Explanation Write the expression for total mass. m t = m L + m R (I) Here, total mass is m t , mass in left tank is m L and mass in right tank is m R . Write the expression for mass in left tank. m L = P L V L R T L (II) Here, pressure in left tank is P L , volume of left tank is V L , the gas constant is R and temperature of left tank is T L . Write the expression for mass in right tank. m R = P R V R R T R (III) Here, the pressure in right tank is P R , volume of right tank is V R and temperature of right tank is T R . Write the expression for internal energy at equilibrium state. u 2 = m L u L + m R u R m t (IV) Here, internal energy at equilibrium is u 2 , the internal energy in left tank is u L and internal energy of right tank is u R . Calculate final pressure. P 2 = m t R T 2 V t (V) Here, final pressure is P 2 , final temperature is T 2 and total volume is V t . Calculate the amount of entropy produced. σ = m L ( s 2 − s L − R ln ( P 2 P L ) ) + m R ( s 2 − s R − R ln ( P 2 P R ) ) (VI) Here, the amount of entropy produced is σ , final entropy of the system is s 2 , entropy of left tank is s L , pressure of left tank is P L , entropy of right tank is s R and pressure of right tank is P R . Conclusion: Convert the left tank temperature from Fahrenheit to Rankine. T L = 40 ° F = ( 40 ° F + 460 ) ° R = 500 ° R Convert the right tank temperature from Fahrenheit to Rankine T R = 200 ° F = ( 200 ° F + 460 ) ° R = 660 ° R The gas constant for air is 53.353 ft ⋅ lbf / lb ⋅ ° R . Substitute 14.7 lbf / in 2 for P L , 12 ft 3 for V L , 53.353 ft ⋅ lbf / lb ⋅ ° R for R and 500 ° R for T L in Equation (II). m L = ( 14.7 lbf / in 2 ) ( 12 ft 3 ) ( 53.353 ft ⋅ lbf / lb ⋅ ° R ) ( 500 ° R ) = ( 176.4 26676.56 lb ⋅ ft 2 / in 2 ) ( 144 in 2 1 ft 2 ) = 0.9522 lb Substitute 50 lbf / in 2 for P R , 10 ft 3 for V R , 53.353 ft ⋅ lbf / lb ⋅ ° R for R and 660 ° R for T R in Equation (III). m R = ( 50 lbf / in 2 ) ( 10 ft 3 ) ( 53.353 ft ⋅ lbf / lb ⋅ ° R ) ( 660 ° R ) = ( 500 35212.98 lb ⋅ ft 2 / in 2 ) ( 144 in 2 1 ft 2 ) = 2.0447 lb Substitute 0.9522 lb for m L and 2.0447 lb for m R in Equation (I). m t = 0.9522 lb + 2.0447 lb = 2 .9969 lb Refer to Table A-22E “Ideal gas properties of Air” to obtain u L as 85.20 Btu / lb corresponding to 500 ° R and u R as 112.67 Btu / lb corresponding to 660 ° R . Substitute 0.9522 lb for m L , 85.20 Btu / lb for u L , 2.0447 lb for m R , 112.67 Btu / lb for u R and 2 .9969 lb for m t in Equation (IV). u 2 = ( 0.9522 lb ) ( 85

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118412930

Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey

Publisher: WILEY

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In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…

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Chapter 6.13, Problem 63P

To determine

The final temperature of air.

The final pressure of air.

The amount of entropy produced.

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Chapter 6.13, Problem 62P

Chapter 6.13, Problem 64P

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The final temperature of air. The final pressure of air. The amount of entropy produced.