exam 1 sample prob 2011

exam 1 sample prob 2011 - Thermodynamics E = Q W = U + KE +...

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UNIVERSITY OF FLORIDA 1 Thermodynamics E = Q – W = U + KE + PE 300 gal 20psig 60degF air 1 Hp motor for 15 min. 500 btu/hr What is the temperature after 15 minutes? Problem 1
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UNIVERSITY OF FLORIDA 2 Thermodynamics Q – Wother -Wmb = U + KE + PE 300 gal 20psig 60degF air 1 Hp motor for 15 min. 500 btu/hr Assumptions Ideal gas KE is zero PE is zero Q= -500*15/60= -125 btu Wmb = zero Wother= -1Hp*2544btu/ hr-Hp*15/60hr=- 636.1btu
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UNIVERSITY OF FLORIDA 3 Thermodynamics -125 – (-636.1)-0= U + 0 + 0 = m*Cv(T2-T1) 300 gal 20psig 60degF air 1 Hp motor for 15 min. 500 btu/hr P1V1=mRT m= P1V1/RT P1= 20+14.7=34.7psia V1=300gal*1 ft3/7.48gal= 40,1ft3 T1= 60+460=520R R = 53.34ft-lbf/lbm- R (pg 840) m= 7.22lbm
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UNIVERSITY OF FLORIDA 4 Thermodynamics -125 – (-636.1)-0= U + 0 + 0 = 7.22*0.171(T2-520) Therefore T2= 933R or 474F 300 gal 20psig 60degF air 1 Hp motor for 15 min. 500 btu/hr Cv = 0.171btu/lbm R Page 840
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UNIVERSITY OF FLORIDA 5 Thermodynamics Q – Wother -Wmb = U + KE + PE 110V 0.3A for 10min Piston cylinder contains 0.2kg steam at 200kPa and 400deg C What is the initial volume and final temp. after 10min.
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This note was uploaded on 08/03/2011 for the course EML 3100 taught by Professor Sherif during the Summer '08 term at University of Florida.

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exam 1 sample prob 2011 - Thermodynamics E = Q W = U + KE +...

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