# Heat and Mass Transfer: Heat Out

• jdawg
In summary: The convective heat transfer resistance (based on the convective heat transfer coefficient and the surface area) will be in parallel and will compete with the conductive heat transfer.In summary, the temperature distribution across a 0.3m thick wall at a certain instant is given by T(x)=a+bx+cx2, with a=200°C, b=-200°C/m, and c=30°C/m2. The wall has a thermal conductivity of 1W/m2K. On a unit surface area basis, the rate of heat transfer into and out of the wall is 200 W/m2 and 182 W/m2, respectively. The rate of change of energy stored by the wall can be determined using the
jdawg

## Homework Statement

The temperature distribution across a wall 0.3m thick at a certain instant of time is T(x)=a+bx+cx2, where T is in degrees Celsius and x is in meters, a = 200°C, b = -200°C/m, and c=30°C/m2. The wall has a thermal conductivity of 1W/m2K.
(a) On a unit surface area basis, determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall.
(b) If the cold surface is exposed to a ﬂuid at 100°C, what is the convection coefﬁcient?

## Homework Equations

q'' conduction= -k[∂T/∂x]
q'' convection= h[Twall-T]

## The Attempt at a Solution

So part A made perfect sense. Assuming 1D conduction, I got the heat transfer into be 200 W/m2 and the heat transfer out to be 182 W/m2.

Part B is where I get a bit confused. So I found the temperature at the cold surface of the wall (at distance L) by just plugging 0.3 m into the given temperature distribution. Then you use the heat transfer equation for convection:

q'' convection= h[Twall-T]

So you know the temperature difference, but how to you know the heat transfer out? Why would it be acceptable to use the heat transfer out due to conduction from part A when we are considering convection in part B? Is it because if you do a surface balance at that surface of the wall, then the heat transfer due to conduction will equal the heat transfer due to convection? But then I suppose your conduction heat transfer would have to be heat in and not heat out in that case... I would greatly appreciate some clarification :)

You assessed this correctly. You calculated the heat transfer rate out as 182 W/m^2. This is in series with the heat transfer rate through the convection boundary layer from the wall to the fluid at 100 C. The 182 is the rate of heat transfer conducted through the wall into the interface.

## What is heat transfer?

Heat transfer is the movement of thermal energy from one object to another due to a difference in temperature. This can occur through three main mechanisms: conduction, convection, and radiation.

## How is heat transferred through conduction?

Conduction is the transfer of heat through a solid material or between two objects in direct contact. It occurs when the molecules in a material vibrate and transfer their energy to neighboring molecules, causing them to vibrate as well.

## What is convection and how does it work?

Convection is the transfer of heat through the movement of fluids (liquids or gases). This occurs when warmer, less dense fluids rise and cooler, denser fluids sink, creating a continuous circulation of heat transfer.

## How is heat transferred through radiation?

Radiation is the transfer of heat through electromagnetic waves, such as infrared radiation. All objects with a temperature above absolute zero emit some form of radiation. This is how the sun's energy reaches the Earth.

## What factors affect the rate of heat transfer?

The rate of heat transfer is affected by several factors, including the temperature difference between objects, the thermal conductivity of the material, the surface area of contact, and the distance between objects. The type of material and its physical properties also play a role in determining the rate of heat transfer.

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