Explanation The system consists of one inlet duct (1) and two exit ducts such as duct (2) and duct (3). Here, the system operates in steady state with one inlet and two exit. Hence, the inlet mass flow rate of the steam and exits flow rate of steam are equal. m ˙ 1 = m ˙ 2 + m ˙ 3 (I) Here, the 20% of inlet mass exits through the exit duct (2). The mass flow rate at the exit duct (2) is as follows. m ˙ 2 = 20 % ( m ˙ 1 ) = ( 20 100 ) m ˙ 1 = 0.2 m ˙ 1 Also, the 80% of inlet mass exits through the exit duct (3). m ˙ 3 = 80 % ( m ˙ 1 ) = ( 80 100 ) m ˙ 1 = 0.8 m ˙ 1 Write the energy rate balance equation for system having multiple inlet and exits ducts. 0 = Q ˙ cv − W ˙ cv + ∑ i m ˙ i ( h i + 1 2 V i 2 + g z i ) − ∑ e m ˙ e ( h e + 1 2 V e 2 + g z e ) (II) It is given that neglect the kinetic and potential energy effects. The Equation (II) becomes as follows. 0 = Q ˙ cv − W ˙ cv + ∑ i m ˙ i ( h i ) − ∑ e m ˙ e ( h e ) (III) Here, the one inlet and two exits present in the system. That is as follows. ∑ i m ˙ i h i = m ˙ 1 h 1 ∑ e m ˙ e h e = m ˙ 2 h 2 + m ˙ 3 h 3 Substitute m ˙ 1 h 1 for ∑ i m ˙ i h i and m ˙ 2 h 2 + m ˙ 3 h 3 for ∑ e m ˙ e h e in Equation (IV). 0 = Q ˙ cv − W ˙ cv + m ˙ 1 h 1 − ( m ˙ 2 h 2 + m ˙ 3 h 3 ) = Q ˙ cv − W ˙ cv + m ˙ 1 h 1 − m ˙ 2 h 2 − m ˙ 3 h 3 (IV) Substitute 0.2 m ˙ 1 for m ˙ 2 and 0.8 m ˙ 1 for m ˙ 3 in Equation (IV). 0 = Q ˙ cv − W ˙ cv + m ˙ 1 h 1 − ( 0.2 m ˙ 1 ) h 2 − ( 0

Fundamentals of Engineering Thermodynamics

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ISBN: 9781118832318

Author: MORAN

Publisher: WILEY

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Chapter 4.12, Problem 51P

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Fundamentals of Engineering Thermodynamics

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The mass flow rate of the steam entering the turbine, in kg/s .

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The mass flow rate of the steam entering the turbine, in kg/s.

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