Question

Steam at \(7 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) enters a two-stage adiabatic turbine at a rate of \(15 \mathrm{kg} / \mathrm{s}\). Ten percent of the steam is extracted at the end of the first stage at a pressure of \(1.8 \mathrm{MPa}\) for other use. The remainder of the steam is further expanded in the second stage and leaves the turbine at 10 kPa. If the turbine has an isentropic efficiency of 88 percent, determine the wasted power potential during this process as a result of irreversibilities. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\).

Short answer

Question: Calculate the wasted power potential due to irreversibilities in a two-stage adiabatic turbine with steam entering at 7 MPa and 400°C. The first stage of the turbine has an isentropic efficiency of 88% and steam is extracted for other use at a pressure of 1.8 MPa. The second stage of the turbine has an isentropic efficiency of 88%, and the final pressure is 10 kPa. The mass flow rate entering the first stage is 15 kg/s, and 90% of mass flow rate is going to the second stage. Answer: The wasted power potential due to irreversibilities is approximately 1849.902 kW.

Step-by-step solution

Calculate the enthalpy and entropy at initial state (state 1)

We will start by calculating the enthalpy and entropy at state 1 where the steam enters the turbine. Using the pressure \(P_1 = 7 \,\text{MPa}\) and temperature \(T_1 = 400 ^\circ\text{C}\), we can get the enthalpy \(h_1\) and entropy \(s_1\) values from steam tables: - \(h_1 \approx 3230.9\, \text{kJ/kg}\) - \(s_1 \approx 6.9211\, \text{kJ/kg}\cdot\text{K}\)

Calculate the enthalpy at isentropic state 2 and actual state 2

We first need to find the isentropic enthalpy at state 2, which is when the steam is extracted for other use. First, we calculate the isentropic entropy as it remains constant with isentropic process: - \(s_2s = s_1 = 6.9211\, \text{kJ/kg}\cdot\text{K}\) Then, we find the corresponding isentropic enthalpy \(h_2s\) using steam table and \(P_2 = 1.8 \, \text{MPa}\). - \(h_{2s} \approx 3168 \, \text{kJ/kg}\) We also need to calculate the actual enthalpy at state 2, which is given by the isentropic efficiency: Isentropic efficiency = \(\frac{h_1 - h_2}{h_1 - h_{2s}}\) Solving for \(h_2\), $$h_2 = h_1 - (\text{Isentropic efficiency}\times(h_1 - h_{2s}))$$ Putting the values, Isentropic efficiency = 0.88 \(h_2 \approx 3185.976\, \text{kJ/kg}\)

Find power output at first stage of turbine

To find the power output at the first stage, we'll use the mass flow rate value given (\(\dot m = 15\, \text{kg/s}\)) and enthalpy values at state 1 and state 2. Power Output = \(\dot m\times(h_1 - h_2)\) $$\text{Power Output}_1 \approx 15\times(3230.9 - 3185.976) \approx 673.86\,\text{kW}$$

Calculate the enthalpy at state 3 and actual state 4

In this step, we calculate the isentropic enthalpy at state 4, the final state of the turbine. First, calculate the isentropic enthalpy \(h_{4s}\) using the pressure value \(P_4 = 10\,\text{kPa}\) and taking isentropic entropy as constant: - \(s_{4s} = s_2 = 6.9211\, \text{kJ/kg}\cdot\text{K}\) - \(h_{4s} \approx 2534 \, \text{kJ/kg}\) Next, we find the actual enthalpy at state 4 using the isentropic efficiency, similar to step 2: $$h_4 = h_2 - (\text{Isentropic efficiency}\times(h_2 - h_{4s}))$$ Putting the values, \(h_4 \approx 2587.084\, \text{kJ/kg}\)

Find power output at the second stage of the turbine

To find the power output at the second stage, we'll again use the mass flow rate value and enthalpy values at state 2 and state 4. Here, the mass flow rate at the second stage is 90% of the initial mass flow rate, due to extraction: $$\dot m_2 = 0.9\times 15 = 13.5\,\text{kg/s}$$ Power Output = \(\dot m_2\times(h_2 - h_4)\) $$\text{Power Output}_2 \approx 13.5\times(3185.976 - 2587.084) \approx 8071.642\,\text{kW}$$

Calculate the wasted power potential

Finally, we compute the wasted power potential, which is the difference between the isentropic power output and the actual power output for both the stages combined: Isentropic Power Output\(_1\) = \( 15\times(h_1 - h_{2s}) \approx 939\,\text{kW}\) Isentropic Power Output\(_2\) = \( 13.5\times(h_2 - h_{4s}) \approx 8682.06\,\text{kW}\) Wasted Power Potential = Total Isentropic Power Output - Total Actual Power Output $$\text{Wasted Power Potential} = (939 + 8682.06) - (673.86 + 8071.642) \approx 875 -1245.502 = 1849.902\,\text{kW}$$ Thus, the wasted power potential due to irreversibilities is approximately \(1849.902\,\text{kW}\).