Answer:
0.12 kW
Explanation:
Given that
The flow rate of air (V)=0.3 ft³/s
V=0.008 m³/s
Pressure, P=14.7 psia
P=1.013529 atm=101.325 kPa
Inlet temperature = 70° F=294.261 K
Exit temperature = 83° F=301.483 K
We know that , specific heat capacity of the air
Cp=1.005 kJ/kg.K
The mass flow rate of air is given as
[tex]\dot{m}=\dfrac{P\times V}{R\times T}\\\dot{m}=\dfrac{101.325\times 0.008}{0.287\times 294.261}\\\dot{m}= 0.0095\ kg/s[/tex]
By using energy conservation
[tex]Electric\ power =\dot{m}\times C_p\times (T_2-T_1)\\Electric\ power =0.0095\times 1.005\times (83-70)=0.12\ kW[/tex]
Therefore electric power dissipate by components will be 0.12 kW.
