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A two-stage compressor takes in air at 1.1 bar and discharge at 20 bar. For max efficiency, the pressure at the end of the I stage is
- a)10.55 bar
- b)7.33 bar
- c)5.5 bar
- d)4.7 bar
Correct answer is option 'D'. Can you explain this answer?
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A two-stage compressor takes in air at 1.1 bar and discharge at 20 ba...
In a two-state Compressor
Max - efficiency is obtained When 
P1= Suction pressure
P2= Intermediate pressure
P1= Delivery pressure
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A two-stage compressor takes in air at 1.1 bar and discharge at 20 ba...
To find the pressure at the end of the I stage of the compressor, we can use the equation for the isentropic compression process. The equation is:
P2 = P1 * (V1/V2)^(γ)
Where:
P1 = initial pressure = 1.1 bar
P2 = final pressure = 20 bar
V1 = initial volume
V2 = final volume
γ = specific heat ratio
The specific heat ratio, γ, for air is approximately 1.4.
We can assume that the specific heat ratio remains constant throughout the compression process, which allows us to rewrite the equation as:
P2/P1 = (V1/V2)^(γ)
To find the pressure at the end of the I stage, we need to find the ratio (V1/V2)^(γ) by rearranging the equation:
(V1/V2)^(γ) = P2/P1
Taking the logarithm of both sides of the equation, we have:
γ * log(V1/V2) = log(P2/P1)
Now, we can solve for (V1/V2)^(γ) using the logarithmic identity:
log(a^b) = b * log(a)
γ * log(V1/V2) = log(P2/P1)
log(V1/V2) = log(P2/P1) / γ
(V1/V2) = 10^(log(P2/P1) / γ)
Now, we substitute the given values into the equation:
(V1/V2) = 10^(log(20/1.1) / 1.4)
(V1/V2) = 10^(1.6812 / 1.4)
(V1/V2) = 10^(1.2009)
(V1/V2) = 15.102
Now, we can rearrange the equation to solve for the final pressure of the I stage:
P2 = P1 * (V1/V2)^(γ)
P2 = 1.1 * (15.102)^(1.4)
P2 = 1.1 * 15.102^(1.4)
P2 ≈ 4.7 bar
Therefore, the pressure at the end of the I stage of the compressor is approximately 4.7 bar. Answer choice D is correct.
P2 = P1 * (V1/V2)^(γ)
Where:
P1 = initial pressure = 1.1 bar
P2 = final pressure = 20 bar
V1 = initial volume
V2 = final volume
γ = specific heat ratio
The specific heat ratio, γ, for air is approximately 1.4.
We can assume that the specific heat ratio remains constant throughout the compression process, which allows us to rewrite the equation as:
P2/P1 = (V1/V2)^(γ)
To find the pressure at the end of the I stage, we need to find the ratio (V1/V2)^(γ) by rearranging the equation:
(V1/V2)^(γ) = P2/P1
Taking the logarithm of both sides of the equation, we have:
γ * log(V1/V2) = log(P2/P1)
Now, we can solve for (V1/V2)^(γ) using the logarithmic identity:
log(a^b) = b * log(a)
γ * log(V1/V2) = log(P2/P1)
log(V1/V2) = log(P2/P1) / γ
(V1/V2) = 10^(log(P2/P1) / γ)
Now, we substitute the given values into the equation:
(V1/V2) = 10^(log(20/1.1) / 1.4)
(V1/V2) = 10^(1.6812 / 1.4)
(V1/V2) = 10^(1.2009)
(V1/V2) = 15.102
Now, we can rearrange the equation to solve for the final pressure of the I stage:
P2 = P1 * (V1/V2)^(γ)
P2 = 1.1 * (15.102)^(1.4)
P2 = 1.1 * 15.102^(1.4)
P2 ≈ 4.7 bar
Therefore, the pressure at the end of the I stage of the compressor is approximately 4.7 bar. Answer choice D is correct.
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A two-stage compressor takes in air at 1.1 bar and discharge at 20 bar. For max efficiency, the pressure at the end of the I stage isa)10.55 barb)7.33 barc)5.5 bard)4.7 barCorrect answer is option 'D'. Can you explain this answer?
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