Consider air as an ideal gas with Cp=1000 J/(kg.K), R=287 J/(kg.K) and γ≈1.4. It is compressed from 1 bar, 300K to 10 bar using a compressor. The compressor can be considered adiabatic but not reversible. The flow rate is 2 kg/s, and the entropy production rate is 200 W/K.

1.What is the change in specific entropy of the gas (units of J/(kg.K)) as it moves from inlet to exit?

2.If the process was reversible apart from being adiabatic, what would have been the final Temperature (in kelvin) of the gas at the exit?

3.What is the actual temperature of the gas at the exit (in kelvin)?

4.What is the power consumed by the compressor (in kW)? Do not enter the sign.

5.What is the isentropic efficiency of the compressor? Enter the value as a fraction.

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Follow the steps to get a numerical answer.

To answer these questions, we need to use thermodynamic equations and principles. Let's go step by step:

1. The change in specific entropy (Δs) can be obtained using the following equation:

Δs = Cp * ln(T2/T1) - R * ln(P2/P1)

where Cp is the specific heat capacity at constant pressure, T1 and T2 are the initial and final temperatures respectively, P1 and P2 are the initial and final pressures respectively, and R is the specific gas constant.

Given: Cp = 1000 J/(kg.K), R = 287 J/(kg.K), T1 = 300 K, T2 (final temperature) is unknown, P1 = 1 bar, P2 = 10 bar.

To find Δs, we need to know the final temperature (T2). Let's move on to question 2 to determine T2.

2. If the process was reversible, the adiabatic process can be assumed to be isentropic. In an isentropic process, the entropy remains constant. Therefore, we can use the equation:

Δs = 0 = Cp * ln(T2/T1) - R * ln(P2/P1)

Rearranging the equation gives:

ln(T2/T1) = R * ln(P2/P1) / Cp

Substituting the given values into the equation, we can solve for T2:

T2 = T1 * exp(R * ln(P2/P1) / Cp)

3. The actual temperature of the gas at the exit can be found by considering the entropy production rate (ηs) given as 200 W/K. The entropy production rate is defined as:

ηs = Δs * mdot

where Δs is the change in specific entropy and mdot is the mass flow rate.

Given: mdot = 2 kg/s, ηs = 200 W/K

We can rearrange the equation to solve for Δs:

Δs = ηs / mdot

Substituting the given values, we can find Δs.

4. The power consumed by the compressor (Pcomp) can be calculated using the equation:

Pcomp = mdot * Δh

where Δh is the change in enthalpy.

Since we know that the process is adiabatic, there is no heat transfer (Q = 0). Therefore, the enthalpy change can be calculated using the equation:

Δh = Cp * (T2 - T1)

Substituting the given values, we can find Pcomp.

5. The isentropic efficiency of the compressor (ηisentropic) can be calculated using the equation:

ηisentropic = (h2s - h1) / (h2 - h1)

Where h2s is the enthalpy at the exit assuming an isentropic process, h2 is the actual enthalpy at the exit, and h1 is the enthalpy at the inlet.

Since we have already found Δh (in question 4), we can use it to calculate h2.

Let's solve the equations to find the answers to each question.