Saygo, Joshua Gabriel C. 2CHE-B THERMODYNAMICS ASSIGNMENT 2.18 Liquid water at 180 C and 1,002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ kg1 and a specific volume of 1.128 cm3 g-1 (a) What is its enthalpy? (b) The water is brought to the vapor state at 300 C and 1,500 kPa, where its internal energy is 2,784.4 kJ kg-1 and its specific volume is 169.7 cm3 g-1. Calculate U and H for the process (a) Enthalpy π» = π + ππ π»1 = 762 + (1002.7 π₯ 1.128 π₯ 10−6 π3 ππ½ ) = 763.131 −3 10 ππ ππ (b) βπ» πππ βπ π»2 = 2784.4 + (1500 π₯ 169.7 π₯ 10−6 π3 ππ½ ) = 3038.99 −3 10 ππ ππ βπ» = π»2 − π»1 = 3038.99 − 763.131 = 2275.86 βπ = βπ2 − βπ1 = 2784.4 − 762 = 2022.4 ππ½ ππ ππ½ ππ 2.23 A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1 of cool water at 25 C with 0.8 kg s-1 of hot water at 75 C. During mixing, the heat is lost to the surroundings at the rate of 30 kJ s-1. What is the temperature of the heat is lost to the surroundings at the rate of 30 kJ s-1. What is the temperature of the warm-stream water? Assume the specific heat of water constant at 4.18 kJ kg-1 K-1. π2 πΆπ βπ = π1 πΆπ βπ + π (0.8)(4.18)(75 − π) = (1)(4.18)(π − 25) + 30 π = 43.23°πΆ 2.25 Water at 28 C flows in a straight horizontal pipe in which there is no exchange of either heat or work with the surroundings. Its velocity is 14 m s-1 in a pipe with an internal diameter of 2.5 cm until it flows into a section where the pipe diameter abruptly increases. What is the temperature change of the water if the downstream diameter is 3.8 cm? If it is 7.5 cm? What is the maximum temperature change for an enlargement in the pipe? π’1 π12 = π’2 π22 Saygo, Joshua Gabriel C. 2CHE-B πΉππ π‘βπ πππ€ππ π‘ππππ ππππππ‘ππ ππ 3.8 ππ: (14)(2.5 π₯ 10−2 )2 = π’2 (3.8 π₯ 10−2 )2 π’2 = 6.06 βπ» + 4.18βπ + π π βπ’2 π’22 − π’12 = 0; πΆπ βπ + =0 2 2 (6.06)2 − (14)2 = 0; βπ = 0.027°πΆ 2 πΉππ πππ€ππ π‘ππππ ππππππ‘ππ ππ 7.5 ππ: (14)(2.5 π₯ 10−2 )2 = π’2 (7.5 π₯ 10−2 )2 π’2 = 1.56 π π (1.56)2 − (14)2 4.18βπ + = 0; βπ = 0.023°πΆ 2 2.26 Fifty (50) kmol per hour of air is compressed from P1 = 1.2 bar to P2 = 6.0 bar in a steady-flow compressor. Delivered mechanical power is 98.8 kW. Temperatures and velocities are: T1 = 300 K T2 = 520 K U1 = 10 m s-1 U2 = 3.5 m s-1 Estimate the rate of heat transfer from the compressor. Assume for air that Cp = 7/2 R and that enthalpy is independent of pressure. πππ π ππππ€ πππ‘π ππ πππ = 50 π₯ 29 = 1450 π (βπ» + ππ 1 βπ ππ π₯ = 0.403 βπ 3600 π π βπ’2 + πβπ§) = π + π 2 ππ ππππ£ππ‘πππ; ∴ βπ§ = 0 π ([πΆπ (π2 − π1 )] + π’22 − π’12 )=π+π 2 (3.5)2 − (10)2 7 8.314 0.403 ([ π₯ π₯(520 − 300)] + ) = π + 98.8 2 29 2 π = −27.52 ππ½ π Saygo, Joshua Gabriel C. 2CHE-B 2.27 Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of 1.5(in). A pressure drop results from flow through a partially opened valve. Just upstream from the valve the pressure is 100(psia), the temperature is 120 F and the average velocity is 20(ft)(s)-1. If the pressure just downstream from the valve is 20(psia), what is the temperature. Assume for nitrogen that PV/T is constant, Cv=(5/2)R, and Cp=(7/2)R. βπ» + βπ’2 πβπ§ = π + π 2 βπ» + βπ’2 =0 2 (20)(100)(π2 ) π’π π’1 π1 π’2 π2 = ππππ π‘πππ‘; = ; π’2 = = 0.311π2 (20)(322) π π! π2 πΆπ βπ + π’22 − π’12 =0 2 (0.311π2 )2 − (20)2 1 7 (6.1324)(π2 − 322) + ( ) (28)(0.0022) = 0 2 2 32.2 π2 = 321.57 πΎ 2.28 Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil the water changes state from liquid at 200 kPa and 80 C to vapor at 100 kPa and 125 C. Its entering velocity is 3 m s-1 and its exit velocity is 200 m s-1. Determine the heat transferred through the coil per unit mass of water. Enthalpies of the inlet and outlet streams are: Inlet: 334.9 kJ kg-1; Outlet: 2726.5 kJ kg-1. βπ» + βπ’2 =π 2 (π»2 − π»1 ) + (2726.5 − 334.9) + π’22 − π’12 =π 2 (200)2 − (3)2 ππ½ π₯10−3 = π; π = 2411.6 2 ππ 2.29 Steam flows at steady state through a converging insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At the nozzle entrance (state 1), the temperature and pressure are 325 C and 700 kPa, and the velocity is 30 m s-1. At the nozzle exit (state 2), the stream temperature and pressure are 240 C and 350 kPa. Property values are: H1 = 3,112.5 kJ kg-1; V1=388.61 cm3 g-1; H2 = 2,945.7 kJ kg-1; V2=667.75 cm3 g-1. What is the velocity of the stream at the nozzle exit, and what is the exit diameter? βπ» + βπ’2 + πβπ§ = π + π 2 Saygo, Joshua Gabriel C. 2CHE-B βπ» + βπ’2 =0 2 (π»2 − π»1 ) + (2945.7 − 3112.5) + π’22 − π’12 =0 2 π’22 − 900 π₯10−3 = 0; π’2 = 578.4 π/π 2 π’1 ππ2 π’2 ππ2 = π1 π2 (30)(102 )(5)2 (578.539)(102 )(ππ2 ) 2 = ; ππ = 1.49 ππ 388.61 667.75 2.34 Carbon dioxide gas enters a water-cooled compressor at conditions P1 = 15(psia) and T1 = 50(F), and is discharged at conditions P2 = 520(psia) and T2 = 200(F). The entering CO2 flows through a 4inch-diameter pipe with a velocity of 20(ft)(s)-1, and is discharged through a 1-inch-diameter pipe. The shaft work supplied to the compressor is 5,360(Btu)(mol)-1. What is the heat-transfer rate from the compressor in (Btu)(hr)-1? H1=307(Btu)(lbm)-1 H2=330(Btu)(lbm)-1 V1=9.25(ft)3(lbm)-1 V2=0.28(ft)3(lbm)-1 π’1 π12 π’2 π22 = π1 π2 (20)(4)2 (π’2 )(1)2 ππ‘ = ; π’2 = 9.69 9.25 0.28 π (π»2 − π»1 ) + (π’22 − π’12 ) =π+π 2 (9.69)2 − (20)2 1 1 ′ π΅π‘π’ (330 − 307) + ( ) = π + (5360 π₯ ) ; π = −104.91 2 25.037 44.01 πππ πππ π ππππ€ πππ‘π = π’π΄ (20)(0.0855) πππ = = 0.185 π 9.25 π π ππ‘π ππ βπππ‘ π‘ππππ πππ = π π₯ π = −104.91 π₯ 0.185 = −19.41 π΅π‘π’ 3600 π π΅π‘π’ π₯ = −69870.06 π 1 βπ βπ Saygo, Joshua Gabriel C. 2CHE-B