Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

Transcription

1 5- Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL OLUMES Conseration of Mass 5-C Mass, energy, momentum, and electric charge are consered, and olume and entropy are not consered durg a process. 5-C Mass flow rate is the amount of mass flowg through a cross-section per unit time whereas the olume flow rate is the amount of olume flowg through a cross-section per unit time. 5-C The amount of mass or energy enterg a control olume does not hae to be equal to the amount of mass or energy leag durg an unsteady-flow process. 5-4C Flow through a control olume is steady when it oles no changes with time at any specified position. 5-5C No, a flow with the same olume flow rate at the let and the exit is not necessarily steady (unless the density is constant). To be steady, the mass flow rate through the deice must rema constant. 5-6E A garden hose is used to fill a water bucket. The olume and mass flow rates of water, the fillg time, and the discharge elocity are to be determed. Assumptions Water is an compressible substance. Flow through the hose is steady. There is no waste of water by splashg. Properties We take the density of water to be 6.4 lbm/ft (Table A-E). Analysis (a) The olume and mass flow rates of water are A ( πd / 4) [ π (/ ft) / 4](8 ft/s) ft /s ρ (6.4 lbm/ft )(0.046 ft /s).7 lbm/s (b) The time it takes to fill a 0-gallon bucket is 0 gal t ft ft 6. s /s gal (c) The aerage discharge elocity of water at the nozzle exit is e A e πd e / ft /s ft/s [ π (0.5 / ft) / 4] Discussion Note that for a gien flow rate, the aerage elocity is ersely proportional to the square of the elocity. Therefore, when the diameter is reduced by half, the elocity quadruples.

2 5-5-7 Air is accelerated a nozzle. The mass flow rate and the exit area of the nozzle are to be determed. Assumptions Flow through the nozzle is steady. Properties The density of air is gien to be. kg/m at the let, and 0.76 kg/m at the exit. Analysis (a) The mass flow rate of air is determed from the let conditions to be m ρ A (. kg/m )(0.009 m )(40 m/s) kg/s (b) There is only one let and one exit, and thus Then the exit area of the nozzle is determed to be ρ m. 40 m/s A 90 cm kg/s m (0.76 kg/m )(80 m/s) A A ρ AIR 58 cm 80 m/s 5-8 Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent crease the elocity of air as it flows through the drier is to be determed. Assumptions Flow through the nozzle is steady. Properties The density of air is gien to be.0 kg/m at the let, and.05 kg/m at the exit. Analysis There is only one let and one exit, and thus. Then, m ρ A ρ A ρ ρ.0 kg/m.05 kg/m.4 (or, and crease of 4% ) Therefore, the air elocity creases 4% as it flows through the hair drier. 5-9E The ducts of an air-conditiong system pass through an open area. The let elocity and the mass flow rate of air are to be determed. Assumptions Flow through the air conditiong duct is steady. Properties The density of air is gien to be lbm/ft at the let. Analysis The let elocity of air and the mass flow rate through the duct are A π D / 4 π 450 ft /m ( 0/ ft) / 4 85 ft/m.8 ft/s 450 ft /m m ρ (0.078 lbm/ft )(450 ft / m) 5. lbm/m lbm/s AIR D 0

3 5-5-0 A rigid tank itially contas air at atmospheric conditions. The tank is connected to a supply le, and air is allowed to enter the tank until the density rises to a specified leel. The mass of air that entered the tank is to be determed. Properties The density of air is gien to be.8 kg/m at the begng, and 7.0 kg/m at the end. Analysis We take the tank as the system, which is a control olume sce mass crosses the boundary. The mass balance for this system can be expressed as Mass balance: Substitutg, m i m m m ρ system mi m m ρ ( ρ ρ ) [( ) kg/m ](m ) 6.0 kg Therefore, 6.0 kg of mass entered the tank. m ρ.8 kg/m 5- The entilatg fan of the bathroom of a buildg runs contuously. The mass of air ented per day is to be determed. Assumptions Flow through the fan is steady. Properties The density of air the buildg is gien to be.0 kg/m. Analysis The mass flow rate of air ented is air ρ air (.0 kg/m )(0.00 m /s) 0.06 kg/s Then the mass of air ented 4 h becomes m air t (0.06 kg/s)(4 600 s) 0 kg Discussion Note that more than tons of air is ented by a bathroom fan one day. 5- A desktop computer is to be cooled by a fan at a high eleation where the air density is low. The mass flow rate of air through the fan and the diameter of the casg for a gien elocity are to be determed. Assumptions Flow through the fan is steady. Properties The density of air at a high eleation is gien to be 0.7 kg/m. Analysis The mass flow rate of air is air ρ air (0.7 kg/m )(0.4 m /m) 0.8 kg/m kg/s If the mean elocity is 0 m/m, the diameter of the casg is πd A 4 D 4 π 4(0.4 m /m) π (0 m/m) 0.06 m Therefore, the diameter of the casg must be at least 6. cm to ensure that the mean elocity does not exceed 0 m/m. Discussion This problem shows that engeerg systems are sized to satisfy certa constrats imposed by certa considerations.

4 A smokg lounge that can accommodate 5 smokers is considered. The required mimum flow rate of air that needs to be supplied to the lounge and the diameter of the duct are to be determed. Assumptions Infiltration of air to the smokg lounge is negligible. Properties The mimum fresh air requirements for a smokg lounge is gien to be 0 L/s per person. Analysis The required mimum flow rate of air that needs to be supplied to the lounge is determed directly from air air per person (No. of persons) (0 L/s person)(5 persons) 450 L/s 0.45 m The olume flow rate of fresh air can be expressed as A ( πd / 4) Solg for the diameter D and substitutg, D 4 π 4(0.45 m /s) π (8 m/s) 0.68 m /s Smokg Lounge 5 smokers 0 L/s person Therefore, the diameter of the fresh air duct should be at least 6.8 cm if the elocity of air is not to exceed 8 m/s. 5-4 The mimum fresh air requirements of a residential buildg is specified to be 0.5 air changes per hour. The size of the fan that needs to be stalled and the diameter of the duct are to be determed. Analysis The olume of the buildg and the required mimum olume flow rate of fresh air are room (.7 m)(00 m ) 540 m ACH (540 m )(0.5/h) 89 m / h 89,000 L/h 50 L/m room The olume flow rate of fresh air can be expressed as A ( πd / 4) Solg for the diameter D and substitutg, D 4 π 4(89 / 600 m π (6 m/s) /s) 0.06 m Therefore, the diameter of the fresh air duct should be at least 0.6 cm if the elocity of air is not to exceed 6 m/s. 0.5 ACH House 00 m

5 Air flows through a pipe. Heat is supplied to the air. The olume flow rates of air at the let and exit, the elocity at the exit, and the mass flow rate are to be determed. Air 00 kpa 0 C 5 m/s Q 80 kpa 40 C Properties The gas constant for air is 0.87 kj/kg.k (Table A-). Analysis (a) (b) The olume flow rate at the let and the mass flow rate are A ρ A c c πd 4 P RT π (0.8 m) 4 πd 4 (5 m/s) m /s (00 kpa) π (0.8 m) (0.87 kj/kg.k)(0 + 7 K) 4 (5 m/s) 0.78 kg/s (c) Notg that mass flow rate is constant, the olume flow rate and the elocity at the exit of the pipe are determed from m ρ A c P RT 0.78 kg/s (80 kpa) (0.87 kj/kg.k)( K) m /s 5.94 m/s π (0.8 m) m /s

6 Refrigerant-4a flows through a pipe. Heat is supplied to R-4a. The olume flow rates of air at the let and exit, the mass flow rate, and the elocity at the exit are to be determed. R-4a 00 kpa 0 C 5 m/s Q 80 kpa 40 C Properties The specific olumes of R-4a at the let and exit are (Table A-) P 00 kpa 0.4 m /kg T 0 C P 80 kpa 0.74 m /kg T 40 C Analysis (a) (b) The olume flow rate at the let and the mass flow rate are A c Ac πd 4 πd 4 π (0.8 m) 4 (5 m/s) m /s π (0.8 m) 0.4 m /kg 4 (5 m/s).696 kg/s (c) Notg that mass flow rate is constant, the olume flow rate and the elocity at the exit of the pipe are determed from (.696 kg/s)(0.74 m /kg) m /s 6.0 m/s A c π (0.8 m) m /s 5-7 Warm water is withdrawn from a solar water storage tank while cold water enters the tank. The amount of water the tank a 0-mute period is to be determed. Properties The density of water is taken to be 000 kg/m for both cold and warm water. Analysis The itial mass the tank is first determed from m ρ tank (000 kg/m )(0. m ) 00 kg The amount of warm water leag the tank durg a 0-m period is m e π (0.0 m) ρa c t (000 kg/m ) 4 Cold water 0 C 5 L/m (0.5 m/s)(0 60 s) 88.5 kg The amount of cold water enterg the tank durg a 0-m period is mi ρ c t (000 kg/m )(0.005 m /m)(0 m) 00 kg The fal mass the tank can be determed from a mass balance as 00 L 45 C m m m m m m + m m kg i e i e Warm water 45 C 0.5 m/s

7 5-7 Flow Work and Energy Transfer by Mass 5-8C Energy can be transferred to or from a control olume as heat, arious forms of work, and by mass. 5-9C Flow energy or flow work is the energy needed to push a fluid to or of a control olume. Fluids at rest do not possess any flow energy. 5-0C Flowg fluids possess flow energy addition to the forms of energy a fluid at rest possesses. The total energy of a fluid at rest consists of ternal, ketic, and potential energies. The total energy of a flowg fluid consists of ternal, ketic, potential, and flow energies. 5-E Steam is leag a pressure cooker at a specified pressure. The elocity, flow rate, the total and flow energies, and the rate of energy transfer by mass are to be determed. Assumptions The flow is steady, and the itial start-up period is disregarded. The ketic and potential energies are negligible, and thus they are not considered. Saturation conditions exist with the cooker at all times so that steam leaes the cooker as a saturated apor at 0 psia. Properties The properties of saturated liquid water and water apor at 0 psia are f ft /lbm, g.749 ft /lbm, u g Btu/lbm, and h g 64. Btu/lbm (Table A-5E). Analysis (a) Saturation conditions exist a pressure cooker at all times after the steady operatg conditions are established. Therefore, the liquid has the properties of saturated liquid and the exitg steam has the properties of saturated apor at the operatg pressure. The amount of liquid that has eaporated, the mass flow rate of the exitg steam, and the exit elocity are m m t g liquid f ρ A c 0.4 gal 0.68 ft ft /lbm gal.45 lbm 45 m g A c lbm/m.65 0 ( lbm - lbm/s lbm/s)(.749 ft /lbm) ft 5.4 ft/s H O Sat. apor P 0 psia (b) Notg that h u + P and that the ketic and potential energies are disregarded, the flow and total energies of the exitg steam are e flow P h u Btu/lbm θ h + ke + pe h 64. Btu/lbm Note that the ketic energy this case is ke / (5.4 ft/s) 7 ft /s Btu/lbm, which is ery small compared to enthalpy. (c) The rate at which energy is leag the cooker by mass is simply the product of the mass flow rate and the total energy of the exitg steam per unit mass, E mass θ (.65 0 lbm/s)(64. Btu/lbm).56 Btu/s Discussion The numerical alue of the energy leag the cooker with steam alone does not mean much sce this alue depends on the reference pot selected for enthalpy (it could een be negatie). The significant quantity is the difference between the enthalpies of the exitg apor and the liquid side (which is h fg ) sce it relates directly to the amount of energy supplied to the cooker. Q

8 Refrigerant-4a enters a compressor as a saturated apor at a specified pressure, and leaes as superheated apor at a specified rate. The rates of energy transfer by mass to and of the compressor are to be determed. Assumptions The flow of the refrigerant through the compressor is steady. The ketic and potential energies are negligible, and thus they are not considered. Properties The enthalpy of refrigerant-4a at the let and the exit are (Tables A- and A-) P 0.8 MPa h h MPa 9.6 kj/kg h 96.8 kj/kg T 60 C Analysis Notg that the total energy of a flowg fluid is equal to its enthalpy when the ketic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass to and of the compressor are θ h (0.06 kg/s)(9.6 kj/kg) 4.5 kj/s 4.5 kw mass, θ h mass, (0.06 kg/s)(96.8 kj/kg) 7.8 kj/s 7.8kW R-4a compressor () 0.4 MPa () 0.8 MPa 60 C Discussion The numerical alues of the energy enterg or leag a deice by mass alone does not mean much sce this alue depends on the reference pot selected for enthalpy (it could een be negatie). The significant quantity here is the difference between the gog and comg energy flow rates, which is E kw mass mass, mass, This quantity represents the rate of energy transfer to the refrigerant the compressor. 5- Warm air a house is forced to leae by the filtratg cold side air at a specified rate. The net energy loss due to mass transfer is to be determed. Assumptions The flow of the air to and of the house through the cracks is steady. The ketic and potential energies are negligible. Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R 0.87 kpa m /kg K (Table A-). The constant pressure specific heat of air at room temperature is c p.005 kj/kg C (Table A-). Analysis The density of air at the door conditions and its mass flow rate are P 0.5 kpa ρ.89 kg/m RT (0.87 kpa m /kg K)(4 + 7)K m ρ (.89 kg/m )(50 m /h) 78.5 kg/h kg/s Notg that the total energy of a flowg fluid is equal to its enthalpy when the ketic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass to and of the house by air are θ h mass, θ h mass, Cold air 5 C Warm air 4 C Warm air 4 C The net energy loss by air filtration is equal to the difference between the gog and comg energy flow rates, which is h h ) mc ( T T ) mass mass, mass, ( p ( kg/s)(.005 kj/kg C)(4-5) C kj/s kw This quantity represents the rate of energy transfer to the refrigerant the compressor. Discussion The rate of energy loss by filtration will be less reality sce some air will leae the house before it is fully heated to 4 C.

9 Air flows steadily 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 determed. Also, the error oled the determation of energy transport by mass is to be determed. Properties The properties of air are R 0.87 kj/kg.k and c p.008 kj/kg.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/kg.k)( K) 0.49 m (00 kpa) (8 / 60 kg/s)(0.49 m 5 m/s 4( m π 00 kpa 77 C /kg /kg) m ) m (b) The rate of flow energy is determed from W flow mp (8/ 60 kg/s)(00 kpa)(0.49 m /kg) 0.4 kw (c) The rate of energy transport by mass is mass + ke) cpt + kj/kg (8/60 kg/s) (.008 kj/kg.k)( K) + (5 m/s) 000 m /s kw (d) If we neglect ketic energy the calculation of energy transport by mass mass mh mc pt (8/60 kg/s)(.005 kj/kg.k)( K) kw Therefore, the error oled if neglect the ketic energy is only 0.09%. Air 5 m/s 8 kg/m

10 5-0 Steady Flow Energy Balance: Nozzles and Diffusers 5-5C A steady-flow system oles no changes with time anywhere with the system or at the system boundaries 5-6C No. 5-7C It is mostly conerted to ternal energy as shown by a rise the fluid temperature. 5-8C The ketic energy of a fluid creases at the expense of the ternal energy as eidenced by a decrease the fluid temperature. 5-9C Heat transfer to the fluid as it flows through a nozzle is desirable sce it will probably crease the ketic energy of the fluid. Heat transfer from the fluid will decrease the exit elocity. 5-0 Air is accelerated a nozzle from 0 m/s to 80 m/s. The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with constant specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-). The specific heat of air at the anticipated aerage temperature of 450 K is c p.0 kj/kg. C (Table A-). Analysis (a) There is only one let and one exit, and thus. Usg the ideal gas relation, the specific olume and the mass flow rate of air are determed to be RT P ( 0.87 kpa m /kg K)(47 K) m /kg 00 kpa A (0.008 m )(0 m/s) kg/s m /kg P 00 kpa T 00 C 0 m/s A 80 cm AIR P 00 kpa 80 m/s (b) We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as 0 (steady) E E Esystem Substitutg, Rate of net energy transfer by heat, work, and mass Rate of change ternal, ketic, potential, etc.energies 0 h + / ) + /) (sce Q W pe 0) h + 0 (.0 kj/kg K)( T 00 It yields T 84.6 C (c) The specific olume of air at the nozzle exit is 0 c p, ae ( T T ) + (80 m/s) (0 m/s) C) + RT P o ( 0.87 kpa m /kg K)( K). m /kg 00 kpa kj/kg 000 m /s m A kg/s A ( 80 m/s) A m 8.7 cm. m /kg

11 5-5- EES Problem 5-0 is reconsidered. The effect of the let area on the mass flow rate, exit elocity, and the exit area as the let area aries from 50 cm to 50 cm is to be estigated, and the fal results are to be plotted agast the let area. Analysis The problem is soled usg EES, and the solution is gien below. Function HCal(WorkFluid$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'Air' WorkFluid$ then HCal:ENTHALPY('Air',TTx) "Ideal gas equ." else HCal:ENTHALPY(WorkFluid$,TTx, PPx)"Real gas equ." endif end HCal "System: control olume for the nozzle" "Property relation: Air is an ideal gas" "Process: Steady state, steady flow, adiabatic, no work" "Knowns - obta from the put diagram" WorkFluid$ 'Air' T[] 00 [C] P[] 00 [kpa] el[] 0 [m/s] P[] 00 [kpa] el[] 80 [m/s] A[]80 [cm^] Am[]A[]*conert(cm^,m^) "Property Data - sce the Enthalpy function has different parameters for ideal gas and real fluids, a function was used to determe h." h[]hcal(workfluid$,t[],p[]) h[]hcal(workfluid$,t[],p[]) "The olume function has the same form for an ideal gas as for a real fluid." []olume(workfluid$,tt[],pp[]) []olume(workfluid$,tt[],pp[]) "Conseration of mass: " m_dot[] m_dot[] "Mass flow rate" m_dot[]am[]*el[]/[] m_dot[] Am[]*el[]/[] "Conseration of Energy - SSSF energy balance" h[]+el[]^/(*000) h[]+el[]^/(*000) "Defition" A_ratioA[]/A[] A[]Am[]*conert(m^,cm^)

12 5- A [cm ] A [cm ] m T m[] A[] [cm^] A[] [cm^] A[] [cm^]

13 5-5- Steam is accelerated a nozzle from a elocity of 80 m/s. The mass flow rate, the exit elocity, and the exit area of the nozzle are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Potential energy changes are negligible. There are no work teractions. Properties From the steam tables (Table A-6) and P 5 MPa m /kg T 400 C h 96.7 kj/kg P MPa T 00 C h 0.55 m /kg 04. kj/kg Steam 0 kj/s Analysis (a) There is only one let and one exit, and thus m. The mass flow rate of steam is 4 A (80 m/s)( m /kg m ) 6.9 kg/s (b) We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass + Q / ) Rate of change ternal, ketic, potential, etc.energies Q h h + + Substitutg, the exit elocity of the steam is determed to be It yields 0 kj/s ( 6.96 kg/s) /) (sce W pe 0) m/s (c) The exit area of the nozzle is determed from A A (80 m/s) kj/kg 000 m /s ( 6.96 kg/s)( 0.55 m /kg) 56.7 m/s m

14 5-4 5-E Air is accelerated a nozzle from 50 ft/s to 900 ft/s. The exit temperature of air and the exit area of the nozzle are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 There are no work teractions. Properties The enthalpy of air at the let is h 4.47 Btu/lbm (Table A-7E). Analysis (a) There is only one let and one exit, and thus. We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as or, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h q + Q + h / ). Btu/lbm Rate of change ternal, ketic, potential, etc.energies Q h h + + /) (900 ft/s) 6.5 Btu/lbm Btu/lbm Thus, from Table A-7E, T 507 R (sce W pe 0) (50 ft/s) (b) The exit area is determed from the conseration of mass relation, A A A ( 508/4.7)( 50 ft/s) A ( )( ) ( ) 0. ft ft 600/ ft/s RT / P A A RT / P Btu/lbm 5,07 ft /s AIR 6.5 Btu/lbm

15 [Also soled by EES on enclosed CD] Steam is accelerated a nozzle from a elocity of 40 m/s to 00 m/s. The exit temperature and the ratio of the let-to-exit area of the nozzle are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Potential energy changes are negligible. There are no work teractions. 4 The deice is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Table A-6), P MPa m /kg T 400 C h.7 kj/kg Analysis (a) There is only one let and one exit, and thus. We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as or, Thus, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h Rate of change ternal, ketic, potential, etc.energies 0 h / ) + /) (sce Q W pe 0) h + P MPa T 400 C 40 m/s kj/kg 000 m /s (00 m/s) (40 m/s) h.7 kj/kg P.5 MPa T 76.6 C h 87.5 kj/kg 0.5 m /kg Steam (b) The ratio of the let to exit area is determed from the conseration of mass relation, A A A A ( m /kg)(00 m/s) 6.46 (0.5 m /kg)(40 m/s) 87.5 kj/kg P.5 MPa 00 m/s

16 Air is accelerated a nozzle from 0 m/s to 80 m/s. The exit temperature and pressure of air are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The enthalpy of air at the let temperature of 500 K is h 50.0 kj/kg (Table A-7). Analysis (a) There is only one let and one exit, and thus. We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as or, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h Rate of change ternal, ketic, potential, etc.energies 0 h / ) + /) (sce Q W pe 0) h + ( 80 m/s) ( 0 m/s) kj/kg 50.0 kj/kg 000 m /s h Then from Table A-7 we read T 46.5 K (b) The exit pressure is determed from the conseration of mass relation, Thus, A A A RT / P RT / P P A AT (46.5 K)(0 m/s) P (600 kpa) 0.8 kpa A T (500 K)(80 m/s) AIR 48.0 kj/kg

17 Air is decelerated 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 determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-). The enthalpy of air at the let temperature of 400 K is h kj/kg (Table A-7). 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 expressed the rate form as or, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h From Table A-7, Rate of change ternal, ketic, potential, etc.energies / ) + /) (sce Q W pe 0), 0 h h + ( 0 m/s) ( 0 m/s) kj/kg 000 m /s h kj/kg T 45.6 K (b) The specific olume of air at the diffuser exit is RT ( kpa m /kg K)( 45.6 K) ( 00 kpa) m /kg P From conseration of mass, A A ( kg/s)(. m 0 m/s /kg) AIR kj/kg m

18 E Air is decelerated a diffuser from 600 ft/s to a low elocity. The exit temperature and the exit elocity of air are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The enthalpy of air at the let temperature of 0 F is h 4.69 Btu/lbm (Table A-7E). 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 expressed the rate form as or, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h From Table A-7E, Rate of change ternal, ketic, potential, etc.energies / ) + /) (sce Q W pe 0), 0 h h + ( 600 ft/s) 0 Btu/lbm 4.69 Btu/lbm 5,07 ft /s h T 50.0 R (b) The exit elocity of air is determed from the conseration of mass relation, Thus, A A A RT / P RT / P A AT P (50 R)( psia) (600 ft/s) 4. ft/s A T P 5 (480 R)(4.5 psia) AIR.88 Btu/lbm

19 CO gas is accelerated a nozzle to 450 m/s. The let elocity and the exit temperature are to be determed. Assumptions This is a steady-flow process sce there is no change with time. CO is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The gas constant and molar mass of CO are kpa.m /kg.k and 44 kg/kmol (Table A-). The enthalpy of CO at 500 C is h 0,797 kj/kmol (Table A-0). Analysis (a) There is only one let and one exit, and thus. Usg the ideal gas relation, the specific olume is determed to be Thus, RT ( kpa m /kg K)( 77 K) m /kg 000 kpa P A A ( 6000/600 kg/s)( 0.46 m /kg) 60.8 m/s 40 0 (b) We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as Substitutg, E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass h Rate of change ternal, ketic, potential, etc.energies 0 h / ) + /) (sce Q W pe 0) h M 6,4 kj/kmol 0,797 kj/kmol h + ( 450 m/s) ( 60.8 m/s) m kj/kg 000 m /s Then the exit temperature of CO from Table A-0 is obtaed to be ( 44 kg/kmol) T K CO

20 R-4a is accelerated a nozzle from a elocity of 0 m/s. The exit elocity of the refrigerant and the ratio of the let-to-exit area of the nozzle are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Potential energy changes are negligible. There are no work teractions. 4 The deice is adiabatic and thus heat transfer is negligible. Properties From the refrigerant tables (Table A-) and P 700 kpa m /kg T 0 C h kj/kg P 400 kpa m /kg T 0 C h kj/kg R-4a Analysis (a) There is only one let and one exit, and thus. We take nozzle as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as Substitutg, It yields E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass 0 Rate of change ternal, ketic, potential, etc.energies 0 h / ) + /) (sce Q W pe 0) ( ) m/s h + kj/kg + ( 0 m/s) kj/kg 000 m /s (b) The ratio of the let to exit area is determed from the conseration of mass relation, A A A A ( m /kg)( m/s) ( m /kg)( 0 m/s) 5.65

21 Air is decelerated a diffuser from 0 m/s. The exit elocity and the exit pressure of air are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Air is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 There are no work teractions. Properties The gas constant of air is 0.87 kpa.m /kg.k (Table A-). The enthalpies are (Table A-7) T T 7 C 00 K h 009. kj / kg 4 C 5 K h 5. 7 kj / kg 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 expressed the rate form as E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass + Q Rate of change ternal, ketic, potential, etc.energies / ) Q 4444 h h /) Substitutg, the exit elocity of the air is determed to be It yields 8 kj/s (.5 kg/s) 6.0 m/s (sce W pe 0) ( ) kj/kg + (0 m/s) 8 kj/s AIR kj/kg 000 m /s (b) The exit pressure of air is determed from the conseration of mass and the ideal gas relations, and P A ( 0.04 m )( 6 m/s) A 0.99 m.5 kg/s RT ( 0.87 kpa m /kg K)( 5 K) RT P 0.99 m /kg /kg 9. kpa

22 5-5-4 Nitrogen is decelerated a diffuser from 00 m/s to a lower elocity. The exit elocity of nitrogen and the ratio of the let-to-exit area are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Nitrogen is an ideal gas with ariable specific heats. Potential energy changes are negligible. 4 The deice is adiabatic and thus heat transfer is negligible. 5 There are no work teractions. Properties The molar mass of nitrogen is M 8 kg/kmol (Table A-). The enthalpies are (Table A-8) T 7 C 80 K T C 95 K h h 84 kj/kmol 8580 kj/kmol 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 expressed the rate form as Substitutg, It yields E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass 0 Rate of change ternal, ketic, potential, etc.energies / ) + /) (sce Q W pe 0) h h, 0 h h + + M ( ) kj/kmol ( 00 m/s) 8 kg/kmol m/s kj/kg 000 m /s N (b) The ratio of the let to exit area is determed from the conseration of mass relation, or, A A A T / P A T / P A A RT / P RT / P ( 80 K/60 kpa)( 9.0 m/s) ( 95 K/85 kpa)( 00 m/s) 0.65

23 5-5-4 EES Problem 5-4 is reconsidered. The effect of the let elocity on the exit elocity and the ratio of the let-to-exit area as the let elocity aries from 80 m/s to 60 m/s is to be estigated. The fal results are to be plotted agast the let elocity. Analysis The problem is soled usg EES, and the solution is gien below. Function HCal(WorkFluid$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'N' WorkFluid$ then HCal:ENTHALPY(WorkFluid$,TTx) "Ideal gas equ." else HCal:ENTHALPY(WorkFluid$,TTx, PPx)"Real gas equ." endif end HCal "System: control olume for the nozzle" "Property relation: Nitrogen is an ideal gas" "Process: Steady state, steady flow, adiabatic, no work" "Knowns" WorkFluid$ 'N' T[] 7 [C] P[] 60 [kpa] {el[] 00 [m/s]} P[] 85 [kpa] T[] [C] "Property Data - sce the Enthalpy function has different parameters for ideal gas and real fluids, a function was used to determe h." h[]hcal(workfluid$,t[],p[]) h[]hcal(workfluid$,t[],p[]) "The olume function has the same form for an ideal gas as for a real fluid." []olume(workfluid$,tt[],pp[]) []olume(workfluid$,tt[],pp[]) "From the defition of mass flow rate, m_dot A*el/ and conseration of mass the area ratio A_Ratio A_/A_ is:" A_Ratio*el[]/[] el[]/[] "Conseration of Energy - SSSF energy balance" h[]+el[]^/(*000) h[]+el[]^/(*000) A Ratio el [m/s] el [m/s]

24 A Ratio el[] [m/s] el[] [m/s] el[] [m/s]

25 R-4a is decelerated a diffuser from a elocity of 0 m/s. The exit elocity of R-4a and the mass flow rate of the R-4a are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Potential energy changes are negligible. There are no work teractions. Properties From the R-4a tables (Tables A- through A-) and P 800 kpa m /kg sat. apor h 67.9 kj/kg P 900 kpa m /kg T 40 C h 74.7 kj/kg kj/s R-4a Analysis (a) There is only one let and one exit, and thus. Then the exit elocity of R-4a is determed from the steady-flow mass balance to be A A (0.075 m /kg) A m/s A.8 (0.056 m /kg) ( 0 ) 60.8 m/s (b) 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 expressed the rate form as E 44 0 (steady) E Esystem 0 Rate of net energy transfer by heat, work, and mass Q + + / ) Q Rate of change ternal, ketic, potential, etc.energies h h /) + (sce W pe 0) Substitutg, the mass flow rate of the refrigerant is determed to be kj/s ( )kJ/kg + ( 60.8 m/s) It yields m.08 kg/s (0 m/s) kj/kg 000 m /s

26 Heat is lost from the steam flowg a nozzle. The elocity and the olume flow rate at the nozzle exit are to be determed. Assumptions This is a steady-flow process sce there is no change with time. Potential energy change is negligible. There are no work teractions. Analysis We take the steam as the system, which is a control olume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as Energy balance: or E E 44 Rate of net energy transfer by heat, work, and mass h + h + Rate of change ternal, ketic, potential, etc.energies h h 0 (steady) Esystem Q Q + 0 The properties of steam at the let and exit are (Table A-6) P 800 kpa m /kg T 400 C h 67.7 kj/kg P 00 kpa.6 m /kg T 00 C h 07. kj/kg The mass flow rate of the steam is Substitutg, A 400 C 800 kpa 0 m/s sce W pe 0) (0.08 m )(0 m/s).08 kg/s m /s (0 m/s) 67.7 kj/kg + kj/kg 000 m /s The olume flow rate at the exit of the nozzle is (.08 kg/s)(.6 m /kg) 07. kj/kg m/s.74 m /s STEAM kj/kg 000 m /s + Q 5 kj/s.08 kg/s 00 C 00 kpa