is first determined to be -5 m /s 5 5 e cr ν (5 0 (.8 0 m /s xcr.6 m V m/s Therefore, both plates are under laminar flow. The eynolds number for the first plate is V ( m/s(m 5 e.8 0 ν m /s Using the relation for Nusselt number, the average heat transfer coefficient and the heat transfer rate are determined to be Nu h 0.66 e k / Pr / 0.66( / ( W/m. C Nu ( W/m. C m /.5 A w ( m(m m ha( Ts T (5.7 W/m. C( m (5 0 C 09 W (b epeating the calculations for the second and third plates, V ( m/s( m 5 e.76 0 ν m /s / / 5 / / Nu 0.66 e Pr 0.66(.76 0 ( k W/m. C h Nu (.7.87 W/m. C m V ( m/s( m 5 e. 0 ν m /s / / 5 / / Nu 0.66 e Pr 0.66(. 0 ( k W/m. C h Nu (85..6 W/m. C m Then h h.6.87 h.7 W/m. C The rate of heat loss from the third plate is ha( T T (.7 W/m. C( m (5 0 C.8 W s POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

58 -5-5 A circuit board is cooled by air. The surface temperatures of the electronic components at the leading edge and the end of the board are to be determined. Assumptions Steady operating conditions exist. The critical eynolds number is e cr adiation effects are negligible. Any heat transfer from the back surface of the board is disregarded. 5 Air is an ideal gas with constant properties. Properties Assuming the film temperature to be approximately 5 C, the properties of air are evaluated at this temperature to be (Table A- k W/m. C ν Pr Analysis (a The convection heat transfer coefficient at the leading edge approaches infinity, and thus the surface temperature there must approach the air temperature, which is 0 C. (b The eynolds number is Vx (6 m/s(0.5 m e x ν m /s m /s which is less than the critical eynolds number but we assume the flow to be turbulent since the electronic components are expected to act as turbulators. Using the Nusselt number uniform heat flux, the local heat transfer coefficient at the end of the board is determined to be hx x 0.8 / 0.8 / Nu x e x Pr 0.008(5.8 0 ( k k x W/m. C hx Nu x ( W/m. C x 0.5 m Then the surface temperature at the end of the board becomes q& hx ( Ts T Ts T q& + h x (0 W/(0.5 m 0 C W/m. C 9.9 C Discussion The heat flux can also be determined approximately using the relation for isothermal surfaces, hx x 0.8 / 0.8 / Nu x e x Pr 0.096(5.8 0 ( k k x W/m. C hx Nu x ( W/m. C x 0.5 m Then the surface temperature at the end of the board becomes q& hx ( Ts T Ts T q& + h Note that the two results are close to each other. x Air 0 C 6 m/s (0 W/(0.5 m 0 C W/m. C Circuit board 0 W 5. C 5 cm 5 cm POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

59 -6-68 A steam pipe is exposed to windy air. The rate of heat loss from the steam is to be determined. Assumptions Steady operating conditions exist. adiation effects are negligible. Air is an ideal gas with constant properties. Properties The properties of air at atm and the film temperature of (T s + T / (90+7/ 8.5 C are (Table A- k 0.07 W/m. C ν.78 0 Pr m Analysis The eynolds number is /s VD [(50 km/h(000 m/km/(600 s/h](0.08 m e ν m /s The Nusselt number corresponding to this eynolds number is Nu hd k e 0.5 Pr [ + ( 0. / Pr ] / 0.6( / (0.7 [ + ( 0. / 0.7 ] / / / e 8,000 / 5 / 8 / ,000 The heat transfer coefficient and the heat transfer rate become h k Nu D A s conv 0.07 W/m. C ( W/m. C 0.08 m πd π (0.08 m(m 0.5 m ha ( T s s T 5 / 8 / 5 Air V 50 km/h T 7 C 59. (5.7 W/m. C(0.5 m (90 7 C 0 W (per m length Pipe D 8 cm T s 90 C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and

60 - -7 The average surface temperature of the head of a person when it is not covered and is subjected to winds is to be determined. Assumptions Steady operating conditions exist. adiation effects are negligible. Air is an ideal gas with constant properties. One-quarter of the heat the person generates is lost from the head. 5 The head can be approximated as a 0-cm-diameter sphere. 6 The local atmospheric pressure is atm. Properties We assume the surface temperature to be 5 C for viscosity. The properties of air at atm pressure and the free stream temperature of 0 C are (Table A- μ C k 0.09 W/m. C ν.6 0 μ Pr Analysis The eynolds number is m /s kg/m.s kg/m.s [(5 000/600 m/s] (0. m 5 VD e.6 0 ν m /s The proper relation for Nusselt number corresponding to this eynolds number is Nu hd k + + The heat transfer coefficient is k h Nu D 0.5 / [ 0. e e ] Pr 0. μ μ s / / / 0. [ 0.( (.6 0 ]( W/m. C (8..0 W/m. C 0. m Then the surface temperature of the head is determined to be As πd π (0. m 0.87 m has ( Ts T Ts T + ha s Air V 5 km/h T 0 C 0 C + (.0 W/m (8/ W C(0.87 m 5 Head Q W D 0. m. C POPIETAY MATEIA. 008 The McGraw-Hill Companies, Inc. imited distribution permitted only to teachers and