# HW4-336 - ME 336 HW PROBLEM SET 4 9.27 A flow of 2 kg/s...

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ME 336 HW PROBLEM SET 4 9.27 A flow of 2 kg/s saturated vapor R-410a at 500 kPa is heated at constant pressure to 60 o C. The heat is supplied by a heat pump that receives heat from the ambient at 300 K and work input, shown in Fig. P9.27. Assume everything is reversible and find the rate of work input. Solution: C.V. Heat exchanger Continuity Eq.: m . 1 = m . 2 ; Energy Eq.: m . 1 h 1 + Q . H = m . 1 h 2 Table B.4.2: h 1 = 274.33 kJ/kg, s 1 = 1.0647 kJ/kg K h 2 = 342.32 kJ/kg, s 2 = 1.2959 kJ/kg K H Q W L Q T L HP 1 2 Notice we can find Q . H but the temperature T H is not constant making it difficult to evaluate the COP of the heat pump. C.V. Total setup and assume everything is reversible and steady state. Energy Eq.: m . 1 h 1 + Q . L + W . in = m . 1 h 2 Entropy Eq.: m . 1 s 1 + Q . L /T L + 0 = m . 1 s 2 (T L is constant, s gen = 0) Q . L = m . 1 T L [s 2 – s 1 ] = 2 × 300 [1. 2959 – 1.0647] = 138.72 kW W . in = m . 1 [h 2 - h 1 ] - Q . L = 2 (342.32 – 274.33) – 138.72 = 2.74 kW 9.34 A reversible steady state device receives a flow of 1 kg/s air at 400 K, 450 kPa and the air leaves at 600 K, 100 kPa. Heat transfer of 800 kW is added from a 1000 K reservoir, 100 kW rejected at 350 K and some heat transfer takes place at 500 K. Find the heat transferred at 500 K and the rate of work produced. Solution: C.V. Device, single inlet and exit flows.

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Energy equation, Eq.6.12: m . h 1 + Q . 3 - Q . 4 + Q . 5 = m . h 2 + W . Entropy equation with zero generation, Eq.9.8: m . s 1 + Q . 3 /T 3 - Q . 4 /T 4 + Q . 5 /T 5 = m . s 2 1 2 T T 500 K 3 3 4 4 Q Q W Q 5 Solve for the unknown heat transfer using Table A.7.1 and Eq. 8.19 for change in s Q . 5 = T 5 [s 2 – s 1 ] m . + T 5 T 4 Q . 4 T 5 T 3 Q . 3
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HW4-336 - ME 336 HW PROBLEM SET 4 9.27 A flow of 2 kg/s...

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