Answer to Question #257376 in Molecular Physics | Thermodynamics for John

Question

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.

Answer 1

(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

h_f = 293.02 KJ/Kg

h_g = 2626.01 KJ/Kg

So h₁ = h_f + x × (h_g- h_f)

= 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

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

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

=96.08 KJ/Kg

(c) Energy equation per unit mass

$h_1 + \frac{v^2_1}{2g} + H_1 + \frac{dQ}{dm} = h_2 + \frac{v^2_2}{2g} + H_2 + \frac{dW}{dm} \\ ΔK \times E = 3.5 \;kg/kg \\ H_1=H_2 \\ \frac{dQ}{dm} = 96.08 \;kJ/kg \\ 2159.41 + 96.08 = 295.04 + 3.5 + \frac{dW}{dm} \\ \frac{dW}{dm} = 1956.95 \;kJ/kg$

(d) System power $= \frac{dW}{dm} \times \frac{dm}{dt}$

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