Answer:
38%
Explanation:
The following data is obtained from water property tables:
[tex]P_{1@satliquid} = 150 KPa\\ h_{1} = 350.54 KJ/kg\\v_{1} = 0.001030 m^3/kg\\\\\\h_{2} = h_{1} + v_{1}*(P_{2} - P_{1})\\h_{2} = 340.54 + 0.001030*(17500-150)\\\\h_{2} = 358.51 KJ/kg\\\\P_{3}= 17500KPa\\T_{3} = 550 C\\s_{3} = 6.430KJ/kgK\\\\h_{3} = 3420 KJ/kg\\\\P_{4} = 4000KPa\\s_{3} = s_{4} = 6.430KJ/kgK\\\\h_{4} = 2840KJ/kg\\\\P_{5} = 2000KPa\\T_{5} = 300 C\\s_{5} = 6.770KJ/kgK\\\\h_{5} = 3020KJ/kg\\\\P_{6} = 50KPa\\s_{5} = s_{6} = 6.770KJ/kgK\\\\h_{6} = 2350KJ/kg\\\\[/tex]
Part a)
[tex]q_{b} = (h_{3}-h_{2})+(h_{5}-h_{4})\\q_{b} = (3420-358.51)+(3020-2840)\\\\q_{b} = 3421.49KJ/kg\\\\q_{c} = h_{6} - h_{1}\\q_{c} = 2350-340.54\\\\q_{c} = 2009.46KJ/kg\\\\u_{th} = 1-\frac{q_{c} }{q_{b} }\\\\u_{th} = 1-\frac{2009.46}{3421.49}\\\\u_{th} = 38%[/tex]