Explanation Write the expression for mass of water. m = V v f (I) Here, the total volume is V and the specific volume of water is v f . Write the expression for entropy balance. Δ s = ∫ ( δ Q T ) b + σ (II) Since there is no heat transfer takes place so δ Q = 0 . Substitute 0 for δ Q in Equation (II). Δ s = ∫ ( 0 T ) b + σ Δ s = σ (III) Here, the entropy generation is σ . Write the expression for change in entropy. Δ s = m c ln ( T 2 T 1 ) (IV) Write the expression for energy balance. Q cv = W cv + m ( h 2 − h 1 ) (V) Since there is no heat transfer, so Q cv = 0 . Substitute 0 for Q cv in Equation (V). 0 = W cv + m ( h 2 − h 1 ) W cv = m ( h 1 − h 2 ) W cv = m c ( T 2 − T 1 ) (VI) Write the expression for increase in exergy stored in water heater. E 2 − E 1 = W ˙ cv − T o ( Δ s ) (VII) Here, the environment temperature is T o . Write the expression for exergetic efficiency. ε = E 2 − E 1 W c v (VIII) Here, the exergetic efficiency is ε . Conclusion: Refer to Table A-2, “properties of saturated water (liquid-vapor): Temperature Table”, to obtain the specific volume at the average temperature 39 ° C by using interpolation method of two variables. 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 (IX) Here, the variables denote by x and y are temperature and specific volume. Show the values of specific volume corresponding to the given temperatures at as in below table: Temperature, ° C specific volume, m 3 / kg 38 ° C 1.0071 × 10 − 3 m 3 / kg 39 ° C ? 40 ° C 1.0078 × 10 − 3 m 3 / kg Substitute 38 ° C for x 1 , 39 ° C for x 2 , 40 ° C for x 3 , 1.0071 × 10 − 3 m 3 / kg for y 1 , and 1.0078 × 10 − 3 m 3 / kg for y 3 in Equation (IX). y 2 = ( 39 ° C − 38 ° C ) [ ( 1.0078 × 10 − 3 m 3 / kg ) − ( 1.0071 × 10 − 3 m 3 / kg ) ] ( 40 ° C − 38 ° C ) + 1

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ISBN: 9781119391630

Author: MORAN

Publisher: WILEY

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Chapter 7.7, Problem 98P

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Chapter 7.7, Problem 97P

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