3Px=y−y2p2first order higher dgree
1 answer:
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Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
a) 180 m³/s
b) 213.4 kg/s
Explanation:
= 1 m²
= 100 kPa
= 180 m/s
Flow rate
Volumetric flow rate = 180 m³/s
Mass flow rate
Mass flow rate = 213.4 kg/s