Question

Water enters a 0.004 m diameter and 21 m long tube at 45ºC with a velocity of 0.5 m/s. What is the required length of the tube, in m, in order for the water to exit the tube at 25ºC if the tube is maintained at a constant temperature of 5ºC?

          Water enters a 0.004 m diameter and 21 m long tube at 45ºC with a velocity of 0.5 m/s. What is the required length of the tube, in m, in order for the water to exit the tube at 25ºC if the tube is maintained at a constant temperature of 5ºC?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Water enters a 0.004 m diameter and 21 m long tube at 45ºC with a velocity of 0.5 m/s. What is the required length of the tube, in m, in order for the water to exit the tube at 25ºC if the tube is maintained at a constant temperature of 5ºC?
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Transcript

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00:01 Hi, in this question we are given for air with the saturation temperature is 298 kelvin.
00:07 So properties of air at this temperature is v is 15 .71 into 10 power minus 6 meter square per second.
00:15 K value is 0 .0261 watt per meter kelvin and we have prandtl number as 0 .71.
00:23 So first we have to calculate reynolds number which is equal to v into divided by v.
00:28 So this on substitution we have 20 into 0 .05 divided by 15 .71 into 10 power minus 6 which is equal to 63653 .72.
00:40 So here this is a turbulent flow.
00:44 So this is a turbulent flow.
00:46 Therefore we have nusselt number is given by it is 0 .023, reynolds number power 0 .8, prandtl number power.
00:58 So prandtl number power 0 .4.
01:00 So we know the value of reynolds number and prandtl number.
01:04 So from that we have nusselt number is so nusselt number can be given by h bar.
01:10 So let us write it as it is h bar into d divided by k which is equal to so this value is about so it is 0 .023 into 63653 .72 whole power 0 .8 into 0 .71 whole power.
01:31 So prandtl number power 0 .4 divided so let it be there.
01:37 So we have to find h bar.
01:38 So h bar will be equal to so this is 0 .023 into 63653 .72 power 0 .8 into 0 .71 whole power 0 .4 divided by diameter is given as 0 .05.
01:55 So 0 .05 into k value is given which is 0 .0261.
02:03 So from this we have heat transfer is equal to 72 .94 watt per meter square kelvin.
02:12 So in part a we have to calculate the length of the tube.
02:15 So we have to use heat equation.
02:17 So we have t s minus so t s minus t mo this is the output temperature divided by t s minus t mi is equal to e power minus pl into h bar divided by m into so m into so m into specific heat of water...
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