Answer:
Explanation:
Given that:
At the inlet:
Temperature [tex]T_1 = 25^0 \ C[/tex]
Pressure [tex]P_1[/tex] = 20bar = 2000 kPa
At the outlet:
[tex]P_2 = 2000 \ kPa[/tex]
a)
Q = 500 kJ/kg
Using the steam tables to determine the properties of water;
[tex]h_1 = 106.6863 \ kJ/kg \\ \\ h_2 -h_1 = 500 \\ \\ h_2 = (500 + 106.6863) \ kJ/kg \\ \\ h_2 = 606.68 \ kJ/kg[/tex]
[tex]at \ p_{sat}= 2000 \ kPa; \ h_f = 908.47 \kJ/kg; \ h_g = 2798.3 \ kJ/kg} \\ \\ h_2< h_f; \text{thus, the soil is compressed or subcooled liquid}[/tex]
b)
[tex]Q = 1500 \ kJ/kg\\ \\ Q= (h_2-h_1) \\ \\ h_2 = 1500 + 106.6863 \\ \\ h_2 = 1606.6863 \ kJ/kg \\ \\ h_f < h_2<hg \ at \ 2000 \ kPa[/tex]
Thus, the outlet steam will be a vapor-liquid mixture at 212.38° C
c)
[tex]Q = 3000 \ kJ/kg \\ \\ Q = h_2 -h_1 \\ \\ h_2 = Q + h_1 \\ \\ h_2 = 3000 + 106.6863 \\ \\ h_2 = 3106.6863 \ kJ/kg\\ \\ h_2 > hg \ at \ 2000 \ kPa[/tex]
Thus, the outlet steam will be a superheated vapor at 380° C
