Question #257376

Steam expands steadily at a constant temperature. The flow rate is 13.6 kg/min with an inlet state of wet steam (water) at 80% quality to a final state of 700C and 2.5 MPa.  The change of kinetic energy across the device is 3.5 kJ/kg, and the heat added is 21.81 kW. Determine (a) the enthalpy of conditions 1 &2, (b) the heat transfer in kJ/kg, (c) the work, and (d) the system power.


1
Expert's answer
2021-10-28T08:53:07-0400

(a) Mass flow rate = 13.6 kg/min

Inlet state quality (x) = 0.8

Final state : temperature T2 = 70 °C

P2 = 2.5 MPa.

Change in kinetic energy = 3.5 KJ/Kg

Heat added (Q) = 21.81 KW

Steam is expanding at constant temperature

Therefore T1 = T2

or T1 = 70 °C

And x = 0.8

at temperature T = 70 °C

hf = 293.02 KJ/Kg

hg = 2626.01 KJ/Kg

So h1 = hf + x × (hg- hf)

= 293.02 + 0.8×(2626.01-293.02)

= 2159.41 KJ/Kg

P2 = 2.5 MPa

T2= 70 °C

At this set of temperature and pressure fluid is in liquid state.

h2 = 295.04 KJ/Kg

(b) Heat transfer in KJ/ Kg

Heat transfer = 21.81 KW

Mass flow rate = 13.6 Kg/ min

=13.660  Kg/s=0.227Kg/s= \frac{13.6 }{ 60} \; Kg/s \\ =0.227 Kg/s

So heat transfer in KJ/Kg =21.810.227= \frac{21.81}{0.227}

=96.08 KJ/Kg

(c) Energy equation per unit mass

h1+v122g+H1+dQdm=h2+v222g+H2+dWdmΔK×E=3.5  kg/kgH1=H2dQdm=96.08  kJ/kg2159.41+96.08=295.04+3.5+dWdmdWdm=1956.95  kJ/kgh_1 + \frac{v^2_1}{2g} + H_1 + \frac{dQ}{dm} = h_2 + \frac{v^2_2}{2g} + H_2 + \frac{dW}{dm} \\ ΔK \times E = 3.5 \;kg/kg \\ H_1=H_2 \\ \frac{dQ}{dm} = 96.08 \;kJ/kg \\ 2159.41 + 96.08 = 295.04 + 3.5 + \frac{dW}{dm} \\ \frac{dW}{dm} = 1956.95 \;kJ/kg

(d) System power =dWdm×dmdt= \frac{dW}{dm} \times \frac{dm}{dt}

=1956.95×0.227=444.23  kW= 1956.95 \times 0.227 \\ = 444.23 \;kW


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