Conduction of heat transfer through hollow cylinder equation
Answers
Answered by
5
It was observed in the previous discussion that for the given plane wall the area for heat transfer was constant along the heat flow direction. The plane solid wall was one of the geometries but if we take some other geometry (tapered plane, cylindrical body, spherical body etc.) in which the area changes in the direction of heat flow. Now we will consider geometrical configuration which will be mathematically simple and also of great engineering importance like hollow cylinder and hollow sphere. In these cases the heat transfer area varies in the radial direction of heat conduction. We will take up both the cases one by one in the following sections.
2.3.1 Cylinder
Consider a hollow cylinder as shown in the fig.2.9a. The inner and outer radius is represented by riand ro , whereas Ti and To (Ti > To ) represent the uniform temperature of the inner and outer wall, respectively.
Consider a very thin hollow cylinder of thicknessdr in the main geometry at a radial distance r. If dr is small enough with respect to r, then the area of the inner and outer surface of the thin cylinder may be considered to be of same area. In other words, for very small dr with respect to r, the lines of heat flow may be considered parallel through the differential element in radial outward direction.
We may ignore the heat flow through the ends if the cylinder is sufficiently large. We may thus eliminate any dependence of the temperature on the axial coordinate and for one dimensional steady state heat conduction, the rate of heat transfer for the thin cylinder,
Where dT is the temperature difference between the inner and outer surface of the thin cylinder considered above and k is the thermal conductivity of the cylinder.
On rearranging,
To get the heat flow through the thick wall cylinder, the above equation can be integrated between the limits
I HOPE IT WILL HELP YOU DEAR
THANKU
2.3.1 Cylinder
Consider a hollow cylinder as shown in the fig.2.9a. The inner and outer radius is represented by riand ro , whereas Ti and To (Ti > To ) represent the uniform temperature of the inner and outer wall, respectively.
Consider a very thin hollow cylinder of thicknessdr in the main geometry at a radial distance r. If dr is small enough with respect to r, then the area of the inner and outer surface of the thin cylinder may be considered to be of same area. In other words, for very small dr with respect to r, the lines of heat flow may be considered parallel through the differential element in radial outward direction.
We may ignore the heat flow through the ends if the cylinder is sufficiently large. We may thus eliminate any dependence of the temperature on the axial coordinate and for one dimensional steady state heat conduction, the rate of heat transfer for the thin cylinder,
Where dT is the temperature difference between the inner and outer surface of the thin cylinder considered above and k is the thermal conductivity of the cylinder.
On rearranging,
To get the heat flow through the thick wall cylinder, the above equation can be integrated between the limits
I HOPE IT WILL HELP YOU DEAR
THANKU
Similar questions