Answer:
Q = 424523.22 kw
Explanation:

k = 48.9 W/m - K
c = 0.115 KJ/kg- K


T_∞ = 35 degree celcius
velocity of air stream = 15 m/s
D = 40 cm
L = 200 cm
mass flow rate
 




solving for h

h = 675.6 kw/m^2K

Q = 675.6*2.513*(285-35)
Q = 424523.22 kw
 
        
             
        
        
        
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:
Jesus is always the answer
 
        
             
        
        
        
Answer: mets 
Explanation: meets are good