EGR 334 Thermodynamics: Homework 08 Problem 3: 49 A closed rigid tank is filled with water. Initially the tank hold 9.9 ft 3 saturated vapor and 0.1 cu. ft saturated liquid, each at 212 deg F. The water is heated until the tank contains only saturated vapor. For the water, determine a) the quality at the initial state, b) the temperature at the final state in deg F, and c) the heat transfer in Btu, Kinetic and potential energy effects can be ignored. ----------------------------------------------------------------------------------------------------------------------------- ------------State 1: V1 = 9.9 ft3 State 2: Saturated vapor. V1liq = 0.1 ft3 T 1 = 212 oF Constant volume process: V1 V2 9.9 0.1 10.0 ft 3 Look up Saturation values at T= 212 deg. F. vf = 0.01672 uf = 180.1 Btu/lbm Mass of saturated liquid: Vliq mliq v f mliq Vliq vf p 1 = 14.7 psi vg = 26.80 ug = 1077.6 Btu/lbm 0.1 ft 3 5.981 lbm 0.01672 ft 3 / lbm Q Mass of saturated vapor: Vvapor mvapor v g mvapor Vvapor vg 9.9 ft 3 0.3694 lbm 26.80 ft 3 / lbm Total mass of system: m mliq mvapor 5.981 0.3694 6.350 lbm Quality of initial state: x1 mvapor mliquid mvapor mvapor mtotal 0.3694 0.05817 5.8% 6.350 Then u1 u f x(ug u f ) 180.1 0.05817(1077.6 180.1) 232.3 Btu / lbm At state 2, all mass is saturated vapor or v2 vsat _ 2 V2 10 ft 3 1.5748 ft 3 / lbm mtotal 6.350 lbm Look up saturated vapor conditions for this specific volume off the saturated vapor table A-2E. Note that at T = 410 vg = 1.6743 ug = 1117.6 T= 420 vg = 1.502 ug = 1118.3 using interpolation: T2 410 1.5748 1.6743 420 410 1.502 1.6743 and T2 415.8 o F u2 1117.6 1.5748 1.6743 1118.3 1117.6 1.502 1.6743 u2 = 1118.0 Btu/lbm For a system which doesn't undergo a change of volume the work done by the system is 0. W 0 1st Law of thermodynamics: U Q W Q U U 2 U1 m(u2 u1 ) 6.350lbm (1118.0 232.5)Btu / lbm 5623.0Btu EGR 334 Thermodynamics: Homework 08 Problem 3:55 A piston cylinder assembly containing water, initially a liquid at 50 deg F. undergoes a process at a constant pressure of 20 psi to a final state where the water is a vapor at 300 deg. F. Kinetic and potential energy effects are negligible. Determine the work and heat transfer, in BTU per lb, for each of three parts of the overall process; a) from the initial liquid state to saturated liquid state, b) from saturated liquid to saturated vapor, and c) from saturated vapor to the final vapor state all at 20 psi. ----------------------------------------------------------------------------------------------------------------------------- -----------State 1: T1 = 50 deg F. State 2: Sat. liq. State 3: sat. vapor State 4: T 2=300 deg F. p1 = 20 psi p2 = 20 psi Find spec. volume and internal energy at each of the four states. State 1: v1 ≈ vf (T=50 deg) State 2: v2 = vf (p=20psi) State 3: v3 = vg(p=20psi) = 0.01602 ft3/lb = 0.01683 ft3/lb = 20.09 ft3/lb u1 ≈ uf (T=50 deg) = 18.06 Btu/lb u2 = uf (p=20psi) =196.19 Btu/lb State 4: v2 = v(T300,p20) = 22.36 ft3/lb u3 = ug(p=20psi) = 1082.0 Btu/lb u2 = u(T300,p20) = 1108.7 Btu/lb State 1 to State 2: U 2 U1 Q12 W1 2 since the liquid is treated as incompressible: V 0 W12 0 Q12 / m (u2 u1 ) 196.19 18.06 178.13 Btu / lbm State 2 to State 3: occurs as a constant pressure process so W23 pdV p (V3 V2 ) or 2 12in 1Btu = 74.3 Btu/lbm W23 / m p(v3 v2 ) (20lb / in )(20.09 0.01683) ft / lbm 1 ft 778.17 ft lb 2 3 Q23 / m (u3 u2 ) W23 / m (1082.0 196.19) 74.3Btu / lbm 960.11Btu / lbm State 3 to State 4: occurs as a constant pressure process so W3 4 pdV p (V4 V3 ) or 2 12in 1Btu = 8.40 Btu/lbm W34 / m p(v4 v3 ) (20lb / in )(22.36 20.09) ft / lbm 1 ft 778.17 ft lb 2 3 Q34 / m (u4 u3 ) W34 / m (1108.7 1082.0) 8.4Btu / lbm 35.1Btu / lbm Summary: W12 / m 0 Q12 / m 178.13 Btu / lbm W23 / m 74.3Btu / lbm W34 / m 8.4 Btu / lbm Q23 / m 960.11Btu / lbm Q34 / m 35.1Btu / lbm EGR 334 Thermodynamics: Homework 08 Problem 3:68 Water contained in a piston cylinder assembly initially at 300 deg F, a quality of 90% and a volume of 6 ft 3 is heated at constant temperature to saturate vapor. If the rate of heat transfer is 0.3 Btu/s, determine the time in minutes for this process of the water to occur. Kinetic and potential energy effects are negligible. ----------------------------------------------------------------------------------------------------------------------------- -----------State 1: T1 = 300 deg F. State 2: sat. vapor x1 = 90% T 2 = 300 deg F. V1 = 6 ft3 Heat Rate: Q dot = 0.3 Btu/s For T = 300 deg. p = 66.98 psi vf = 0.01745 ft3/lbm uf = 269.5 Btu/lbm vg = 6.472 ft3/lbm ug = 1100.0 Btu/lbm Find the internal energy at each state: State 1: v1 v f x(vg v f ) 0.01745 0.90(6.472 0.01745) 5.827 ft 3 / lb u1 u f x(u g u f ) 269.5 0.9(1100 269.5) 1016.95 Btu / lbm State 2: v2 6.472 ft 3 / lb u2 1100 Btu / lbm Find amount of mass: m V1 6 ft 3 1.0297lbm v1 5.827 ft 3 / lb Volume of State 2: 1st Law of Thermo: U Q W where: U 1.0297lbm (1100 1016.95)Btu / lbm 85.5Btu Q Q t (0.3Btu / s) t W pdV p(V2 V1 ) p m(v2 v1 ) 2 12in 1Btu (66.98lb / in )(1.0297lbm)(6.472 5.827) ft / lbm 8.23Btu 1 ft 778.17 ft lb 2 therefore: U Q W 85.5Btu (0.3Btu / s )t 8.23Btu (85.5 8.23) Btu t 312.4s 5.21min 0.3Btu / s 3 EGR 334 Thermodynamics: Homework 08 Problem 3:78 One lbm of water contained in a piston cylinder assembly undergoes the power cycle shown in Fig. 3.78. For each of the four processes, evaluate the work and heat transfer each in Btu. For the overall cycle, evaluate the thermal efficiency. ----------------------------------------------------------------------------------------------------------------------------- --------------mass = 1 lbm State 1: p1 = 700 psi saturated liquid. State 2: p 2=700 psi saturated vapor State 3: p3= 70 psi mixture v3 = v2 State 4: p4 = 70 psi v4 = v1 p mixture 1 700 psi Start by looking up intensive properties at the four states: State 1: T1= Tsat at p = 700 psi. from table A-3E: T1= 503.23 oF v1 = vf(p=700) = 0.02051 ft3/lbm u1 = uf(p=700) = 488.9 Btu/lbm 70 psi 4 State 2: T2= Tsat at p = 700 psi. from table A-3E: T1= 503.23 oF v2 = vg(p=700) = 0.656 ft3/lbm u2 = ug(p=700) = 1117.0 Btu/lbm v1=v4 State 3: p3= 70 psi ( from table A-3E: p=70 psi then vf=0.01748 ft3/lbm vg= 6.209 ft3/lbm uf = 272.6 Btu/lbm ug = 1100.6 Btu/lbm v2 = v3 = 0.656 ft3/lbm therefore quality is x3 v3 v f vg v f 0.656 0.01748 0.1031 10.31% 6.209 0.01748 and then u3 u f x(ug u f ) 272.6 0.1031(1100.6 272.6) 358.0Btu / lbm State 4: v4 = v1 = 0.02051 ft3/lbm therefore quality is x4 v4 v f vg v f 0.02051 0.01748 0.000489 0.0489% 6.209 0.01748 and then u4 u f x4 (u g u f ) 272.6 0.000489(1100.6 272.6) 273.0 Btu / lbm Process 1 to 2: U Q W where: U12 m(u2 u1 ) 1lbm(1117 488.9) Btu / lbm 628.1Btu W1 2 pdV p1 (V2 V1 ) p1m(v2 v1 ) 2 (700lb f / in2 )(1lbm)(0.656 0.02051) ft 3 / lbm 12in 1Btu 82.3Btu 1 ft 778.17lb f ft Q12 U W 628.1 82.3 710.4Btu Process 2 to 3: U Q W where: U 23 m(u3 u2 ) 1lbm(358.0 1117) Btu / lbm 759Btu W23 0 Q23 U W 759 0 759Btu 2 1 3 v2=v3 v Process 3 to 4: U Q W where: U 34 m(u4 u3 ) 1lbm(273 358) Btu / lbm 85Btu W3 4 pdV p3 (V4 V3 ) p3 m(v4 v3 ) 2 12in 1Btu (70lb f / in )(1lbm)(0.02051 0.656) ft / lbm 8.23Btu 1 ft 778.17lb f ft 2 3 Q34 U W 85 8.23 93.23Btu Process 4 to 1: U Q W where: U 41 m(u1 u4 ) 1lbm(488.9 273) Btu / lbm 215.9 Btu W41 0 Q41 U W 215.9 0 215.9Btu Heat input into the system: Qin Q12 Q41 710.4 215.9 926.3Btu Heat dumped to the environment: Qout Q23 Q34 759 93.23 852.2Btu Work done by system: Wcycle 82.3 8.23 74.07 Btu thermal efficiency: Wcycle Qin 74.07 0.07996 8.0% 926.3