Answer:
work done= 2.12 kJ
Explanation:
Given
N=2500 rpm
T=4.5 N.m
Period ,t= 30 s
P=
P=70,685W
P=70.685 KW
power=
work done = power * time
= 70.685*30=2120.55J
= 2.12 kJ
Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each
1 answer:
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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
<u>a) Calculate for the temp of sheet metal when it leaves the oil bath </u>
<em>first step : find the Biot number </em>
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 )
<em>next : find the time constant </em>
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
<em>Determine the elapsed time </em>
T = L / V = 9/20 = 0.45 min
∴<u> temp of sheet metal when it leaves the oil bath </u>
= (T(t) - 45 ) / (820 - 45) = e^-(0.10092 * 27 )
T∞ = 45°C
Ti = 820°C
hence : T(t) = 95.80°C
<u>b) Calculate the required rate of heat removal form the oil </u>
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
