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2 answers:
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Answer:
(a)
Plastic section modulus Z =92.72 in^3
Plastic moment Mp = 4636 kip-in
(b)
Elastic section modulus S = 80.88 in^3
Yield moment My = 4044 kip-in
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached files.i hope my explanation will help you in understanding this particular question.
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.