For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p
1 answer:
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Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as
W=2 MW
For Carnot heat pump
2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C
Answer:
The heater load =35 KJ/kg
Explanation:
Given that
At initial condition
Temperature= 15°C
RH=80%
At final condition
Temperature= 50°C
We know that in sensible heating process humidity ratio remain constant.
Now from chart
At temperature= 15°C and RH=80%
At temperature= 50°C
The heater load = 73 - 38 KJ/kg
The heater load =35 KJ/kg
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu