HEAT TRANSFER OPERATIONS

General Heat Conduction Equation

Dr. Muhammad Rizwan Assistant Professor

COMSATS

GENERAL HEAT CONDUCTION EQ: For rectangular coordinates

It reduces to the following forms under specified conditions

GENERAL HEAT CONDUCTION EQ:

Example 2.2 (Fundamentals of heat and mass transfer by Incropera ):

GENERAL HEAT CONDUCTION EQ FOR CYLINDRICAL COORDINATE SYSTEM:

Based upon basic energy balance equations

Figure: Differential control volume, dr.dz.rdΦ, for conduction analysis in cylindrical coordinates (r, Φ, z).

After lengthy manipulations, we obtain

GENERAL HEAT CONDUCTION EQ FOR CYLINDRICAL COORDINATE SYSTEM:

Based upon basic energy balance equations

Figure: Differential control volume, dr.rSinθdΦ.rdθ, for conduction analysis in spherical coordinates (r, Φ, θ).

After lengthy manipulations, we obtain

HOME WORK: Problem 2.23 (Fundamentals of heat and mass transfer by Incropera ):

The steady-state temperature distribution in a one dimensional wall of thermal conductivity 50 W/m. K and thickness 50 mm is observed to be T (°C) = a + bx2, where a = 200 °C, b = -2000 °C/m2, and x is in meters.

(a) What is the heat generation rate ‘q’ in the wall?

(b) Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate?

ONE DIMENSIONAL STEADY-STATE CONDUCTION:

“One Dimensional” – Only one coordinate considered Hence temperature gradient exists along only a single

coordinate direction. At steady state conditions, temperature at each point is

independent to time. One dimensional steady state models are simple but still

provide accurate representation of numerous engineering systems.

The objective is to determine expressions for the temp distribution and heat transfer rate in common (planar, cylindrical and spherical) geometries.

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Temperature is a function of the ‘x’ coordinate only and hence heat is transferred exclusively in this direction.

Figure: Heat transfer through a plane wall (a) Temperature distribution (b) Equivalent thermal circuit.

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Temperature distribution can be determined by solving the heat equation.

Temperature Distribution

Steady state, no heat generation, one dimensional coordinates.

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Just as an electrical resistance is associated with the conduction of electricity, a thermal resistance may be associated with the conduction of heat.

Precisely, it is a ratio of driving potential to the corresponding transfer rate.

Thermal Resistance

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Thermal resistance is also associated with the convection heat transfer.

Newton’s law of cooling is: Then thermal resistance for convection will be:

Thermal Resistance

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

The Composite Wall

Figure: Equivalent thermal circuit for a series composite wall

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

The one dimensional heat transfer rate for this wall may be expressed as:

The Composite wall

Alternatively, the heat transfer rate can be related to the temperature difference and resistance associated with each element:

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

For composite systems it is often convenient to use overall heat transfer coefficient, U:

The Composite wall

Where,

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Contact Resistance

Figure: Temperature drop due to the thermal contact resistance

Problem: A furnace is constructed with 0.2m of fire bricks, 0.1m of insulated bricks and 0.2m of building bricks. The inside temperature of the furnace is 1200 K and outside temperature is 330 K. If the thermal conductivities are 1.4, 0.21, 0.7 W/m2.K, respectively, Determine the heat loss per unit area and the temperature at the junction of fire bricks and insulated bricks?

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

Problem: A flat furnace wall is constructed of a 114mm layer of silicon bricks with a thermal conductivity of 0.138 W/m.K backed by a 229mm layer of common bricks of thermal conductivity 1.38 W/m.K. The temperature of the inner surface of the wall is 1033 K and that of the outer surface is 349.6K.1. What is the heat loss to the wall? 2. What is the temperature of the interface b/w

refractory and common bricks? 3. Supposing that the contact between the two bricks

layer is poor and that is contact resistance of 0.088 k.m2 is present. What would be the heat loss?

ONE DIMENSIONAL STEADY-STATE CONDUCTION THROUGH PLANE WALL:

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