The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.
<h3>What is the rate of heat transfer?</h3>
Rate of heat transfer is the power rating of the machine.
Work done and changes in potential and kinetic energy are neglected since it is a steady state process.
The specific heat in terms of specific heat capacity and temperature change is given as:
The rate of heat transfer, is then determined as follows:
- Qout = flow rate × specific heat
Qout = 0.75 × 20.08 = 15.06 kW
Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.
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Explanation:
For true Strain:
step 1:
E true = Ln(1 + 0.5 ) = 0.40
Step 2:
E true = Ln(1 + 0.33 ) = 0.29
By single step process:
E true = Ln(1 + 1 ) = 0.69
total strain of step process = 0.40 + 0.29 = 0.69 units
SO TRUE STRAIN IS ADDITIVE.
Solution :
Given :
The number of blows is given as :
0 - 6 inch = 4 blows
6 - 12 inch = 6 blows
12 - 18 inch = 6 blows
The vertical effective stress 
Now,
corrected N - value of overburden
effective stress at level of test
0 - 6 inch, 
= 9.86
6 - 12 inch, 
= 14.8
12 - 18 inch,
= 14.8
= 13.14
= 13
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
