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Answer:
(a) The mass flow rate of the steam is approximately 1.803×10⁶ lb/h
(b) The rate of heat transfer is approximately 2.52×10⁹ BTU/h
(c) The thermal efficiency is approximately 39.68%
(d) The mass flow rate of cooling water is approximately 9.478 × 10⁷ lb/h
Explanation:
(a) The parameters are;
T₁ = 1000 F, P₁ = 1400 psi
By using an online application, we have;
h₁ = 1494 BTU/lb = 3,475 kJ/kg
s₁ = 1.61 BTU/(lb·R) = 6.741 kJ/(kg·K)
Therefore, due to isentropic expansion from state 1 to state 2, we have;
s₁ = s₂ = 1.61 BTU/(lb·R)
P₂ = 2 psi
T₂ =
= 0.17498 BTU/(lb·R)
= 94.02 BTU/(lb)
= 1116 BTU/lb
= 1.919 BTU/(lb·R)
We have;
x₂ = (1.61 - 0.17498)/(1.919 - 0.17498) ≈ 0.823
h₂ = + x₂×(
-
P₃ = P₂ = 2 psi
h₃ = = 94.02 BTU/(lb)
v₃ = 0.01605 ft³/lb
h₄ = h₃ + v₃ × (P₄ - P₃)
h₄ = 94.02 + 0.01605 × (1400 - 2) ×144/778 = 98.17 BTU/lb
The mass flow rate of the steam, =
/((h₁ - h₂) - (h₄ - h₃)) = 1 * 10^9/((1494 -935.11) - (98.17 -94.02)) ≈ 1.803×10⁶ lb/h
≈ 1.803×10⁶ lb/h
The mass flow rate of the steam ≈ 1.803×10⁶ lb/h
(b)The rate of heat transfer, =
× (h₁ - h₄) = 1.803×10⁶×(1494 -98.17) ≈ 2.52×10⁹ BTU/h
≈ 2.52×10⁹ BTU/h
(c) The thermal efficiency = /
= 1×10⁹/(2.52×10⁹) = 0.3968 ≈ 39.68%
The thermal efficiency ≈ 39.68%
(d) The mass flow rate of cooling water =
(h₂- h₃)/(
)
= 1 BTU/(lb·°F)
= 1.803×10⁶ (935.11- 94.02)/(1 * (76 - 60)) ≈ 9.478 × 10⁷ lb/h
≈ 9.478 × 10⁷ lb/h
The mass flow rate of cooling water ≈ 9.478 × 10⁷ lb/h.