300 kpa 77 C. (d) If we neglect kinetic energy in the calculation of energy transport by mass

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Milton Hunter
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1 6-6- Air flows steadily a ie at a secified state. The diameter of the ie, the rate of flow energy, and the rate of energy transort by mass are to be determed. Also, the error oled the determation of energy transort by mass is to be determed. Proerties The roerties of air are R 0.87 kj.k and c.008 kj.k (at 50 K from Table A-b) Analysis (a) The diameter is determed as follows RT P m A D 4A π 0.87 kj.k)(77 7 K) 0.49 m (00 kpa) ( (8 / 60 kg/s)(0.49 m 5 m/s 4( m π ) m ) m (b) The rate of flow energy is determed from W flow mp (8/ 60 kg/s)(00 kpa)(0.49 m ) 0.4 kw (c) The rate of energy transort by mass is mass ( h ke) ct 00 kpa 77 C kj (8/60 kg/s) (.008 kj.k)(77 7 K) (5 m/s) 000 m /s kw (d) If we neglect ketic energy the calculation of energy transort by mass mass mh mc T (8/60 kg/s)(.005 kj.k)(77 7 K) kw Therefore, the error oled if neglect the ketic energy is only 0.09%. Air 5 m/s 8 kg/m PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

2 6-6-7 Air is decelerated a diffuser from 0 m/s to 0 m/s. The exit temerature of air and the exit area of the diffuser are to be determed. Assumtions This is a steady-flow rocess sce there is no change with time. Air is an ideal gas with ariable secific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Proerties The gas constant of air is 0.87 kpa.m.k (Table A-). The enthaly of air at the let temerature of 400 K is h kj (Table A-). Analysis (a) There is only one let and one exit, and thus. We take diffuser as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be exressed the rate form as or, E 44 E ΔEsystem 0 h ( h From Table A-, Rate of change ternal, ketic, otential, etc.energies 4444 / ) ( h /) (sce W Δe, 0 h h ( 0 m/s) ( 0 m/s) kj 000 m /s h kj T 45.6 K (b) The secific olume of air at the diffuser exit is RT ( kpa m K)( 45.6 K) ( 00 kpa) m P From conseration of mass, AIR kj A A ( kg/s)(. m 0 m/s ) m PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

3 Steam exands a turbe. The change ketic energy, the ower ut, and the turbe let area are to be determed. Assumtions This is a steady-flow rocess sce there is no change with time. Potential energy changes are negligible. The deice is adiabatic and thus heat transfer is negligible. Proerties From the steam tables (Tables A-4 through 6) and P 0 MPa m T 450 C h 4.4 kj P 0 MPa T 450 C 80 m/s P 0 kpa h x 0.9 h x f h fg kj Analysis (a) The change ketic energy is determed from Δke ( 50 m/s) (80 m/s) kj 000 m /s.95 kj (b) There is only one let and one exit, and thus. We take the turbe as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be exressed the rate form as E 44 E ΔEsystem 0 ( h / ) W Rate of change ternal, ketic, otential, etc.energies W h 4444 ( h h /) (sce Δe Then the ower ut of the turbe is determed by substitution to be W ( kg/s)( ) kj 0. MW (c) The let area of the turbe is determed from the mass flow rate relation, STEAM m kg/s P 0 kpa x m/s W A A ( kg/s)( m 80 m/s ) m PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

4 Air is comressed an adiabatic comressor. The mass flow rate of the air and the ower ut are to be determed. Assumtions This is a steady-flow rocess sce there is no change with time. The comressor is adiabatic. Air is an ideal gas with constant secific heats. Proerties The constant ressure secific heat of air at the aerage temerature of (0400)/0 C48 K is c.06 kj K (Table A-b). The gas constant of air is R 0.87 kpa m K (Table A-). Analysis (a) There is only one let and one exit, and thus m. We take the comressor as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steadyflow system can be exressed the rate form as E 44 E ΔEsystem 0 h W W Rate of change ternal, ketic, otential, etc. energies h h h c ( T T ) The secific olume of air at the let and the mass flow rate are RT ( 0.87 kpa m K)(0 7 K) m P 00 kpa.8 MPa 400 C Comressor 00 kpa 0 C 0 m/s A (0.5 m )(0 m/s) 5.5kg/s m Similarly at the let, RT ( 0.87 kpa m K)(400 7 K) 0.07 m P 800 kpa m (5.5 kg/s)(0.07 m ) A 0.08 m (b) Substitutg to the energy balance equation gies 7.77 m/s W c ( T T ) (7.77 m/s) (0 m/s) (5.5 kg/s) (.06 kj K)(400 0)K 084 kw kj 000 m /s PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

5 A hot water stream is mixed with a cold water stream. For a secified mixture temerature, the mass flow rate of cold water is to be determed. Assumtions Steady oeratg conditions exist. The mixg chamber is well-sulated so that heat loss to the surroundgs is negligible. Changes the ketic and otential energies of fluid streams are negligible. 4 Fluid roerties are constant. 5 There are no work teractions. Proerties Notg that T < T 50 kpa 7.4 C, the water all three streams exists as a comressed liquid, which can be aroximated as a saturated liquid at the gien temerature. Thus, h h 80 C 5.0 kj h h 0 C 8.95 kj h h 4 C kj Analysis We take the mixg chamber as the system, which is a control olume. The mass and energy balances for this steady-flow system can be exressed the rate form as Mass balance: Energy balance: Δ system 0 m E E 44 h h Rate of change ternal, ketic, otential, etc. energies h 0 ΔEsystem (sce W Δke Δe Combg the two relations and solg for m gies ( ) m h m h h h h h h m Substitutg, the mass flow rate of cold water stream is determed to be ( ) kj ( 0.5 kg/s) kg/s kj ( ) T 80 C m 0.5 kg/s T 0 C m H O (P 50 kpa) T 4 C PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

6 Oil is to be cooled by water a th-walled heat exchanger. The rate of heat transfer the heat exchanger and the exit temerature of water is to be determed. Assumtions Steady oeratg conditions exist. The heat exchanger is well-sulated so that heat loss to the surroundgs is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Changes the ketic and otential energies of fluid streams are negligible. 4 Fluid roerties are constant. Proerties The secific heats of water and oil are gien to be 4.8 and.0 kj. C, resectiely. Analysis We take the oil tubes as the system, which is a control olume. The energy balance for this steady-flow system can be exressed the rate form as E E 44 mh 0 ΔEsystem 4444 Rate of change ternal, ketic, otential, etc.energies 0 mh (sce Δke Δe mc ( T T ) Then the rate of heat transfer from the oil becomes [ mc ( T T )] oil ( kg/s)(. kj. C)(50 C 40 C) 484 kw Notg that the heat lost by the oil is gaed by the water, the let temerature of the water is determed from 484 kj/s [ mc ( T T )] water T T C 99. C c (.5 kg/s)(4.8 kj. C) water Cold water C.5 kg/s 40 C Hot oil 50 C kg/s PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

7 The mass flow rate of a comressed air le is diided to two equal streams by a T-fittg the le. The elocity of the air at the lets and the rate of change of flow energy (flow ower) across the T- fittg are to be determed. Assumtions Air is an ideal gas with constant secific heats. The flow is steady. Sce the lets are identical, it is resumed that the flow diides eenly between the two. Proerties The gas constant of air is R 0.87 kpa m K (Table A-). Analysis The secific olumes of air at the let and lets are RT ( 0.87 kpa m K)(40 7 K) m P 600 kpa RT ( 0.87 kpa m K)(6 7 K) m P 400 kpa Assumg an een diision of the let flow rate, the mass balance can be written as A A A m/s A The mass flow rate at the let is A πd π (0.05 m) 50 m/s m 0.47 kg/s.6 MPa 40 C 50 m/s while that at the lets is 0.47 kg/s 0.86 kg/s Substitutg the aboe results to the flow ower exression roduces W flow P P (0.86 kg/s)(400 kpa)(0.065 m ) (0.47 kg/s)(600 kpa)( m ) kw.4 MPa 6 C.4 MPa 6 C PROPRIETARY MATERIAL. 008 The McGraw-Hill Comanies, Inc. Limited distribution ermitted only to teachers and educators for course rearation. If you are a student usg this Manual, you are usg it with ermission.

The average velocity of water in the tube and the Reynolds number are Hot R-134a

The average velocity of water in the tube and the Reynolds number are Hot R-134a hater 0:, 8, 4, 47, 50, 5, 55, 7, 75, 77, 8 and 85. 0- Refrigerant-4a is cooled by water a double-ie heat exchanger. he overall heat transfer coefficient is to be determed. Assumtions he thermal resistance