Question

PROBLEM 4 A feedwater heater operates at steady state with liquid water entering at inlet 1 at 7 bar, 40°C, and a mass flow rate of 70 kg/s. A separate stream of water enters at inlet 2 as a two-phase liquid-vapor mixture at 7 bar with a quality of 97%. Saturated liquid at 7 bar exits the feedwater heater at 3. Ignoring heat transfer with the surroundings and neglecting kinetic and potential energy effects, determine the mass flow rate, in kg/s, at inlet 2.

Answer 1

P2=7bar X2= 0.97 .- - feed water heater - Py = Ibar Ti= 40°c - - - - - - - P2=7bar - - - mg = 70kgls Engineering model d. the accompanying figure The is control volume shown on at steady state 9, Heat transfer and kinetic and potential energy effects are neglected. Wcv-o since a feed water heater has no work associated with it..

The Steady-state mass rate balance gives &mi= { me. mg + miza miz e Steady State energy balance gives o= 8cr-Wert & milhit V,?r 9z;) -şme (het ve² + gze) Neglecting heat transfer and kinetic and potential energy effects and recognizing no work is associated with a

feedwater heater, the energy balance simplifies to . O= {mini- & mehe = 0 = m hit maha mizha Substituting for mig from the mass rate balance, 0 = minit maha - (mit mia) hz Solving for me m2 = ... (i) m (ha-hi) The-ha)

At inlet to the water is compressed liquid. Pi= Ibar, Ti= 40°c - ha 168. 154 kilkg At inlet 2, two-phase liquid vapor mixture. BO MIX P2= Ibar

hf= 697 kJ/kg, bg = 2762.8 kg/kg | b = họ + 3 (h-hg) = 697 + 0-97 ( 2762-8-697) - 2700.826 kilky. So, hz= 2700- 826 kJ 1kg. At liquid. exit 3, the water is saturated I Pa = zbar. ha= hq = 697 kilkg. from substituting the values of hi, ha and ean in ha., we and get

m2 = 70kgls ( 697 - 168.154) kilka (2700-826-697) kJ/kg. » m2 = 18.474 ugls . Hence, the mass flow rate at inlet 2, m2 = 18.424 ke515