Question

Steam enters an adiabatic turbine at \(6 \mathrm{MPa}, 600^{\circ} \mathrm{C}\) and \(80 \mathrm{m} / \mathrm{s}\) and leaves at \(50 \mathrm{kPa}, 100^{\circ} \mathrm{C},\) and \(140 \mathrm{m} / \mathrm{s}\). If the power output of the turbine is \(5 \mathrm{MW}\), determine \((a)\) the reversible power output and ( \(b\) ) the second-law efficiency of the turbine. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\).

Short answer

Question: Determine the reversible power output and the second-law efficiency of an adiabatic turbine with the initial conditions: inlet pressure \(P_1 = 6 \mathrm{MPa}\), inlet temperature \(T_1 = 600^{\circ} \mathrm{C}\), outlet pressure \(P_2=50 \mathrm{kPa}\), and outlet temperature \(T_2 = 100^{\circ} \mathrm{C}\). The initial velocity is \(\nu_1=80 \mathrm{m} / \mathrm{s}\), the final velocity is \(\nu_2=140 \mathrm{m} / \mathrm{s}\), and the power output of the turbine is \(5 \mathrm{MW}\). Answer: To determine the reversible power output and the second-law efficiency of the adiabatic turbine, we must first obtain the specific enthalpies \(h_1\) and \(h_2\) using the given conditions. Then, we use the expressions derived in the step-by-step solution: the reversible power output is given by \(W_{rev} = -(h_2 - h_1)\) (Step 5), and the second-law efficiency is given by \(\eta_{II} = \frac{-\frac{1}{2} (140^2 - 80^2) - (h_2 - h_1)}{-(h_2 - h_1)}\) (Step 6). Plug in the values of \(h_1\) and \(h_2\) to obtain the numerical values for the reversible power output and the second-law efficiency.

Step-by-step solution

(Step 1: Use the first law of thermodynamics for an adiabatic turbine)

The first law of thermodynamics for a steady-flow system can be written as: \(Q - W = \Delta KE + \Delta U + \Delta PE\) Since the turbine is adiabatic, there is no heat transfer \((Q=0)\). Due to the assumption of sufficient height difference being negligible, changes in potential energy \((\Delta PE = 0)\). Thus, we can rewrite the energy equation as: \(-W = \Delta KE + \Delta U\) To determine the work output of the turbine, we need to calculate the changes in kinetic energy \((\Delta KE)\) and internal energy \((\Delta U)\).

(Step 2: Calculate changes in kinetic energy)

The change in kinetic energy is given by: \(\Delta KE = \frac{1}{2} m (\nu_2^2 - \nu_1^2)\) where \(m\) is mass flow rate of the steam, \(\nu_1\) and \(\nu_2\) are the initial and final velocities of the steam, respectively. We are given \(\nu_1=80 \mathrm{m} / \mathrm{s}\) and \(\nu_2=140 \mathrm{m} / \mathrm{s}\). We will leave the change in kinetic energy in terms of mass flow rate, as the mass flow rate will cancel out when we calculate work output. \(\Delta KE = \frac{1}{2} m (140^2 - 80^2)\)

(Step 3: Calculate changes in internal energy)

The change in internal energy is given by the specific enthalpy difference at the inlet and outlet of the turbine. Using the given properties of steam (inlet pressure \(P_1 = 6 \mathrm{MPa}\), inlet temperature \(T_1 = 600^{\circ} \mathrm{C}\), outlet pressure \(P_2=50 \mathrm{kPa}\), and outlet temperature \(T_2 = 100^{\circ} \mathrm{C}\)), we can determine specific enthalpy \(h_1\) and \(h_2\) by referring to steam tables or using software tools. The change in internal energy is then: \(\Delta U = m (h_2 - h_1)\)

(Step 4: Calculate work output of the turbine)

Now, we can determine the work output of the turbine using the first law of thermodynamics: \(-W = \Delta KE + \Delta U\) \(W = - \frac{1}{2} m (140^2 - 80^2) - m (h_2 - h_1)\) The problem states that the power output of the turbine is \(5 \mathrm{MW}\). Power is the work output per unit time, so we can write: \(5 \times 10^6 = - \frac{1}{2} (\frac{m}{\Delta t}) (140^2 - 80^2) - (\frac{m}{\Delta t}) (h_2 - h_1)\) Dividing both sides by \(\frac{m}{\Delta t}\), the mass flow rate, gives the work output per unit mass: \(\frac{W}{m} = -\frac{1}{2} (140^2 - 80^2) - (h_2 - h_1)\)

(Step 5: Calculate the reversible power output)

For a reversible adiabatic turbine, the reversible work output, \(W_{rev}\), is the difference in specific enthalpy at the inlet and outlet: \(W_{rev} = -(h_2 - h_1)\)

(Step 6: Calculate the second-law efficiency of the turbine)

The second-law efficiency of the turbine, \(\eta_{II}\), can be calculated as the ratio of actual work output to reversible work output: \(\eta_{II} = \frac{W}{W_{rev}}\) Substitute the values of \(W\) and \(W_{rev}\) from steps 4 and 5: \(\eta_{II} = \frac{-\frac{1}{2} (140^2 - 80^2) - (h_2 - h_1)}{-(h_2 - h_1)}\) This is the final expression to determine the second-law efficiency of the turbine. Once we obtain specific enthalpies \(h_1\) and \(h_2\) from the steam tables, we can plug their values into this equation to get the numerical value for \(\eta_{II}\). In conclusion, we have formulated the expressions to calculate the reversible power output (Step 5) and the second-law efficiency of the turbine (Step 6) using the given conditions and energy equations.