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Consider a steady-flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 60 deg C. The quality of the water is 0.891 at the beginning of the heat-rejection process and 0.1 at the end. Show the cycle on a T-s diagramrelative to the saturation lines, and determine a) the thermal efficiency, b) the pressure at the turbine inlet, and c) the net work output. (ANSWER = a) 0.465, b) 1.4 MPa, c) 1623 kJ/kg)

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tochinwachukwu33

Answer:

A. Thermal efficiency = 0.465 (46.5%)

B. The pressure at the turbine inlet = 1.4 MPa

C. The net work output = 1623 KJ/kg

Explanation:

The thermal efficiency of the cycle can be simply determined with only the maximum and minimum temperatures;

[tex]T_{max} = 350C = 273+ 350 = 623K[/tex]

[tex]T_{min} = 60C = 273+ 60 = 333K[/tex]

The thermal efficiency, [tex]\mu _{t} =1-\frac{T_{max} }{T_{min}}[/tex]

[tex]\mu _{t} =1-\frac{333k }{623K}= 0.465[/tex]

∴Thermal efficiency = 46.5%

B. To get the pressure at the turbine inlet, we need to go to the thermodynamic table with temperature [tex]T_{max} = 350C[/tex] and the specific entropy of the steam [tex]s_{2}[/tex] .

for the isentropic process, [tex]s_{3}=s_{2}[/tex] =[tex]s_{f}+ x_{2}s_{fg}[/tex]

[tex]s_{2}=0.8313 + 0.891\times 7.0769= 7.1369KJ/kg.K[/tex]

From tables A- 6, using values for [tex]T_{2}[/tex] and [tex]s_{2}[/tex] we can read off [tex]P_{2}[/tex] as

[tex]P_{2}= 1.4 MPa[/tex]

Pressure = 1.4 MPa

C.  We determine the net work output by calculating the enclosed area on the TS diagram attached below.

[tex]s_{4}= s_{f}+ x_{4}s_{fg}[/tex]

[tex]s_{4}= 0.8313+ 0.1\times 7.0769=1.539KJ/Kg.K[/tex]

The Net work output is then calculated as

[tex]w=(T_{max}-T_{min})\times(s_{3}-s_{4})[/tex]

[tex]w=(623K-333K)(7.140-1.540)= 1623KJ/Kg[/tex]

∴ Net work output of the turbine =  1623 KJ/Kg

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A) The thermal efficiency is; μ_t = 0.465

B) The pressure at the turbine inlet is; P₂ = 1.4 MPa

C) The net work output is; W = 1623 kJ/kg

A) We are given;

Maximum temperature; T_max = 350°C = 623 K

Minimum Temperature; T_min = 60°C = 333K

Formula for thermal efficiency is;

μ_t = 1 - (T_min/T_max)

Thus;

μ_t = 1 - (333/623)

μ_t = 0.465

B) I have drawn and attached a T-S diagram of this cycle relative to the saturation lines.

From the attached diagram, we can use the formula;

s₃ = s₂ = s_f + x₂*s_fg

At T_min = 60°C , looking at the attached table A4, we can see that;

s_f = 0.8313 KJ/Kg.K

s_fg = 7.0769 KJ/Kg.K

We are given the quality of water as; x₂ = 0.891. Thus;

Entropy at stage 2 is;

s₂ = 0.8313 + (0.891 × 7.0769)

s₂ = 7.1368 KJ/Kg.K

From table A6 attached, at s = 7.1368 KJ/Kg.K and T = 350°C, we have;

Pressure at the turbine inlet; P₂ = 1.4 MPa

C) We are told the quality of the water at the end of the heat rejection process is 0.1. Thus;

s₄ = s_f + x₄*s_fg

s₄ = 0.8313 + (0.1 × 7.0769)

s₄ = 1.53899 KJ/Kg.K

The net work output will be gotten from the formula;

W = (T_max - T_min) × (s₃ - s₄)

Thus;

W = (623 - 333) × (7.1368 - 1.53899)

W ≈ 1623 KJ/Kg

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