Water is the working fluid in an ideal Rankine cycle. Dry saturated vapour enters the turbine at 16 MPa, and the condenser pressure is 8 kPa. The mass flow rate of steam entering the turbine is 120 kg/s, and the expansion is isentropic.
Determine the;
i. power developed (in kW),
ii. rate of heat transfer to the steam passing though the boiler,
iii. thermal efficiency, and
iv. mass flow rate of condenser cooling water, if the cooling water undergoes a temperature increase of 18 OC with negligible pressure change in passing through the condenser. The specific heat capacity of cooling water is 4.179 kJ/kg.
The Correct Answer and Explanation is :
Let’s solve each part of the problem step by step for the ideal Rankine cycle:
Given Data:
- Turbine inlet pressure (P₁) = 16 MPa = 16000 kPa
- Condenser pressure (P₂) = 8 kPa
- Mass flow rate of steam (ṁ) = 120 kg/s
- Isentropic expansion (no entropy change in the turbine)
We will use the steam tables to find the necessary thermodynamic properties at the given pressures.
Step 1: Power Developed by the Turbine (Part i)
The turbine power (P) is calculated using:
[
P = ṁ \cdot (h₁ – h₂)
]
Where:
- h₁ = enthalpy at turbine inlet (saturated steam at 16 MPa)
- h₂ = enthalpy at turbine outlet (saturated liquid at 8 kPa)
From the steam tables:
- At P₁ = 16 MPa (saturated steam):
- h₁ = 3640 kJ/kg
- At P₂ = 8 kPa (saturated liquid):
- h₂ = 204 kJ/kg
Thus:
[
P = 120 \, \text{kg/s} \cdot (3640 – 204) \, \text{kJ/kg} = 120 \cdot 3436 \, \text{kJ/s} = 412,320 \, \text{kW}
]
Power developed by the turbine = 412.32 MW
Step 2: Rate of Heat Transfer to the Steam (Part ii)
The heat added to the steam in the boiler is the difference in enthalpies between the saturated liquid at P₂ and the saturated steam at P₁:
[
Q_{\text{in}} = ṁ \cdot (h₁ – h₄)
]
Where:
- h₄ = enthalpy of saturated liquid at P₂ (8 kPa)
From the steam tables:
- h₄ = 191.81 kJ/kg
So:
[
Q_{\text{in}} = 120 \, \text{kg/s} \cdot (3640 – 191.81) \, \text{kJ/kg} = 120 \cdot 3448.19 \, \text{kJ/s} = 413,782.8 \, \text{kW}
]
Rate of heat transfer to the steam = 413.78 MW
Step 3: Thermal Efficiency (Part iii)
The thermal efficiency of the Rankine cycle is given by:
[
\eta = \frac{P_{\text{turbine}}}{Q_{\text{in}}} = \frac{412,320}{413,782.8} \approx 0.997 \, \text{or} \, 99.7\%
]
Thermal efficiency = 99.7%
Step 4: Mass Flow Rate of Condenser Cooling Water (Part iv)
The rate of heat rejected in the condenser is:
[
Q_{\text{out}} = ṁ \cdot (h₃ – h₂)
]
Where:
- h₃ = enthalpy at the condenser inlet (saturated vapor at 8 kPa)
- h₂ = enthalpy at the condenser outlet (saturated liquid at 8 kPa)
From the steam tables:
- h₃ = 2676 kJ/kg
- h₂ = 204 kJ/kg
Thus:
[
Q_{\text{out}} = 120 \cdot (2676 – 204) = 120 \cdot 2472 = 296,640 \, \text{kJ/s} = 296.64 MW
]
Now, the rate of heat rejected to the cooling water is:
[
Q_{\text{out}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T
]
Where:
- mₓ is the mass flow rate of the cooling water
- cₓ = specific heat capacity of water = 4.179 kJ/kg°C
- ΔT = temperature rise of water = 18°C
Rearranging the equation:
[
m_{\text{water}} = \frac{Q_{\text{out}}}{c_{\text{water}} \cdot \Delta T} = \frac{296,640}{4.179 \cdot 18} = 3,930.5 \, \text{kg/s}
]
Mass flow rate of condenser cooling water = 3,930.5 kg/s
Summary of Results:
- Power developed = 412.32 MW
- Rate of heat transfer to the steam = 413.78 MW
- Thermal efficiency = 99.7%
- Mass flow rate of condenser cooling water = 3,930.5 kg/s
These results reflect the ideal Rankine cycle, assuming no losses and perfect performance in the turbine.