Explanation Write the equation of mass balance. m in − m e = Δ m system (I) Here, the inlet mass is m in , the exit mass is m e and the change in mass of the system is Δ m system . The change in mass of the system for the control volume is expressed as, Δ m system = ( m 2 − m 1 ) cv Here, the suffixes 1 and 2 indicates the initial and final states of the system. Consider the given bottle as the control volume. The bottle has inlet only and no exit, i.e. m e = 0 . Initially the bottle is evacuated, i.e. m 1 = 0 . Rewrite the Equation (I) as follows. m in − 0 = ( m 2 − 0 ) cv m in = m 2 (II) Write the energy balance equation. E i n − E o u t = Δ E s y s t e m { [ Q in + W in + m in ( h + ke + pe ) in ] − [ Q e + W e + m e ( h + ke + pe ) e ] } = [ m 2 ( u + ke + pe ) 2 − m 1 ( u + ke + pe ) 1 ] system (III) Here, the heat transfer is Q , the work transfer is W , the enthalpy is h , the internal energy is u , the kinetic energy is ke , the potential energy is pe and the change in net energy of the system is Δ E system ; the suffixes 1 and 2 indicates the inlet and outlet of the system. Since the bottle is not insulated, the heat transfer occurs through the bottle wall. In control volume, there is no work transfer, i.e. ( W in = W e = 0 ) . Neglect the kinetic and potential energy changes i.e. ( Δ ke = Δ pe = 0 ) . There is no exit for the bottle, the exit mass is neglected i.e. ( m e = 0 ) . At initial state, the bottle is evacuated i.e. ( m 1 = 0 ) . The Equation (III) reduced as follows. Q in + m in h in = m 2 u 2 (IV) Substitute m 2 for m in in Equation (IV). Q in + ( m 2 ) h in = m 2 u 2 Q in = m 2 u 2 − m 2 h in = m 2 ( u 2 − h in ) (V) Write the formula for mass of air ( m ) when the bottle is fully filled

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Chapter 5.5, Problem 115P

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The net heat transfer through the wall of the bottle during this filling process.

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Chapter 5.5, Problem 113P

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Chapter 5 Solutions

THERMODYNAMICS: ENG APPROACH LOOSELEAF

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The net heat transfer through the wall of the bottle during this filling process.

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