Answer:
the turbine work in MW is [tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
Explanation:
Given that:
the inlet temperature = 500°C
pressure = 600 kPa
exit temperature = 45°С
mass flow rate = 11 kg/s
The objective is to find the turbine work in MW given that the mass flow rate of the steam is at 11 kg/s.
From the steam tables, the data obtained for the enthalpies at the inlet temperature of 500°C and a pressure of 600 kPa is:
[tex]\mathtt{h_1 : 3482.7 \ kJ/kg}[/tex]
The turbine expansion process is also the isentropic process, as such:
Inlet entropy [tex]\mathbf{s_1}[/tex] = Exit entropy [tex]\mathbf{s_2}[/tex]
The data obtained from the steam table for [tex]\mathbf{s_1}[/tex] = 8.003 kJ/kg.K
[tex]s_2 = s_{f2}+ x_2 *s_{fg2}[/tex]
The data obtained from the steam tables for this entities are as follows:
[tex]\mathsf{s_{f2} = 0.638 \ kJ/Kg/K }[/tex]
[tex]\mathtt{s_{fg2}=7.528 \ kJ/kg/K}[/tex]
since [tex]\mathbf{s_1}[/tex] = [tex]\mathbf{s_2}[/tex] = 8.003 kJ/kg.K
Therefore;
[tex]8.003 = 0.638 + x_2 * 7.528[/tex]
[tex]8.003 -0.638 = 7.528x_2[/tex]
[tex]7.365= 7.528x_2[/tex]
[tex]x_ 2= \dfrac{ 7.365}{7.528}[/tex]
[tex]x_2 = 0.978[/tex]
From the steam tables, the data obtained for the enthalpies at the exit temperature of 45°C and a pressure of 600 kPa is:
[tex]h_{f2}[/tex] = 188.4
[tex]h_{fg2[/tex] =2394.9
Thus;
[tex]h_ 2= 188.4 +0.978*2394.9[/tex]
[tex]h_ 2=2530.61[/tex] kJ/kg
The workdone for the turbine process can be computed as:
[tex]\mathsf{W_t = m(h1-h_2)}[/tex]
[tex]\mathsf{W_t = 11(3482.7-2530.61)}[/tex]
[tex]\mathsf{W_t = 11(952.09)}[/tex]
[tex]\mathsf{W_t = 10472.99}[/tex] kW
To MW, we have
[tex]\mathbf{W_t = 10.47299 \ kW}[/tex]
