Calculate power in air compression.

[MrNathan]

Homework Statement

In an air compressor air flows steadily at a rate of 15 kg per minute. The air
enters the compressor at 5 m/s with a pressure of 1 bar and a specific volume
of 0.5 m3/kg. The air leaves the compressor at 7.5 m/s with a pressure of 7
bar and a specific volume of 0.15 m3/kg. The internal energy of air leaving the
compressor is 165 kJ/kg greater than that of the air entering the compressor.
The cooling water in the compressor jacket absorbs heat from the air at a rate
of 125 kJ/s. Calculate the power required to drive the compressor.

Homework Equations

Q+W=m (mass flow rate)(u2+p2V2+0.5c2^2)-(u1+p1v1+0.5c1^2)

The Attempt at a Solution

Not sure if this is the right equation?

the 1s and 2s should be subtext. after putting i the variables, i got a negative number. should the mass flow rate be a factor of both the brackets or just the first?

thanks

 

[mark726]

for your question and for providing all the relevant information. Here is my attempt at solving the problem:

First, let's define all the variables given in the problem:

m = 15 kg/min (mass flow rate)
v1 = 5 m/s (inlet velocity)
p1 = 1 bar (inlet pressure)
u1 = 0.5 m3/kg (specific volume at inlet)
v2 = 7.5 m/s (outlet velocity)
p2 = 7 bar (outlet pressure)
u2 = 0.15 m3/kg (specific volume at outlet)
Δu = 165 kJ/kg (change in internal energy)
Q = 125 kJ/s (heat absorbed by cooling water)

To calculate the power required to drive the compressor, we need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added, and W is the work done.

We can rearrange this equation to solve for W:

W = Q - ΔU

Now, let's plug in the values given in the problem:

W = (125 kJ/s) - (165 kJ/kg * 15 kg/min) = 125 kJ/s - 2475 kJ/min = -2350 kJ/min

Note that the units for Q and ΔU are both in kJ, so we can simply subtract them to get the work done in kJ.

Finally, to convert this to power, we need to divide by the time interval. Since we are given the mass flow rate in kg/min, we can convert the time interval to minutes as well:

P = (-2350 kJ/min) / (1 min) = -2350 kJ/min * (1 kW / 60 kJ/min) = -39.17 kW

The negative sign indicates that the work is being done on the compressor, rather than being done by the compressor. This means that the compressor is using 39.17 kW of power to compress the air.

I hope this helps! Let me know if you have any further questions.