Question

Taking the convection heat transfer coefficient on both sides of the plate to be 860 W/m2 ·K, deter- mine the temperature of the sheet metal when it leaves the oil bath. Also, determine the required rate of heat removal from the oil to keep its temperature constant at 45°C.

Answer 1

Answer:

Hello your question is incomplete attached below is the complete question

answer :

a) 95.80°C

b) 8.23 MW

Explanation:

Convection heat transfer coefficient = 860 W/m^2 . k

a) Calculate for the temp of sheet metal when it leaves the oil bath

first step : find the Biot number

Bi = hLc / K  ------- ( 1 )

where : h = 860 W/m^2 , Lc = 0.0025 m ,  K = 60.5 W/m°C

Input values into equation 1 above

Bi = 0.036 which is < 1  ( hence lumped parameter analysis can be applied )

next : find the time constant

t ( time constant ) = h / P*Cp *Lc  --------- ( 2 )

where : p = 7854 kg/m^3 , Lc = 0.0025 m , h = 860 W/m^2, Cp = 434 J/kg°C

Input values into equation 2 above

t ( time constant ) = 0.10092 s^-1

Determine the elapsed time

T = L / V = 9/20 = 0.45 min

∴   temp of sheet metal when it leaves the oil bath

= (T(t) - 45 ) / (820 - 45)  = e^-(0.10092 * 27 )

T∞ =  45°C

Ti = 820°C

hence : T(t) = 95.80°C

b) Calculate the required rate of heat removal form the oil

Q = mCp ( Ti - T(t) ) ------------ ( 3 )

m = ( 7854 *2 * 0.005 * 20 ) = 26.173 kg/s

Cp = 434 J/kg°C

Ti =  820°C

T(t) = 95.80°C

Input values into equation 3 above

Q = 8.23 MW