Question

Methane at \(27^{\circ} \mathrm{C}, 10\) MPa enters a turbine operating at steady state, expands adiabatically through a \(5: 1\) pressure ratio, and exits at \(-48^{\circ} \mathrm{C}\). Kinetic and potential energy effects are negligible. If \(\bar{c}_{p b}=35 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}\), determine the work developed per \(\mathrm{kg}\) of methane flowing through the turbine. Compare with the value obtained using the ideal gas model.

Short answer

-163.57 kJ/kg. The real gas result closely matches the ideal gas result.

Step-by-step solution

- Identify Given Data

Given data: Methane enters the turbine at \(27^{\text{°}} \text{C} = 300 \text{K}\) and \(10 \text{ MPa}\). It exits at \(-48^{\text{°}} \text{C} = 225 \text{K}\). The pressure ratio is \(5:1\). The specific heat at constant pressure \(\bar{c}_{p b}=35 \text{ kJ/kmol} \text{·} \text{K}\).

- Establish the Energy Equation

Since the process is adiabatic and neglecting kinetic and potential energy changes, apply the energy equation for an adiabatic process: \( \frac{dh}{dt} = \frac{{dW}}{dt} \), where \(h\) is the specific enthalpy.

- Calculate Enthalpy Change

For adiabatic expansion, the change in enthalpy \( \bar{h_2} - \bar{h_1} = -W_{turbine} \) is used. Using \(\bar{h} = \bar{c}_{p b} (T_2 - T_1) \) where \(T_1 = 300 \text{ K}\) and \(T_2 = 225 \text{ K}\): \( \bar{h_2} - \bar{h_1} = 35 \text{ kJ/kmol·K} (225 \text{ K} - 300 \text{ K}) = 35 \text{ kJ/kmol·K}(-75 \text{ K}) = -2625 \text{ kJ/kmol} \).

- Convert Work to per kg basis using molar mass

The molar mass of methane \(CH_4\) is 16.04 kg/kmol. Therefore, the work developed per kg of methane, \(w = \frac{{-2625 \text{kJ/kmol}}}{16.04 \text{kg/kmol}} = -163.57 \text{kJ/kg} \).

- Compare with Ideal Gas Model

For an ideal gas undergoing an adiabatic process, \(P_1 V_1^{\theta} = P_2 V_2^{\theta}\) where \(\theta = \frac{c_p}{c_v}\). Given the pressure ratio and temperature relation: \( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\theta-1}{\theta}}\). Using \(c_p = 2.23 R\) and \(c_v = 1.23 R\), \(\theta = 1.81\). Calculate \(T_2\): \[ \frac{225}{300} = \left(\frac{1}{5}\right)^{0.44} \to T_2 \approx 225 \text{K} \]. Therefore the ideal gas result should be close to real gas result.

Key concepts

adiabatic process

An adiabatic process is a crucial concept in thermodynamics where no heat is exchanged between the system and its surroundings. This means any change in internal energy of the system is due solely to work done by or on the system. For a turbine, this implies that all the energy extracted from the gas is converted into work.
In the exercise, methane undergoes adiabatic expansion, requiring us to use the energy equation for adiabatic processes. This can be written as:
\(\frac{dh}{dt} = \frac{dW}{dt}\)
where \(h\) refers to specific enthalpy and \(W\) is work. Importantly, if we ignore kinetic and potential energy changes, the only consideration remains the change in enthalpy as the gas expands.

enthalpy calculation

Calculating enthalpy is vital in this exercise. To find the work produced in the turbine, we need to evaluate the difference in specific enthalpy between the initial and final states of the methane gas. This is done using the equation:
\bar{h} = \bar{c}_{pb} (T_2 - T_1)\
Given \(T_1 = 300 \text{K}\) and \(T_2 = 225 \text{K}\), and with the specific heat at constant pressure \( \bar{c}_{pb} = 35 \text{kJ/kmol·K}\), we can calculate the change in enthalpy as follows:
\ \bar{h_2} - \bar{h_1} = \bar{c}_{pb} \times (T_2 - T_1) = 35 \text{kJ/kmol·K} \times (225 \text{K} - 300 \text{K}) = -2625 \text{kJ/kmol}\
This negative value indicates a decrease in enthalpy, implying the gas has done work on the surroundings. To get the work per kilogram of methane, we convert the result using the molar mass of methane \(16.04 \text{kg/kmol}\):
\ w = \frac{-2625 \text{kJ/kmol}}{16.04 \text{kg/kmol}} = -163.57 \text{kJ/kg}\

ideal gas comparison

For comparison with an ideal gas model, it's essential to understand how real gases like methane behave differently from ideal gases under certain conditions. Ideal gases follow the equation \(P V = nRT\) and have simplified relations during adiabatic processes. For an ideal gas, the relationship between pressure and temperature during adiabatic processes is:
\frac{T_2}{T_1} = \bigg(\frac{P_2}{P_1}\bigg)^{\frac{\theta-1}{\theta}}\
where \(\theta\) is the ratio of specific heats \( c_p/c_v \). Given \( c_p = 2.23 R \) and \( c_v = 1.23 R \), we have \(\theta = 1.81\). Considering the exercise's pressure ratio \(5:1\) and initial temperature \(300 \text{K}\), we approximate the final temperature \(T_2\) as:
\frac{225}{300} = \bigg(\frac{1}{5}\bigg)^{0.44}\rightarrow T_2 \approx 225 \text{K}\
The result indicates that the temperature drop aligns closely between the ideal and real gas assumptions, confirming that the calculated turbine work value of \(-163.57 \text{kJ/kg}\) is reliable.