A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast
ing and oil can be modeled as incompressible with specific heats 0.10 Btu/lbm o R, and 0.45 Btu/lbm o R, respectively. For the iron casting and oil as the system, determine: a) The final equilibrium temperature (o F) b) The total entropy change for this process (Btu/ o R) (Hint: Total entropy change is the sum of entropy change of iron casting and oil.)
1 answer:
You might be interested in
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 )