(a) Explanation Write the expression for the ideal gas equation at initial state. p 1 v 1 = n R T 1 T 1 = p 1 v 1 n R (I) Here, the initial pressure is p 1 , the number of moles of the gas is n , the initial temperature is T 1 , the initial specific volume of the gas is v 1 and the universal gas constant is R . Write the expression for the electric work transfer. W i n = W ˙ e l e c × Δ t (II) Here, the electric work transfer is W i n , the energy transfer through the resister is W ˙ e l e c and the time for the transfer is Δ t . Write the expression for the internal energy at state 2 using first law of the thermodynamics. d Q = d U + d W (III) Here, the change in heat transfer is d Q , the change in internal energy is d U and the change in work transfer is d W . Substitute 0 for d Q , u 2 − u 1 for d U and − W n for d W in Equation (III). d Q = d U + d W 0 = ( u 2 − u 1 ) − W n ( u 2 − u 1 ) = W n (IV) Here, the internal energy at initial state is u 1 , the work done is W and the internal energy at final state is u 2 . Write the expression for the ideal gas equation at final state. p 2 v 2 = n R T 2 p 2 = n R T 2 v 2 (V) Here, the final pressure is p 2 , the final specific volume is v 2 and the temperature at the final state is T 2 . Write the expression for the change in exergy of the gas. Δ e = m [ ( u 2 − u 1 ) + p o ( v 2 − v 1 ) − T o ( s o ( T 2 ) − s o ( T 1 ) − R ln ( p 2 p 1 ) ) ] (VI) Here, the change in exergy of the gas is Δ e , the ambient temperature is T o and the absolute entropy of the gas is s o . Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (VII) Here, the variables denote by x and y are temperature and entropy. Conclusion: Convert the ambient temperature from Fahrenheit to degree Rankin. T 0 ° R = T ° F + 460 = 70 ° F + 460 = 530 ° R Substitute 5 atm for p 1 , 90 ft 3 for v 1 , 1 lb ⋅ mol for n and 1545 lbf ⋅ ft / lb ⋅ mol ⋅ °R for R in Equation (I). T 1 = ( 5 atm ) ( 90 ft 3 ) ( 1 lb ⋅ mol ) ( 1545 lbf ⋅ ft / lb ⋅ mol ⋅ °R ) = 450 atm ⋅ ft 3 ( 14 .7 lbf / in 2 1 atm ) ( 144 in 2 1 ft 2 ) 1545 lbf ⋅ ft / ⋅ °R = 616.54 °R Substitute 10 Btu / s for W ˙ e l e c and 2 min for Δ t in (II). W i n = 10 Btu / s × 2 min = 20 Btu ⋅ min / s ( 60 s 1 min ) = 1200 Btu Refer to table A-23E “the ideal gas properties of selected gases” to obtain the initial internal energy at temperature T 1 = 616.54 °R as 3056.3 Btu / lb ⋅ mol . Substitute 3056.3 Btu / lb ⋅ mol for u 1 , 1200 Btu for W and 1 lb ⋅ mol for n in Equation (IV). u 2 = 1200 Btu 1 lb ⋅ mol + ( 3056.3 Btu / lb ⋅ mol ) = 4256.3 Btu / lb ⋅ mol Refer to table A-23E “the ideal gas properties of selected gases” to obtain final temperature at final internal energy u 2 = 3056.3 Btu / lb ⋅ mol as 854 .5°R . Substitute 854 .5°R for T 2 90 ft 3 for v 2 , 1 lb ⋅ mol for n and 1545 lbf ⋅ ft / lb ⋅ mol ⋅ °R for R in Equation (V) (b) To determine The electric work of the system. (c) To determine The exergy destruction for system.

Fundamentals of Engineering Thermodynamics

8th Edition

ISBN: 9781118832301

Author: SHAPIRO

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 7.7, Problem 34P

(a)

To determine

The change in the exergy of the gas.

(b)

To determine

The electric work of the system.

(c)

To determine

The exergy destruction for system.

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Chapter 7.7, Problem 33P

Chapter 7.7, Problem 35P

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