FirsrLawSolns - 1 ME 201 Thermodynamics Solutions First Law...

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Unformatted text preview: 1 ME 201 Thermodynamics Solutions First Law Practice Problems 1. Consider a balloon that has been blown up inside a building and has been allowed to come to equilibrium with the inside temperature of 25°C and inside pressure of 100 kPa. The diameter of the balloon is measured and found to be 0.13 m. The balloon is then taken outside and allowed to come to equilibrium with the outside temperature of -5°C and outside pressure of 100 kPa. Determine the boundary work, heat transfer, and final balloon diameter. Solution: System Type: Closed System Substance Type: Ideal gas Process: Isobaric Initial State: Fixed Final State: Fixed Q = UNKNOWN W sh = 0 W bnd = UNKNOWN Conservation of Mass: m 1 = m 2 1st Law: W- Q = ) u- m(u bnd 1 2 Boundary Work: P(v- v ) 2 1 State 1 State 2 T 1 = 25°C = 298 K T 2 = -5°C = 268 K P 1 = 100 kPa P 2 = 100 kPa D 1 = 0.13 m D 2 = 0.1255 m V 1 = 1.150 x 10-3 m 3 V 2 = 1.034 x 10-3 m 3 m 1 = 1.345 x 10-3 kg m 2 = 1.345 x 10-3 kg u 1 = 209.06 kJ/kg u 2 = 191.17 kJ/kg φ 1 = 1.6783 kJ/(kg K) φ 2 = 1.588 kJ/(kg K) v 1 = 0.855 m3/kg v 2 = 0.769 m3/kg Italicized values are from ideal gas relations. Bold values are calculated. Approach: We begin by evaluating the properties at state 1 and state 2 by using the air tables and the ideal gas law. We can then use our boundary work equation to calculate the boundary work. Finally, we use the conservation of energy to determine the heat transfer. ME 201 Thermodynamics 2 We start by determining our total volume at state 1. The volume of a sphere is given by m 10 x 1.150 R 3 4 V 3 3- 3 1 = π = Next we determine our mass. Using the ideal gas law we have kg 10 x 1.345 ) 298 )( 287 . ( ) 10 x (100)(1.15 RT V P m 3--3 1 1 1 1 = = = /kg m 0.855 ) 10 x 1.345 ( ) 10 x (1.150 m V v 3 3--3 1 1 1 = = = We can go to the air tables and find u 1 = 209.06 kJ/kg φ 1 = 1.6783 kJ/(kg ⋅ K) u 2 = 191.17 kJ/kg φ 2 = 1.5888 kJ/(kg ⋅ K) The specific volume at state 2 must be given by the ideal gas law, so that /kg m 0.769 ) 100 ( ) 268 )( 287 . ( P RT v 3 2 2 2 = = = The total final volume is then V 2 = m 2 v 2 = (1.345 x 10-3 )(0.769) = 1.034 x 10-3 m 3 The final diameter is m 0.1255 4 3V 2 D 1/3 2 2 = π = The boundary work is calculated kJ 0.0116- ) 10 x 1.150- 10 x 4 (100)(1.03 ) V- P(V W-3-3 1 2 bnd = = = and the heat transfer is kJ 0.0357- (-0.0116) 209.06)- )(191.17 10 x (1.345 W ) u- m(u Q-3 bnd 1 2 = + = + = ME 201 Thermodynamics 3 2. Air at 1800 K and 800 kPa enters an ideal turbine at 2.3 kg/s. The power output required of this turbine is 700 kW. Determine the exhaust temperature and pressure. Solution: System Type: Control Volume System Substance Type: Ideal gas Process: Isentropic Initial State: Fixed Final State: Unknown Q = 0 W sh = 700 kW Conservation of Mass: kg/s 2.3 m = m 1 2 = & & 1st Law: ( 29 W- h- h m sh 1 2 & & = State 1 State 2 T 1 = 1800 K T 2 = 1552 K P 1 = 800 kPa P 2 = 424 kPa h 1 = 2003.3 kJ/kg h 2 = 1698.95 kJ/kg φ 1 = 3.6684 kJ/(kg K) φ 2 =...
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.

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FirsrLawSolns - 1 ME 201 Thermodynamics Solutions First Law...

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