In a quench hardening process, steal rods ( rho = 7832 kg/m3, C rho = 434 kg mid

Question

In a quench hardening process, steal rods ( rho = 7832 kg/m3, C rho = 434 kg middot K, k = 63.9 w/m middot K) are heated in a furnace to 850 degree C and then cooled in a water bath to 95 degree C. The water bath has a uniform temperature of 40 degree C and convective heat transfer coefficient of 450 W/m middot K. if the steel rods have a diameter of 50 mm and length of 2 m determine the time required to cool the rod from 850 degree C to 95 degree C. rho = 7832 kg/m3, C rho = 434 kg middot K, k = 63.9 w/m middot K h = 450 W/m2 middot K T infinity = 40 degree C

Explanation / Answer

Biot Number, Bi = h(V/A)/k

Here, V = Volume, A = surface area, k = Thermal conductivity, h = Heat transfer coefficient.

We have V = d2L/4, and A dL.

So V/A = d/4 = 50/4 = 12.5 mm = 0.0125 m

So, Bi = 450*(0.0125)/63.9 = 0.088

Since Bi << 1, we can use lumped mass assumption.

Temperature after time t is given by

T = Tatm + (Tinit - Tatm) * e(-hAt/CpV)

Putting T = 95 deg C, Tatm = 40 deg C and Tinit = 850 deg C

95 = 40 + (850 - 40) * e(-450*t/(7832*434*0.0125))

Solving this gives, t = 253.96 sec 4 min 14 sec

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