Answer:
Explanation:
Given that:
Superheated vapor enters the turbine at 10 MPa, 480°C,
From the tables of superheated steam tables; the following values are obtained
[tex]h_1 = 3322.02 \ kJ/kg\\\\ s_1 = 6.52846 \ kJ/kg.K[/tex]
Also; from the system, the isentropic line is 1-2 in which s_2 is in wet state
[tex]s_2 = s_{f \ 6 kpa} +xs_{fg \ 6 kpa}[/tex]
[tex]s_2 =0.51624 + x(7.82)[/tex]
[tex]s_2 =0.51624 + 7.82x[/tex]
From the values obtained;
[tex]s_1 =s_2= 6.52846 \ kJ/kg.K[/tex]
Therefore;
6.52846 = 0.51624+7.82x
6.52846 - 0.51624 = 7.82 x
6.01222 = 7.82 x
x = 6.01222/7.82
x = 0.7688
The enthalpy for this process at state (s_2) can be determined as follows:
[tex]h_2 = h _f +xh_{fg} \\ \\ h_2 = 150.15 +(0.77 \times 2415.92) \\ \\ h_2 =150.15 +( 1629.2584 ) \\ \\ h_2 =2010.4084 \ kJ/kg[/tex]
The actual enthalpy at s_2 by using the isentropic efficiency of the turbine can determined by using the expression:
[tex]n_T = \dfrac{h_1-h_{2a}}{h_1-h_2}[/tex]
[tex]0.8 = \dfrac{3322.02-h_{2a}}{3322.02-2010.4084}[/tex]
[tex]0.8 = \dfrac{3322.02-h_{2a}}{1311.6116}[/tex]
[tex]0.8 * {1311.6116}= {3322.02-h_{2a}[/tex]
[tex]1049.28928= {3322.02-h_{2a}[/tex]
[tex]h_{2a}= {3322.02- 1049.28928[/tex]
[tex]h_{2a}= 2272.73072[/tex] kJ/kg
The work pump is calculated by applying the formula:
[tex]w_p = v_{f \ 6 kpa} (p_4-p_3)[/tex]
[tex]w_p = 0.0010062 * (10000-6)[/tex]
[tex]w_p = 0.0010062 *9994[/tex]
[tex]w_p = 10.0559628 \ kJ/kg[/tex]
However;
[tex]w_p = h_4 -h_3[/tex]
From the process;
[tex]h_3 = h_{f(6 kpa)} = 150.15 \ kJ/kg[/tex]
[tex]10.0559628 = h_4 - 150.15[/tex]
[tex]10.0559628+ 150.15 = h_4[/tex]
[tex]160.2059628= h_4[/tex]
[tex]h_4= 160.2059628 \ kJ/kg[/tex]
The actual enthalpy at s_4 by using the isentropic efficiency of the turbine can determined by using the expression:
[tex]n_P = \dfrac{h_4-h_{3}}{h_{4a}-h_3}[/tex]
