Question

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)

Answer 1

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;

$T_{max} = 350C = 273+ 350 = 623K$

$T_{min} = 60C = 273+ 60 = 333K$

The thermal efficiency, $\mu _{t} =1-\frac{T_{max} }{T_{min}}$

$\mu _{t} =1-\frac{333k }{623K}= 0.465$

∴Thermal efficiency = 46.5%

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

for the isentropic process, $s_{3}=s_{2}$ =$s_{f}+ x_{2}s_{fg}$

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

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

$P_{2}= 1.4 MPa$

Pressure = 1.4 MPa

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

$s_{4}= s_{f}+ x_{4}s_{fg}$

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

The Net work output is then calculated as

$w=(T_{max}-T_{min})\times(s_{3}-s_{4})$

$w=(623K-333K)(7.140-1.540)= 1623KJ/Kg$

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