Explanation Write the initial mass of the air in rigid tank from the ideal gas equation. m i = p i V R T i (I) Here, the initial pressure is p i , the volume of tank is V , the gas constant is R and the initial temperature of air in tank is T i . Write the expression for gas constant of air. R = R u M Here, the universal gas constant is R u and the molar mass of the air is M . Substitute R u M for R in Equation (I). m i = p i V ( R u M ) T i = p i V M ( R u ) T i (II) Write the final mass of the air in rigid tank from the ideal gas equation. m f = p f V R T f (IIII) Here, the final pressure is p f , the volume of tank is V , the gas constant is R and the final temperature of air in tank is T f . Substitute R u M for R in Equation (III). m f = p f V ( R u M ) T f = p f V M ( R u ) T f (IV) Write the mass balance for the control volume. m i + m i n = m f + m e x (V) Here, the mass of air at the inlet of the control volume is m i n and the mass of air at the exit of the control volume is m e x . There is no mass leaving the hence the mass at exit is 0 , that is m e x = 0 . Write the energy balance equation. Δ U + W = Q + Δ H Δ U = Q − W + Δ H (VI) Here, the heat transfer from the control volume is Q , the enthalpy change of the control volume is Δ H , the change of internal energy is Δ U and the work flow through the control volume is W . Write the enthalpy change for the control volume. Δ H = m i n h i n Here, the specific enthalpy at inlet is h i n . There is no work transfer hence rate of work flow through the control volume is zero, that is W = 0 . Write the change in internal energy. Δ U = m f u 2 − m i u 1 (VII) Here, the specific internal energy at initial state is u 1 and the specific internal energy at final state is u 2 . Substitute m f u 2 − m i u 1 for Δ U , 0 for W and m i n h i n for Δ H in Equation (VI). m f u 2 − m i u 1 = Q − 0 + m i n h i n Q = m f u 2 − m i u 1 − m i n h i n (VIII) Conclusion: Substitute 28.97 g / mol for M , 8.314 J / mol ⋅ K for R u , 1 bar for p i , 2 m 3 for V and 295 K for T i in Equation (II). m i = ( 1 bar ) ( 2 m 3 ) ( 28.97 g / mol ) ( 8.314 J / mol ⋅ K ) ( 295 K ) = ( 1 bar ) ( 10 5 Pa 1 bar ) ( 1 N / m 2 1 Pa ) ( 2 m 3 ) ( 28.97 g / mol ) ( 1 kg 1000 g ) ( 8

Fundamentals of Engineering Thermodynamics

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ISBN: 9781118412930

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

Publisher: WILEY

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Fundamentals of Engineering Thermodynamics

Ch. 4.12 - Prob. 1E Ch. 4.12 - 2. When a drip coffeemaker on-off switch is turned...Ch. 4.12 - Prob. 3E Ch. 4.12 - Prob. 4E Ch. 4.12 - Prob. 5E Ch. 4.12 - Prob. 6E Ch. 4.12 - Prob. 7E Ch. 4.12 - Prob. 8E Ch. 4.12 - Prob. 9E Ch. 4.12 - 10. How does the operator of a pumper-tanker fire...

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The heat transfer between the tank contents and the surrounding in kJ .

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