Finite slab Flashcards

1
Q

What are the 6 steps involved in solving a transport phenomena problem?

A

1) Draw a diagram of the system and define the boundary conditions
2) Draw the control volume and indicate inlet and outlet fluxes
3) Construct the energy balance
4) Simplify the differential equation by non-dimensionalising it and re-writing the boundary conditions
5) Solve the differential equation
6) Apply the boundary conditions to find the constraints
7) Simplify the final equation

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2
Q

Draw a diagram of the system and define the boundary conditions for heat transfer in a finite rod.

A

See footnote 1.

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3
Q

Draw the control volume and indicate the inlet and outlet fluxes for heat transfer in a finite rod.
State the assumptions.

A

See footnote 2.

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4
Q

Construct the energy balance for heat transfer in a finite rod.

A

See footnote 3.

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5
Q

Define thermal diffusivity

A

alpha=k/(rho x c)

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6
Q

Simplify the differential equation given by the energy balance for heat transfer in a finite rod and rewrite the boundary conditions

A

See footnote 4.

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7
Q

What is the range of the non-dimensional variables for temperature and space?

A

0 to 1

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8
Q

Why is the non-dimensional variable for time different?

A

There is no ‘final time’ so it must be made up from other variables.

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9
Q

Solve the simplified differential equation for the heat transfer through a finite rod.

A

See footnote 5.

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10
Q

Apply the non-dimensionalised boundary conditions to find the new constraints for heat transfer through a finite rod.

A

See footnote 6.

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11
Q

Simplify the final equation from the equation for the non-dimensionalised variable of temperature.

A

See footnote 7.

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12
Q

What type of series is the solution for the heat transfer through an infinite rod?

A

An infinite series

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13
Q

What happens to the temperature as the time tends towards infinity?

A

It tends towards T1

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14
Q

What happens to the series given by heat transfer through an infinite rod when t increases?

A

The higher frequency terms become less important because the exponential decay of these terms is so fast. Therefore the approximation using the first few terms becomes more accurate as time increases.

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