# hw4 - MEEN 315, SECTION 503 PRINCIPLES OF THERMODYNAMICS...

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MEEN 315, SECTION 503 PRINCIPLES OF THERMODYNAMICS Fall 2009 Homework #4 DUE September 29, 2009 1. The following table lists temperature and specific volumes of ammonia vapor at two pressures: P = 50 psia P = 60 psia T (F) v (ft 3 /lb m ) T (F) v (ft 3 /lb m ) 100 6.836 100 5.659 120 7.110 120 5.891 140 7.380 140 6.120 Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided below and the table above, estimate: (a) the specific volume (ft 3 /lb m ) at T = 120F, P = 54 psia. (b) the temperature (F) at P = 60 psia, v = 5.982 ft 3 /lb m . (c) the specific volume (ft 3 /lb m ) at T = 110F, P = 58 psia. Solution:

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2. For each case, determine the specific volume at the indicated state. Locate the state on a sketch of the T- v diagram. (a) water at P = 14.7 psia, T = 100F. Find v , in ft 3 /lb m . (b) water at P = 10 psia, T = 320F. Find v in ft 3 /lb m . (c) R-134a, T = 15F, x = 0.5. Find v in ft 3 /lb m . Solution: (a) At P = 14.7 psia, T sat = 211.95 F (Table A-5E). Since T < T sat , water at P = 14.7 psia and T = 100 F is a liquid . The compressed liquid tables (Table A-7E) do not list properties until P = 500 psia. In this situation, assume water has the same properties as saturated liquid at T sat = T. Thus, v = v f,T=100F = 0.01613 ft 3 /lb m . 10 -2 10 -1 10 0 10 1 10 2 10 3 0 100 200 300 400 500 600 700 800 v [ft 3 /lb m ] T [°F] 1600 psia 95 psia 14.7 psia Water
(b) At P = 10 psia, T sat = 193.16 F (Table A-5E). Since T > T sat , water at P = 10 psia and T = 320 F is a superheated vapor . Use Table A-6E. From Table A-6E at P = 10 psia and T = 320 F,

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## This note was uploaded on 12/03/2009 for the course MEEN 315 taught by Professor Ramussen during the Spring '07 term at Texas A&M.

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hw4 - MEEN 315, SECTION 503 PRINCIPLES OF THERMODYNAMICS...

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