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.
1) from 2nd law:
S_gen/m = 200/2 = 100 J/kg-K
s_gen = 0.10 kJ/kg-K
s_gen=s2-s1 = .10kJ/kg-K
look in table for ideal gas for air to find s1
=> s1 = 6.86926 kJ/kg-k
s_gen + s1 = s2
=> 0.10 + 6.86926 = s_gen
3) actual temperature of gas at exit:
apply change in entropy law for ideal gas:
formula can be found in book:
s2-s1=C_p*ln(T2/T1) - R*ln(P2/P1)
all values are given except T2, solve for T2 (be careful about units)
2) if reversible + adiabatic:
0=C_p*ln(T2/T1) - R*ln(P2/P1)
find T2
4) Apply first law w/out Q, potential and kinetic energy
dQ-dW=mC_p*deltaT
W = (2kg/s) * (0.10J/kg-K)*(T2-300)
T2 here is same as T2 found in number 3
5) isentropic efficiency of compressor:
n= W_isen / W_actual
Donho, can you please send the final answer
day after the submission date
To solve these questions, we need to use the following equations:
1. The change in specific entropy (Δs) of a gas can be calculated using the 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, and P1 and P2 are the initial and final pressures.
2. If the process is reversible, we can use the isentropic relation to find the final temperature (T2s): T2s = T1 * (P2/P1)^((γ-1)/γ), where γ is the adiabatic index.
3. Since in this case the process is adiabatic but not reversible, we can't directly use the above equation. Instead, we'll use the formula: T2 = T1 * (P2/P1)^((γ-1)/γ), where γ is still the adiabatic index.
4. The power consumed by the compressor can be calculated using the equation: Power = m_dot * Cp * (T2 - T1), where m_dot is the mass flow rate.
5. The isentropic efficiency (η) of the compressor is given by: η = (actual work done on the gas) / (work done on the gas if the process was reversible).
Let's solve each question one by one:
1. To find the change in specific entropy (Δs):
Δs = Cp * ln(T2/T1) - R * ln(P2/P1)
= 1000 * ln(T2/300) - 287 * ln(10/1)
2. If the process was reversible apart from being adiabatic, the final temperature (T2s) can be found using the isentropic relation:
T2s = 300 * (10/1)^((1.4-1)/1.4)
3. The actual temperature of the gas at the exit (T2) can be found using the adiabatic relation:
T2 = 300 * (10/1)^((1.4-1)/1.4)
4. The power consumed by the compressor can be calculated using the equation:
Power = 2 * 1000 * (T2 - 300)
5. The isentropic efficiency (η) of the compressor can be found using the formula:
η = (actual work done on the gas) / (work done on the gas if the process was reversible)
Now, plug in the values provided into these equations to get the respective answers.