Air Compressor Problem Analysis

This problem involves the application of the First Law of Thermodynamics for a steady-flow open system, specifically an air compressor. We need to determine the rate of work done on the air as it passes through the compressor. The steady-flow energy equation will be our primary tool, considering changes in enthalpy, kinetic energy, and heat transfer.

Given Parameters for the Air Compressor

Let's list the properties of the air at the inlet (state 1) and outlet (state 2) of the air compressor, along with the mass flow rate and heat transfer rate:

Since heat is absorbed by cooling water from the air, it means heat is rejected from the air (our system). Therefore, the heat transfer rate for the air compressor is negative: \(\dot{Q} = -100 \text{ kW}\).

Thermodynamic Principles Applied to Air Compressor

For a steady-flow system like an air compressor, the First Law of Thermodynamics (Steady-Flow Energy Equation) is given by:

\[ \dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{(V_2^2 - V_1^2)}{2} + g(z_2 - z_1) \right] \]

Where:

• \(\dot{Q}\) is the rate of heat transfer to the system.
• \(\dot{W}\) is the rate of work done by the system.
• \(\dot{m}\) is the mass flow rate.
• \(h\) is specific enthalpy.
• \(V\) is velocity.
• \(g\) is acceleration due to gravity.
• \(z\) is elevation.

For a horizontal air compressor, the change in potential energy is negligible, so \((z_2 - z_1) = 0\).

The equation simplifies to:

\[ \dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{(V_2^2 - V_1^2)}{2} \right] \]

Calculating Enthalpy Change for the Air

The specific enthalpy (\(h\)) is related to specific internal energy (\(u\)), pressure (\(P\)), and specific volume (\(v\)) by the equation: \(h = u + Pv\).

Therefore, the change in specific enthalpy is:

\[ (h_2 - h_1) = (u_2 - u_1) + (P_2 v_2 - P_1 v_1) \]

Let's calculate the \(Pv\) terms:

• For the inlet: \[ P_1 v_1 = 100 \text{ kPa} \times 1 \text{ m}^3/\text{kg} = 100 \text{ kPa} \cdot \text{m}^3/\text{kg} \] Since \(1 \text{ kPa} \cdot \text{m}^3 = 1 \text{ kJ}\), \[ P_1 v_1 = 100 \text{ kJ/kg} \]
• For the outlet: \[ P_2 v_2 = 800 \text{ kPa} \times 0.2 \text{ m}^3/\text{kg} = 160 \text{ kPa} \cdot \text{m}^3/\text{kg} \] \[ P_2 v_2 = 160 \text{ kJ/kg} \]

Now, substitute these values and the given internal energy change into the enthalpy change equation:

\[ (h_2 - h_1) = 100 \text{ kJ/kg} + (160 \text{ kJ/kg} - 100 \text{ kJ/kg}) \]

\[ (h_2 - h_1) = 100 \text{ kJ/kg} + 60 \text{ kJ/kg} = 160 \text{ kJ/kg} \]

Calculating Kinetic Energy Change for the Air

The change in specific kinetic energy is:

\[ \frac{(V_2^2 - V_1^2)}{2} = \frac{(6 \text{ m/s})^2 - (10 \text{ m/s})^2}{2} \]

\[ \frac{(V_2^2 - V_1^2)}{2} = \frac{36 \text{ m}^2/\text{s}^2 - 100 \text{ m}^2/\text{s}^2}{2} \]

\[ \frac{(V_2^2 - V_1^2)}{2} = \frac{-64 \text{ m}^2/\text{s}^2}{2} = -32 \text{ m}^2/\text{s}^2 \]

Since \(1 \text{ J/kg} = 1 \text{ m}^2/\text{s}^2\), we have \(-32 \text{ J/kg}\). To convert to kJ/kg:

\[ \frac{(V_2^2 - V_1^2)}{2} = -32 \text{ J/kg} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = -0.032 \text{ kJ/kg} \]

Determining the Rate of Work Done by the Air Compressor

Now, we substitute all calculated values and given parameters into the simplified steady-flow energy equation:

\[ \dot{Q} - \dot{W} = \dot{m} \left[ (h_2 - h_1) + \frac{(V_2^2 - V_1^2)}{2} \right] \]

Given: \(\dot{m} = 1 \text{ kg/s}\) and \(\dot{Q} = -100 \text{ kW}\).

\[ -100 \text{ kW} - \dot{W} = (1 \text{ kg/s}) \left[ 160 \text{ kJ/kg} + (-0.032 \text{ kJ/kg}) \right] \]

\[ -100 \text{ kW} - \dot{W} = (1 \text{ kg/s}) \left[ 159.968 \text{ kJ/kg} \right] \]

\[ -100 \text{ kW} - \dot{W} = 159.968 \text{ kW} \]

Now, solve for \(\dot{W}\):

\[ -\dot{W} = 159.968 \text{ kW} + 100 \text{ kW} \]

\[ -\dot{W} = 259.968 \text{ kW} \]

\[ \dot{W} = -259.968 \text{ kW} \]

The negative sign indicates that work is done on the air (the system), which is expected for an air compressor. The magnitude of the rate of work done is approximately 260 kW.

Comparing this result with the given options, 260 kW matches one of the choices.