CALCULATE THE FOLLOWING:We performed the experiment to measureMATERIAL 1 - BRASSthe thermal conductivity of 2 materials (Brass& Steel) in the laboratory and measured thefollowing tabulated values:Calculation for BrassQuantitiesValuesMaterial 1- BRASSCalculatedPower(Q') W(Diameter = 25mm)Area of cross section (A)m2PowerDifference in TemperatureTemperature (°C)between two points (AT) °CDifference in distancebetween two points (Ax) mQ'1 2 3(W)Thermal conductivity ofbrass (kp) W/m°C45689.14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6MATERIAL 2 - STEELCalculation for SteelMaterial 2 - STEEL%3DQuantitiesValuesCalculated(Diameter = 25mm)%3DPower(Q') WPowerArea of cross section (A)m2Temperature (°C)Difference in Temperaturebetween two points (AT) °CQ'1|(W)Difference in distancebetween two points (Ax) m237814.2 88.6 87.5 85 33.9 33.6 32.4Thermal conductivity ofsteel (kg) W/m°C

Material 1 Brassd = 25mm, 0.025mQuantity calculatedPower ( θ') W = 14.6WArea of cross section A =…

We performed the experiment to measure CALCULATE THE FOLLOWING: MATERIAL 1 - BRASS the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Calculation for Brass Quantities Values Material 1- BRASS Calculated Power (Q') W (Diameter = 25mm) Area of cross section (A) m2 Power Difference in Temperature Temperature (°C) between two points (AT) °C Difference in distance between two points (Ax) m 12 3 4 5 6 7 8 9 Thermal conductivity of brass (kb) W/m°C (W) 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6 MATERIAL 2 - STEEL Calculation for Steel Material 2 - STEEL Quantities Values Calculated (Diameter = 25mm) Power (Q') W Power Area of cross section (A) m2 Temperature (°C) Difference in Temperature between two points (AT) °C Q' 1 (W) Difference in distance between two points (Ax) m 2 3 8 14.2 88.6 87.5 85 33.9 33.6 32.4 Thermal conductivity of steel (kg) W/m°C

We performed the experiment to measure CALCULATE THE FOLLOWING: MATERIAL 1 - BRASS the thermal conductivity of 2 materials (Brass & Steel) in the laboratory and measured the following tabulated values: Calculation for Brass Quantities Values Material 1- BRASS Calculated Power (Q') W (Diameter = 25mm) Area of cross section (A) m2 Power Difference in Temperature Temperature (°C) between two points (AT) °C Difference in distance between two points (Ax) m 12 3 4 5 6 7 8 9 Thermal conductivity of brass (kb) W/m°C (W) 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 33.6 MATERIAL 2 - STEEL Calculation for Steel Material 2 - STEEL Quantities Values Calculated (Diameter = 25mm) Power (Q') W Power Area of cross section (A) m2 Temperature (°C) Difference in Temperature between two points (AT) °C Q' 1 (W) Difference in distance between two points (Ax) m 2 3 8 14.2 88.6 87.5 85 33.9 33.6 32.4 Thermal conductivity of steel (kg) W/m°C

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter5: Analysis Of Convection Heat Transfer

Section: Chapter Questions

Problem 5.10P: 5.10 Experiments have been performed on the temperature distribution in a homogeneous long cylinder...

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1.4 To measure thermal conductivity, two similar 1-cm-thick specimens are placed in the apparatus shown in the accompanying sketch. Electric current is supplied to the guard heater, and a wattmeter shows that the power dissipation is 10 W. Thermocouples attached to the warmer and to the cooler surfaces show temperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of the material at the mean temperature in W/m K. Problem 1.4
5.10 Experiments have been performed on the temperature distribution in a homogeneous long cylinder (0.1 m diameter, thermal conductivity of 0.2 W/m K) with uniform internal heat generation. By dimensional analysis, determine the relation between the steady-state temperature at the center of the cylinder , the diameter, the thermal conductivity, and the rate of heat generation. Take the temperature at the surface as your datum. What is the equation for the center temperature if the difference between center and surface temperature is when the heat generation is ?
A square silicon chip 7mm7mm in size and 0.5-mm thick is mounted on a plastic substrate as shown in the sketch below. The top surface of the chip is cooled by a synthetic liquid flowing over it. Electronic circuits on the bottom of the chip generate heat at a rate of 5 W that must be transferred through the chip. Estimate the steady-state temperature difference between the front and back surfaces of the chip. The thermal conductivity of silicon is 150 W/m K. Problem 1.6
1.3 A furnace wall is to be constructed of brick having standard dimensions of Two kinds of material are available. One has a maximum usable temperature of 1040°C and a thermal conductivity of 1.7 W/(m K), and the other has a maximum temperature limit of 870°C and a thermal conductivity of 0.85 W/(m K). The bricks have the same cost and are laid in any manner, but we wish to design the most economical wall for a furnace with a temperature of 1040°C on the hot side and 200°C on the cold side. If the maximum amount of heat transfer permissible is 950 , determine the most economical arrangement using the available bricks.
A section of a composite wall with the dimensions shown below has uniform temperatures of 200C and 50C over the left and right surfaces, respectively. If the thermal conductivities of the wall materials are: kA=70W/mK,kB=60W/mK, kC=40W/mK, and kD=20W/mK, determine the rate of heat transfer through this section of the wall and the temperatures at the interfaces.
One end of a 0.3-m-long steel rod is connected to a wall at 204C. The other end is connected to a wall that is maintained at 93C. Air is blown across the rod so that a heat transfer coefficient of 17W/m2 K is maintained over the entire surface. If the diameter of the rod is 5 cm and the temperature of the air is 38C, what is the net rate of heat loss to the air?
A plane wall of thickness 2L has internal heat sources whose strength varies according to qG=qocos(ax) Where qo is the heat generated per unit volume at the center of the wall (x=0) and a is a constant. If both sides of the wall are maintained at a constant temperature of Tw, derive an expression for the total heat loss from the wall per unit surface area.
Using Table 1.4 as a guide, prepare a similar table showing the orders of magnitude of the thermal resistances of a unit area for convection between a surface and various fluids.