Respuesta :
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 =
[tex]s_f[/tex] = 0.17498 BTU/(lb·R)
[tex]h_{f2}[/tex] = 94.02 BTU/(lb)
[tex]h_g[/tex] = 1116 BTU/lb
[tex]s_g[/tex] = 1.919 BTU/(lb·R)
We have;
x₂ = (1.61 - 0.17498)/(1.919 - 0.17498) ≈ 0.823
h₂ = [tex]h_f[/tex] + x₂×([tex]h_g[/tex] -
P₃ = P₂ = 2 psi
h₃ = [tex]h_{f2}[/tex] = 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, [tex]\dot m[/tex] = [tex]\dot W[/tex]/((h₁ - h₂) - (h₄ - h₃)) = 1 * 10^9/((1494 -935.11) - (98.17 -94.02)) ≈ 1.803×10⁶ lb/h
[tex]\dot m[/tex] ≈ 1.803×10⁶ lb/h
The mass flow rate of the steam ≈ 1.803×10⁶ lb/h
(b)The rate of heat transfer, [tex]\dot Q_{in}[/tex] = [tex]\dot m[/tex] × (h₁ - h₄) = 1.803×10⁶×(1494 -98.17) ≈ 2.52×10⁹ BTU/h
[tex]\dot Q_{in}[/tex] ≈ 2.52×10⁹ BTU/h
(c) The thermal efficiency = [tex]\dot W_{cyc}[/tex]/[tex]\dot Q_{in}[/tex] = 1×10⁹/(2.52×10⁹) = 0.3968 ≈ 39.68%
The thermal efficiency ≈ 39.68%
(d) The mass flow rate of cooling water [tex]\dot m_w[/tex] = [tex]\dot m[/tex](h₂- h₃)/([tex]c_w \Delta T[/tex])
[tex]c_w[/tex] = 1 BTU/(lb·°F)
[tex]\dot m_w[/tex] = 1.803×10⁶ (935.11- 94.02)/(1 * (76 - 60)) ≈ 9.478 × 10⁷ lb/h
[tex]\dot m_w[/tex] ≈ 9.478 × 10⁷ lb/h
The mass flow rate of cooling water ≈ 9.478 × 10⁷ lb/h.
A) The mass flow rate of steam, in lb/h is; 1.803 × 10⁶ lb/h
B) The rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator is; 2.52 × 10⁹ btu/h
C) The thermal efficiency is; 39.68%
D) The mass flow rate of cooling water, in lb/h is; 9.478 × 10⁷ lb/h
Thermal Efficiency and Mass Flow rate
We are given;
- Inlet Temperature; T₁ = 1000 F
- Inlet Pressure; P₁ = 1400 psi
- Power Output; W' = 1 * 10⁹ btu/hr
From online thermodynamic property table, we have;
enthalpy at inlet; h₁ = 1494 Btu/lbm
entropy at inlet; s₁ = 1.61 Btu/lbm·R
Now, since there is an isentropic expansion from state 1 to state 2, then we can say that;
s₁ = s₂ = 1.61 Btu/lbm·R
Also at s₂ = 1.61 Btu/lbm·R, from thermodynamics steam property table online, we have;
P₂ = 2 psi
Entropy of saturated Liquid; s_f = 0.17498 Btu/lbm·R
Enthalpy of saturated liquid; h_f₂ = 94.02 Btu/lbm
Enthalpy of evaporation liquid; h_g = 1116 Btu/lbm
Entropy of saturation vapour; s_g = 1.919 Btu/lbm.R
A) Thus, quality of steam at second state is calculated from;
x₂ = (s₂ - s_f)/(s_g - s_f)
x₂ = (1.61 - 0.17498)/(1.919 - 0.17498)
x₂ = 0.823
formula for enthalpy at second state;
h₂ = h_f + x₂(h_g - h_f)
h₂ = 94.02 + 0.823(1116 - 94.02)
h₂ = 935.11 Btu/lbm.R
Similar to the first to second state, we can equally say that;
P₃ = P₂ = 2 psi
h₃ = h_f₂ = 94.02 Btu/lbm
From online thermodynamic steam tables, for those values at state 3, we have;
Specific Volume; ν₃ = 0.01605 ft³/lb = 0.0029707 btu/lb
Enthalpy at state 4 is;
h₄ = h₃ + (v₃(P₄ - P₃))
h₄ = 94.02 + 0.0029707(1400 - 2)
h₄ = 98.17 Btu/lbm
Mass flow rate of the steam is calculated from;
m' = W'/[(h₁ - h₂) - (h₄ - h₃)]
m' = (1 × 10⁹)/((1494 -935.11) - (98.17 -94.02))
m' = 1.803 × 10⁶ lb/h
B) Rate of the heat transfer is calculated from;
Q' = m'(h₁ - h₄)
Q' = 1.803 × 10⁶ ×(1494 - 98.17)
Q' = 2.52 × 10⁹ btu/h
C) Thermal efficiency is gotten from the formula;
η = W'/Q'
η = (1 × 10⁹)/(2.52×10⁹)
η = 39.68%
D) The mass flow rate of cooling water is calculated from;
m'_w = m'(h₂- h₃)/(c_w * Δt)
c_w is specific heat capacity of cool water = 1 Btu/(lb·°F)
Thus;
m'_w = 1.803 × 10⁶(935.11 - 94.02)/(1 × (76 - 60))
m'_w ≈ 9.478 × 10⁷ lb/h
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