Sure, here's a problem that incorporates advanced calculus within the context of chemical engineering thermodynamics. This problem involves the application of the first and second laws of thermodynamics to an open system with steady-state flow, which is a typical topic for third-year chemical engineering students.

Problem:

A steady-state, adiabatic flow process involves a gas entering a turbine operating at steady state. The gas enters the turbine at a pressure of 2 MPa, a temperature of 500 K, and a velocity of 100 m/s. The gas exits the turbine at a pressure of 100 kPa, a temperature of 300 K, and a velocity of 200 m/s. The mass flow rate of the gas is 2 kg/s. The specific heat capacity at constant pressure $C_p$ is 1.005 kJ/kg·K, and the specific heat capacity at constant volume $C_v$ is 0.718 kJ/kg·K. Assume the gas behaves as an ideal gas with $R = 0.287$ kJ/kg·K.

Determine the power output of the turbine.

Solution:

Step 1: Apply the First Law of Thermodynamics (Energy Balance)

For a steady-state process in an open system, the first law of thermodynamics can be expressed as:

$\dot{Q} - \dot{W} = \dot{m} (h_2 - h_1 + \frac{1}{2} (v_2^2 - v_1^2) + g(z_2 - z_1))$

Since the process is adiabatic (no heat transfer):

$\dot{Q} = 0$

And since we can neglect potential energy changes (elevation change $z_1 = z_2$):

$\dot{W} = \dot{m} (h_1 - h_2 + \frac{1}{2} (v_2^2 - v_1^2))$

Step 2: Calculate the Specific Enthalpies

The specific enthalpy for an ideal gas is given by:

$h = C_p T$

Therefore:

$h_1 = C_p T_1 = 1.005 \, \text{kJ/kg·K} \times 500 \, \text{K} = 502.5 \, \text{kJ/kg}$

$h_2 = C_p T_2 = 1.005 \, \text{kJ/kg·K} \times 300 \, \text{K} = 301.5 \, \text{kJ/kg}$

Step 3: Calculate the Kinetic Energy Change

The kinetic energy change per unit mass is:

$\frac{1}{2} (v_2^2 - v_1^2) = \frac{1}{2} (200^2 - 100^2) \, \text{m}^2/\text{s}^2 = \frac{1}{2} (40000 - 10000) \, \text{m}^2/\text{s}^2 = \frac{1}{2} \times 30000 \, \text{m}^2/\text{s}^2 = 15000 \, \text{m}^2/\text{s}^2$

Since 1 m$^2$/s$^2$ = 0.001 kJ/kg,

$\frac{1}{2} (v_2^2 - v_1^2) = 15 \, \text{kJ/kg}$

Step 4: Calculate the Power Output

Now we can plug the values into the energy balance equation:

$\dot{W} = \dot{m} \left( (h_1 - h_2) + \frac{1}{2} (v_2^2 - v_1^2) \right)$

$\dot{W} = 2 \, \text{kg/s} \left( (502.5 - 301.5) \, \text{kJ/kg} + 15 \, \text{kJ/kg} \right)$

$\dot{W} = 2 \, \text{kg/s} \left( 201 + 15 \right) \, \text{kJ/kg}$

$\dot{W} = 2 \, \text{kg/s} \times 216 \, \text{kJ/kg}$

$\dot{W} = 432 \, \text{kW}$

Conclusion:

The power output of the turbine is 432 kW.

Would you like any more details or have any questions? Here are 5 questions you might ask:

1. Can you explain the significance of the adiabatic assumption in this problem?
2. How would the solution change if the gas did not behave as an ideal gas?
3. What if the process was not at steady state? How would the approach change?
4. Can you provide an example of a similar problem with heat transfer?
5. How do the values of $C_p$ and $C_v$ affect the specific enthalpy calculations?

Tip: In thermodynamic problems, always start by writing down the relevant laws and assumptions to clarify the problem's constraints and scope.