00:01
Hello, first we will determine the mass flow rate of air.
00:08
So, we will use the ideal gas law which follows pv is equal to mrt.
00:15
Here p is equal to 101 .325 kpa, t is equal to 15 degree celsius plus 273 .15 which means 288 .15 kelvin, volume is 90 meter cube per hour and r is equal to 0 .287 kilojoule per kg kelvin.
00:47
Now, we solve for m, then m will be equal to p multiplied by v divided by r multiplied by t.
00:54
So, from here we will get m is equal to 101 .325 multiplied by 90 whole divided by 0 .287 multiplied by 288 .15 which means m is equal to 11 .13 kilogram per second.
01:15
This is mass flow rate of air.
01:18
Now we will determine the work done by compressor.
01:27
So it can be calculated using the equation w is equal to mrat2 minus rat1 whole divided by n minus mcp multiplied by t2 minus t1.
01:48
So, here t1 is equal to 288 .15 kelvin, t2 is equal to 735 kilo pascal divided by 101 .325 kilo pascal multiplied by 288 .15 kelvin which means t2 comes out to be 2075 .5 kelvin.
02:16
The value of cp will be 1 .005 kilojoule per kg kelvin.
02:25
Cp is specific heat capacity of air at constant pressure and n is equal to 1 .22 for polytropic efficiency.
02:37
Polytropic efficiency n is equal to 1 .22.
02:40
Now if we put the values then w is equal to 11 .13 multiplied by 0 .287 multiplied by 2075 .5 minus 0 .287 multiplied by 288 .15 whole divided by n which is 1 .22 minus 11 .13 multiplied by 1 .005 multiplied by 2075 .5 minus 288 .15.
03:26
So from here w is equal to 104 .6 kilo watt.
03:36
Now we will calculate the power needed to drive the compressor.
03:41
So power needed this should be equal to w divided by mechanical efficiency...