Explanation Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal. The mass flow rate of steam ( m ˙ s ) is as follows. m ˙ 1 = m ˙ 2 = m ˙ s The mass flow rate of feed water ( m ˙ f w ) is as follows. m ˙ 3 = m ˙ 4 = m ˙ f w 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 Neglect the work transfer, heat transfer to the surrounding, potential and kinetic energies. The Equations (I) reduced as follows for steam water. [ 0 + 0 + m ˙ 1 ( h 1 + 0 + 0 ) ] − [ 0 + 0 + m ˙ 2 ( h 2 + 0 + 0 ) ] = 0 m ˙ 1 h 1 − m ˙ 2 h 2 = 0 m ˙ 1 h 1 = m ˙ 2 h 2 (II) The Equations (I) reduced as follows for feed water. [ 0 + 0 + m ˙ 3 ( h 3 + 0 + 0 ) ] − [ 0 + 0 + m ˙ 4 ( h 4 + 0 + 0 ) ] = 0 m ˙ 3 h 3 − m ˙ 4 h 4 = 0 m ˙ 3 h 3 = m ˙ 4 h 4 (III) Combining Equation (II) and (III). m ˙ 1 h 1 + m ˙ 3 h 3 = m ˙ 2 h 2 + m ˙ 4 h 4 m ˙ 2 h 2 − m ˙ 1 h 1 = m ˙ 3 h 3 − m ˙ 4 h 4 Substitute m ˙ s for m ˙ 1 , m ˙ 2 and m ˙ f w for m ˙ 3 , m ˙ 4

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Author: CENGEL

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Chapter 5.5, Problem 152RP

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The ratio of the mass flow rates of extracted steam and the feed water.

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Chapter 5.5, Problem 151RP

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