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Expansion of Liquids

Steam enters ...

Question

Steam enters a well insulated turbine and expands isentropically throughout. At an intermediate pressure, $20$ percent of the mass is extracted for process heating and the remaining steam expands isentropically to $9$ kPa.
Inlet to turbine; P = $14$ MPa, T = $560^{o} C$ , h = $3486$ kJ/kg, s $= 6.6 k J / (k g K)$
Intermediate state; h = $276$ kJ/kg
Exit of turbine: P $= 9$ kPa, $h_{f} = 174$ kJ\kg.
$h_{g} = 2574$ kJ/kg, $s_{f} = 0.6$ kJ/(kgK)
$s_{g} = 8.1$ kJ(kgK)
If the flow rate of steam entering the turbine is $100$ kg/s, then the work output (in MW) is
125.56

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Solution

The correct option is A 125.56

Given data:
At turbine inlet:
$h_{1} = 3486$ kJ/kg
$s_{1} = 6.6$ kJ/kgK
$m_{1} = 100$ kg/s
At intermediate stage:
$h_{2} = 2776$ kJ/kg
$m_{2} = 0.2 m_{1} = 0.2 \times 100 = 20$ kg/s
At turbine exit:
$P_{3} = 9$ kPa
$m_{3} = 0.8 m_{1} = 0.8 \times 100 = 80$ kg/s
At $p_{3} = 9$ kPa
$h_{f} = 174$ kJ/kg
$h_{g} = 2574$ kJ/kgK
$s_{f} = 0.6$ kJ/kgK
$s_{g} = 8.1$ kJ/kgK
$s_{3} = s_{1} = s_{f} + x_{3} (s_{g} - s_{f})$
$6.6 = 0.6 + x_{3} (8.1 - 0.6)$
$6 = 7.5 x_{3}$
or $x_{3} = 0.8$
$h_{3} = h_{f} + x_{3} (h_{g} - h_{f})$
$174 + 0.8 (2574 - 174)$
$2094$ kJ/kg
Work output,
$W = m_{1} (h_{1} - h_{2}) + m_{3} (h_{2} - h_{3})$
$= 100 (3486 - 2776)$
$+ 80 (2776 - 2094)$
$71000 + 54560$
$125560$ kW
$125.56$ MW

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