Explanation Write the expression for the mass balance. m i n − m o u t = Δ m Here, mass of the water entering into the system is m i n , mass of water leaving out of the system is m o u t and change in mass is Δ m . Rewrite the mass balance equation for the mixing in chamber. m ˙ 1 + m ˙ 2 = m ˙ 3 (I) Here, mass flow rate of liquid water entering into the chamber is m ˙ 1 , mass flow rate of superheated steam entering into the chamber is m ˙ 2 and mass flow rate of water leaving out the chamber is m ˙ 3 . Write the energy rate balance equation for a control volume. E ˙ in − E ˙ out = d E ˙ system / d t = 0 ( For steady state ) Here, total energy rate at inlet is E ˙ in , total energy rate at outlet is E ˙ out and change in net energy rate is d E ˙ system / d t . Rewrite the energy balance equation for the mixing of water in chamber neglecting the potential and kinetic energies. E ˙ in = E ˙ out m ˙ 1 h 1 + m ˙ 2 h 2 = m ˙ 3 h 3 (II) Here, specific enthalpy of liquid water entering into the chamber is h 1 , the specific enthalpy of the superheated steam entering into the chamber is h 2 and specific enthalpy of water leaving out of chamber is h 3 . Conclusion: Refer Table A-4,“Saturated water - Temperature”, obtain the specific enthalpy ( h 1 ) of liquid water at temperature ( T 1 ) of 20 ° C . h 1 = h f = 83 .91 kJ / kg . Refer Table A-6, “Superheated water”, obtain the specific enthalpy ( h 2 ) of superheated steam at pressure ( P 2 ) of 300 kPa and temperature ( T 2 ) of 300 ° C as 3069

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ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 5.5, Problem 71P

To determine

The mass flow rate of the superheated steam.

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Chapter 5.5, Problem 70P

Chapter 5.5, Problem 72P

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