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A solar-powered steam plant uses a solar collector to heat the water in place of the typical boiler. For a solar-powered steam plant operating on an ideal Rankine cycle, saturated vapor leaves the solar collector at 1 MPa and the pressure in the condenser is 10 kPa. Determine the net work per unit mass and the thermal efficiency of this cycle.

User Trekkie
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1 Answer

4 votes

Answer:

The net work per unit mass
W_(net) = 692.14 kJ/kg

The thermal efficiency for the cycle =
\mathbf{26.78 \ \%}}

Step-by-step explanation:

For an Ideal Rankine cycle:

The following properties were obtained from the steam tables at 1 MPa.


h_1 = h_g = 2777.1 \ kJ/kg


s_1 = 6.585 \ kJ/kg.k

The properties obtained at 10 kPa are:


h_f = 191.81 \ kJ/kg\\ \\ h_f_g = 2392.1 \ kJ/kg \\ \\ s_f= 0.6492 \ kJ/kg .k \\ \\ s_f_g = 7.4996 \ kJ/kg/k \\ \\ v_f = 0.001010 \ m^3 /kg

For 1 - 2 isentropic expansion:


s_1 = s_2


s_1 = (s_f+x_2 * 7.4996)

6.585 = 0.6492 + x₂ × 7.4996

6.585 - 0.6492 = 7.4996x₂

5.9358 = 7.4996x₂

x₂ = 5.9358/7.4996

x₂ = 0.791


h_2 = h_f + x_2 \ hfg


h_2 = 191.81 + 0.791 * 2392.1


h_2 = 2083.96 \ kJ/kg

At 10 kPa;


h_3= h_f = 191.81 \ kJ//kg

The pump work for the process:


h_4 - h_3 = vf(\Delta \ P)


h_4 - h_3 = 0.001010(1000 -10)


h_4 - h_3 = 0.9999 kJ/kg


h_4 - 191.81 \ kJ/kg = 0.9999 kJ/kg


h_4 = 191.81 \ kJ/kg + 0.9999 kJ/kg


h_4 = 192.8 \ kJ/kg

However, the turbine work
W_T can be computed by using the formula:


W_T = h_1 -h_2


W_T = ( 2777.1 - 2083.96 ) kJ/kg


W_T = 693.14 \ kJ/kg

Thus, the net work
W_(net) can be determined as:


W_(net) =
W_T - W_p


W_(net) = 693.14 - 0.9999


W_(net) = 692.14 kJ/kg

Similarly, to determine the thermal efficiency of this cycle, we need to first know the heat addition
Q_s= h_1 - h_4


Q_s= (2777.1 -192.8 ) \ kJ/kg


Q_s= 2584.3 \ kJ/kg

Finally, the thermal efficiency can be calculated by using the formula:


n_(th) = (W_(net))/(Q_s)


n_(th) = (692.14)/(2584.3)


n_(th) = 0.2678

The thermal efficiency for the cycle =
\mathbf{26.78 \ \%}}

User Emil
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