A specialized heavy-duty equipment draws 550 A current from the power supply at
ID: 2079361 • Letter: A
Question
A specialized heavy-duty equipment draws 550 A current from the power supply at a steady rate. The power cables are made of 1/0 AWG wire, which consists of a 8.25-mm-diameter specially-designed composite wire-strand inside a 2-mm-thick PVC insulation. The resistivity of the wire is 0.0003 /m. The composite wire-strand has thermal conductivity of 70 W/m_C. PVC insulation has thermal conductivity of 0.2 W/m_C, and starts to decompose at 140_C. A forced convection system is employed to cool the power cable. The convection heat transfer coeffcient between the insulation and cooling uid is 80 W/m2_C, and ambient temperature of the cooling fluid is 20_C. Clearly specify all boundary conditions for both parts of the problem. (a) Determine if the temperature of the insulation is safe (i.e., PVC does not decompose). (b) Find out the temperature at the center of the wire.
Explanation / Answer
At thermal equilibrium condition the rate of heat transfer will be constant and that is equal heat loss of power cable
Q/t = I2*R=5502 *0.0003=90.75 W/m
Outer temperature of insulation can be found by using below relation
Q/t=convection heat transfer coefficient X area of outer insulation x(Tinsout-20)
90.75=80*(pi*(8.25+4)*10^-3*1)* (Tinsout-20)
By solving above equation
We get Tinsout=49.5 0C
inner temperature of insulation can be found by using below relation
Q/t=(thermal conductivity X area of inner insulation x(Tinsin-49.5))/(thickness of insulation)
90.75=0.2*(pi*(8.25)*10^-3*1)* (Tinsin-49.5)/(2*10^-3)
By solving above equation
We get Tinsin=84.5 0C
Which is less than the starts to decompose limit so insulation is safe .
b. center temperature of can be found as below
Q/t=(thermal conductivity X area of outer cunductor x(Tinnercond-159.5))/(radius of conductor)
90.75=70*(pi*(8.25)*10^-3*1)* (Tinsin-84.5)/(4.125*10^-3)
By solving above equation
We get Tinnercond-=84.7 0C
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