MAE 110A. Homework 6: Solutions 11/9/2017

Transcription

1 MAE 110A Hoework 6: Solutions 11/9/2017

2 H6.1: Two kg of H2O contained in a piston-cylinder assebly, initially at 1.0 bar and 140 C undergoes an internally ersible, isotheral copression to 25 bar. Given Model p = 1.0 bar Water is the closed syste p = 25 bar Isotheral, T = T T = T = 140 C Internally ersible = 2.0 kg Ignore potential and kinetic energy effects Deterine a) Sketch p v and T s diagras b) Heat transfer, in kj c) Work, in kj Basic Equations ds = δq T int ΔKE ΔPE ΔU = W DE Q HIJ W HIJ Analysis b) ds = δqekj DE T int δqekj DE = TdS QEKJ DE = TdS = T S S = T s s Deterine s fro the superheated vapor, A-4, using T = 140 C and p = 1.0 bar s = kj/kg K Deterine s fro the copressed liquid tables, A-5, using T = 140 C and p = 25 s = kj/kg K QEKJ DE QEKJ DE = 2 kg 140 C 273 K kj/kg K kj/kg K = kj is negative heat transfer is out of the syste Q HIJ = kj c) ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = u u = W DE Q HIJ W DE = u u Q HIJ Deterine u fro the superheated vapor, A-4, using T = 140 C and p = 1.0 bar u = kj/kg Deterine u fro the copressed liquid tables, A-5, using T = 140 C and p = 25 u = kj/kg W DE = 2 kg kj/kg kj/kg kj = kj 1

3 H6.2: One kg of air contained in a piston-cylinder assebly initially at 1.0 MPa and 960 K expands in an internally ersible adiabatic process to 200 kpa. Assue: a) variable specific heats (ideal gas table) b) constant specific heats Given p = 1.0 MPa p = 200 kpa T = 960 K = 1.0 kg Model Air is the closed syste Air is an ideal gas Internally ersible Ignore potential and kinetic energy effects Adiabatic Deterine T, the final teperature, in K W, the work of the process, in kj Basic Equations ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ds = δq T int Tds = dh vdp pv = RT c = c n R Analysis ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = W HIJ u u W HIJ = u u ds = δq δq S = = 0 S T int T = S pv = RT v = RT p Tds = dh vdp = c n dt RT p dp ds = c n T dt R p dp s = c n T dt R p dp = c n T dt R ln p p = 0 1 = W HIJ a) Variable specific heats Equation 1 can be expressed as: s = s H T s H T R ln p = 0 s H T p = s H T R ln p p Deterine s H T fro the ideal gas tables, A-22, using T s H T s H T = kj/kg K = kj/kg K J/ol K 29.0 g/ol 200 kpa ln = kj/kg K 1000 kpa 2

4 Deterine T fro the ideal gas tables, A-22, using s H T T = 630 K Deterine u and u fro the ideal gas tables using T and T u = kj/kg u = kj/kg W HIJ = 1.0 kg kj/kg kj/kg = kj b) Constant specific heats Equation 1 can be integrated directly: s = c n ln T R ln p = 0 ln T = R ln p T = p T p T c n p T p Deterine c n fro ideal gas tables, A-20, using T c n = kj/kg K x.yz J/ol K {. g/ol.yz kj/kg K 200 kpa T = 960 K = 639 K 1000 kpa W HIJ = u u = c T T J/ol K c = c n R = kj/kg K = kj/kg K 29.0 g/ol W HIJ = 1.0 kg kj/kg K 960 K 639 K = kj u v w T = T p p u v w 3

5 H6.3: Consider a Carnot power cycle with water contained in a piston cylinder assebly (recall Fig in text but states are nubered differently here). At the start of the heat addition process, the water is saturated liquid at p 1 = 50 bar. The water is heated isotherally until it is saturated vapor (state 2). The water then expands adiabatically until the pressure is 5 bar. Given p = 50 bar p y = 5 bar Model Water is the closed syste Ignore potential and kinetic energy effects All processes are internally ersible Carnot cycle Deterine a) Sketch p v and T s diagras b) Q, heat transfer, and W, work, for each process, in kj/kg c) η, theral efficiency using part b d) η, theral efficiency using T H and T L Basic Equations ΔKE ΔPE ΔU = W DE Q HIJ W HIJ W = ds = δq T pd int Tds = du pdv η = WEKJ HIJ η ƒ = 1 T T u = u x u u Analysis b) Property Evaluations: State (1) saturated liquid at p = 50 bar Deterine u, v, s, and T fro the saturated water tables, A-3, using p u = u = kj/kg v = v = y 3 /kg s = s = kj/kg K T = T Œ J = T = C = 537 K State (2) saturated vapor T = T = T also p = p Œ J = p = 50 bar Deterine u, v, and s fro the saturated water tables, A-3, using p u = u = kj/kg 4

6 v = v = /kg s = s = kj/kg K State (3) ersible, adiabatic expansion fro state (2) to p y = 5 bar ds = δq = 0 S = s = 0 s T y = s int Deterine s y, s y, u y, u y, v y, v y, and T y fro saturated water tables, A-3, using p y s y = kj/kg K, s y = kj/kg K u y = kj/kg, u y = kj/kg v y = y 3 /kg, v y = /kg T y = T Œ J = T = C = K x y = s y s y kj/kg K kj/kg K = s y s y kj/kg K kj/kg K = 0.83 u y = u y x y u y u y u y = kj/kg kj/kg kj/kg = kj/kg v y = v y x y v y v y v y = y 3 /kg /kg y 3 /kg = /kg State (4) ersible, isotheral process fro state (3) so T z = T y, also process 4-1 is ersible, adiabatic like process 2-3, therefore, s z = s Fro T-s diagra state 3 saturation quantities are the sae as state 4 s z = s y, s z = s y, u z = u y, u z = u y, v z = v y, v z = v y x z = s z s z kj/kg K kj/kg K = s z s z kj/kg K kj/kg K = 0.21 u z = u z x z u z u z u z = kj/kg kj/kg kj/kg = kj/kg v z = v z x z v z v z v z = y 3 /kg /kg y 3 /kg = /kg Process 1-2: ersible, isotheral expansion T = T = T ds = δq T T ds = δq T ds = δq δq int = T ds Q = T s s = Q = T s s = 537 K W = pd W = δq kj/kg K kj/kg K = kj/kg pdv = p v v = T ds W = 50 bar 100 kpa/bar /kg y 3 /kg = kj/kg 5

7 W > 0 expansion so work is directed out of the syste: W HIJ = kj/kg Alternative: Use 1 st Law ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = u u = W HIJ Process 2-3: ersible, adiabatic expansion = Q HIJ = 0 ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = W HIJ W HIJ = u y u W HIJ = u u y W HIJ = kj/kg kj/kg = kj/kg Process 3-4: ersible, isotheral copression T y = T = T ds = δq T T ds = δq T ds = δq δq int = T ds Q yz = T s z s y Q HIJ z δq y z = T ds y = Q yz = T s z s y Q HIJ = K kj/kg K kj/kg K = kj/kg W = z pd y W = z pdv y = p y v z v y W = 5 bar 100 kpa/bar /kg /kg = kj/kg W < 0 copressible so work is directed into the syste: W DE = kj/kg Alternative: Use 1 st Law ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = u z u y = W HIJ Process 4-1: ersible, adiabatic copression = Q HIJ = 0 ΔKE ΔPE ΔU = W DE Q HIJ W HIJ ΔU = W DE W DE = u u z W DE = u u z W DE c) η = η = = kj/kg kj/kg = kj/kg WEKJ HIJ = W HIJ W DE = W HIJ W HIJ W DE W DE kj/kg kj/kg kj/kg kj/kg kj/kg =

8 d) η ƒ = 1 T = K = T 537 K 7

9 6.33: Air in a piston cylinder assebly undergoes a Carnot power cycle. The isotheral expansion and copression processes occur at 1400 K and 350 K, respectively. The pressures at the beginning and end of the isotheral copression are 100 kpa and 500 kpa, respectively. Assue the ideal gas odel with cp = kj/kg*k. Given T, = 1400 K T y,z = 350 K p y = 100 kpa p z = 500 kpa c n = kj/kg K Model Air is the closed syste Ignore potential and kinetic energy effects Carnot cycle Air is an ideal gas w/ constant specific heats Deterine a) p and p, the pressures at the beginning and end of the isotheral expansion, in kpa b) Q, heat transfer, and W, work, for each process, in kj/kg c) η, theral efficiency Basic Equations ΔKE ΔPE ΔU = W DE Q HIJ W HIJ P 2 ds = δq T int Tds = dh vdp η = WEKJ HIJ η ƒ = 1 T T pv = RT c = c n R Analysis a) Tds = dh vdp = c n dt RT p dp ds = c n T dt R p dp c n s = T dt R p dp = c n ln T R ln p T p because this is the Carnot cycle it is known that process 2-3 and 4-1 are isentropic s y s = c n ln T y T R ln p y p = 0 p = p y T T y p = 100 kpa 1400 K 350 K. œ œž Ÿ v w u x.yz J/ol K {. g/ol = kpa 8

10 s s z = c n ln T T z R ln p p z = 0 p = p z T T z p = 500 kpa 1400 K 350 K. œ œž Ÿ v w u x.yz J/ol K {. g/ol = kpa b) Process 1-2: ersible, isotheral expansion T = T = T ds = δq T Q δq = 1400 K T ds = δq T ds = δq δq int = T ds = T ds Q = T s = T c n ln T T R ln p p = T R ln p J/ol K 29.0 g/ol kpa ln kpa = kj/kg Constant teperature ideal gas eans ΔU = u T u T = 0 ΔKE ΔPE ΔU = W DE Q HIJ W HIJ = W HIJ = W HIJ W HIJ = kj/kg Process 2-3: ersible, adiabatic expansion = Q HIJ = 0 = 0 ΔKE ΔPE ΔU = W DE Q HIJ W HIJ W HIJ = ΔU = u y u = c T y T W HIJ = c T y T J/ol K c = c n R = kj/kg K = kj/kg K 29.0 g/ol W HIJ = kj/kg K 350 K 1400 K = kj/kg p Process 3-4: ersible, isotheral copression T y = T z = T Q yz = T R ln p z J/ol K 500 kpa = 350 K ln p y 29.0 g/ol 100 kpa Q HIJ = kj/kg ΔU = u z T z u y T y = 0 = W DE Q HIJ W DE = Q HIJ = kj/kg = kj/kg 9

11 Process 4-1: ersible, adiabatic copression = Q HIJ = 0 Q HIJ = 0 W DE = Δu = c T T z = kj/kg K 1400 K 350 K = kj/kg c) η = WEKJ HIJ = W HIJ W HIJ W DE W DE kj/kg kj/kg kj/kg kj/kg η = = kj/kg Alternatively, because this is a Carnot cycle η = η ƒ η ƒ = 1 T = K = T 1400 K 10

12 6.42: A rigid, insulated container fitted with a paddle wheel contains 5 lb of water, initially at 260 F and a quality of 60%. The water is stirred until the teperature is 350 F. Given Model T = 260 F Water is the closed syste T = 350 F Ignore potential and kinetic energy effects x = 0.6 Process is adiabatic = 5 lb Constant volue, v = v Deterine a) W, work, in Btu b) σ, entropy produced, in Btu/ R Basic Equations δq S = σ T v = v x v v ΔKE ΔPE ΔU = W DE Q HIJ W HIJ Analysis a) ΔKE ΔPE ΔU = W DE Q HIJ W HIJ W DE = ΔU = u u State (1): T = 260 F, x = 0.6 u = u x u u v = v x v v Deterine u, u, v, and v fro the saturated water tables, A-2E, using T = 260 F u = Btu/lb u = Btu/lb v = ft 3 /lb v = ft 3 /lb u = Btu/lb Btu/lb Btu/lb = Btu/lb v = ft y /lb ft 3 /lb ft y /lb = ft 3 /lb Deterine u fro the superheated water vapor tables, A-4E, using T = 350 F and v = v u = Btu/lb W DE = 5 lb Btu/lb Btu/lb = Btu/lb b) δq S = σ σ = s T s s = s x s s Deterine s and s fro the saturated water tables, A-2E, using T = 260 F s = Btu/lb R s = Btu/lb R s = Btu/lb R Btu/lb R Btu/lb R s = Btu/lb R Deterine s fro the superheated water vapor tables, A-4E, using T = 350 F and v = v 11

13 s = Btu/lb R σ = 5 lb Btu/lb R Btu/lb R = Btu/ R 12

14 6.50: Air as an ideal gas contained within a piston cylinder assebly is copressed between two specified states. Given (a) Given (b) Model p = 0.1 MPa p = 3 at Air is the closed syste p = 0.5 MPa p = 10 at Ignore potential and kinetic energy effects T = 27 C T = 80 F Air is an ideal gas T = 207 C T = 240 F c n = Btu/lb R Deterine For each of the two sets of givens, deterine if the process can occur adiabatically. If yes, deterine the work for an adiabatic process between the states. If no, deterine the direction of the heat transfer. Basic Equations ΔKE ΔPE ΔU = W DE Q HIJ W HIJ δq S = σ T Tds = dh vdp s s = s H T s H T R ln p p Analysis To deterine if the process can occur adiabatically, start with the closed syste entropy balance: δq S = σ S = σ, σ 0 S 0 for an adiabatic process T a) s s = s H T s H T R ln p p Deterine s H T and s H T fro the ideal gas tables, A-22, using T and T respectively s H T = kj/kg K s H T = kj/kg K s = kj/kg K kj/kg K J/ol K 29.0 g/ol 0.5 MPa ln 0.1 MPa = kj/kg K s > 0 the process can occur adiabatically ΔKE ΔPE ΔU = W DE Q HIJ W HIJ W DE = ΔU = u u Deterine u and u fro the ideal gas tables, A-22, using T and T respectively u = kj/kg u = kj/kg W DE = u u = kj/kg kj/kg = kj/kg 13

15 (a) b) Tds = dh vdp = c n dt RT p dp ds = c n T dt R p dp c n s = T dt R p dp = c n ln T R ln p T s = Btu/lb R ln 240 F 460 R 80 F 460 R s = kj/kg K Fro the entropy balance equation: δq S = T S < 0 only when p Btu/ol R lb/ol 10 at ln 3 at σ, since σ cannot be negative, process cannot occur adiabtically 2 δq 1 T < 0 the heat transfer is out of the syste (b) 14