A steam power plant working on Carnot cycle operates between 320-C- 10 Mpa and 31-C and 6 kpa- The following data table is availablePT-ChfkJ-kghgkJ-kgsfkJ-kgKsgkJ-kgK10 MPa320-C140727253-355-616kPa31-C13625600-488-40 Back work ratio of the cycle is -

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Standard X

Physics

Electric Power

A steam power...

Question

A steam power plant working on Carnot cycle operates between $320^{\circ} C$ , 10 Mpa and $31^{\circ} C$ and 6 kpa. The following data table is available

P

$T (^{\circ} C)$

$h_{f} (k J / k g)$

$h_{g} (k J / k g)$

$s_{f} (k J / k g K)$

$s_{g} (k J / k g K)$

10 MPa

$320^{\circ} C$

1407

2725

3.35

5.61

6kPa

$31^{\circ} C$

136

2560

0.48

8.40

Back work ratio of the cycle is ______.

0.37

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Solution

The correct option is A 0.37

$r_{1} = x - \frac{T_{L}}{T_{H}} = 1 - (\frac{31 + 273}{320 + 273}) = 0.487$

$s_{1} = s_{2}$

5.61 = 0.48 + x(8.40 – 0.48)

X = 0.647

$h_{1} = h_{f} + x h_{f g}$

=136 + 0.647 x (2500 – 136) = 1706.1 kJ/kg

$W_{T} = h_{1} - h_{2}$

= 2725 – 1706.1 = 1018.9 kJ/kg

$Q_{s} = h_{1} - h_{4} = 2725 - 1407 = 1318 k J / k g$

$n = \frac{W_{n e t}}{Q_{s}}$

$\Rightarrow 0.487 = \frac{W_{n e t}}{1318}$

$W_{n e t} = 641.866 k J / k g$

Work ration = $\frac{W_{n e t}}{W_{T}} = \frac{641.866}{1018.9}$

$0.6299 ≃ 0.63$

Back work ratio = 1 – work ratio = 0.37

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P	$T (^{\circ} C)$	$h_{f} (k J / k g)$	$h_{g} (k J / k g)$	$s_{f} (k J / k g K)$	$s_{g} (k J / k g K)$
10 MPa	$320^{\circ} C$	1407	2725	3.35	5.61
6kPa	$31^{\circ} C$	136	2560	0.48	8.40

A steam power plant working on Carnot cycle operates between 320∘C, 10 Mpa and 31∘C and 6 kpa. The following data table is available P T(∘C) hf(kJ/kg) hg(kJ/kg) sf(kJ/kgK) sg(kJ/kgK) 10 MPa 320∘C 1407 2725 3.35 5.61 6kPa 31∘C 136 2560 0.48 8.40 Back work ratio of the cycle is ______. 0.37