Question

A compressor operating at steady state takes in \(45 \mathrm{~kg} / \mathrm{min}\) of methane gas \(\left(\mathrm{CH}_{4}\right.\) ) at 1 bar, \(25^{\circ} \mathrm{C}, 15 \mathrm{~m} / \mathrm{s}\), and compresses it with negligible heat transfer to 2 bar, \(90 \mathrm{~m} / \mathrm{s}\) at the exit. The power input to the compressor is \(110 \mathrm{~kW}\). Potential energy effects are negligible. Using the ideal gas model, determine the temperature of the gas at the exit, in \(\mathrm{K}\).

Short answer

The temperature of the gas at the exit is approximately 362.13 K.

Step-by-step solution

Identify the Given Data

Extract and list all known quantities from the problem statement:- Mass flow rate, \(\frac{dm}{dt} = 45 \text{ kg/min} = 0.75 \text{ kg/s}\)- Inlet pressure, \(P_1 = 1 \text{ bar} = 100 \text{ kPa} \)- Inlet temperature, \(T_1 = 25^\text{C} = 298 \text{ K} \)- Inlet velocity, \(v_1 = 15 \text{ m/s} \)- Outlet pressure, \(P_2 = 2 \text{ bar} = 200 \text{ kPa} \)- Outlet velocity, \(v_2 = 90 \text{ m/s} \)- Power input, \(W_{\text{in}} = 110 \text{ kW} \)

Use the Steady-Flow Energy Equation

Apply the Steady-Flow Energy Equation (SFEE) for an ideal gas considering the given data: \[ \frac{dE_{\text{system}}}{dt} = \frac{dE_{\text{in}}}{dt} - \frac{dE_{\text{out}}}{dt} + \frac{dQ}{dt} + \frac{dW}{dt} \] For this problem, heat transfer (Q̇) and potential energy changes are negligible. For steady-state conditions, \[ W_{\text{in}} = \frac{dm}{dt} \times \bigg[ \bigg( h_2 + \frac{v_2^2}{2} \bigg) - \bigg( h_1 + \frac{v_1^2}{2} \bigg) \bigg] \]where \( h = c_p T \).

Solve for the Outlet Enthalpy

First, rewrite the SFEE to solve for the outlet enthalpy, \( h_2 \):\[ W_{\text{in}} = \frac{dm}{dt} \times \bigg[ c_p T_2 + \frac{v_2^2}{2} - c_p T_1 - \frac{v_1^2}{2} \bigg] \] Rearrange to isolate \( T_2 \):\[ T_2 = \frac{ \bigg[ W_{\text{in}} + \frac{dm}{dt} \times \bigg( c_p T_1 + \frac{v_1^2}{2} \bigg) \bigg] - \frac{dm}{dt} \times \frac{v_2^2}{2} }{ \frac{dm}{dt} \times c_p } \]Specify \( c_p \) for methane, approximately \(2.22 \text{ kJ/kg.K} \)

Substitute Known Values

Insert the known values into the equation:\[ T_2 = \frac{ \bigg[ 110 \text{ kW} + (0.75 \text{ kg/s}) \times \bigg( (2.22 \text{ kJ/kg.K}) \times 298 \text{ K} + \frac{(15 \text{ m/s})^2}{2 \times 1000} \bigg) \bigg] - (0.75 \text{ kg/s}) \times \frac{(90 \text{ m/s})^2}{2 \times 1000} }{ (0.75 \text{ kg/s}) \times (2.22 \text{ kJ/kg.K}) } \]

Calculate the Outlet Temperature

After solving the above equation, first calculate intermediary terms:\[ \frac{v_1^2}{2 \times 1000} = \frac{(15)^2}{2000} = 0.1125 \text{ kJ/kg} \]Insert in the equation:\[ T_2 = \frac{ 110 \text{ kW} + 0.75 \times (661.56 + 0.1125) - 0.75 \times 4.05 }{ 0.75 \times 2.22 } \] Perform calculations step by step for clarity.

Perform Final Calculations

Calculate each term: \( (661.56 + 0.1125) = 661.6725 \text{ kJ/kg} \); \( 0.75 \times 661.6725 = 496.254 \text{ kJ/s} \); \( 0.75 \times 4.05 = 3.0375 \text{ kJ/s} \); \[ T_2 = \frac{ 110 + 496.254 - 3.0375 }{ 1.665 } = \frac{ 603.2165 }{ 1.665 } = 362.13 \text{ K} \] Checking units and calculated values.

Key concepts

Ideal Gas Model

The ideal gas model is a simplified way of understanding the behavior of gases under various conditions. This model assumes that gases consist of many small particles that move independently and occupy no volume themselves. The model also assumes no intermolecular forces act between these particles. This makes the calculations straightforward, as opposed to more complex models that consider intermolecular forces and molecular volume.
The behavior of an ideal gas can be described by the ideal gas law: \[ PV = nRT \] where:

• P is the pressure
• V is the volume
• n is the number of moles of gas
• R is the universal gas constant
• T is the temperature in Kelvin.

In the context of our problem, we use the ideal gas model to simplify the calculations involved in determining the temperature at the compressor's outlet.

Compressor Work

A compressor is a device that increases the pressure of a gas by reducing its volume. The energy needed for this operation is directly related to the work done on the gas. The Steady-Flow Energy Equation (SFEE) helps us analyze the energy changes in such processes:
\[ \frac{dE_{\text{system}}}{dt} = \frac{dE_{\text{in}}}{dt} - \frac{dE_{\text{out}}}{dt} + \frac{dQ}{dt} + \frac{dW}{dt} \]
For our problem, we consider a steady state where the heat transfer effect is negligible, simplifying our analysis to:
\[ W_{\text{in}} = \frac{dm}{dt} \times \bigg[ \bigg( h_2 + \frac{v_2^2}{2} \bigg) - \bigg( h_1 + \frac{v_1^2}{2} \bigg) \bigg] \]
where:

• W_in is the power input
• \frac{dm}{dt} is the mass flow rate
• h_1 and h_2 are the specific enthalpies at the inlet and outlet, respectively
• v_1 and v_2 are the gas velocities at the inlet and outlet.

By substituting the known quantities into this equation and solving it step-by-step, we can determine the output temperature of the gas.

Enthalpy Change

Enthalpy is a measure of the total energy in a thermodynamic system, including both internal energy and energy required to make room for it by displacing its surroundings. It is often represented as:

• h = c_p T

In the context of compressing an ideal gas, we are interested in the change in enthalpy (\text{Δ}h) which can be calculated when we know specific heat capacity (\text{c_p}) and the temperature difference:
C_p is a property that indicates the amount of heat added or removed per unit mass per degree temperature change at constant pressure.
Using the SFEE, we rearrange terms to isolate the outlet temperature (T_2):
\[ T_2 = \frac{ \bigg[ W_{\text{in}} + \frac{dm}{dt} \times \bigg( c_p T_1 + \frac{v_1^2}{2} \bigg) \bigg] - \frac{dm}{dt} \times \frac{v_2^2}{2} }{ \frac{dm}{dt} \times c_p } \]
where:

• W_in is the power input
• \frac{dm}{dt} is the mass flow rate
• c_p is the specific heat capacity
• T_1 and T_2 are the inlet and outlet temperatures respectively
• v_1 and v_2 are the inlet and outlet velocities.

By dealing with the intermediate calculations one by one, we arrive at the output temperature T_2, encapsulating the change in enthalpy.