Water is the working fluid in an ideal Rankine cycle. Steam enters the turbine at 1400 lbf/in2 and 1000°F. The net power output of the cycle is 1 x 109 Btu/hr. Cooling water experiences a temperature increase from 60°F to 76°F, with negligible pressure drop, as it passes through the condenser. a) The mass flow rate of steam, in lb/h
b) The rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator
c) The thermal efficiency
d) The mass flow rate of cooling water, in lb/h

(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.

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