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EML 3100 Homework No 2 soln 2010

# EML 3100 Homework No 2 soln 2010 - EML 3100 Homework No 2...

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EML 3100 Homework No. 2 Due Wednesday June 9 1. A piston cylinder device is to be analyzed. At the initial state, the cylinder holds 25 grams of saturated water vapor at 300 kPa. A resistance heater is installed in the walls of the cylinder and is turned on for 5 minutes. During this time 0.2 amps of current flows through the heater from a 120V power source. The cylinder is fairly well insulated but loses during this time period a total amount of heat equal to 3.7kJ. Assume that during the heating that the piston is allowed to move so that pressure is constant. a. show that for this closed system that the boundary work Wb and the change in internal energy delta-U, in the first law relation can be combined into one term, delta-H for the constant pressure process. b. Determine the final temperature of the steam State the assumptions that you should make for this analysis. Soln. Tank is stationary so KE and PE changes are zero Therefore delta-E = delta-U Neglect the energy changes in the wires and cylinder Q- Wother –Wb = U2 – U1 Wb = P (V2-V1) for a constant pressure process Therefore Q – Wother –P(V2-V1) = U2 – U1 P1 = P2 = P Therefore Q – Wother = (U2+P2V2) – (U1+ P1V1) Or Q – Wother = H2 – H1 Wother = V*I*delta-t = 120 V * 0.2A * 300 seconds *(1kJ/s/1000VA) = 7.2kJ At state point 1 Satd water vapor at 300 kPa, h = 2724.9kJ/kg

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EML 3100 Homework No 2 soln 2010 - EML 3100 Homework No 2...

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