Question

Under steady state conditions, superheated steam enters the turbine with enthalpy, $h_1 = 3200$ kJ/kg and wet steam leaves the turbine at pressure $p_2 = 0.1$ bar. The heat loss is 100 kJ/kg and work output is 1000 kJ/kg. Kinetic and potential energies for inflow and outflow are neglected. At pressure 0.1 bar, the enthalpy of saturated liquid is 200 kJ/kg and the enthalpy of vaporization is 2400 kJ/kg. The dryness fraction of the steam at the exit of the turbine is ................. (round off to 2 decimal places).

• A. 0.75
• B. 0.80
• C. 0.85
• D. 0.90

Hint:

To calculate the dryness fraction, use the enthalpy at the exit of the turbine and solve using the formula $h_2 = h_f + x \cdot h_{fg}$.

Answer 1

The Correct Option is A

Given:
Enthalpy of steam at inlet, $h_1 = 3200$ kJ/kg
Heat loss = 100 kJ/kg
Work output = 1000 kJ/kg
Enthalpy of saturated liquid at $p_2 = 0.1$ bar = 200 kJ/kg
Enthalpy of vaporization at $p_2 = 0.1$ bar = 2400 kJ/kg
Dryness fraction = ?
The enthalpy at the exit of the turbine, $h_2$, is given by: \[ h_2 = h_1 - \text{heat loss} - \text{work output} \] \[ h_2 = 3200 - 100 - 1000 = 2100 \, \text{kJ/kg} \] Now, the dryness fraction $x$ is given by: \[ h_2 = h_f + x \cdot h_{fg} \] where:
- $h_f$ is the enthalpy of the saturated liquid = 200 kJ/kg
- $h_{fg}$ is the enthalpy of vaporization = 2400 kJ/kg
Substitute the values: \[ 2100 = 200 + x \cdot 2400 \] \[ 2100 - 200 = x \cdot 2400 \] \[ 1900 = x \cdot 2400 \] \[ x = \frac{1900}{2400} = 0.79 \] Thus, the dryness fraction is approximately 0.79.