Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.901 and ex

Question

Steam at 1 bar enters an adiabatic heat exchanger with a quality of 0.901 and exits as a saturated liquid. The steam is being cooled by a 947 kg/s stream of water which enters the heat exchanger at 20oC and exits at 40oC. The two flows do not mix. Determine the volumetric flow rate (m3/s) of the steam entering the heat exchanger.

Correct answer: 59.41 ± 2%

Determine the power output (kW) of a power cycle given the following information:

Efficiency = 39.4%

Qcold = 30.1 kW

Explanation / Answer

enthalpy of saturated liquid at 1 bar,Hf = 418.5 kj/kg

enthalpy of saturated vapour at 1 bar,Hg = 2675.5 kj/kg

since quqlity of steam,x = 0.901

so enthalpy of steam entering the heat exchanger Hi= (1-x)*Hf+x*Hg

hi = (1-0.901)*418.5+0.901*2675.5 = 2452.057 kj/kg

since steam leaving the heat exchanger is saturated liquid,so Ho= Hf

enthalpy of steam leaving the heat exchanger Ho = 418.5 kj/kg

energy released by steam per kg of steam = Hi-Ho

= 2452.057-418.5 = 2033.557 kj/kg

now energy taken by water = m*Cp*(To-Ti) = 947*4.2*(40-20)

= 79548 kj

let m' be the flow rate of steam

by energy conservation

energy given by steam = energy taken by water

m'*2033.557 = 79548

so, m' = 39.12 kg/s

specific volume = (1-x)*Vf+x*Vg

= (1-0.901)*0.001044+0.901*1.6729 = 1.507386 m^3 / kg

so, volumetric flow rate = 39.12*1.507386 = 58.97 m^3/s

we know that efficiency of a power cycle = (Qhot-Qcold)/Qhot

0.394 = 1- Qcold/Qhot

Qcold/Qhot =1-0.394 = 0.606

Qhot = 30.1/0.606 = 49.669 kw

power output = Qhot-Qcold

49.669-30.1 = 19.569 kw

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