# hw5 - HOMEWORK SET 5 5.55 A cylinder having a piston...

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HOMEWORK SET 5 5.55 A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m) contains 0.5 kg of saturated vapor water at 120 ° C, as shown in Fig. P5.55. Heat is transferred to the water, causing the piston to rise. If the piston cross- sectional area is 0.05 m 2 , and the pressure varies linearly with volume until a final pressure of 500 kPa is reached. Find the final temperature in the cylinder and the heat transfer for the process. Solution: C.V. Water in cylinder. Continuity: m 2 = m 1 = m ; Energy Eq.5.11: m(u 2 - u 1 ) = 1 Q 2 - 1 W 2 State 1: (T, x) Table B.1.1 => v 1 = 0.89186 m 3 /kg, u 1 = 2529.2 kJ/kg Process: P 2 = P 1 + k s m A p 2 (v 2 - v 1 ) = 198.5 + 15 × 0.5 (0.05) 2 (v 2 - 0.89186) State 2: P 2 = 500 kPa and on the process curve (see above equation). => v 2 = 0.89186 + (500 - 198.5) × (0.05 2 /7.5) = 0.9924 m 3 /kg (P, v) Table B.1.3 => T 2 = 803 ° C ; u 2 = 3668 kJ/kg The process equation allows us to evaluate the work 1 W 2 = PdV = P 1 + P 2 2 m(v 2 - v 1 ) = 198.5 + 500 2 × 0.5 × (0.9924 - 0.89186) = 17.56 kJ Substitute the work into the energy equation and solve for the heat transfer 1 Q 2 = m(u 2 - u 1 ) + 1 W 2 = 0.5 × (3668 - 2529.2) + 17.56 = 587 kJ P v 1 2 T v 1 2 k m s A p 2

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5.61 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.61. Room A is at 200 kPa, v = 0.5 m 3 /kg, V A = 1 m 3 , and room B contains 3.5 kg at 0.5 MPa, 400 ° C. The membrane now ruptures and heat transfer takes place so the water comes to a uniform state at 100 ° C. Find the heat transfer during the process. Solution: C.V.: Both rooms A and B in tank. B A Continuity Eq.: m 2 = m A1 + m B1 ; Energy Eq.: m 2 u 2 - m A1 u A1 - m B1 u B1 = 1 Q 2 - 1 W 2 State 1A: (P, v) Table B.1.2, m A1 = V A /v A1 = 1/0.5 = 2 kg x A1 = v – v f v fg = 0.5 - 0.001061 0.88467 = 0.564 u A1 = u f + x u fg = 504.47 + 0.564 × 2025.02 = 1646.6 kJ/kg State 1B: Table B.1.3, v B1 = 0.6173, u B1 = 2963.2, V B = m B1 v B1 = 2.16 m 3 Process constant total volume: V tot = V A + V B = 3.16 m 3 and 1 W 2 = 0 / m 2 = m A1 + m B1 = 5.5 kg => v 2 = V tot /m 2 = 0.5746 m 3 /kg State 2: T 2 , v 2 Table B.1.1
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## hw5 - HOMEWORK SET 5 5.55 A cylinder having a piston...

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