Steam flows through a turbine stage at the rate of 4 500 kg/h. The steam velocities at inlet and outlet are 15 m/s and 180 m/s respectively. The rate of heat energy flow from the turbine casing to the surroundings is 23 kJ/kg of steam flowing. If specific enthalpy of the steam decreases by 420 kJ/kg in passing through the turbine stage. Calculate the power developed. (476 kW)
The correct answer and explanation is :
To calculate the power developed by the steam turbine, we need to apply the First Law of Thermodynamics and incorporate the given data: mass flow rate, velocity, heat loss, and change in enthalpy. We will use the energy equation for a control volume:
[
\text{Power} = \frac{\dot{m}}{1000} \times (\Delta h + \Delta KE + \text{Heat loss})
]
Where:
- ( \dot{m} ) = mass flow rate (kg/s)
- ( \Delta h ) = change in specific enthalpy (kJ/kg)
- ( \Delta KE ) = change in kinetic energy (kJ/kg)
- Heat loss = heat energy lost to surroundings (kJ/kg)
Step 1: Convert Mass Flow Rate
The mass flow rate is given as 4,500 kg/h. First, convert this to kg/s:
[
\dot{m} = \frac{4500}{3600} = 1.25 \, \text{kg/s}
]
Step 2: Change in Enthalpy (( \Delta h ))
The problem states that the specific enthalpy of the steam decreases by 420 kJ/kg. Thus, ( \Delta h = -420 \, \text{kJ/kg} ).
Step 3: Change in Kinetic Energy (( \Delta KE ))
The change in kinetic energy per unit mass of the steam can be calculated using the equation for kinetic energy:
[
\Delta KE = \frac{v_2^2 – v_1^2}{2}
]
Where:
- ( v_2 = 180 \, \text{m/s} ) (outlet velocity)
- ( v_1 = 15 \, \text{m/s} ) (inlet velocity)
Substitute the velocities into the equation:
[
\Delta KE = \frac{180^2 – 15^2}{2} = \frac{32400 – 225}{2} = \frac{32175}{2} = 16,587.5 \, \text{J/kg} = 16.59 \, \text{kJ/kg}
]
Step 4: Heat Loss to Surroundings
The heat loss is given as 23 kJ/kg.
Step 5: Total Energy per Unit Mass
The total energy per unit mass of steam is the sum of the enthalpy change, kinetic energy change, and heat loss:
[
\Delta E = \Delta h + \Delta KE + \text{Heat loss} = -420 + 16.59 + 23 = -380.41 \, \text{kJ/kg}
]
Step 6: Power Developed
Now, calculate the power developed by multiplying the total energy change per unit mass by the mass flow rate and converting from kJ/s to kW:
[
\text{Power} = \dot{m} \times \Delta E = 1.25 \times (-380.41) = -475.51 \, \text{kW}
]
Since the negative sign indicates energy output, the magnitude of power developed by the turbine is:
[
\text{Power} = 476 \, \text{kW}
]
Conclusion:
The power developed by the turbine is 476 kW. This value represents the mechanical power extracted from the steam, considering the heat loss and changes in velocity and enthalpy.
