# hw4 - HOMEWORK SET 4 4.41 Ammonia(0.5 kg is in a piston...

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HOMEWORK SET 4 4.41 Ammonia (0.5 kg) is in a piston cylinder at 200 kPa, -10 o C is heated in a process where the pressure varies linear with the volume to a state of 120 o C, 300 kPa. Find the work the ammonia gives out in the process. Solution: Take CV as the Ammonia, constant mass. Continuity Eq.: m 2 = m 1 = m ; Process: P = A + BV (linear in V) State 1: Superheated vapor v 1 = 0.6193 m 3 /kg State 2: Superheated vapor v 2 = 0.63276 m 3 /kg Work is done while piston moves at increasing pressure, so we get 1 W 2 = P dV = area = P avg (V 2 V 1 ) = 1 2 (P 1 + P 2 )m(v 2 v 1 ) = ½(200 + 300) × 0.5 (0.63276 0.6193) = 1.683 kJ 4.49 Helium gas expands from 125 kPa, 350 K and 0.25 m 3 to 100 kPa in a polytropic process with n = 1.667. How much work does it give out? Solution: Process equation: PV n = constant = P 1 V n 1 = P 2 V n 2 Solve for the volume at state 2 V 2 = V 1 (P 1 /P 2 ) 1/n = 0.25 × 125 100 0.6 = 0.2852 m 3 Work from Eq.4.4 1 W 2 = P 2 V 2 - P 1 V 1 1-n = 100 × 0.2852 - 125 × 0.25 1 - 1.667 kPa m 3 = 4.09 kJ P P 2 P v 1 1 2

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The actual process is on a steeper curve than n = 1. 4.67
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## This note was uploaded on 03/18/2010 for the course IOE 201 taught by Professor Dennisblumenfield during the Fall '09 term at University of Michigan-Dearborn.

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hw4 - HOMEWORK SET 4 4.41 Ammonia(0.5 kg is in a piston...

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