ch06.pdf

• | outros

SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 6 SONNTAG BORGNAKKE VAN WYLEN FUNDAMENTALS of Thermodynamics Sixth Edition Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION Correspondence table Concept-Study guide problems Continuity equation and flow rates Single flow, single-device processes Nozzles, diffusers Throttle flow Turbines, expanders Compressors, fans Heaters, coolers Pumps, pipe and channel flows Multiple flow, single-device processes Turbines, compressors, expanders Heat exchangers Mixing processes Multiple devices, cycle processes Transient processes Review Problems Heat Transfer Problems English Unit Problems PROB NO. 1- 21 22-29 30-39 40-47 48-54 55-62 63-70 71-77 78-83 84-90 91-98 99-107 108-123 124-134 135-138 139-175 Sonntag, Borgnakke and van Wylen Correspondence List This chapter 6 homework problem set corresponds to the 5th edition chapter 6 as follows. Problems 1-21 are all new. New 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 5th 1 2 new 4 5 new 6 new 18 new 19 new new new new 20 new 22 23 new new 24 new 26 mod new 25 new 30 New 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 5th new 33 new 32 new new 36 new new 38 mod new new new 9 new new 10 12 new new new new 35 new 11 mod new new new 29 31 New 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 5th 37 mod 34 41 new 14 new 13 new new new new new 8 new 16 new 27 new new 39a 39b 40a 42 43 new 46 47 new 58 mod 53 New 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 5th new 49 new 48 51 new new new 50 52 new 57 59 new 15 new new new 44 45 55 56 62 63 65 new new new new Sonntag, Borgnakke and van Wylen CONCEPT-STUDY GUIDE PROBLEMS 6.1 A mass flow rate into a control volume requires a normal velocity component. Why? The tangential velocity component does not bring any substance across the control volume surface as it flows parallel to it, the normal component of velocity brings substance in or out of the control volume according to its sign. The normal component must be into the control volume to bring mass in, just like when you enter a bus (it does not help that you run parallel with the bus side). V Vnormal Vtangential 6.2 .A temperature difference drives a heat transfer. Does a similar concept apply to m? Yes. A pressure difference drives the flow. The fluid is accelerated in the direction of a lower pressure as it is being pushed harder behind it than in front of it. This also means a higher pressure in front can decelerate the flow to a lower velocity which happens at a stagnation on a wall. F1 =P1 A F2 =P 2 A 6.3 Can a steady state device have boundary work? No. Any change in size of the control volume would require either a change in mass inside or a change in state inside, neither of which is possible in a steady-state process. Sonntag, Borgnakke and van Wylen 6.4 .Can you say something about changes in m and V through a steady flow device? The continuity equation expresses the conservation of mass, so the total .amount of m entering must be equal to the total amount leaving. For a single flow device the mass flow rate is constant through it, so you have the same mass flow rate across any total cross-section of the device from the inlet to the exit. The volume flow rate is related to the mass flow rate as .V=vm so it can vary if the state changes (then v changes) for a constant mass flow rate. This also means that the velocity can change (influenced by the area as V =VA) and the flow can experience an acceleration (like in a nozzle) or a deceleration (as in a diffuser). Sonntag, Borgnakke and van Wylen 6.5 How does a nozzle or sprayhead generate kinetic energy? By accelerating the fluid from a high pressure towards the lower pressure, which is outside the nozzle. The higher pressure pushes harder than the lower pressure so there is a net force on any mass element to accelerate it. 6.6 Liquid water at 15oC flows out of a nozzle straight up 15 m. What is nozzle Vexit? Energy Eq.6.13: hexit +2 Vexit +gHexit =h2 +2 V2 +gH2 If the water can flow 15 m up it has specific potential energy of gH2 which must equal the specific kinetic energy out of the nozzle Vexit/2. The water does not change P or T so h is the same. Vexit/2 =g(H2 Hexit) =gH Vexit =2gH =2 2 1 2 1 2 =2 9.807 15 m2/s2 =17.15 m/s 6.7 What is the difference between a nozzle flow and a throttle process? In both processes a flow moves from a higher to a lower pressure. In the nozzle the pressure drop generates kinetic energy, whereas that does not take place in the throttle process. The pressure drop in the throttle is due to a flow restriction and represents a loss. Sonntag, Borgnakke and van Wylen 6.8 If you throttle a saturated liquid what happens to the fluid state? If it is an ideal gas? The throttle process is approximated as a constant enthalpy process. Changing the state from saturated liquid to a lower pressure with the same h gives a two-phase state so some of the liquid will vaporize and it becomes colder. 1 2 P 1 2 h=C T h=C v If the same process happens in an ideal gas then same h gives the same temperature (h a function of T only) at the lower pressure. 6.9 R-134a at 30oC, 800 kPa is throttled so it becomes cold at 10oC. What is exit P? State 1 is slightly compressed liquid so Table B.5.1: h =hf =241.79 kJ/kg At the lower temperature it becomes two-phase since the throttle flow has constant h and at 10oC: hg =392.28 kJ/kg P =Psat =210.7 kPa 6.10 Air at 500 K, 500 kPa is expanded to 100 kPa in two steady flow cases. Case one is a throttle and case two is a turbine. Which has the highest exit T? Why? 1. Throttle. In the throttle flow no work is taken out, no kinetic energy is generated and we assume no heat transfer takes place and no potential energy change. The energy equation becomes constant h, which gives constant T since it is an ideal gas. 2. Turbine. In the turbine work is taken out on a shaft so the fluid expands and P and T drops. Sonntag, Borgnakke and van Wylen 6.11 A turbine at the bottom of a dam has a flow of liquid water through it. How does that produce power? Which terms in the energy equation are important? The water at the bottom of the dam in the turbine inlet is at a high pressure. It runs through a nozzle generating kinetic energy as the pressure drops. This high kinetic energy flow impacts a set of rotating blades or buckets which converts the kinetic energy to power on the shaft so the flow leaves at low pressure and low velocity. Lake DAM T H Sonntag, Borgnakke and van Wylen 6.12 A windmill takes a fraction of the wind kinetic energy out as power on a shaft. In what manner does the temperature and wind velocity influence the power? Hint: write the power as mass flow rate times specific work. The work as a fraction f of the flow of kinetic energy becomes .1 2 1 2 W =mw =m f 2 Vin =AVin f 2 Vin so the power is proportional to the velocity cubed. The temperature enters through the density, so assuming air as ideal gas =1/v =P/RT and the power is inversely proportional to temperature. Sonntag, Borgnakke and van Wylen 6.13 If you compress air the temperature goes up, why? When the hot air, high P flows in long pipes it eventually cools to ambient T. How does that change the flow? As the air is compressed, volume decreases so work is done on a mass element, its energy and hence temperature goes up. If it flows at nearly constant P and cools its density increases (v decreases) so it slows down .for same mass flow rate ( m =AV ) and flow area. 6.14 In a boiler you vaporize some liquid water at 100 kPa flowing at 1 m/s. What is the velocity of the saturated vapor at 100 kPa if the pipe size is the same? Can the flow then be constant P? The continuity equation with average values is written .mi =me =m =AV =AV/v =AVi/vi =AVe/ve From Table B.1.2 at 100 kPa we get vf =0.001043 m3/kg; vg =1.694 m3/kg 1.694 Ve =Vi ve/vi =1 0.001043 =1624 m/s To accelerate the flow up to that speed you need a large force ( PA ) so a large pressure drop is needed. Pe Pi Pi cb Sonntag, Borgnakke and van Wylen 6.15 A mixing chamber has all flows at the same P, neglecting losses. A heat exchanger has separate flows exchanging energy, but they do not mix. Why have both kinds? You might allow mixing when you can use the resulting output mixture, say it is the same substance. You may also allow it if you definitely want the outgoing mixture, like water out of a faucet where you mix hot and cold water. Even if it is different substances it may be desirable, say you add water to dry air to make it more moist, typical for a winter time air-conditioning set-up. In other cases it is different substances that flow at different pressures with one flow heating or cooling the other flow. This could be hot combustion gases heating a flow of water or a primary fluid flow around a nuclear reactor heating a transfer fluid flow. Here the fluid being heated should stay pure so it does not absorb gases or radioactive particles and becomes contaminated. Even when the two flows have the same substance there may be a reason to keep them at separate pressures. 1 MIXING 2 cb 1 2 4 cb 3 CHAMBER 3 Sonntag, Borgnakke and van Wylen 6.16 In a co-flowing (same direction) heat exchanger 1 kg/s air at 500 K flows into one channel and 2 kg/s air flows into the neighboring channel at 300 K. If it is infinitely long what is the exit temperature? Sketch the variation of T in the two flows. C.V. mixing section (no W, Q) Continuity Eq.:Energy Eq.6.10: Same exit T: m1 =m3 and m2 =m4 .m1h1 +m2h2 =m1h3 +m2h4 .h3 =h4 =[m1h1 +m2h2] / [m1 +m2] .m1 .m1 +m2 .m2 Using conctant specific heat T3 =T4 =T1 +T2 =3 500 +3 300 =367 K .m1 +m2 T 1 x 2 cb 1 2 3 500 300 T1 T2 x 4 Sonntag, Borgnakke and van Wylen 6.17 Air at 600 K flows with 3 kg/s into a heat exchanger and out at 100oC. How much (kg/s) water coming in at 100 kPa, 20oC can the air heat to the boiling point? C.V. Total heat exchanger. The flows are not mixed so the two flowrates are constant through the device. No external heat transfer and no work. Energy Eq.6.10: mairhair in +mwaterhwater in =mairhair out +mwaterhwater out .mair[hair in -hair out] =mwater[hwater out hwater in] Table B.1.2: hwater out hwater in =2675.46 83.94 =2591.5 kJ/kg Table A.7.1: hair in -hair out =607.32 374.14 =233.18 kJ/kg Solve for the flow rate of water from the energy equation hair in -hair out 233.18 .mwater =mair h -hwater in =3 2591.5 =0.27 kg/s water out Air out Air in cb Sonntag, Borgnakke and van Wylen 6.18 Steam at 500 kPa, 300oC is used to heat cold water at 15oC to 75oC for domestic hot water supply. How much steam per kg liquid water is needed if the steam should not condense? Solution: C.V. Each line separately. No work but there is heat transfer out of the steam flow and into the liquid water flow. Water line energy Eq.:mliqhi +Q =mliqhe Q =mliq(he hi) For the liquid water look in Table B.1.1 hliq =he hi =313.91 62.98 =250.93 kJ/kg ( Cp T =4.18 (75 15) =250.8 kJ/kg ) Steam line energy has the same heat transfer but it goes out .Steam Energy Eq.:msteamhi =Q +msteamhe Q =msteam(hi he) For the steam look in Table B.1.3 at 500 kPa hsteam =hi he =3064.2 2748.67 =315.53 kJ/kg Now the heat transfer for the steam is substituted into the energy equation for the water to give .250.93 msteam / mliq =hliq / hsteam =315.53 =0.795 Hot Steam out cb water out Steam in Cold water in Sonntag, Borgnakke and van Wylen 6.19 Air at 20 m/s, 260 K, 75 kPa with 5 kg/s flows into a jet engine and it flows out at 500 m/s, 800 K, 75 kPa. What is the change (power) in flow of kinetic energy? 1 2 2 m KE =m 2 (Ve Vi ) 1 1 =5 kg/s 2 (5002 202) (m/s)2 1000 (kW/W) =624 kW cb 6.20 An initially empty cylinder is filled with air from 20oC, 100 kPa until it is full. Assuming no heat transfer is the final temperature larger, equal to or smaller than 20oC? Does the final T depend on the size of the cylinder? This is a transient problem with no heat transfer and no work. The balance equations for the tank as C.V. become Continuity Eq.:m2 0 =mi Energy Eq.:m2u2 0 =mihi +Q W =mihi +0 0 Final state: u2 =hi & P2 =Pi T2 Ti and it does not depend on V Sonntag, Borgnakke and van Wylen 6.21 A cylinder has 0.1 kg air at 25oC, 200 kPa with a 5 kg piston on top. A valve at the bottom is opened to let the air out and the piston drops 0.25 m towards the bottom. What is the work involved in this process? What happens to the energy? If we neglect acceleration of piston then P =C =Pequilibrium W =P V To get the volume change from the height we need the cylinder area. The force balance on the piston gives mpg mpg 5 9.807 P =Po +A A =P -P =100 1000 =0.000 49 m2 o V =AH =0.000 49 0.25 =0.000 1225 m3 W =P V =200 kPa (-0.000 1225) m3 =0.0245 kJ The air that remains inside has not changed state and therefore not energy. The work leaves as flow work Pv m. m cb g e AIR Pcyl Sonntag, Borgnakke and van Wylen Continuity equation and flow rates 6.22 Air at 35 C, 105 kPa, flows in a 100 mm 150 mm rectangular duct in a heating system. The volumetric flow rate is 0.015 m3/s. What is the velocity of the air flowing in the duct and what is the mass flow rate? Solution: Assume a constant velocity across the duct area with A =100 150 10-6 m2 =0.015 m2 and the volumetric flow rate from Eq.6.3, V =mv =AV .V 0.015 m3/s =1.0 m/s V=A= 0.015 m2 Ideal gas so note: RT 0.287 308.2 v= P =0.8424 m3/kg 105 .V 0.015 m =v =0.8424 =0.0178 kg/s Sonntag, Borgnakke and van Wylen 6.23 A boiler receives a constant flow of 5000 kg/h liquid water at 5 MPa, 20 C and it heats the flow such that the exit state is 450 C with a pressure of 4.5 MPa. Determine the necessary minimum pipe flow area in both the inlet and exit pipe(s) if there should be no velocities larger than 20 m/s. Solution: Mass flow rate from Eq.6.3, both V 20 m/s .1 mi =me =(AV/v) i =(AV/v) e =5000 3600 kg/s Table B.1.4 Table B.1.3 vi =0.001 m3/kg, ve =(0.08003 +0.00633)/2 =0.07166 m3/kg, 5000 Ai vi m/Vi =0.001 3600 / 20 =6.94 10-5 m2 =0.69 cm2 .5000 Ae ve m/Ve =0.07166 3600 / 20 =4.98 10-3 m2 =50 cm2 Super heater Q Q boiler Exit Superheated vapor vapor Inlet liquid i cb e Sonntag, Borgnakke and van Wylen 6.24 An empty bathtub has its drain closed and is being filled with water from the faucet at a rate of 10 kg/min. After 10 minutes the drain is opened and 4 kg/min flows out and at the same time the inlet flow is reduced to 2 kg/min. Plot the mass of the water in the bathtub versus time and determine the time from the very beginning when the tub will be empty. Solution: During the first 10 minutes we have dmcv .mi =10 kg/min ,m =m t1 =10 10 =100 kg dt So we end up with 100 kg after 10 min. For the remaining period we have dmcv .mi -me= 2 4 =2 kg/min dt .m m2 =mnet t2 t2 =100/-2 =50 min. mnet So it will take an additional 50 min. to empty ttot =t1 +t2 =10 +50 =60 min. kg 100 m 10 .m 0 t 0 10 20 min 0 -2 t 0 10 min Sonntag, Borgnakke and van Wylen 6.25 Nitrogen gas flowing in a 50-mm diameter pipe at 15 C, 200 kPa, at the rate of 0.05 kg/s, encounters a partially closed valve. If there is a pressure drop of 30 kPa across the valve and essentially no temperature change, what are the velocities upstream and downstream of the valve? Solution: Same inlet and exit area: A =4 (0.050)2 =0.001963 m2 RTi 0.2968 288.2 Ideal gas: vi =P =0.4277 m3/kg 200 i From Eq.6.3, mvi 0.05 0.4277 Vi =A =0.001963 =10.9 m/s RTe 0.2968 288.2 Ideal gas: ve =P =0.5032 m3/kg 170 e .mve 0.05 0.5032 Ve =A =0.001963 =12.8 m/s Sonntag, Borgnakke and van Wylen 6.26 Saturated vapor R-134a leaves the evaporator in a heat pump system at 10 C, with a steady mass flow rate of 0.1 kg/s. What is the smallest diameter tubing that can be used at this location if the velocity of the refrigerant is not to exceed 7 m/s? Solution: Mass flow rate Eq.6.3: m =V/v =AV/v =v =vg =0.04945 m3/kg Exit state Table B.5.1: (T =10 C, x =1) .The minimum area is associated with the maximum velocity for given m .mvg 0.1 kg/s 0.04945 m3/kg 2 =0.000706 m2 =4 DMIN AMIN =7 m/s VMAX DMIN =0.03 m =30 mm Exit cb Sonntag, Borgnakke and van Wylen 6.27 A hot air home heating system takes 0.25 m3/s air at 100 kPa, 17oC into a furnace and heats it to 52oC and delivers the flow to a square duct 0.2 m by 0.2 m at 110 kPa. What is the velocity in the duct? Solution: The inflate flow is given by a mi .Continuity Eq.:mi =Vi / vi =me =AeVe/ve RTi 0.287 290 m3 =0.8323 kg Ideal gas: vi =P =100 i RTe 0.287 (52 +273) ve =P =110 e =0.8479 m3/ kg .mi =Vi/vi =0.25/0.8323 =0.30 kg/s .0.3 0.8479 m3/s Ve =m ve/ Ae =6.36 m/s 0.2 0.2 m2 Sonntag, Borgnakke and van Wylen 6.28 Steam at 3 MPa, 400 C enters a turbine with a volume flow rate of 5 m3/s. An extraction of 15% of the inlet mass flow rate exits at 600 kPa, 200 C. The rest exits the turbine at 20 kPa with a quality of 90%,and a velocity of 20 m/s. Determine the volume flow rate of the extraction flow and the diameter of the final exit pipe. Solution: Inlet flow :mi =V/v =5/0.09936 =50.32 kg/ s .me =0.15 mi =7.55 kg/ s; (Table B.1.3) v =0.35202 m3/kg Extraction flow :Vex =mev =7.55 0.35202 =2.658 m3/ s .Exit flow :m =0.85 mi =42.77 kg /s Table B.1.2 .m =AV/v v =0.001017 +0.9 7.64835 =6.8845 m3/kg .A =(/4) D2 =m v/V =42.77 6.8845/20 =14.723 m2 D =4.33 m Inlet flow 1 2 Extraction flow WT HP section Exit flow LP section 3 Sonntag, Borgnakke and van Wylen 6.29 A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at 105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the inlet velocity and the outgoing volume flow rate in m3/s? Solution: Continuity Eq. Ideal gas .mi =me =AV/ v v =RT/P Area :A =4 D 2 =4 0.752 =0.442 m2 .Ve =AVe =0.442 1.5 =0.6627 m3/s RTe 0.287 (23 +273) =0.8091 m3/kg ve =P =105 e .mi =Ve/ve =0.6627/0.8091 =0.819 kg/s .AVi /vi =mi =AVe / ve RTi 0.287 (22 +273) Vi =Ve (vi / ve) =Ve P v =1.5 =1.6 m/s 98 0.8091 i e Sonntag, Borgnakke and van Wylen Single flow single device processes Nozzles, diffusers 6.30 Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated find the exit velocity. Solution: C.V. Nozzle steady state one inlet and exit flow, insulated so it is adiabatic. Inlet Exit Low V Hi P, A Energy Eq.6.13: 2 cb Hi V Low P, A 1 2 h1 +h2 +2 V2 V2 =2 ( h1 -h2 ) 2 CPN2 (T1 T2 ) =2 1.042 (400 330) =145.88 kJ/kg =145 880 J/kg V2 =381.9 m/s Sonntag, Borgnakke and van Wylen 6.31 A nozzle receives 0.1 kg/s steam at 1 MPa, 400oC with negligible kinetic energy. The exit is at 500 kPa, 350oC and the flow is adiabatic. Find the nozzle exit velocity and the exit area. Solution: Energy Eq.6.13: h1+ 2 V1 +gZ1 =h2 +2 V2 +gZ2 Process: State 1: State 2: Z1 =Z2 V1 =0 ,Table B.1.3 Table B.1.3 h1 =3263.88 kJ/kg h2 =3167.65 kJ/kg 1 2 1 2 Then from the energy equation 1 2 2 V2 =h1 h2 =3263.88 3167.65 =96.23 kJ/kg V2 =2(h1 -h2) =2 96.23 1000 =438.7 m/s The mass flow rate from Eq.6.3 .m =AV =AV/v .2 2 A =mv/V =0.1 0.57012 / 438.7 =0.00013 m =1.3 cm Inlet Exit Low V Hi P, A cb Hi V Low P, A Sonntag, Borgnakke and van Wylen 6.32 Superheated vapor ammonia enters an insulated nozzle at 20 C, 800 kPa, shown in Fig. P6.32, with a low velocity and at the steady rate of 0.01 kg/s. The ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature (or quality, if saturated) and the exit area of the nozzle. Solution: C.V. Nozzle, steady state, 1 inlet and 1 exit flow, insulated so no heat transfer. 2 2 Energy Eq.6.13: q +hi +Vi /2 =he +Ve /2, Process: Table B.2.2: Table B.2.1: q =0, Vi =0 he =1363.6 kJ/kg Sat. state at -9.2 C :hi =1464.9 =he +4502/(2 1000) Pe =300 kPa he =1363.6 =138.0 +xe 1293.8, xe =0.947, ve =0.001536 +xe 0.4064 =0.3864 m3/kg .Ae =meve/Ve =0.01 0.3864 / 450 =8.56 10-6 m2 Inlet Exit Low V Hi P, A cb Hi V Low P, A Sonntag, Borgnakke and van Wylen 6.33 In a jet engine a flow of air at 1000 K, 200 kPa and 30 m/s enters a nozzle, as shown in Fig. P6.33, where the air exits at 850 K, 90 kPa. What is the exit velocity assuming no heat loss? Solution: C.V. nozzle. No work, no heat transfer .Continuity mi =me =m .Energy :m (hi +Vi2) =m(he+ Ve2) Due to high T take h from table A.7.1 Ve2 =Vi2 +hi -he 1 =2000 (30)2 +1046.22 877.4 =0.45 +168.82 =169.27 kJ/kg Ve =(2000 169.27)1/2 =581.8 m/s Sonntag, Borgnakke and van Wylen 6.34 In a jet engine a flow of air at 1000 K, 200 kPa and 40 m/s enters a nozzle where the air exits at 500 m/s, 90 kPa. What is the exit temperature assuming no heat loss? Solution: C.V. nozzle, no work, no heat transfer .Continuity mi= me =m .Energy :m (hi +Vi2) =m(he+ Ve2) Due to the high T we take the h value from Table A.7.1 he =hi +Vi2 -Ve2 =1046.22 +0.5 (402 5002) (1/1000) =1046.22 124.2 =922.02 kJ/kg Interpolation in Table A.7.1 922.02 -877.4 Te =850 +50 933.15 -877.4 =890 K 40 m/s 200 kPa 500 m/s 90 kPa Sonntag, Borgnakke and van Wylen 6.35 A sluice gate dams water up 5 m. There is a small hole at the bottom of the gate so liquid water at 20oC comes out of a 1 cm diameter hole. Neglect any changes in internal energy and find the exit velocity and mass flow rate. Solution: Energy Eq.6.13: h1+ 2 V1 +gZ1 =h2 +2 V2 +gZ2 Process: h1 =h2 both at P =1 atm Z1 =Z2 +5 m V1 =0 1 2 1 2 Water 5m 1 2 2 V2 =g (Z1 -Z2) 2g(Z1 -Z2) =2 9.806 5 =9.902 m/s .m =V =AV/v =4 D2 (V2/v) V2 =4 (0.01)2 (9.902 / 0.001002) =0.776 kg/s Sonntag, Borgnakke and van Wylen 6.36 A diffuser, shown in Fig. P6.36, has air entering at 100 kPa, 300 K, with a velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit pressure and temperature of the air. Solution: Continuity Eq.6.3: mi =AiVi/vi =me =AeVe/ve, hi +2Vi2 =he +2Ve2 1 1 Energy Eq.(per unit mass flow)6.13: 1 1 he -hi =2 2002/1000 -2 202/1000 =19.8 kJ/kg Te =Ti +(he -hi)/Cp =300 +19.8/1.004 =319.72 K Now use the continuity equation and the ideal gas law AeVe AeVe ve =vi =(RTi/Pi) A V =RTe/Pe AiVi i i Te AiVi 319.72 100 200 Pe =Pi T =123.92 kPa =100 300 860 20 i AeVe Inlet Exit Hi V Low P, A Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.37 A diffuser receives an ideal gas flow at 100 kPa, 300 K with a velocity of 250 m/s and the exit velocity is 25 m/s. Determine the exit temperature if the gas is argon, helium or nitrogen. Solution: C.V. Diffuser: Energy Eq.6.13: mi =me 1 2 1 & assume no heat transfer 2 1 2 1 2 hi +2 Vi =2 Ve +he he =hi +2 Vi -2Ve 1 2 2 1 he hi Cp ( Te Ti ) =2 ( Vi -Ve ) =2 ( 2502 252 ) =30937.5 J/kg =30.938 kJ/kg Specific heats for ideal gases are from table A.5 Argon Helium Nitrogen Cp =0.52 kJ/kg K; 30.938 T =0.52 =59.5 Te =359.5 K Te =306 K Te =330 K Exit 30.938 Cp =5.913 kJ/kg K; T =5.193 =5.96 30.938 Cp =1.042 kJ/kg K; T =1.042 =29.7 Inlet Hi V Low P, A Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.38 Air flows into a diffuser at 300 m/s, 300 K and 100 kPa. At the exit the velocity is very small but the pressure is high. Find the exit temperature assuming zero heat transfer. Solution: Energy Eq.:Process: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 Z1 =Z2 and V2 =0 h2 =h1 +2 V1 T2 =T1 +2 (V1 / Cp) =300 +2 3002 / (1.004 1000) =344.8K 1 1 2 1 2 1 2 1 2 Inlet Exit Hi V Low P, A Low V Hi P, A Sonntag, Borgnakke and van Wylen 6.39 The front of a jet engine acts as a diffuser receiving air at 900 km/h, 5 C, 50 kPa, bringing it to 80 m/s relative to the engine before entering the compressor. If the flow area is reduced to 80% of the inlet area find the temperature and pressure in the compressor inlet. Solution: C.V. Diffuser, Steady state, 1 inlet, 1 exit flow, no q, no w. Continuity Eq.6.3: mi =me =(AV/v) .1 2 1 2 Energy Eq.6.12: m ( hi +2 Vi ) =m ( 2 Ve +he ) he hi =Cp ( Te Ti ) =2 Vi -2 Ve =2 =28050 J/kg =28.05 kJ/kg T =28.05/1.004 =27.9 Now use the continuity eq.:AiVi /vi =AeVe /ve ve =v i Ideal gas: AeVe ve =v i AiVi Te =5 +27.9 =22.9 C 1 2 1 2 1 900 1000 2 ( 3600 ) -2 (80)2 1 0.8 80 =vi 0.256 1 250 Pv =RT =ve =RTe/Pe =RT i 0.256/Pi Pe =Pi (Te/T i)/0.256 =50 296/268 0.256 =215.7 kPa Fan Sonntag, Borgnakke and van Wylen Throttle flow 6.40 Helium is throttled from 1.2 MPa, 20 C, to a pressure of 100 kPa. The diameter of the exit pipe is so much larger than the inlet pipe that the inlet and exit velocities are equal. Find the exit temperature of the helium and the ratio of the pipe diameters. Solution: C.V. Throttle. Steady state, Process with: Energy Eq.6.13: AV m =RT/P q =w =0; and hi =he, Vi =Ve, Zi =Ze Ti =Te =20 C PiAi =PeAe Ideal gas =But m, V, T are constant De Pi 1/2 1.21/2 =0.1 =3.464 Di =Pe Sonntag, Borgnakke and van Wylen 6.41 Saturated vapor R-134a at 500 kPa is throttled to 200 kPa in a steady flow through a valve. The kinetic energy in the inlet and exit flow is the same. What is the exit temperature? Solution: Steady throttle flow .Continuity mi =me =m Energy Eq.6.13: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 and V2 =V1 Process: Z1 =Z2 h2 =h1 =407.45 kJ/kg from Table B.5.2 superheated vapor State 2: P2 & h2 Interpolate between 0oC and 10oC in table B.5.2 in the 200 kPa subtable 407.45 400.91 T2 =0 +10 409.5 400.91 =7.6oC i e T 500 kPa cb 1 2 1 2 i h=C 200 e v Sonntag, Borgnakke and van Wylen 6.42 Saturated liquid R-12 at 25oC is throttled to 150.9 kPa in your refrigerator. What is the exit temperature? Find the percent increase in the volume flow rate. Solution: Steady throttle flow. Assume no heat transfer and no change in kinetic or potential energy. he =hi =hf 25oC =59.70 kJ/kg =hf e +xe hfg e From table B.3.1 we get Te =Tsat ( 150.9 kPa ) =20oC he hf e 59.7 17.82 xe =h =160.92 =0.26025 fg e ve =vf +xe vfg =0.000685 +xe 0.10816 =0.0288336 m3/kg vi =vf 25oC =0.000763 m3/kg .V =mv so the ratio becomes .Ve mve ve 0.0288336 =v =0.000763 =37.79 .i mvi Vi So the increase is 36.79 times or 3679 %e cb at 150.70 kPa i T i e h=C v Sonntag, Borgnakke and van Wylen 6.43 Water flowing in a line at 400 kPa, saturated vapor, is taken out through a valve to 100 kPa. What is the temperature as it leaves the valve assuming no changes in kinetic energy and no heat transfer? Solution: C.V. Valve. Steady state, single inlet and exit flow Continuity Eq.:Energy Eq.6.12: m1 =m2 .m1h1 +Q =m2h2 +W Process: Throttling .Small surface area: Q =0; No shaft: W=0 1 2 Table B.1.2: h2 =h1 =2738.6 kJ/kg T2 =131.1 C Sonntag, Borgnakke and van Wylen 6.44 Liquid water at 180oC, 2000 kPa is throttled into a flash evaporator chamber having a pressure of 500 kPa. Neglect any change in the kinetic energy. What is the fraction of liquid and vapor in the chamber? Solution: Energy Eq.6.13: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 and V2 =V1 Process: Z1 =Z2 h2 =h1 =763.71 kJ/kg from Table B.1.4 2 phase State 2: P2 & h2 h2 =hf +x2 hfg 763.71 640.21 x2 =(h2 -hf ) / hfg= 0.0586 2108.47 Fraction of Vapor: x2 =0.0586 Liquid: 1 -x2 =0.941 (5.86 %) (94.1 %) Two-phase out of the valve. The liquid drops to the bottom. 1 2 1 2 Sonntag, Borgnakke and van Wylen 6.45 Water at 1.5 MPa, 150 C, is throttled adiabatically through a valve to 200 kPa. The inlet velocity is 5 m/s, and the inlet and exit pipe diameters are the same. Determine the state (neglecting kinetic energy in the energy equation) and the velocity of the water at the exit. Solution: CV: valve. m =const, Ve =Vi(ve/vi) Energy Eq.6.13: 1 2 ve2 (he -hi) +2 Vi v -1 =0 i Now neglect the kinetic energy terms (relatively small) from table B.1.1 we have the compressed liquid approximated with saturated liquid same T 1 2 1 2 A =const hi +2 Vi =2 Ve +he or he =hi =632.18 kJ/kg ;Table B.1.2: Substituting and solving, xe =0.0579 vi =0.001090 m3/kg he =504.68 +xe 2201.96, ve =0.001061 +xe 0.88467 =0.052286 m3/kg Ve =Vi(ve/vi) =5 m/s (0.052286 / 0.00109) =240 m/s Sonntag, Borgnakke and van Wylen 6.46 R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of exit pipe diameter to that of the inlet pipe (Dex/Din) so the velocity stays constant. Solution: Energy Eq.6.13: Process: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 and V2 =V1 Z1 =Z2 1 2 1 2 State 1, Table B.5.1: h1 =234.59 kJ/kg, v1 =vf =0.000829 m3/kg Use energy eq.:h2 =h1 =234.59 kJ/kg 2 phase and T2 =Tsat (165 kPa) =15 C State 2: P2 & h2 h2 =hf +x2 hfg =234.59 kJ/kg x2 =(h2 -hf ) / hfg= (234.59 180.19) / 209 =0.2603 v2 =vf +x2 vfg =0.000746 +0.2603 0.11932 =0.0318 m3/kg Now the continuity equation with V2 =V1 gives, from Eq.6.3, m =V =AV/v =A1V1/v1 =(A2 V1) / v2 (A2 / A1) =v2 / v1 =(D2 / D1) (D2/D1) =(v2 / v1) 0.5 2 0.5 =(0.0318 / 0.000829) =6.19 Sonntag, Borgnakke and van Wylen 6.47 Methane at 3 MPa, 300 K is throttled through a valve to 100 kPa. Calculate the exit temperature assuming no changes in the kinetic energy and ideal gas behavior. Repeat the answer for real-gas behavior. C.V. Throttle (valve, restriction), Steady flow, 1 inlet and exit, no q, w Energy Eq.:Real gas :hi =he =Ideal gas Ti =Te =300 K hi =he =598.71 Table B.7 Pe =0.1 MPa Te =13.85 C ( =287 K) Sonntag, Borgnakke and van Wylen Turbines, Expanders 6.48 A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350oC and velocity of 15 m/s. The exit is at 100 kPa, x =1 and very low velocity. Find the specific work and the power produced. Solution: Energy Eq.6.13: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 +wT and V2 =0 Process: Z1 =Z2 Table B.1.3: h1 =3157.65 kJ/kg, h2 =2675.46 kJ/kg wT =h1 +2 V1 h2 =3157.65 +15 / 2000 2675.46 =482.3 kJ/kg .WT =m wT =2 482.3 =964.6 kW 1 WT 2 1 2 1 2 1 2 2 Sonntag, Borgnakke and van Wylen 6.49 A small, high-speed turbine operating on compressed air produces a power output of 100 W. The inlet state is 400 kPa, 50 C, and the exit state is 150 kPa, 30 C. Assuming the velocities to be low and the process to be adiabatic, find the required mass flow rate of air through the turbine. Solution: C.V. Turbine, no heat transfer, no KE, no PE Energy Eq.6.13: hin =hex +wT Ideal gas so use constant specific heat from Table A.5 wT =hin -hex Cp(Tin -Tex) =1.004 (kJ/kg K) [50 -(-30)] K =80.3 kJ/kg .W =mwT .m =W/wT =0.1/80.3 =0.00125 kg/s The dentist's drill has a small air flow and is not really adiabatic. Sonntag, Borgnakke and van Wylen 6.50 A liquid water turbine receives 2 kg/s water at 2000 kPa, 20oC and velocity of 15 m/s. The exit is at 100 kPa, 20oC and very low velocity. Find the specific work and the power produced. Solution: Energy Eq.6.13: h1 +2 V1 +gZ1 =h2 +2 V2 +gZ2 +wT Process: State 1: State 2: 1 2 1 2 1 2 Z1 =Z2 Table B.1.4 Table B.1.1 and V2 =0 (which is at 2.3 kPa so we h1 =85.82 kJ/kg h2 =83.94 2 should add Pv =97.7 0.001 to this) wT =h1 +2 V1 -h2 =85.82 +15 /2000 83.94 =1.99 kJ/kg .WT =m wT =2 1.9925 =3.985 kW Notice how insignificant the specific kinetic energy is. Sonntag, Borgnakke and van Wylen 6.51 Hoover Dam across the Colorado River dams up Lake Mead 200 m higher than the river downstream. The electric generators driven by water-powered turbines deliver 1300 MW of power. If the water is 17.5 C, find the minimum amount of water running through the turbines. Solution: C.V.:H2O pipe +turbines, Lake Mead DAM T H Continuity: min =mex; (h+ V2/2 +gz)in =(h+ V2/2 +gz)ex +wT vin vex hin hex ;Energy Eq.6.13: Water states: Now the specific turbine work becomes wT =gzin -gzex =9.807 200/1000 =1.961 kJ/kg .1300 103 kW m =WT/wT =1.961 kJ/kg =6.63 105 kg/s .V =mv =6.63 105 0.001001 =664 m3/s Sonntag, Borgnakke and van Wylen 6.52 A windmill with rotor diameter of 30 m takes 40% of the kinetic energy out as shaft work on a day with 20oC and wind speed of 30 km/h. What power is produced? Solution: Continuity Eq. Energy .mi =me =m .m (hi +Vi2 +gZi) =m(he+ Ve2 +gZe) +W .Process information: W =m Vi2 0.4 .m =AV =AVi /vi A =4 D 2 =4 302 =706.85 m2 0.287 293 vi =RTi/Pi =0.8301 m3/kg 101.3 30 1000 Vi =30 km/h =3600 =8.3333 m/s .706.85 8.3333 m =AVi /vi =7096 kg/s 0.8301 Vi2 =8.33332 m2/s2 =34.722 J/kg .W =0.4 m Vi2 =0.4 7096 34.722 =98 555 W =98.56 kW Sonntag, Borgnakke and van Wylen 6.53 A small turbine, shown in Fig. P 6.53, is operated at part load by throttling a 0.25 kg/s steam supply at 1.4 MPa, 250 C down to 1.1 MPa before it enters the turbine and the exhaust is at 10 kPa. If the turbine produces 110 kW, find the exhaust temperature (and quality if saturated). Solution: C.V. Throttle, Steady, q =0 and w =0. No change in kinetic or potential energy. The energy equation then reduces to Energy Eq.6.13: h1 =h2 =2927.2 kJ/kg from Table B.1.3 110 C.V. Turbine, Steady, no heat transfer, specific work: w =0.25 =440 kJ/kg Energy Eq.:h1 =h2 =h3 +w =2927.2 kJ/kg h3 =2927.2 -440 =2487.2 kJ/kg State 3: (P, h) Table B.1.2 h hg 2487.2 =191.83 +x3 2392.8 T =45.8 C ,x3 =0.959 T 1 2 3 v Sonntag, Borgnakke and van Wylen 6.54 A small expander (a turbine with heat transfer) has 0.05 kg/s helium entering at 1000 kPa, 550 K and it leaves at 250 kPa, 300 K. The power output on the shaft is measured to 55 kW. Find the rate of heat transfer neglecting kinetic energies. Solution: C.V. Expander. Steady operation Cont. Energy .mi= me =m .mhi +Q =mhe +W cb i Q WT e .Q =m (he-hi) +W Use heat capacity from tbl A.5: Cp He =5.193 kJ/kg K .Q =mCp (Te-Ti) +W =0.05 5.193 (300 -550) +55 =64.91 +55 =9.9 kW Sonntag, Borgnakke and van Wylen Compressors, fans 6.55 A compressor in a commercial refrigerator receives R-22 at -25oC, x =1. The exit is at 800 kPa, 40oC. Neglect kinetic energies and find the specific work. Solution: i C.V. Compressor, steady state, single inlet and exit flow. For this device we also assume no heat transfer and Z1 =Z2 From Table B.4.1 :From Table B.4.2 :h1 =239.92 kJ/kg h2 =274.24 kJ/kg e cb -WC Energy Eq.6.13 reduces to wc =h1 -h2 =(239.92 274.24) =34.3 kJ/kg Sonntag, Borgnakke and van Wylen 6.56 The compressor of a large gas turbine receives air from the ambient at 95 kPa, 20 C, with a low velocity. At the compressor discharge, air exits at 1.52 MPa, 430 C, with velocity of 90 m/s. The power input to the compressor is 5000 kW. Determine the mass flow rate of air through the unit. Solution: C.V. Compressor, steady state, single inlet and exit flow. Energy Eq.6.13: q +hi +Vi2/2 =he +Ve2/2 +w Here we assume q 0 and Vi 0 so using constant CPo from A.5 (90)2 -w =CPo(Te -Ti) +Ve2/2 =1.004(430 -20) +2 1000 =415.5 kJ/kg Notice the kinetic energy is 1% of the work and can be neglected in most cases. The mass flow rate is then from the power and the specific work .Wc .5000 m =w =415.5 =12.0 kg/s Sonntag, Borgnakke and van Wylen 6.57 A compressor brings R-134a from 150 kPa, 10oC to 1200 kPa, 50oC. It is water cooled with a heat loss estimated as 40 kW and the shaft work input is measured to be 150 kW. How much is the mass flow rate through the compressor? Solution: C.V Compressor. Steady flow. Neglect kinetic and potential energies. Energy :m hi +Q =mhe +W .m =(Q -W)/(he -hi) Look in table B.5.2 hi =393.84 kJ/kg, he =426.84 kJ/kg -40 (-150) .m =426.84 393.84 =3.333 kg/s 1 Compressor -Wc 2 Q cool Sonntag, Borgnakke and van Wylen 6.58 An ordinary portable fan blows 0.2 kg/s room air with a velocity of 18 m/s. What is the minimum power electric motor that can drive it? Hint: Are there any changes in P or T? Solution: C.V. Fan plus space out to near stagnant inlet room air. Energy Eq.6.13: q +hi +Vi2/2 =he +Ve2/2 +w Here q 0, Vi 0 and hi =he same P and T -w =Ve2/2 =182/2000 =0.162 kJ/kg .W =mw =0.2 kg/s 0.162 kJ/kg =0.032 kW Sonntag, Borgnakke and van Wylen 6.59 An air compressor takes in air at 100 kPa, 17 C and delivers it at 1 MPa, 600 K to a constant-pressure cooler, which it exits at 300 K. Find the specific compressor work and the specific heat transfer in the cooler. Solution C.V. air compressor q =0 .Continuity Eq.:m2 =m1 Energy Eq.6.13: h1 +wc =h2 Q cool 1 Compressor -W c Compressor section Table A.7: 2 3 Cooler section wc in =h2 -h1 =607.02 -290.17 =316.85 kJ/kg C.V. cooler w =0 / Continuity Eq.:Energy Eq.6.13: m3 =m1 h2 =qout +h3 qout =h2 -h3 =607.02 -300.19 =306.83 kJ/kg Sonntag, Borgnakke and van Wylen 6.60 A 4 kg/s steady flow of ammonia runs through a device where it goes through a polytropic process. The inlet state is 150 kPa, 20oC and the exit state is 400 kPa, 80oC, where all kinetic and potential energies can be neglected. The specific work input has been found to be given as [n/(n-1)] (Pv). a) Find the polytropic exponent n b) Find the specific work and the specific heat transfer. Solution: C.V. Steady state device. Single inlet and single exit flows. Energy Eq.6.13: h1 +2 V1 +gZ1 +q =h2 +2 V2 +gZ2 +w Process: State 1: State 2: Pvn =constant Table B.2.2 Table B.2.2 and Z1 =Z2 ,V1 =V2 =0 v1 =0.79774, h1 =1422.9 v2 =0.4216, h2 =1636.7 1 2 1 2 From the polytropic process equation and the two states we can find the exponent n: P2 v1 400 0.79774 n =ln P / ln v =ln 150 / ln 0.4216 =1.538 1 2 Before we can do the heat transfer we need the work term w=-n (P v P1v1) =2.8587(400 0.4216 150 0.79774) n-1 2 2 =140.0 kJ/kg q =h2 +w -h1 =1636.7 140.0 1422.9 =73.8 kJ/kg Sonntag, Borgnakke and van Wylen 6.61 An exhaust fan in a building should be able to move 2.5 kg/s air at 98 kPa, 20oC through a 0.4 m diameter vent hole. How high a velocity must it generate and how much power is required to do that? Solution: C.V. Fan and vent hole. Steady state with uniform velocity out. Continuity Eq.:Ideal gas :m =constant =V =AV / v =AVP/RT Pv =RT, and area is A =4 D2 Now the velocity is found .V =m RT/(4 D2P) =2.5 0.287 293.15 / ( 4 0.42 98) =17.1 m/s The kinetic energy out is 1 2 1 2 2 V2 =2 17.1 / 1000 =0.146 kJ/kg which is provided by the work (only two terms in energy equation that does not cancel, we assume V1 =0) .1 2 Win =m 2 V2 =2.5 0.146 =0.366 kW Sonntag, Borgnakke and van Wylen 6.62 How much power is needed to run the fan in Problem 6.29? A household fan of diameter 0.75 m takes air in at 98 kPa, 22oC and delivers it at 105 kPa, 23oC with a velocity of 1.5 m/s. What are the mass flow rate (kg/s), the inlet velocity and the outgoing volume flow rate in m3/s? Solution: Continuity Eq. mi =me =AV/ v Ideal gas v =RT/P Area :A =4 D2 =4 0.752 =0.442 m2 .Ve =AVe =0.442 1.5 =0.6627 m3/s RTe 0.287 296 =0.8091m3/kg ve =P =105 e .mi =Ve/ve =0.6627/0.8091 =0.819 kg/s .AVi /vi =mi =AVe / ve Vi =Ve (vi / ve) =Ve (RTi)/(Pive) =1.5 0.287 (22 +273) =1.6 m/s 98 0.8091 .m (hi +Vi2) =m(he+ Ve2) +W .W =m(hi +Vi2 he Ve2 ) =m [Cp (Ti-Te) +Vi2 Ve2 ] =0.819 [ 1.004 (-1) +0.81 kW 1.62 -1.52 2000 ] =0.819 [ -1.004 +0.000155] Sonntag, Borgnakke and van Wylen Heaters/Coolers 6.63 Carbon dioxide enters a steady-state, steady-flow heater at 300 kPa, 15oC, and exits at 275 kPa, 1200oC, as shown in Fig. P6.63. Changes in kinetic and potential energies are negligible. Calculate the required heat transfer per kilogram of carbon dioxide flowing through the heater. Solution: C.V. Heater Steady state single inlet and exit flow. Energy Eq.6.13: q +h i =he e i Q Table A.8: q =he -hi =1579.2 204.6 =1374.6 kJ/kg q 0.842(1200 -15) =997.8 kJ/kg) (If we use Cp0 from A.5 then Too large T, Tave to use Cp0 at room temperature. Sonntag, Borgnakke and van Wylen 6.64 A condenser (cooler) receives 0.05 kg/s R-22 at 800 kPa, 40oC and cools it to 15o C. There is a small pressure drop so the exit state is saturated liquid. What cooling capacity (kW) must the condenser have? Solution: C.V. R-22 condenser. Steady state single flow, heat transfer out and no work. Energy Eq.6.12: Inlet state: Table B.4.2 Exit state: Table B.4.1 .m h1 =m h2 +Qout h1 =274.24 kJ/kg, h2 =62.52 kJ/kg Process: Neglect kinetic and potential energy changes. Cooling capacity is taken as the heat transfer out i.e. positive out so .Qout =m ( h1- h2) =0.05 kg/s (274.24 62.52) kJ/kg =10.586 kW =10.6 kW Q cool 1 2 Sonntag, Borgnakke and van Wylen 6.65 A chiller cools liquid water for air-conditioning purposes. Assume 2.5 kg/s water at 20oC, 100 kPa is cooled to 5oC in a chiller. How much heat transfer (kW) is needed? Solution: C.V. Chiller. Steady state single flow with heat transfer. Neglect changes in kinetic and potential energy and no work term. Energy Eq.6.13: qout =hi he Properties from Table B.1.1: hi =83.94 kJ/kg and he =20.98 kJ/kg Now the energy equation gives qout =83.94 20.98 =62.96 kJ/kg .Qout =m qout =2.5 62.96 =157.4 kW Alternative property treatment since single phase and small T If we take constant heat capacity for the liquid from Table A.4 qout =hi he Cp (Ti -Te ) =4.18 (20 5) =62.7 kJ/kg .Qout =m qout =2.5 62.7 =156.75 kW Q cool 1 2 Sonntag, Borgnakke and van Wylen 6.66 Saturated liquid nitrogen at 500 kPa enters a boiler at a rate of 0.005 kg/s and exits as saturated vapor. It then flows into a super heater also at 500 kPa where it exits at 500 kPa, 275 K. Find the rate of heat transfer in the boiler and the super heater. Solution: C.V.:boiler steady single inlet and exit flow, neglect KE, PE energies in flow Continuity Eq.:m1 =m2 =m3 Super heater Q Q boiler Table B.6.1: h1 =87.095 kJ/kg, h2 =86.15 kJ/kg, Table B.6.2: h3 =284.06 kJ/kg Energy Eq.6.13: qboiler =h2 h1 =86.15 -(- 87.095) =173.25 kJ/kg .Qboiler =m1qboiler =0.005 173.25 =0.866 kW C.V. Superheater (same approximations as for boiler) Energy Eq.6.13: qsup heater =h3 h2 =284.06 86.15 =197.9 kJ/kg .Qsup heater =m2qsup heater =0.005 197.9 =0.99 kW P 3 500 1 2 3 T 1 2 v v 3 1 cb vapor 2 Sonntag, Borgnakke and van Wylen 6.67 In a steam generator, compressed liquid water at 10 MPa, 30 C, enters a 30-mm diameter tube at the rate of 3 L/s. Steam at 9 MPa, 400 C exits the tube. Find the rate of heat transfer to the water. Solution: C.V. Steam generator. Steady state single inlet and exit flow. Constant diameter tube: Table B.1.4 Ai =Ae =4 (0.03)2 =0.0007068 m2 .m =Vi/vi =0.003/0.0010003 =3.0 kg/s .Vi =Vi/Ai =0.003/0.0007068 =4.24 m/s Exit state properties from Table B.1.3 Ve =Vi ve/vi =4.24 0.02993/0.0010003 =126.86 m/s The energy equation Eq.6.12 is solved for the heat transfer as .Q =m (he -hi) +Ve2 -Vi2 /2 126.862 -4.242 =3.0 3117.8 -134.86 +2 1000 =8973 kW ( ) Steam exit gas out Typically hot combustion gas in cb liquid water in Sonntag, Borgnakke and van Wylen 6.68 The air conditioner in a house or a car has a cooler that brings atmospheric air from 30oC to 10oC both states at 101 kPa. If the flow rate is 0.5 kg/s find the rate of heat transfer. Solution: CV. Cooler. Steady state single flow with heat transfer. Neglect changes in kinetic and potential energy and no work term. Energy Eq.6.13: qout =hi he Use constant heat capacity from Table A.5 (T is around 300 K) qout =hi -he =Cp (Ti -Te) kJ =1.004 kg K (30 10) K =20.1 kJ/kg .Qout .m qout =0.5 20.1 =10 kW Sonntag, Borgnakke and van Wylen 6.69 A flow of liquid glycerine flows around an engine, cooling it as it absorbs energy. The glycerine enters the engine at 60oC and receives 9 kW of heat transfer. What is the required mass flow rate if the glycerine should come out at maximum 95o C? Solution: C.V. Liquid flow (glycerine is the coolant), steady flow. no work. Energy Eq.:mhi +Q =mhe .Q m =Q/( he -hi) =C (T -T ) gly e i From table A.4 Cgly =2.42 kJ/kg-K .9 m =2.42 (95 60) =0.106 kg/s Air intake filter Shaft power cb Fan Radiator Atm. air Exhaust flow Coolant flow Sonntag, Borgnakke and van Wylen 6.70 A cryogenic fluid as liquid nitrogen at 90 K, 400 kPa flows into a probe used in cryogenic surgery. In the return line the nitrogen is then at 160 K, 400 kPa. Find the specific heat transfer to the nitrogen. If the return line has a cross sectional area 100 times larger than the inlet line what is the ratio of the return velocity to the inlet velocity? Solution: C.V line with nitrogen. No kinetic or potential energy changes Continuity Eq.:Energy Eq.6.13: m =constant =me =mi =AeVe/ve =AiVi/vi q =he -hi State i, Table B.6.1: hi =95.58 kJ/kg, vi =0.001343 m3/kg State e, Table B.6.2: he =162.96 kJ/kg, ve =0.11647 m3/kg From the energy equation q =he -hi =162.96 (-95.58) =258.5 kJ/kg From the continuity equation 1 0.11647 Ve/Vi =Ai/Ae (ve/vi) =100 0.001343 =0.867 Sonntag, Borgnakke and van Wylen Pumps, pipe and channel flows 6.71 A small stream with 20oC water runs out over a cliff creating a 100 m tall waterfall. Estimate the downstream temperature when you neglect the horizontal flow velocities upstream and downstream from the waterfall. How fast was the water dropping just before it splashed into the pool at the bottom of the waterfall? Solution: CV. Waterfall, steady state. Assume no Q nor W 1 Energy Eq.6.13: h +2V2 +gZ =const. State 1: At the top zero velocity Z1 =100 m State 2: At the bottom just before impact, Z2 =0 State 3: At the bottom after impact in the pool. 1 2 h1 +0 +gZ1 =h2 +2 V2 +0 =h3 +0 +0 Properties: h1 h2 same T, P 1 2 =2 V2 =gZ1 V2 =2gZ1 =2 9.806 100 =44.3 m/s Energy equation from state 1 to state 3 h3 =h1 +gZ1 use h =Cp T with value from Table A.4 (liquid water) T3 =T1 +gZ1 / Cp =20 +9.806 100 /4180 =20.23 C Sonntag, Borgnakke and van Wylen 6.72 A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. The exit line enters a pipe that goes up to an elevation 20 m above the pump and river, where the water runs into an open channel. Assume the process is adiabatic and that the water stays at 10oC. Find the required pump work. Solution: C.V. pump +pipe. Steady state ,1 inlet, 1 exit flow. Assume same velocity in and out, no heat transfer. Continuity Eq.:Energy Eq.6.12: m(hin +(1/2)Vin2 +gzin) =m(hex +(1/2) Vex2 +gzex) +W States: hin =hex same (T, P) i .min =mex =m e H cb .W =m g(zin -zex) =5 9.807 (0 -20)/1000 =0.98 kW I.E. 0.98 kW required input Sonntag, Borgnakke and van Wylen 6.73 A steam pipe for a 300-m tall building receives superheated steam at 200 kPa at ground level. At the top floor the pressure is 125 kPa and the heat loss in the pipe is 110 kJ/kg. What should the inlet temperature be so that no water will condense inside the pipe? Solution: C.V. Pipe from 0 to 300 m, no KE, steady state, single inlet and exit flow. Neglect any changes in kinetic energy. Energy Eq.6.13: q +hi =he +gZe No condensation means: Table B.1.2, hi =he +gZe -q =2685.4 +he =hg at 125 kPa =2685.4 kJ/kg 9.807 300 -(-110) =2810.1 kJ/kg 1000 At 200 kPa: T 170oC Table B.1.3 Sonntag, Borgnakke and van Wylen 6.74 The main waterline into a tall building has a pressure of 600 kPa at 5 m below ground level. A pump brings the pressure up so the water can be delivered at 200 kPa at the top floor 150 m above ground level. Assume a flow rate of 10 kg/s liquid water at 10oC and neglect any difference in kinetic energy and internal energy u. Find the pump work. Solution: C.V. Pipe from inlet at -5 m up to exit at +150 m, 200 kPa. 1 1 Energy Eq.6.13: hi +2Vi2 +gZi =he +2Ve2 +gZe +w With the same u the difference in h's are the Pv terms 1 w =hi -he +2 (Vi2 -Ve2) +g (Zi- Ze) =Pivi Peve +g (Zi Ze) =600 0.001 200 0.001 +9.806 (-5-150)/1000 =0.4 1.52 =1.12 kJ/kg .W =mw =10 (-1.12) =11.2 kW Sonntag, Borgnakke and van Wylen 6.75 Consider a water pump that receives liquid water at 15oC, 100 kPa and delivers it to a same diameter short pipe having a nozzle with exit diameter of 1 cm (0.01 m) to the atmosphere 100 kPa. Neglect the kinetic energy in the pipes and assume constant u for the water. Find the exit velocity and the mass flow rate if the pump draws a power of 1 kW. Solution: 2 .Continuity Eq.:mi =me =AV/v ;A =4 De =4 0.01 2 =7.854 10 -5 1 2 1 2 Energy Eq.6.13: hi +2Vi +gZi =he +2Ve +gZe +w Properties: hi =ui +Pivi =he =ue +Peve ;Pi =Pe ;vi =ve .1 3 1 2 .1 2 -W =m (2 Ve ) =A 2 Ve /ve w =2 Ve .2 W ve 1/3 2 1000 0.001001 1/3 Ve =29.43 m/s A 7.854 10 -5 ( ) ( ) .m =AVe/ve =7.854 10 -5 29.43 / 0.001001 =2.31 kg/s Sonntag,