Explanation Consider the air flows at steady state through one inlet and one exit system (pipe and duct flow). 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 air flows at steady state through the system. Hence, the rate of change in net energy of the system becomes zero. Δ E ˙ system = 0 The heat is added to the air. Neglect the heat transfer at the outlet, work transfer, kinetic and potential energies. The Equations (I) reduced as follows. [ 0 + Q ˙ 1 + m ˙ ( h 1 + 0 + 0 ) ] − [ 0 + 0 + m ˙ ( h 2 + 0 + 0 ) ] = 0 Q ˙ 1 + m ˙ h 1 − m ˙ h 2 = 0 Q ˙ 1 = m ˙ h 2 − m ˙ h 1 Q ˙ 1 = m ˙ ( h 2 − h 1 ) (II) The change in enthalpy is expresses as follow. h 2 − h 1 = c p ( T 2 − T 1 ) Here, the specific heat is c p , the exit temperature is T 2 and the inlet temperature is T 1 . Substitute c p ( T 2 − T 1 ) for h 2 − h 1 in Equation (II). Q ˙ 1 = m ˙ [ c p ( T 2 − T 1 ) ] = m ˙ c p ( T 2 − T 1 ) (III) Rearrange the Equation (III) to obtain the exit temperature ( T 2 ) . Q ˙ 1 m ˙ c p = T 2 − T 1 T 2 = Q ˙ 1 m ˙ c p + T 1 (IV) Write the formula for mass flow rate of air ( m ˙ )

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Author: CENGEL

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Chapter 5.5, Problem 96P

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The average temperature at which the air leaves the hollow core.

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The average temperature at which the air leaves the hollow core.

Solution Summary: The author calculates the energy rate balance equation for one inlet and one outlet system.

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The average temperature at which the air leaves the hollow core.

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