Calculate the mass flow rate of steam entering the low-pressure turbine
To find the mass flow rate of steam entering the low-pressure turbine, we can use the mass balance equation:
\(m_{in} = m_{extracted} + m_{low}\)
From the problem, we know that \(m_{extracted}=1000\,\text{kg/min}\), and we can convert it to kg/s by dividing it by 60:
\(m_{extracted} = \frac{1000}{60}\,\text{kg/s} \approx 16.67\,\text{kg/s}\)
Now we can find \(m_{low}\) from the power produced by the low-pressure turbine and the isentropic efficiency:
\(P_{low} = m_{low}h_{low,in} - m_{low}h_{low,out}\)
First, we can find the enthalpy (\(h_{low,in}\)) at the entrance of the low-pressure turbine from the steam tables, which is the saturation vapor value at \(1.4\,\text{MPa}\):
\(h_{low,in} \approx 2799.5\,\text{kJ/kg}\)
Next, we can calculate the enthalpy at the exit of the low-pressure turbine using the isentropic efficiency:
\(\eta_{low} = \frac{h_{low,in} - h_{low,out}}{h_{low,in} - h'_{low,out}}\)
Where \(h'_{low,out}\) is the enthalpy of the isentropic expansion to \(10\,\text{kPa}\). We can find the corresponding enthalpy from the steam tables:
\(h'_{low,out} \approx 191.83\,\text{kJ/kg}\)
Now we can find \(m_{low}\) from the power produced by the low-pressure turbine and the isentropic efficiency:
\(P_{low} = m_{low}(h_{low,in} - h_{low,out})\)
Rearranging for \(m_{low}\):
\(m_{low}=\frac{P_{low}}{h_{low,in}-h_{low,out}}\)
Now use the isentropic efficiency to find \(h_{low,out}\):
\(h_{low,out} = h_{low,in} - \eta_{low}(h_{low,in} - h'_{low,out})\)
\(h_{low,out}=2799.5-0.8(2799.5-191.83) \approx 2229.17\,\text{kJ/kg}\)
Now we can plug in the values and solve for \(m_{low}\):
\(m_{low}=\frac{1.5\,\text{MW}}{2799.5\,\text{kJ/kg}-2229.17\,\text{kJ/kg}}=13.93\,\text{kg/s}\)
Now we can find the mass flow rate of steam entering the high-pressure turbine using the mass balance equation:
\(m_{in} = m_{extracted} + m_{low}\)
\(m_{in} = 16.67\,\text{kg/s} + 13.93\,\text{kg/s} = 30.6\,\text{kg/s}\)
