(a) Explanation The helium gas flows steadily through pipe (one inlet and one exit system). Hence, the inlet and exit mass flow rates are equal. m ˙ 1 = m ˙ 2 = m ˙ Write the energy rate balance equation for one inlet and one outlet system. [ Q ˙ 1 + W ˙ 1 + m ˙ ( h 1 + V 1 2 2 + g z 1 ) ] − [ Q ˙ 2 + W ˙ 2 + m ˙ ( h 2 + V 2 2 2 + g z 2 ) ] = Δ E ˙ system (I) Here, the rate of heat transfer is Q ˙ , the rate of work transfer is W ˙ , the enthalpy is h and the velocity is V , the gravitational acceleration is g , the elevation from the datum is z and the rate of change in net energy of the system is Δ E ˙ system ; the suffixes 1 and 2 indicates the inlet and outlet of the system. The system is at steady state. Hence, the rate of change in net energy of the system becomes zero. Δ E ˙ system = 0 In this system, only the heat is removed ( Q ˙ 2 ) from the helium gas. Neglect work transfer, heat transfer at inlet, kinetic and potential energies. The Equations (I) reduced as follows. [ 0 + 0 + m ˙ ( h 1 + 0 + 0 ) ] − [ Q ˙ 2 + 0 + m ˙ ( h 2 + 0 + 0 ) ] = 0 m ˙ h 1 − Q ˙ 1 − m ˙ h 2 = 0 Q ˙ 2 = m ˙ h 1 − m ˙ h 2 Q ˙ 2 = m ˙ ( h 1 − h 2 ) (II) The change in enthalpy is expresses as follow. h 1 − h 2 = c p ( T 1 − T 2 ) Here, the specific heat is c p , the exit temperature is T 2 and the inlet temperature is T 1 (b) To determine The helium volume flow rate at the pipe exit, in m 3 /s .

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Chapter 5.5, Problem 138RP

(a)

To determine

The heat transfer for the process, in kW.

(b)

To determine

The helium volume flow rate at the pipe exit, in m3/s.

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Chapter 5.5, Problem 137RP

Chapter 5.5, Problem 139RP

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Chapter 5 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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The heat transfer for the process, in kW .

Solution Summary: The author explains the heat transfer for the process, in 16616.32kW. The inlet and exit mass flow rates are equal.

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The heat transfer for the process, in kW.

The helium volume flow rate at the pipe exit, in m3/s.

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