Answer to Question #221987 in Mechanical Engineering for Omelatum

Question

With 3% moisture, 2.5 kg of steam has an enthalpy of 6600 kJ. It is heated at constant pressure to a final condition of 90

Answer 1

$3 \%$ of moisture, $x= 1-0.03=0.97 \\ h= \frac{H}{m}=\frac{6600}{2.5}=2640 kJ/kg\\ At \space x = 0.97, h= 2640 kJ/kg; P=0.2MPa\\$

From the steam table

$P= 0.2 MPa\\ T_{sat}=120.21^0C\\ h_f= 504.7 kJ/kg\\ h_{fg}=2201.5 kJ/kg\\ h= h_f+ xh_{fg}= 504.7+0.97*2201.5 =2640.155 kJ/kg\\$

Degree of super heat $=T_2-T_{sat}$

$80^0C=T_2 -120.21 ^0C \implies T_2 = 200.21^0C$

$At \space P= 0.2MPa\\ T= 200.12 ^0C\\ h= 2871.2 kJ/kg\\ H_2= m*h_2= 2.5*2871.2 = 7178 kJ\\$

Heat received is given as H₂-H₁

$7178-6600=578 kJ$