Baixe exercicios resolv fund termo van wylen cap 6 e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity! SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 6 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study guide problems 1- 21 Continuity equation and flow rates 22-29 Single flow, single-device processes Nozzles, diffusers 30-39 Throttle flow 40-47 Turbines, expanders 48-54 Compressors, fans 55-62 Heaters, coolers 63-70 Pumps, pipe and channel flows 71-77 Multiple flow, single-device processes Turbines, compressors, expanders 78-83 Heat exchangers 84-90 Mixing processes 91-98 Multiple devices, cycle processes 99-107 Transient processes 108-123 Review Problems 124-134 Heat Transfer Problems 135-138 English Unit Problems 139-175 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 . = v m . 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 + 1 2 V 2 exit + gHexit = h2 + 1 2 V 2 2 + 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 V 2 exit/2. The water does not change P or T so h is the same. V 2 exit/2 = g(H2 – Hexit) = gH => Vexit = 2gH = 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 2 P v 1 T h = C h = C 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 6.13 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 m . i = m . e = m . = "AV = AV/v = AVi/vi = AVe/ve From Table B.1.2 at 100 kPa we get vf = 0.001043 m 3/kg; vg = 1.694 m 3/kg Ve = Vi ve/vi = 1 1.694 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. Pi cb Pe < Pi SOLUTION MANUAL SI UNIT PROBLEMS CHAPTER 6 FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN Sonntag, Borgnakke and van Wylen CONTENT SUBSECTION PROB NO. Correspondence table Concept-Study guide problems 1- 21 Continuity equation and flow rates 22-29 Single flow, single-device processes Nozzles, diffusers 30-39 Throttle flow 40-47 Turbines, expanders 48-54 Compressors, fans 55-62 Heaters, coolers 63-70 Pumps, pipe and channel flows 71-77 Multiple flow, single-device processes Turbines, compressors, expanders 78-83 Heat exchangers 84-90 Mixing processes 91-98 Multiple devices, cycle processes 99-107 Transient processes 108-123 Review Problems 124-134 Heat Transfer Problems 135-138 English Unit Problems 139-175 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 . = v m . 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 + 1 2 V 2 exit + gHexit = h2 + 1 2 V 2 2 + 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 V 2 exit/2. The water does not change P or T so h is the same. V 2 exit/2 = g(H2 – Hexit) = gH => Vexit = 2gH = 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 2 P v 1 T h = C h = C 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 6.13 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 m . i = m . e = m . = "AV = AV/v = AVi/vi = AVe/ve From Table B.1.2 at 100 kPa we get vf = 0.001043 m 3/kg; vg = 1.694 m 3/kg Ve = Vi ve/vi = 1 1.694 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. Pi cb Pe < Pi 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 2 3MIXING CHAMBER cb 1 2 3 cb 4 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.: m . 1 = m . 3 and m . 2 = m . 4 Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 1h3 + m . 2h4 Same exit T: h3 = h4 = [m . 1h1 + m . 2h2] / [m . 1 + m . 2] Using conctant specific heat T3 = T4 = m . 1 m . 1 + m . 2 T1 + m . 2 m . 1 + m . 2 T2 = 1 3 ! 500 + 2 3 ! 300 = 367 K x cb 3 4 1 2 T x300 500 1T 2T 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? m . #KE = m . 1 2 (V 2 e – V 2 i ) = 5 kg/s ! 1 2 (500 2 – 202) (m/s)2 1 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 P = Po + mpg A $ A = mpg P - P o = 5 × 9.807 100 × 1000 = 0.000 49 m2 #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. Pcyl AIR e cb m g 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 . = m . v = AV V = V . A = 0.015 m3/s 0.015 m2 = 1.0 m/s Ideal gas so note: v = RT P = 0.287 × 308.2 105 = 0.8424 m3/kg m . = V . v = 0.015 0.8424 = 0.0178 kg/s 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 Ideal gas: vi = RTi Pi = 0.2968 × 288.2 200 = 0.4277 m3/kg From Eq.6.3, Vi = m . vi A = 0.05 × 0.4277 0.001963 = 10.9 m/s Ideal gas: ve = RTe Pe = 0.2968 × 288.2 170 = 0.5032 m3/kg Ve = m . ve A = 0.05 × 0.5032 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 Exit state Table B.5.1: (T = 10&C, x =1) => v = vg = 0.04945 m 3/kg The minimum area is associated with the maximum velocity for given m . AMIN = m . vg VMAX = 0.1 kg/s × 0.04945 m3/kg 7 m/s = 0.000706 m2 = * 4 D 2 MIN DMIN = 0.03 m = 30 mm cb Exit 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 m . i Continuity Eq.: m . i = V . i / vi = m . e = AeVe/ve Ideal gas: vi = RTi Pi = 0.287 ! 290 100 = 0.8323 m3 kg ve = RTe Pe = 0.287 ! (52 + 273) 110 = 0.8479 m3/ kg m . i = V . i/vi = 0.25/0.8323 = 0.30 kg/s Ve = m . ve/ Ae = 0.3 ! 0.8479 0.2 ! 0.2 m3/s m2 = 6.36 m/s 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 Low V Exit Hi V Hi P, A Low P, Acb Energy Eq.6.13: h1 + , = h2 + 1 2 V 2 2 V 2 2 = 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+ 1 2 V 2 1 + gZ1 = h2 + 1 2 V 2 2 + gZ2 Process: Z1 = Z2 State 1: V1 = 0 , Table B.1.3 h1 = 3263.88 kJ/kg State 2: Table B.1.3 h2 = 3167.65 kJ/kg Then from the energy equation 1 2 V 2 2 = 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 A = m . v/V = 0.1 × 0.57012 / 438.7 = 0.00013 m 2 = 1.3 cm 2 Inlet Low V Exit Hi V Hi P, A Low P, Acb 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. Energy Eq.6.13: q + hi + V 2 i /2 = he + V 2 e /2, Process: q = 0, Vi = 0 Table B.2.2: hi = 1464.9 = he + 450 2/(2×1000) $ he = 1363.6 kJ/kg Table B.2.1: Pe = 300 kPa Sat. state at -9.2°C : he = 1363.6 = 138.0 + xe × 1293.8, => xe = 0.947, ve = 0.001536 + xe × 0.4064 = 0.3864 m 3/kg Ae = m . eve/Ve = 0.01 × 0.3864 / 450 = 8.56 × 10 -6 m2 Inlet Low V Exit Hi V Hi P, A Low P, Acb 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+ 1 2 V 2 1 + gZ1 = h2 + 1 2 V 2 2 + gZ2 Process: h1 = h2 both at P = 1 atm V1 = 0 Z1 = Z2 + 5 m Water 5 m 1 2 V 2 2 = g (Z1 - Z2) V2 = 2g(Z1 - Z2) = 2 ! 9.806 ! 5 = 9.902 m/s m . = ".V = AV/v = * 4 D 2 × (V2/v) = * 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: m . i = AiVi/vi = m . e = AeVe/ve, Energy Eq.(per unit mass flow)6.13: hi + 1 2Vi 2 = he + 1 2Ve 2 he - hi = 1 2 ×200 2/1000 - 1 2 ×20 2/1000 = 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 ve = vi / 0 1 2 3 4AeVe AiVi = (RTi/Pi) / 0 1 2 3 4AeVe AiVi = RTe/Pe Pe = Pi / 0 1 2 3 4Te Ti / 0 1 2 3 4AiVi AeVe = 100 /0 1 23 4319.72 300 / 0 1 2 3 4100'×'200 860'×'20 = 123.92 kPa Inlet Low V Exit Hi V Hi P, ALow 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: m . i = m . e & assume no heat transfer $ Energy Eq.6.13: hi + 1 2 V 2 i = 1 2 V 2 e + he $ he = hi + 1 2 V 2 i - 1 2V 2 e he – hi 5 Cp ( Te – Ti ) = 1 2 ( V 2 i - V 2 e ) = 1 2 ( 250 2 – 252 ) = 30937.5 J/kg = 30.938 kJ/kg Specific heats for ideal gases are from table A.5 Argon Cp = 0.52 kJ/kg K; #T = 30.938 0.52 = 59.5 Te = 359.5 K Helium Cp = 5.913 kJ/kg K; #T = 30.938 5.193 = 5.96 Te = 306 K Nitrogen Cp = 1.042 kJ/kg K; #T = 30.938 1.042 = 29.7 Te = 330 K Inlet Low V Exit Hi V Hi P, ALow P, A 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: q = w = 0; and Vi = Ve, Zi = Ze Energy Eq.6.13: hi = he, Ideal gas => Ti = Te = 20°C m . = AV RT/P But m . , V, T are constant => PiAi = PeAe $ De Di = / 0 1 2 3 4Pi Pe 1/2 = /0 1 23 41.2 0.1 1/2 = 3.464 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 m . i = m . e = m . Energy Eq.6.13: h1 + 1 2 V 2 1 + gZ1 = h2 + 1 2 V 2 2 + gZ2 Process: Z1 = Z2 and V2 = V1 $''''h2 = h1 = 407.45 kJ/kg from Table B.5.2 State 2: P2 & h2 $' superheated vapor Interpolate between 0oC and 10oC in table B.5.2 in the 200 kPa subtable T2 = 0 + 10 407.45 – 400.91 409.5 – 400.91 = 7.6oC i e cb e T v i 500 kPa 200 h = C 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 at 150.70 kPa From table B.3.1 we get Te = Tsat ( 150.9 kPa ) = -20 oC xe = he – hf e hfg e = 59.7 – 17.82 160.92 = 0.26025 ve = vf + xe vfg = 0.000685 + xe 0.10816 = 0.0288336 m 3/kg vi = vf 25oC = 0.000763 m 3/kg V . = m . v so the ratio becomes V . e V . i = m . ve m . vi = ve vi = 0.0288336 0.000763 = 37.79 So the increase is 36.79 times or 3679 % i e cb e T v i h = C 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, A = const $ Ve = Vi(ve/vi) Energy Eq.6.13: hi + 1 2 V 2 i = 1 2 V 2 e + he or (he - hi) + 1 2 V 2 i 6 7 8 9 : ; / 0 1 2 3 4ve vi 2 - 1 = 0 Now neglect the kinetic energy terms (relatively small) from table B.1.1 we have the compressed liquid approximated with saturated liquid same T he = hi = 632.18 kJ/kg ; vi = 0.001090 m 3/kg Table B.1.2: he = 504.68 + xe'×'2201.96, Substituting and solving, xe = 0.0579 ve = 0.001061 + xe'×'0.88467 = 0.052286 m 3/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: h1 + 1 2 V 2 1 + gZ1 = h2 + 1 2 V 2 2 + gZ2 Process: Z1 = Z2 and V2 = V1 State 1, Table B.5.1: h1 = 234.59 kJ/kg, v1 = vf = 0.000829 m 3/kg Use energy eq.: $''''h2 = h1 = 234.59 kJ/kg State 2: P2 & h2 $''''''2 – phase and T2 = Tsat (165 kPa) = -15&C 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 m 3/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) 2 (D2/D1) = (v2 / v1) 0.5 = (0.0318 / 0.000829) 0.5 = 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.: hi = he => Ideal gas Ti = Te = 300 K Real gas : < = >hi = he = 598.71 Pe = 0.1 MPa Table B.7 Te = 13.85°C ( = 287 K) 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 + 1 2 V 2 1 + gZ1 = h2 + 1 2 V 2 2 + gZ2 + wT Process: Z1 = Z2 and V2 = 0 State 1: Table B.1.4 h1 = 85.82 kJ/kg State 2: Table B.1.1 h2 = 83.94 (which is at 2.3 kPa so we should add #Pv = 97.7 × 0.001 to this) wT = h1 + 1 2 V 2 1 - h2 = 85.82 + 15 2 /2000 – 83.94 = 1.99 kJ/kg W . T = 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, T H DAM Lake Mead Continuity: m . in = m . ex; Energy Eq.6.13: (h+ V2/2 + gz)in = (h+ V 2/2 + gz)ex + wT Water states: hin % hex ; vin % vex Now the specific turbine work becomes wT = gzin - gzex = 9.807'×'200/1000 = 1.961 kJ/kg m . = W . T/wT = 1300×103 kW 1.961 kJ/kg = 6.63 ×105 kg/s V . = m . v = 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. m . i = m . e = m . Energy m . (hi + ½Vi 2 + gZi) = m . (he+ ½Ve 2 + gZe) + W . Process information: W . = m . ½Vi 2 ! 0.4 m . = "AV =AVi /vi A = * 4 D 2 = * 4 302 = 706.85 m2 vi = RTi/Pi = 0.287 ! 293 101.3 = 0.8301 m3/kg Vi = 30 km/h = 30 ! 1000 3600 = 8.3333 m/s m . = AVi /vi = 706.85 ! 8.3333 0.8301 = 7096 kg/s ½ Vi 2 = ½ 8.33332 m2/s2 = 34.722 J/kg W . = 0.4 m . ½ Vi 2 = 0.4 !7096 ! 34.722 = 98 555 W = 98.56 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: C.V. Compressor, steady state, single inlet and exit flow. For this device we also assume no heat transfer and Z1 = Z2 WC i e cb - From Table B.4.1 : h1 = 239.92 kJ/kg From Table B.4.2 : h2 = 274.24 kJ/kg 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 + Vi 2/2 = he + Ve 2/2 + w Here we assume q % 0 and Vi % 0 so using constant CPo from A.5 -w = CPo(Te - Ti) + Ve 2/2 = 1.004(430 - 20) + (90)2 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 m . = W . c -w = 5000 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 . = m . he + W . m . = (Q . - W . )/(he - hi) 1 2 Qcool Compressor -Wc Look in table B.5.2 hi = 393.84 kJ/kg, he = 426.84 kJ/kg m . = -40 – (-150) 426.84 – 393.84 = 3.333 kg/s 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 + 1 2 V 2 1 + gZ1 + q = h2 + 1 2 V 2 2 + gZ2 + w Process: Pvn = constant and Z1 = Z2 , V1 = V2 = 0 State 1: Table B.2.2 v1 = 0.79774, h1 = 1422.9 State 2: Table B.2.2 v2 = 0.4216, h2 = 1636.7 From the polytropic process equation and the two states we can find the exponent n: n = ln P2 P1 / ln v1 v2 = ln 400 150 / ln 0.79774 0.4216 = 1.538 Before we can do the heat transfer we need the work term w = - n n - 1 (P2v2 – P1v1) = -2.8587(400!0.4216 – 150!0.79774) = -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.: m . = constant = ".V = AV / v =AVP/RT Ideal gas : 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 !'@AB2 !'98C'D'EFAE'm/s The kinetic energy out is 1 2 V 2 2 = 1 2 ! 17.1 2 / 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) W . in = m . 1 2 V 2 2 = 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. m . i = m . e = AV/ v Ideal gas v = RT/P Area : A = * 4 D2 = * 4 ! 0.752 = 0.442 m2 V . e = AVe = 0.442 !1.5 = 0.6627 m 3/s ve = RTe Pe = 0.287 ! 296 105 = 0.8091m3/kg m . i = V . e/ve = 0.6627/0.8091 = 0.819 kg/s AVi /vi = m . i = AVe / ve Vi = Ve ! (vi / ve) = Ve ! (RTi)/(Pive) = 1.5 ! 0.287 ! (22 + 273) 98 ! 0.8091 = 1.6 m/s m . (hi + ½Vi 2) = m . (he+ ½Ve 2) +W . W . = m . (hi + ½Vi 2 – he – ½Ve 2 ) = m . [Cp (Ti-Te) + ½ Vi 2 – ½Ve 2 ] = 0.819 [ 1.004 (-1) + 1.62 - 1.52 2000 ] = 0.819 [ -1.004 + 0.000155] = - 0.81 kW 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 Q . out = 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 Q . out = m . qout = 2.5 !'62.7 = 156.75 kW 1 2 Q cool 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.: m . 1 = m . 2 = m . 3 1 2 3 Q Q boiler Super heater vapor cb 500 P 1 2 3 v T 1 2 3 v 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 Q . boiler = m . 1qboiler = 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 Q . sup heater = m . 2qsup heater = 0.005 ! 197.9 = 0.99 kW 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: Ai = Ae = * 4 (0.03) 2 = 0.0007068 m2 Table B.1.4 m . = V . i/vi = 0.003/0.0010003 = 3.0 kg/s Vi = V . i/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 . 68 9;(he - hi) + ( )Ve2 - Vi2 /2 = 3.0 6 7 8 9 : ; 3117.8 - 134.86 + 126.862 - 4.242 2 × 1000 = 8973 kW Typically hot combustion gas in Steam exit cb liquid water in gas out 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.: m . = constant = m . e = m . i = AeVe/ve = AiVi/vi Energy Eq.6.13: q = he - hi State i, Table B.6.1: hi = -95.58 kJ/kg, vi = 0.001343 m 3/kg State e, Table B.6.2: he = 162.96 kJ/kg, ve = 0.11647 m 3/kg From the energy equation q = he - hi = 162.96 – (-95.58) = 258.5 kJ/kg From the continuity equation Ve/Vi = Ai/Ae (ve/vi) = 1 100 0.11647 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 . Energy Eq.6.13: h + 1 2 V2 + 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. h1 + 0 + gZ1 = h2 + 1 2 V 2 2 + 0 = h3 + 0 + 0 Properties: h1 % h2 same T, P => 1 2 V 2 2 = 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.: m . in = m . ex = m . Energy Eq.6.12: m . (hin + (1/2)Vin 2 + gzin) = m . (hex + (1/2)'Vex 2 + gzex) + W . States: hin = hex same (T, P) i 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 6.75 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: Continuity Eq.: m . i = m . e = AV/v ; A = * 4 D 2 e = * 4 !'0.01 2 = 7.854!'10 -5 Energy Eq.6.13: hi + 1 2 V 2 i + gZi = he + 1 2 V 2 e + gZe + w Properties: hi = ui + Pivi = he = ue + Peve ; Pi = Pe ; vi = ve w = - 1 2 V 2 e $ -W . = m . ( 1 2 V 2 e ) = A !' 1 2 V 3 e /ve Ve = ( -2 W . ve A ) 1/3 = ( 2 !'1000 !'0.001001 7.854!10 -5 ) 1/3 = 29.43 m/s m . = AVe/ve = 7.854!'10 -5 !'29.43 / 0.001001 = 2.31 kg/s Sonntag, Borgnakke and van Wylen 6.76 A cutting tool uses a nozzle that generates a high speed jet of liquid water. Assume an exit velocity of 1000 m/s of 20oC liquid water with a jet diameter of 2 mm (0.002 m). How much mass flow rate is this? What size (power) pump is needed to generate this from a steady supply of 20oC liquid water at 200 kPa? Solution: C.V. Nozzle. Steady state, single flow. Continuity equation with a uniform velocity across A m . = AV/v = * 4 D2 V / v = * 4 0.0022 !'1000 / 0.001002 = 3.135 kg/s Assume Zi = Ze = Ø, ue = ui and Vi = 0 Pe = 100 kPa (atmospheric) Energy Eq.6.13: hi + Ø + Ø = he + 1 2V 2 e + Ø + w w = hi - he - 1 2V 2 e = ui - ue + Pi vi - Pe ve - 1 2V 2 e = (Pi - Pe) vi - 1 2V 2 e = 0.001002 !'(200 – 100) – 0.5 !'(10002 / 1000) = 0.1002 – 500 %'-500 kJ/kg W . = m . w = 3.135 (-500) = -1567.5 kW Sonntag, Borgnakke and van Wylen 6.77 A pipe flows water at 15oC from one building to another. In the winter time the pipe loses an estimated 500 W of heat transfer. What is the minimum required mass flow rate that will ensure that the water does not freeze (i.e. reach 0oC)? Solution: Energy Eq.: m . hi + Q . = m . he Assume saturated liquid at given T from table B.1.1 m . = Q . he - hi = -500 ! 10-3 0 - 62.98 = 0.5 62.98 = 0.007 94 kg/s 1 2 -Q . Sonntag, Borgnakke and van Wylen 6.80 Two steady flows of air enters a control volume, shown in Fig. P6.80. One is 0.025 kg/s flow at 350 kPa, 150&C, state 1, and the other enters at 450 kPa, 15&C, both flows with low velocity. A single flow of air exits at 100 kPa, -40&C, state 3. The control volume rejects 1 kW heat to the surroundings and produces 4 kW of power. Neglect kinetic energies and determine the mass flow rate at state 2. Solution: C.V. Steady device with two inlet and one exit flows, we neglect kinetic energies. Notice here the Q is rejected so it goes out. 1 2 3 Engine Q . loss W . Continuity Eq.6.9: m . 1 + m . 2 = m . 3 = 0.025 + m . 2 Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3h3 + W . CV + Q . loss Substitute the work and heat transfer into the energy equation and use constant heat capacity 0.025'×'1.004'×'B23.2 + m . 2'×'1.004'×'288.2 = (0.025'+'m . 2) 1.004 × 233.2 +'4.0 + 1.0 Now solve for m . 2. m . 2 = 4.0 + 1.0 + 0.025 × 1.004 × (233.2 – 423.2) 1.004 (288.2 - 233.2) Solving, m . 2 = 0.0042 kg/s Sonntag, Borgnakke and van Wylen 6.81 A large expansion engine has two low velocity flows of water entering. High pressure steam enters at point 1 with 2.0 kg/s at 2 MPa, 500&C and 0.5 kg/s cooling water at 120 kPa, 30&C enters at point 2. A single flow exits at point 3 with 150 kPa, 80% quality, through a 0.15 m diameter exhaust pipe. There is a heat loss of 300 kW. Find the exhaust velocity and the power output of the engine. Solution: C.V. : Engine (Steady state) Constant rates of flow, Q . loss and W . State 1: Table B.1.3: h1 = 3467.6 kJ/kg State 2: Table B.1.1: h2 = 125.77 kJ/kg h3 = 467.1 + 0.8 !'GGGHAI'D'2248.3 kJ/kg 1 2 3 Engine Q . loss W . v3 = 0.00105 + 0.8 ! 1.15825 = 0.92765 m 3/kg Continuity Eq.6.9: m . 1+ m . 2 = m . 3 = 2 + 0.5= 2.5 kg/s = (AV/v) = (*/4)D 2V/v Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3(h3 + 0.5 V 2) + Q . loss + W . V = m . 3v3 / [ * 4 D2] = 2.5 ! 0.92765 / (0.7854 ! 0.152 ) = 131.2 m/s 0.5 V2 = 0.5 !'131.22 /1000 = 8.6 kJ/kg ( remember units factor 1000) W . = 2 !3467.6 + 0.5 !125.77 – 2.5 (2248.3 + 8.6) – 300 = 1056 kW Sonntag, Borgnakke and van Wylen 6.82 Cogeneration is often used where a steam supply is needed for industrial process energy. Assume a supply of 5 kg/s steam at 0.5 MPa is needed. Rather than generating this from a pump and boiler, the setup in Fig. P6.82 is used so the supply is extracted from the high-pressure turbine. Find the power the turbine now cogenerates in this process. Solution: C.V. Turbine, steady state, 1 inlet and 2 exit flows, assume adiabatic, Q . CV = 0 Continuity Eq.6.9: m . 1 = m . 2 + m . 3 Energy Eq.6.10: Q . CV + m . 1h1 = m . 2h2 + m . 3h3 + W . T ; Supply state 1: 20 kg/s at 10 MPa, 500&C Process steam 2: 5 kg/s, 0.5 MPa, 155&C, Exit state 3: 20 kPa, x = 0.9 Table B.1.3: h1 = 3373.7, h2 = 2755.9 kJ/kg, Table B.1.2: h3 = 251.4 + 0.9 × 2358.3 = 2373.9 kJ/kg WT 1 2 3 HP LP W . T = 20'×'3373.7 - 5'×'2755.9 - 15'×'2373.9 = 18.084 MW Sonntag, Borgnakke and van Wylen 6.85 A cooler in an air conditioner brings 0.5 kg/s air at 35oC to 5oC, both at 101 kPa and it then mix the output with a flow of 0.25 kg/s air at 20oC, 101 kPa sending the combined flow into a duct. Find the total heat transfer in the cooler and the temperature in the duct flow. Solution: 1 2 3 4 Q cool Cooler section Mixing section C.V. Cooler section (no W . ) Energy Eq.6.12: m . h1 = m . h2 + Q . cool Q . cool = m . (h1 - h2) = m . Cp (T1 - T2) = 0.5 !'1.004 ! (35-5) = 15.06 kW C.V. mixing section (no W . , Q . ) Continuity Eq.: m . 2 + m . 3 = m . 4 Energy Eq.6.10: m . 2h2 + m . 3h3 = m . 4h4 m . 4 = m . 2 + m . 3 = 0.5 + 0.25 = 0.75 kg/s m . 4h4 = (m . 2 + m . 3)h4 = m . 2h2 + m . 3h3 m . 2 (h4 - h2) + m . 3 (h4 - h3) = Ø m . 2 Cp (T4 - T2) + m . 3 Cp (T4 - T3) = Ø T4 = (m . 2 / m . 4) T2 + (m . 3 / m . 4) T3 = 5(0.5/0.75) + 20(0.25/0.75) = 10&C Sonntag, Borgnakke and van Wylen 6.86 A heat exchanger, shown in Fig. P6.86, is used to cool an air flow from 800 K to 360 K, both states at 1 MPa. The coolant is a water flow at 15&C, 0.1 MPa. If the water leaves as saturated vapor, find the ratio of the flow rates m . H2O /m . air Solution: C.V. Heat exchanger, steady flow 1 inlet and 1 exit for air and water each. The two flows exchange energy with no heat transfer to/from the outside. 3 water 1 air 4 2 Continuity Eqs.: Each line has a constant flow rate through it. Energy Eq.6.10: m . airh1 + m . H2O h3 = m . airh2 + m . H2O h4 Process: Each line has a constant pressure. Air states, Table A.7.1: h1 = 822.20 kJ/kg, h2 = 360.86 kJ/kg Water states, Table B.1.1: h3 = 62.98 kJ/kg (at 15&C), Table B.1.2: h4 = 2675.5 kJ/kg (at 100 kPa) m . H2O /m . air = h1 - h2 h 4 - h3 = 822.20 - 360.86 2675.5 - 62.99 = 0.1766 Sonntag, Borgnakke and van Wylen 6.87 A superheater brings 2.5 kg/s saturated water vapor at 2 MPa to 450oC. The energy is provided by hot air at 1200 K flowing outside the steam tube in the opposite direction as the water, which is a counter flowing heat exchanger. Find the smallest possible mass flow rate of the air so the air exit temperature is 20oC larger than the incoming water temperature (so it can heat it). Solution: C.V. Superheater. Steady state with no external Q . or any W . the two flows exchanges energy inside the box. Neglect kinetic and potential energies at all states. 3 water 1 air 4 2 Energy Eq.6.10: m . H2O h3 + m . air h1 = m . H2O h4 + m . air h2 Process: Constant pressure in each line. State 1: Table B.1.2 T3 = 212.42&C, h3 = 2799.51 kJ/kg State 2: Table B.1.3 h4 = 3357.48 kJ/kg State 3: Table A.7 h1 = 1277.81 kJ/kg State 4: T2 = T3 + 20 = 232.42&C = 505.57 K A.7 : h2 = 503.36 + 5.57 20 (523.98 – 503.36) = 509.1 kJ/kg From the energy equation we get m . air / m . H2O = (h4 - h3)/(h1 - h2) = 2.5 (3357.48 – 2799.51) / (1277.81 – 509.1) = 1.815 kg/s Sonntag, Borgnakke and van Wylen 6.90 A copper wire has been heat treated to 1000 K and is now pulled into a cooling chamber that has 1.5 kg/s air coming in at 20oC; the air leaves the other end at 60oC. If the wire moves 0.25 kg/s copper, how hot is the copper as it comes out? Solution: C.V. Total chamber, no external heat transfer Energy eq.: m . cu h icu + m . air hi air = m . cu he cu + m . air he air m . cu ( he – hi )cu = m . air( hi – he )air m . cu Ccu ( Te – Ti )cu = m . air Cp air( Te – Ti )air Heat capacities from A.3 for copper and A.5 for air ( Te – Ti )cu = m . airCp air m . cuCcu ( Te – Ti )air = 1.5 !1.004 0.25 ! 0.42 (20 - 60) = - 573.7 K Te = Ti – 573.7 = 1000 - 573.7 = 426.3 K Air Air Cu Sonntag, Borgnakke and van Wylen Mixing processes 6.91 An open feedwater heater in a powerplant heats 4 kg/s water at 45oC, 100 kPa by mixing it with steam from the turbine at 100 kPa, 250oC. Assume the exit flow is saturated liquid at the given pressure and find the mass flow rate from the turbine. Solution: C.V. Feedwater heater. No external Q . or W . 1 2 3MIXING CHAMBER cb Continuity Eq.6.9: m . 1 + m . 2 = m . 3 Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3h3 = (m . 1+ m . 2)h3 State 1: Table B.1.1 h = hf = 188.42 kJ/kg at 45 oC State 2: Table B.1.3 h2 = 2974.33 kJ/kg State 3: Table B.1.2 h3 = hf = 417.44 kJ/kg at 100 kPa m . 2 = m . 1× h1 - h3 h3 - h2 = 4 ! 188.42 – 417.44 417.44 – 2974.33 = 0.358 kg/s T v1 23 100 kPa 2 P v 31 Sonntag, Borgnakke and van Wylen 6.92 A desuperheater mixes superheated water vapor with liquid water in a ratio that produces saturated water vapor as output without any external heat transfer. A flow of 0.5 kg/s superheated vapor at 5 MPa, 400&C and a flow of liquid water at 5 MPa, 40&C enter a desuperheater. If saturated water vapor at 4.5 MPa is produced, determine the flow rate of the liquid water. Solution: 1 2 3 CV . Sat. vapor Q = 0 LIQ VAP Continuity Eq.: m . 1 + m . 2 = m . 3 Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3h3 Table B.1 0.5'×'3195.7 + m . 2'×'171.97 = (0.5 + m . 2) 2797.9 => m . 2 = 0.0757 kg/s Sonntag, Borgnakke and van Wylen 6.95 Two flows are mixed to form a single flow. Flow at state 1 is 1.5 kg/s water at 400 kPa, 200oC and flow at state 2 is 500 kPa, 100oC. Which mass flow rate at state 2 will produce an exit T3 = 150 oC if the exit pressure is kept at 300 kPa? Solution: C.V. Mixing chamber and valves. Steady state no heat transfer or work terms. Continuity Eq.6.9: m . 1 + m . 2 = m . 3 Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3h3 = (m . 1+ m . 2)h3 1 2 3MIXING CHAMBER 2 P v 3 1 Properties Table B.1.3: h1 = 2860.51 kJ/kg; h3 = 2760.95 kJ/kg Table B.1.4: h2 = 419.32 kJ/kg m . 2 = m . 1× h1 - h3 h3 - h2 = 1.5 × 2860.51 – 2760.95 2760.95 – 419.32 = 0.0638 kg/s Sonntag, Borgnakke and van Wylen 6.96 An insulated mixing chamber receives 2 kg/s R-134a at 1 MPa, 100&C in a line with low velocity. Another line with R-134a as saturated liquid 60&C flows through a valve to the mixing chamber at 1 MPa after the valve. The exit flow is saturated vapor at 1 MPa flowing at 20 m/s. Find the flow rate for the second line. Solution: C.V. Mixing chamber. Steady state, 2 inlets and 1 exit flow. Insulated q = 0, No shaft or boundary motion w = 0. Continuity Eq.6.9: m . 1 + m . 2 = m . 3 ; Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3( h3 + 1 2 V 2 3 ) m . 2 (h2 – h3 – 1 2 V 2 3 ) = m . 1 ( h3 + 1 2 V 2 3 – h1 ) 1: Table B.5.2: 1 MPa, 100&C, h1 = 483.36 kJ/kg 2: Table B.5.1: x = ,, 60&C, h2 = 287.79 kJ/kg 3: Table B.5.1: x = 1, 1 MPa, 20 m/s, h3 = 419.54 kJ/kg Now solve the energy equation for m . 2 m . 2 = 2 ! [419.54 + 1 2 20 2 !' 1 1000 – 483.36] / [287.79 – 419.54 – 1 2 202 1000 ] = 2 ! ( -63.82 + 0.2) / ( -131.75 - 0.2) = 0.964 kg/s Notice how kinetic energy was insignificant. 1 2 3MIXING CHAMBER cb 2 P v 3 1 Sonntag, Borgnakke and van Wylen 6.97 To keep a jet engine cool some intake air bypasses the combustion chamber. Assume 2 kg/s hot air at 2000 K, 500 kPa is mixed with 1.5 kg/s air 500 K, 500 kPa without any external heat transfer. Find the exit temperature by using constant heat capacity from Table A.5. Solution: C.V. Mixing Section Continuity Eq.6.9: m . 1 + m . 2 = m . 3 => m . 3 = 2 + 1.5 = 3.5 kg/s Energy Eq.6.10: m . 1h1 + m . 2h2 = m . 3h3 h3 = (m . 1h1 + m . 2h2) / m . 3 ; For a constant specific heat divide the equation for h3 with Cp to get T3 = m . 1 m . 3 T1 + m . 2 m . 3 T2 = 2 3.5 2000 + 1.5 3.5 500 = 1357 K 1 2 3 Mixing section Sonntag, Borgnakke and van Wylen 6.100 For the same steam power plant as shown in Fig. P6.99 and Problem 6.99, assume the cooling water comes from a lake at 15&C and is returned at 25&C. Determine the rate of heat transfer in the condenser and the mass flow rate of cooling water from the lake. Solution: Condenser A7 = (*/4)(0.075) 2 = 0.004 418 m2, v7 = 0.001 008 m 3/kg V7 = m . v7/A7 = 25 × 0.001 008 / 0.004 418 = 5.7 m/s h6 = 191.83 + 0.92'×'2392.8 = 2393.2 kJ/kg qCOND = h7 - h6 + 1 2 ( V 2 7 - V 2 6 ) = 168 - 2393.2 + (5.72 - 2002 )/(2×1000) = -2245.2 kJ/kg Q . COND = 25'×'(-2245.2) = -56 130 kW This rate of heat transfer is carried away by the cooling water so -Q . COND = m . H2O (hout - hin)H2O = 56 130 kW => m . H2O = 56 130 104.9 - 63.0 = 1339.6 kg/s Sonntag, Borgnakke and van Wylen 6.101 For the same steam power plant as shown in Fig. P6.99 and Problem 6.99, determine the rate of heat transfer in the economizer, which is a low temperature heat exchanger. Find also the rate of heat transfer needed in the steam generator. Solution: Economizer A7 = *D 2 7/4 = 0.004 418 m 2, v7 = 0.001 008 m 3/kg V2 = V7 = m . v7/A7 = 25 × 0.001 008/0.004 418 = 5.7 m/s, V3 = (v3/v2)V2 = (0.001 118 / 0.001 008) 5.7 = 6.3 m/s 5 V2 so kinetic energy change unimportant qECON = h3 - h2 = 744 - 194 = 550.0 kJ/kg Q . ECON = m . qECON = 25 (550.0) = 13 750 kW Generator A4 = *D 2 4/4 = 0.031 42 m 2, v4 = 0.060 23 m 3/kg V4 = m . v4/A4 = 25 × 0.060 23/0.031 42 = 47.9 m/s qGEN = 3426 - 744 + (47.9 2 - 6.32)/(2×1000) = 2683 kJ/kg Q . GEN = m . qGEN = 25'×'(2683) = 67 075 kW Sonntag, Borgnakke and van Wylen 6.102 A somewhat simplified flow diagram for a nuclear power plant shown in Fig. 1.4 is given in Fig. P6.102. Mass flow rates and the various states in the cycle are shown in the accompanying table. The cycle includes a number of heaters in which heat is transferred from steam, taken out of the turbine at some intermediate pressure, to liquid water pumped from the condenser on its way to the steam drum. The heat exchanger in the reactor supplies 157 MW, and it may be assumed that there is no heat transfer in the turbines. a. Assume the moisture separator has no heat transfer between the two turbinesections, determine the enthalpy and quality (h4, x4). b. Determine the power output of the low-pressure turbine. c. Determine the power output of the high-pressure turbine. d. Find the ratio of the total power output of the two turbines to the total power delivered by the reactor. Solution: HP W 2 3 17 12 moisture separator WLP 9 4 5 8 a) Moisture Separator, steady state, no heat transfer, no work Mass: m . 3 = m . 4 + m . 9, Energy: m . 3h3 = m . 4h4 + m . 9h9 ; 62.874'×'2517 = 58.212'×'h4 + 4.662'×'558 $ h4 = 2673.9 kJ/kg h4 = 2673.9 = 566.18 + x4'×'2160.6 => x4 = 0.9755 b) Low Pressure Turbine, steady state no heat transfer Energy Eq.: m . 4h4 = m . 5h5 + m . 8h8+ W . CV(LP) W . CV(LP) = m . 4h4 - m . 5h5 - m . 8h8 = 58.212'×'2673.9 - 55.44'×'2279 - 2.772'×'2459 = 22 489 kW = 22.489 MW c) High Pressure Turbine, steady state no heat transfer Energy Eq.: m . 2h2 = m . 3h3 + m . 12h12 + m . 17h17 + W . CV(HP) W . CV(HP) = m . 2h2 - m . 3h3 - m . 12h12 - m . 17h17 = 75.6 × 2765 - 62.874 × 2517 - 8.064 × 2517 - 4.662 × 2593 = 18 394 kW = 18.394 MW d) J = (W . HP + W . LP)/Q . REACT = 40.883/157 = 0.26