Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure =
= 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ
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Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate = 
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.

Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is

As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%
Answer:A
Explanation:
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