2- Steady State Heat Conduction (3).pdf - Solution of UPSC Papers Of Heat Transfer Steady State Heat Conduction Prepared By Brij Bhooshan Asst Professor

2- Steady State Heat Conduction (3).pdf - Solution of UPSC...

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For more information log on Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India) Copyright by Brij Bhooshan @ 2013 Page 1 Solution of UPSC Papers Of Heat Transfer Steady State Heat Conduction Prepared By Brij Bhooshan Asst. Professor B. S. A. College of Engg. And Technology Mathura, Uttar Pradesh, (India) Supported By: Purvi Bhooshan Please welcome for any correction or misprint in the entire manuscript and your valuable suggestions kindly mail us [email protected] 1985 Problem: Develop an expression for the heat flux passing radially through a long composite cylinder of length L made up of three concentric layers of different materials and thicknesses in terms of r 1 the inside radius of the inner cylinder; r 2 , r 3 and r 4 , the outside radii of the inner, middle and the outer cylinders respectively; their corresponding thermal conductivities k l , k 2 and k 3 ; the temperature T 1 at the inner surface of the inner cylinder and the temperature T 4 at the outer surface of the outer, cylinder, T 1 being greater than T 4 . If the inside heat transfer coefficient of a fluid at temperature T 1 flowing through this cylinder is h i and the outside heat transfer coefficient of another fluid at temperature T 2 flow in the outside is h 0 , determine the overall heat transfer coefficient referred to the inside surface. [Engg. Services-1985] Solution: Now according to problem electrical network as shown in diagram. Now thermal resistance due to fluid is T 2 T 1 R ci R 1 R 2 R 3 R c0
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For more information log on Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India) Copyright by Brij Bhooshan @ 2013 Page 2 2 Solution of UPSC Papers of Steady State Heat Conduction Now thermal resistance due to conduction is Now net thermal resistance is R th = R ci + R 1 + R 2 + R 3 + R c0 Now we know that heat flux Also we know that Using equation (1) and (2), we have where U 0 is overall heat transfer Then overall heat transfer is 1991 Problem: Starting from basic differential equation, derive an expression for the temperature distribution in a cylindrical rod in which a uniform heat generation occurs and the surface temperature of the rod is T S . Mention the assumptions. [IAS-1991] Solution: Let us consider a long solid cylinder of radius R with internal heat generation as shown in above diagram, such as an electric coil in which heat is generated as a result of the electric current in the wire or a cylindrical nuclear fuel element in which heat is generated by nuclear fission. The one-dimensional heat conduction equation in cylindrical coordinates is On integration 0
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For more information log on Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India) Copyright by Brij Bhooshan @ 2013 Page 3 3 Solution of UPSC Papers of Heat Transfer By Brij Bhooshan Again integration where A and B are the arbitrary constants.
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