Question

During a quenching process, a 0.01 diameter and 0.2m long cylindrical rod made of 1080 Steel is heated to 750°C and then placed in water at 80°C. Considering a convection heat transfer of 250 W/m^2·K, how much time will the rod be at 300°C?

          During a quenching process, a 0.01 diameter and 0.2m long cylindrical rod made of 1080 Steel is heated to 750°C and then placed in water at 80°C. Considering a convection heat transfer of 250 W/m^2·K, how much time will the rod be at 300°C?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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During a quenching process, a 0.01 diameter and 0.2m long cylindrical rod made of 1080 Steel is heated to 750°C and then placed in water at 80°C. Considering a convection heat transfer of 250 W/m^2·K, how much time will the rod be at 300°C?
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Transcript

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00:01 Hi there, so for this problem we have a 0 .6 centimeters diameter for this so let's label this as capital d.
00:12 0 .006 this will be in meters and this at a temperature that we're going to call just simply the temperature t sub zero that is equal to 38 and it is immersed in a liquid that will be the temperature infinity is equal to 93.
00:42 And what we want to determine is the time required for the road to warm to a temperature that is equal to 88.
00:57 And we are given some information for this.
01:01 And that will be that the thermal conductivity is equal to 43 in units of waps, per meter per kelvin and also the specific heat that is 473 joules per kilogram per kelvin and the density that is 7 ,801 in units of kilograms per cubic meter and finally a thermof diffusivity that is 1 .172 times 10 to the minus 5 meters square per second.
01:56 So the first thing that we need to determine is the biot number.
02:02 So we can check if the terminal, if the internal resistance is negligible.
02:07 So that biot number is just the product between hc, which is the value that we are given for this.
02:14 That is 110.
02:17 And this times the diameter, this divided by four times the thermal conductivity.
02:24 So that will be 110 times the diameter that is 0 .006.
02:30 That it meters, this divided by four times thermal conductivity, that is 43.
02:37 Then from this, we obtain a value of 0 .0038.
02:46 That is much less than 0 .1.
02:51 And so with this, we can conclude that internal resistance of the rod is negligible.
02:59 So the temperature time history of the rot can be calculated through the following equation.
03:04 That is the temperature minus the temperature infinity.
03:07 This divided by the temperature sub -0 minus the temperature infinity is equal to the exponential of hc times, times the area per this divided by the specific heat, times the density, times the volume, times the at the time in here...
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