Explanation Given information: Flashpoint of ethanol is 16.6 ° C , inside diameter is 3 cm , wall thickness is 3 mm , average temperature of ethanol is 10 ° C and the heat transfer coefficient for convection is 50 W / m 2 K . Write the expression for the Fourier’s law of heat conduction in the radial direction. q s = − k d T d r ...... (I) Here, thermal conductivity of material is k and the temperature gradient is d T d r . The heat conducted to the outer surface of walls of pipe is lost through convection. Write the expression for the heat equation for the pipe. h [ T − T ∞ ] = − k d T d r ...... (II) Here, the heat transfer coefficient is h , temperature of the surrounding medium is T and the temperature of the atmosphere is T ∞ and the thermal conductivity of the material is k Write the expression for the differential equation for heat conduction in cylindrical coordinate. d d r ( r d T d r ) = 0 ...... (III) Here, radius of the pipe is r . Write the expression for the inner radius of pipe. r 1 = d 1 2 ...... (IV) Here, the inner diameter of pipe is d 1 . Write the expression for the outerradius of pipe. r 2 = r 1 + 3 mm ......(V) Calculation: Integrate the Equation (III). ∫ d d r ( r d T d r ) = ∫ 0 r d T d r = C 1 d T d r = C 1 r ...... (VI) Integrate the Equation (VI). ∫ d T d r = ∫ C 1 r T ( r ) = C 1 ln r + C 2 ...... (VII) Apply the boundary conditions, that is, at r = r 1 , T = T ( r 1 ) . Substitute T ( r 1 ) for T ( r ) and r 1 for r in Equation (VII). T ( r 1 ) = C 1 ln r 1 + C 2 ...... Apply the boundary conditions, that is, at r = r 2 , T = T ( r 2 ) . Substitute T ( r 2 ) for T ( r ) and r 2 for r in Equation (I). q s = − k d T ( r 2 ) d r ...... (VIII) Apply the boundary conditions, that is, at r = r 2 , T = T ( r 2 ) . Substitute T ( r 2 ) for T ( r ) and r 2 for r in Equation (VI). d T ( r 2 ) d r = C 1 r 2 Substitute C 1 r 2 for d T ( r 2 ) d r in Equation (VIII). q s = − k C 1 r 2 C 1 = − q s r 2 k Apply the boundary conditions, that is, at r = r 2 , T = T ( r 2 ) . Substitute T ( r 1 ) for T ( r ) and r 1 for r in Equation (VI). d T ( r 1 ) d r = C 1 r 1 Apply the boundary conditions, that is, at r = r 1 , T = T ( r 1 ) . Substitute T ( r 1 ) for T ( r ) , r 1 for r , C 1 r 1 for d T ( r 1 ) d r and C 1 ln r 1 + C 2 for T ( r 1 ) in Equation (II). h [ C 1 ln r 1 + C 2 − T ∞ ] = − k C 1 r 1 C 2 = − k C 1 h r 1 + T ∞ − C 1 ln r 1 Substitute ( − k C 1 h r 1 + T ∞ − C 1 ln r 1 ) for C 2 and ( − q s r 2 k ) for C 1 in Equation (VII). T ( r ) = ( − q s r 2 k ) ln r + ( − k C 1 h r 1 + T ∞ − C 1 ln r 1 ) = ( − q s r 2 k ) ln r + ( − k ( − q s r 2 k ) h r 1 + T ∞ − ( − q s r 2 k ) ln r 1 ) = ( q s r 2 k ) ( − ln r + k h r 1 + ln r 1 ) + T ∞ = ( q s r 2 k ) ( k h r 1 + ln r 1 r ) + T ∞

6th Edition

ISBN: 9781260440027

Author: CENGEL

Publisher: MCG

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Question

Chapter 2, Problem 74P

To determine

The variation of temperature in the pipe wall.

The temperatures on the inner and the outer surfaces of the pipe.

Whether both surfaces are safely below the flashpoint of liquid ethanol.

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Chapter 2, Problem 73P

Chapter 2, Problem 75P

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The variation of temperature in the pipe wall. The temperatures on the inner and the outer surfaces of the pipe. Whether both surfaces are safely below the flashpoint of liquid ethanol.

The variation of temperature in the pipe wall. The temperatures on the inner and the outer surfaces of the pipe. Whether both surfaces are safely below the flashpoint of liquid ethanol.

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Chapter 2 Solutions

The variation of temperature in the pipe wall.

The temperatures on the inner and the outer surfaces of the pipe.

Whether both surfaces are safely below the flashpoint of liquid ethanol.