Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
Answer:
I think Microsoft Corporation
All of the above. Answer.
Answer:
Explanation:
solution:
from this below equation (1)
σ/2εo
...........(1)
we obtain:
