Free Air Delivered and Relative humidity

[Keeeen]

Homework Statement

Air is drawn into a compressor at normal temperature and pressure (NTP) and compressed to a pressure of 6 bar gauge. After compression the air is delivered at 1.2m^3/min and cooled to a temperature of 30C, at which point condensate is collected at the rate of 2 litres per hour. Estimate the FAD (NTP) of the compressor, and the relative humidity of the air entering the compressor.

Homework Equations

P1 V1 = P2 V2
V1 = P2 V2 / P1

The Attempt at a Solution

P1 = 1.013 bar
V1 =
T1 = 15C
P2 = 6 bar
V2 = 1.2m^3/min
T2 = 30C

V1 = 6 * 1.2 / 1.013
V1 = 7.12 m^3/min

Does this look correct so far? I'm struggling abit with calculating the relative humidity if anyone can help

EDIT -
Noticed I haven't taken the temperature into account so after recalculating with temperature -
V1 = 6 * 1.2 * 30 / 1.013 * 15
V1 = 14.22

And this is my attempt at calculating the relative humidity -

inlet volume = 14.22 * 60 = 853m^3/hour

dew point @ 15C @ 1.013bar on the chart = 12.5g

inlet saturation quantity = 12.5 * 853
= 10663

outlet volume = 1.2 * 60 = 72m^3/hour

dew point @ 30C @ 6bar on the chart = 4g

outlet saturation quantity = 4 * 72
=288g/hour

2L = 2000g
absolute humidity = 288 + 2000
2288g/hour

Relative humidity = 2288/10665 * 100
= 21.45%

how does this look for an answer? I was thinking I might have made a mistake at the calculation of the FAD which would have thrown off the calculation of the relative humidity?

 

[KataKoniK]

Your calculation for the FAD (free air delivery) looks correct. However, your calculation for the relative humidity is not quite accurate.

To calculate the relative humidity, you need to first find the actual water vapor content in the inlet air. This can be done using the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Using the given values, we can calculate the number of moles of air in the inlet:

n = (P1*V1)/(RT1)
= (1.013*14.22)/(0.082*288)
= 0.5922 moles

Next, we can calculate the partial pressure of water vapor in the inlet air using Dalton's law of partial pressures:

Pv = nRTv
= (0.5922*0.082*288)/(14.22/1000)
= 0.0118 bar

Now, we can calculate the actual water vapor content in the inlet air using the ideal gas law again:

n = (Pv*V1)/(RT1)
= (0.0118*14.22)/(0.082*288)
= 0.0019 moles

Finally, we can calculate the relative humidity using the formula:

Relative humidity = (actual water vapor content/inlet saturation quantity) * 100
= (0.0019/12.5) * 100
= 0.0152 * 100
= 1.52%

So the relative humidity of the inlet air is approximately 1.52%.

Note: The calculation of the inlet saturation quantity is incorrect in your attempt. It should be calculated as follows:

Inlet saturation quantity = (12.5/1000) * 14.22 * 60
= 0.008928 * 853
= 7.615 g/hour

This value is used in the calculation of the relative humidity.