HW2-336 - ME 336 HW SET 2, C. BORGNAKKE, ME 5.46 A 10-L...

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ME 336 HW SET 2, C. BORGNAKKE, ME 5.46 A 10-L rigid tank contains R-410a at 10 ° C, 80% quality. A 10-A electric current (from a 6-V battery) is passed through a resistor inside the tank for 10 min, after which the R-410a temperature is 40 ° C. What was the heat transfer to or from the tank during this process? Solution: C.V. R-410a in tank. Control mass at constant V. Continuity Eq.: m 2 = m 1 = m ; Energy Eq.: m(u 2 - u 1 ) = 1 Q 2 - 1 W 2 Process: Constant V v 2 = v 1 => no boundary work, but electrical work v P 1 2 State 1 from table B.4.1 v 1 = 0.000827 + 0.8 × 0.04470 = 0.03659 m 3 /kg u 1 = 42.32 + 0.8 × 207.36 = 208.21 kJ/kg m = V/v = 0.010/0.03659 = 0.2733 kg State 2: Table B.4.2 at 40 ° C and v 2 = v 1 = 0.03659 m 3 /kg => superheated vapor, so use linear interpolation to get P 2 = 800 + 200 × (0.03659 – 0.04074)/(0.03170 – 0.04074) = 800 + 200 × 0.45907 = 892 kPa u 2 = 286.83 + 0.45907 × (284.35 – 286.83) = 285.69 kJ/kg 1 W 2 elec = –power × t = –Amp × volts × t = – 10 × 6 × 10 × 60 1000 = –36 kJ 1 Q 2 = m(u 2 – u 1 ) + 1 W 2 = 0.2733 ( 285.69 – 208.21) – 36 = – 14.8 kJ 5.66 A piston cylinder has a water volume separated in V A = 0.2 m 3 and V B = 0.3 m 3 by a stiff membrane. The initial state in A is 1000 kPa, x = 0.75 and in B it is 1600 kPa and 250°C. Now the membrane ruptures and the water comes to a uniform state at 200°C. What is the final pressure? Find the work and the heat transfer in the process. Take the water in A and B as CV. Continuity: m 2 - m 1A - m 1B = 0
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Energy: m 2 u 2 - m 1A u 1A - m 1B u 1B = 1 Q 2 - 1 W 2 Process: P 2 = P eq = constant = P 1A as piston floats and m p , P o do not change State 1A: Two phase. Table B.1.2 v 1A = 0.001127 + 0.75 × 0.19332 = 0.146117 m
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HW2-336 - ME 336 HW SET 2, C. BORGNAKKE, ME 5.46 A 10-L...

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