(a) Explanation Write the expression for the vapor pressure. p v 1 = ϕ p g (I) Here, relative humidity of the air at the state-1is ϕ , saturates pressure of the water pressure at state-1 p g . Write the expression for the ideal gas equation for the vapor. p v 1 v v 1 = R T 1 v v 1 = R T 1 p v 1 (II) Here, pressure of the of water vapor at state-1is p v 1 , gas constant for the air is R , specific volume at state-1 is v v 1 and temperature is T 1 . Write the expression for the pressure of dry air in the state-1. p a 1 = p 1 − p v 1 (III) Write the expression for the ideal gas equation for the dry air. p a 1 V = m a R ¯ M a T 1 m a = p a 1 V M a R ¯ T 1 (IV) Here, mass of the air is m a , universal gas constant is R ¯ , and molar mass of the air is M a . Substitute p 1 − p v 1 for p a 1 in the Equation (IV) m a = ( p 1 − p v 1 ) V M a R ¯ T 1 (V) Write the expression for the internal energy at the state-1. U 1 = m a 1 u a 1 + m v 1 u v 1 (VI) Here, internal energy of dry air at the state-1 is u a 1 and internal energy of water vapor at the state-1. u v 1 . Write the expression for the internal energy at the state-2. U 2 = m a 1 u a 2 + m v 1 u v 2 (VII) Here, internal energy of dry air at the state-2 is u a 2 and internal energy of water vapor at the state-2 is u v 2 . Write the expression for the heat transfer. Q = U 2 − U 1 (VIII) Conclusion: Refer Table A-2 “Properties of Saturated Water (Liquid–Vapor)” to obtain p g at temperature 125 ° C 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) Show the temperature at 120 ° C and 130 ° C as in Table (1). Temperature, ° C specific enthalpy, bar 120 1.985 125 y 2 130 2.701 Substitute 120 ° C for x 1 , 125 ° C for x 2 , 130 ° C for x 3 , 1.985 bar for y 1 , and 2.701 bar for y 3 in the Equation (IX). y 2 = ( 125 ° C − 120 ° C ) ( 2.701 bar − 1.985 bar ) ( 130 ° C − 120 ° C ) + 1.985 bar = 0.335 bar + 1.985 bar = 2 .32 bar From above calculation the pressureat temperature 125 ° C is 2 .32 bar . Substitute 0.5 for ϕ , 2.32 bar for p g in the Equation (I). p v 1 = 0.5 ( 2.32 bar ) = 1.16 bar ( 100 kPa 1 bar ) = 116 kPa Substitute 0.461 kJ / kg ⋅ K for R , 125 ° C for T 1 , and 116 kPa for p v 1 in the Equation (II). v v 1 = ( 0.461 kJ / kg ⋅ K ) ( 125 ° C ) 116 kPa = ( 0.461 kJ / kg ⋅ K ) ( 125 ° C + 273 ) K 116 kPa ( 1 kN / m 2 1 kPa ) ( 10 3 Pa 1 kPa ) = 183.478 kJ / kg ( 1 kN ⋅ m 1 kJ ) 116 kN / m 2 = 1.548 m 3 / kg Substitute 4 bar for p 1 , 1.16 bar for p v 1 , 2.5 m 3 for V , 28.97 kg / mol for M a , 8314 J / mol ⋅ K for R ¯ , and 125 ° C for T 1 in theEquation (V). m a = ( 4 bar − 1.16 bar ) ( 2.5 m 3 ) ( 28.97 kg / mol ) ( 8314 J / mol ⋅ K ) ( 125 ° C ) = ( 4 bar ( 10 5 N / m 2 1 bar ) − 1.16 bar ( 10 5 N / m 2 1 bar ) ) ( 2.5 m 3 ) ( 28.97 kg / mol ) ( 8314 J / mol ⋅ K ) ( 125 ° C + 273 ) K = ( 2.84 × 10 5 N / m 2 ) ( 2.5 m 3 ) ( 28.97 kg / mol ) 3308972 J / mol = ( 7.1 × 10 5 N ⋅ m ) ( 1 J 1 N ⋅ m ) ( 28.97 kg / mol ) 3308972 J / mol m a = 6 (b) To determine The heat transfer in the tank at the temperature 30 ° C .

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Chapter 12.9, Problem 58P

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The heat transfer in the tank at the temperature 110°C.

(b)

To determine

The heat transfer in the tank at the temperature 30°C.

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Chapter 12.9, Problem 57P

Chapter 12.9, Problem 59P

Chapter 12 Solutions

FUNDAMENTALS OF ENGINEERING THERMODYNAM

Ch. 12.9 - Prob. 1E Ch. 12.9 - Prob. 2E Ch. 12.9 - Prob. 3E Ch. 12.9 - Prob. 4E Ch. 12.9 - Prob. 5E Ch. 12.9 - Prob. 6E Ch. 12.9 - Prob. 7E Ch. 12.9 - Prob. 8E Ch. 12.9 - Prob. 9E Ch. 12.9 - Prob. 10E

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The heat transfer in the tank at the temperature 110 ° C .

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The heat transfer in the tank at the temperature 110°C.

The heat transfer in the tank at the temperature 30°C.

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