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TransientSolns - ME 201 Thermodynamics Solutions to...

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1 ME 201 Thermodynamics Solutions to Transient System Practice Problems 1. A balloon initially contains 5 m 3 CO 2 at 100 kPa and 22 ° C. It is connected to a CO 2 gas line that provides CO 2 at 170 kPa and 30 ° C. The balloon is then filled to a pressure of 170 kPa. During this process the balloon is insulated and it is observed that the pressure and total volume are related by PV -1 = constant Determine the final temperature and volume of the balloon. Solution: We begin by setting up our table. System Type: Transeint System Substance Type: Ideal gas Process: Polytropic with n=-1 State 1: Fixed State 2: Unknown State In: Fixed W sh = 0 Q = 0 W bnd = n 1 V P - V P 1 1 2 2 - Conservation of mass: m 2 - m 1 = m in 1 st Law: m 2 u 2 - m 1 u 1 = m in h in -W bnd State 1 State In State 2 P 1 = 100 kPa P in = 170 kPa P 2 = 170 kPa T 1 = 295 K T in = 303 K T 2 = 324 K V 1 = 5 m 3 V 2 = 8.5 m 3 m 1 = 8.973 kg m in = 14.71 kg m 2 = 23.6 kg u 1 = 154.49 kJ/kg h in = 216.87 kJ/kg u 2 = 173.89 kJ/kg Italicized values are obtained from N 2 tables or ideal gas equation. Bold values are calculated. At state 1 we know both the temperature and pressure, so the state is fixed. We can calculate the mass from kg 8.973 = 95) (0.1889)(2 (100)(5) = RT V P = m 1 1 1 1
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ME 201 Thermodynamics 2 The internal energy can be read from the CO 2 tables, but will have to be converted to kJ/kg or kJ/kg 154.49 = 44.01 6799 = MW u = u 2 N 1 1 At the inlet state we can find the enthalpy from the CO 2 tables or h in = 216.87 kJ/kg To begin to fix our final state, we use the polytropic relation or V P = V P -1 1 1 -1 2 2 or m 8.5 = 100 170 (5) = P P V = V 3 1 2 1 2 The boundary work can now be calculated kJ 472.5 = (-1) - 1 (100)(5) - (170)(8.5) = n 1 V P - V P = W 1 1 2 2 bnd -
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