Consider a steam power plant operating on an ideal reheat Rankine cycle. The work input to the pump is 20 kJ/kg. The work output from the high pressure turbine is 750 kJ/kg. The work output from the low pressure turbine is 1500 kJ/kg. The thermal efficiency of the cycle is 50 %. The enthalpy of saturated liquid and saturated vapour at condenser pressure are 200 kJ/kg and 2600 kJ/kg, respectively. The quality of steam at the exit of the low pressure turbine is _______________ % (round off to the nearest integer).

The correct answer is 93.

Here is the step-by-step educational explanation to determine the quality of steam at the exit of the low-pressure turbine:

1. Identify the given parameters:
- Pump work input, W_p = 20 kJ/kg
- High-pressure turbine work output, W_T = 750 kJ/kg (in the source derivation, a value of 750 is given, but note that 700 + 1500 - 20 = 2230 is calculated, meaning W_T is treated as 750 kJ/kg, but with 750 - 50 = 700 or a slight numerical adjustment. Let's trace: W_net = W_T + W_T' - W_p = 750 + 1500 - 20 = 2230 kJ/kg. The source contains a slight typo where it writes "700 + 1500 - 20" instead of "750 + 1500 - 20", but the sum is correctly computed as 2230 kJ/kg)
- Low-pressure turbine work output, W_T' = 1500 kJ/kg
- Thermal efficiency of the cycle, η = 50% = 0.5
- Enthalpy of saturated liquid at condenser pressure, h_f = 200 kJ/kg
- Enthalpy of saturated vapour at condenser pressure, h_g = 2600 kJ/kg

2. Calculate the net work output (W_net):
The net work output is the total work produced by the turbines minus the work consumed by the pump:
$$W_{\text{net}}=W_{\text{T}}+W_{\text{T}}^{��}-W_{\text{p}}$$
Substituting the given values:
$$W_{\text{net}}=750+1500-20=2230\text{ kJ/kg}$$

3. Calculate the heat supplied (Q_s) and heat rejected (Q_R):
Thermal efficiency (η) is defined as the ratio of net work done to the heat supplied:
$$\eta =\frac{W_{\text{net}}}{Q_{\text{s}}}$$
Rearranging to find the heat supplied (Q_s):
$$Q_{\text{s}}=\frac{2230}{0.5}=4460\text{ kJ/kg}$$
Using the efficiency relation with heat rejection:
$$\eta =1-\frac{Q_{\text{R}}}{Q_{\text{s}}}\Rightarrow Q_{\text{R}}=(1-0.5)\times 4460=2230\text{ kJ/kg}$$

4. Determine the enthalpy at the turbine exit:
The heat rejected in the condenser can be represented in terms of enthalpies at the entry and exit of the condenser:
$$Q_{\text{R}}=h_{\text{4}}-h_{\text{5}}$$
Here, h₄ is the enthalpy of the steam leaving the low-pressure turbine (entering the condenser), and h₅ is the enthalpy of the saturated liquid leaving the condenser (h₅ = h_f = 200 kJ/kg).
$$2230=h_{\text{4}}-200\Rightarrow h_{\text{4}}=2430\text{ kJ/kg}$$

5. Calculate the quality of steam (x) at the exit of the low-pressure turbine:
Using the mixture enthalpy formula:
$$h_{\text{4}}=h_{\text{f}}+x(h_{\text{g}}-h_{\text{f}})$$
Substitute the values:
$$2430=200+x(2600-200)$$
$$2230=2400x$$
$$x=\frac{2230}{2400}\approx 0.92917$$
Converting to a percentage and rounding off to the nearest integer:
$$x=92.91\mathrm{\%}\approx 93\mathrm{\%}$$