Question

Air enters an approximately frictionless duct with \(V_{1}=70 \mathrm{m} / \mathrm{s}, T_{1}=600 \mathrm{K},\) and \(P_{1}=350\) kPa. Letting the exit temperature \(T_{2}\) vary from 600 to 5000 \(\mathrm{K},\) evaluate the entropy change at intervals of \(200 \mathrm{K},\) and plot the Rayleigh line on a \(T\) -s diagram.

Short answer

Question: Calculate the change in entropy of air inside a frictionless duct for different exit temperatures and plot the Rayleigh line on a T-s diagram. Given the initial velocity \(V_1 = 70~m/s\), initial temperature \(T_1 = 600~K\), initial pressure \(P_1 = 350~kPa\) and exit temperature range from \(600~K\) to \(5000~K\) with intervals of \(200~K\). Answer: To find the change in entropy, we used the isentropic relations and calculated \(\Delta s\) for each exit temperature in the given range and intervals. Then, we plotted the Rayleigh line on a T-s diagram to represent the relationship between exit temperature and entropy change.

Step-by-step solution

Understand the given data

The given data in the problem are: Initial velocity: \(V_1 = 70~m/s\) Initial temperature: \(T_1 = 600~K\) Initial pressure: \(P_1 = 350~kPa\) Exit temperature: \(T_2\) varying from \(600~K\) to \(5000~K\) with intervals of \(200~K\)

Calculate entropy change

To find the entropy change for each exit temperature \(T_2\) in the given range, we can use the following isentropic relation between temperature, pressure, and specific heat ratios: $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^\frac{\gamma - 1}{\gamma}$$ Where \(\gamma\) is the specific heat ratio, which for air is approximately \(1.4\). We can rearrange this equation to solve for \(P_2\) and substitute \(P_1\) and \(T_1\) values: $$P_2 = P_1 \left(\frac{T_2}{T_1}\right)^\frac{\gamma}{\gamma-1}$$ Now, we calculate the entropy change as follows: $$\Delta s = c_p \ln\frac{T_2}{T_1} - R \ln\frac{P_2}{P_1}$$ Where \(c_p\) is the specific heat at constant pressure and \(R\) is the specific gas constant. For air, \(c_p = 1.005~kJ/(kg~K)\) and \(R = 0.287~kJ/(kg~K)\). Now, calculate the entropy change \(\Delta s\) for each exit temperature \(T_2\) in the given range and intervals.

Plot the Rayleigh line

Using the calculated values of entropy change and exit temperature, plot the Rayleigh line on a T-s diagram. This plot represents the relation between the exit temperature \(T_2\) and entropy change \(\Delta s\). In conclusion, to solve this exercise, we have derived the isentropic relation to find the entropy change for different exit temperatures. We have calculated the entropy change for each given exit temperature in the specified range and intervals. Finally, we plotted the Rayleigh line on a T-s diagram to represent the relationship between exit temperature and entropy change.