Question

Nitrogen gas enters a turbine operating at steady state with a velocity of \(60 \mathrm{~m} / \mathrm{s}\), a pressure of \(0.345 \mathrm{Mpa}\), and a temperature of \(700 \mathrm{~K}\). At the exit, the velocity is \(0.6 \mathrm{~m} / \mathrm{s}\), the pressure is \(0.14 \mathrm{Mpa}\), and the temperature is \(390 \mathrm{~K}\). Heat transfer from the surface of the turbine to the surroundings occurs at a rate of \(36 \mathrm{~kJ}\) per \(\mathrm{kg}\) of nitrogen flowing. Neglecting potential energy effects and using the ideal gas model, determine the power developed by the turbine, in \(\mathrm{kW}\).

Short answer

The power developed by the turbine is 282.89 kW.

Step-by-step solution

Apply the Energy Balance Equation

Use the steady-flow energy equation for the turbine. Since potential energy changes are negligible, the energy balance equation is: \[ \frac{dE_{system}}{dt} = \frac{dm}{dt}(h_1 + \frac{v_1^2}{2}) - \frac{dm}{dt}(h_2 + \frac{v_2^2}{2}) + Q - W \] where: - \frac{dE_{system}}{dt} is the rate of change of the system energy. - h_1 and h_2 are the specific enthalpies at the inlet and exit respectively. - v_1 and v_2 are the velocities at the inlet and exit respectively. - Q is the heat transfer rate (per kg). - W is the power developed by the turbine.

Define the Known Values

Enter all known values from the problem statement: - Velocity at the inlet, \(v_1 = 60 \, m/s\). - Pressure at the inlet, \(P_1 = 0.345 \, MPa\). - Temperature at the inlet, \(T_1 = 700 \, K\). - Velocity at the exit, \(v_2 = 0.6 \, m/s\). - Pressure at the exit, \(P_2 = 0.14 \, MPa\). - Temperature at the exit, \(T_2 = 390 \, K\). - Heat transfer rate, \(Q = -36 \, kJ/kg\) (negative since heat is transferred from the system).

Determine Enthalpies at Inlet and Exit

Use the ideal gas model to determine the specific enthalpies: For an ideal gas, specific enthalpy, \(h\), is given by: \[ h = c_p T \] where: - \(c_p\) is the specific heat at constant pressure. For Nitrogen, \(c_p = 1.039 \, kJ/(kg\cdot K)\). So, - \(h_1 = c_p \, T_1 = 1.039 \, kJ/(kg\cdot K) \cdot 700 \, K = 727.3 \, kJ/kg\) - \(h_2 = c_p \, T_2 = 1.039 \, kJ/(kg\cdot K) \cdot 390 \, K = 405.21 \, kJ/kg\)

Convert Units of the Kinetic Energies

The kinetic energy term is given by: \[ \frac{v^2}{2} \] Convert velocities from m/s to kJ/kg: - Inlet: \( \frac{60^2}{2} = 1800 \, (m^2/s^2) \times 0.001 \, kJ/(kg\cdot m^2/s^2) = 1.8 \, kJ/kg\) - Exit: \( \frac{0.6^2}{2} = 0.18 \, (m^2/s^2) \times 0.001 \, kJ/(kg\cdot m^2/s^2) = 0.00018 \, kJ/kg\)

Calculate the Power Developed by the Turbine

Substitute the known values back into the energy balance equation to find the power developed by the turbine per unit mass flow rate: \[ 0 = (727.3 + 1.8) - (405.21 + 0.00018) + (-36) - W\] Solve for \(W\): \[ W = 727.3 + 1.8 - 405.21 - 0.00018 - 36\] \[ W = 724.1 - 441.21 = 282.89 \, kJ/kg\] The power developed by the turbine: \[ P = \frac{282.89 \, kJ/kg}{1 \, kg/s} = 282.89 \, kW\]

Key concepts

Steady-Flow Energy Equation

The steady-flow energy equation is a fundamental concept used in analyzing the energy transformations in fluid systems, such as turbines. It allows us to consider the energy changes due to work, heat transfer, and changes in kinetic and potential energy. The equation can be written as: The equation can be simplified by neglecting any potential energy changes and focusing on the effects of heat transfer (Q) and work (W). In this problem, kinetic energy changes are considered due to the differing velocities at the inlet and exit of the turbine. This equation helps us understand how the energy input and output affect the power developed by the turbine. Including all energy conversions ensures we have a comprehensive view of the system's energy balance. The ideal gas model is a simplified representation of the behavior of gases. It's particularly useful in thermodynamic calculations, such as those involving turbines, to estimate properties like pressure, volume, and temperature. The model assumes that:

• Gas molecules do not exert attractive or repulsive forces on each other.
• Gas molecules occupy negligible space compared to the container volume.

These assumptions allow us to use straightforward equations to relate different properties of the gas. When conducting energy balance calculations involving specific enthalpy and kinetic energy, treating nitrogen as an ideal gas simplifies our work. Step 3 of the solution requires us to calculate specific enthalpy. Specific enthalpy is a measure of the total energy of a gas and is given by the equation:

Using Specific Heat Capacity

where:

Because specific heat capacity for nitrogen is known, we can easily calculate the specific enthalpy at both the inlet and exit states of the turbine: Understanding specific enthalpy lets us quantify the internal energy content of the gas before and after it passes through the turbine. One crucial part of the energy balance in our problem is the kinetic energy change due to the differing velocities at the turbine's inlet and exit. Kinetic energy, which represents the energy due to motion, is given by the formula: In the solution, we converted the kinetic energy from m/s to kJ/kg to keep the units consistent with other parts of the energy equation. The differences in inlet and exit velocities (e.g., 60 m/s vs. 0.6 m/s) play a significant role in the overall energy transformation, impacting the calculated power developed by the turbine. Paying attention to both the kinetic and specific enthalpy changes provides a complete picture of the energy exchanges in the turbine and ensures our final power calculation is accurate.