Using the Heisler charts in solving non-steady state heat transfer problems

Heisler Charts

Heisler Zoom 7

Heisler Zoom 5

Heisler charts are a tool used in unsteady-state heat transfer problems, and they relate the temperature to the time of an object under transient heat conduction. They are useful for finding the temperature in an object at a specified position and time, or if given a temperature and position, finding the time at which that temperature occurs.

How Heisler Charts work

Dimensionless temperature’s dependence on time throughout a given geometry is the same. This means that if the dimensionless center temperature drops by 10 percent at a specified time, the dimensionless temperature anywhere else in the geometry also drops by 10 percent. This dependence creates ratios that will build the y-axis of most Heisler charts.

How to use Heisler Charts:

There are a variety of different ways these charts can be used, a few of which are detailed below.

If the time is given and temperature is needed, the Fourier's number must be calculated to find the x-axis coordinate. Then, the inverse of the Biot number must be solved to determine which of the red curves to use as illustrated above. The intersection of the x-axis position and the curve will return the value of θ0/θi, or the temperature ratio on the y-axis, which is used to calculate the temperature at the given time.

If the current, initial, and final temperatures are given and the time at which the current temperature occurs is needed, the procedure follows the same as above, however instead of calculating the Fourier's number to ultimately solve for the temperature ratio, the temperature ratio and inverse Biot number must be calculated to find the Fourier's number, which can then be rearranged to solve for the time.

If the current, initial, and final temperatures and a position somewhere between the centerpoint and characteristic length of the solid are given and the heat transfer coefficient is needed, a new chart must be used where the inverse of the Biot number is on the x-axis and the curve is categorized by the length given over the characteristic length (x/L for a slab, or r/r0 for infinite cylinder or sphere). The y-axis on these charts is categorized by θ/θ0. To solve this, the temperature ratio must be calculated on the y-axis, then the length ratio for the curve, and use the intersection of these two to find the value of the inverse Biot number on the x-axis. The Biot number can be rearranged to solve for h.

The charts are specific to the volume that is being observed and must be carefully chosen based on this geometry (sphere, infinite cylinder, or slab). Note that the characteristic length for each of these geometries will change.

Fourier's number

$F=\left(\frac{\alpha t}{x^2}\right)$

$\alpha$ is the thermal diffusivity in $\left(\frac{m^2}{s}\right)$, t is the time in seconds, and x is the length through which conduction occurs.

Biot's number

$B^{-}1=m=\left(\frac{k}{hx}\right)$

k is the thermal conductivity of the material, h is the heat transfer coefficient, and x is the length of which is being considered.

Transient Heat Transfer Equations

Cartesian Coordinates

$\left(\frac{{\partial }^{2}T(x,t)}{\partial x^2}\right)=\left(\frac{1}{\alpha }\frac{\partial T(x,t)}{\partial t}\right)$

(1)

Cylindrical Coordinates

$\left(\frac{1}{r}\frac{\partial }{\partial r}r\frac{\partial T}{\partial r}\right)=\left(\frac{1}{\alpha }\frac{\partial T}{\partial t}\right)$

(2)

Spherical Coordinates

$\left(\frac{{\partial }^{2}T}{\partial r^2}\right)+\left(\frac{2}{r}\frac{\partial t}{\partial r}\right)=\left(\frac{1}{\alpha }\frac{\partial T}{\partial t}\right)$

(3)

Heating a Sphere

A sphere of radius 1.5cm initially at ambient temperature is heated to 200°C, after 3 minutes what is the temperature at the center of the sphere?

Given:

Ti=25°C = 298.15K

T∞=200°C = 473.15K

t= 3 min = 180 s

$k=1.52\left(\frac{W}{mK}\right)$

r0=1.5cm = .015m

$h=110\left(\frac{W}{m^{2}K}\right)$

$\alpha =9.5*10^{-}7\left(\frac{m^2}{s}\right)$

Use Heisler Chart for: center temperature as a function of time in a sphere of radius r0

Steps

1. Plug in values of α, t, and r0 into the Fourier's number equation. This will be your x-axis value. This should be a dimensionless number.

Fo=.76

2. Plug in values of k, h, and r0 into the inverse Biot number equation. This will determine which curve you use. This should also be a dimensionless number.

$B{i}^{-}1=.921$

3. Use x=.76 and the curve for Bi-1=.921 to solve for the y-axis value

y≈.18

4. The y-axis value is equal to θ0/θi which is T0-T∞ / Ti-T∞ Plug in T∞ = 200°C and Ti=25°C and rearrange to solve for T0

T0=168.5°C = 441.65 K

5. Check: Does this answer make sense?

The temperature should fall between the initial and final temperature, which it does.

Temperature within a sphere

Given the sphere detailed above, at the same time, what is the temperature at a distance of 8mm radially away from the center?

For this problem, use Heisler charts for: temperature distribution in a sphere of radius r0

Steps

1. Calculate the inverse Biot number. Because h, k, and r0 have not changed in the system, it will be equal to the previous example. This value is now the x-axis value on this chart.

$B{i}^{-}1=.921$

2. Calculate the characteristic length ratio by dividing the given radius, 8mm, over the total radius, 1.5cm (r/r0). This is the curve you will use on the chart.

$\left(\frac{r}{r0}\right)=.5333$

3. Observing the chart, find the intersection of the x-axis value and the curve using the numbers calculated. This is the y-axis value, or θ/θ0

y≈.888

4. $\left(\frac{\theta }{\theta 0}\right)$ is equal to T-T∞/T0-T∞. Note that T0 is at the time given - this means you must use the T0 that you calculated in the previous problem in order to solve for T at the new radius.

T=172.028 °C = 445.178 K

5. Does it make sense?

The outside of the sphere is hotter than the center, so as the radius increases, the temperature should get closer to T∞.

Time estimation using Lumped-Sum model for a sphere

The Lumped-Sum model is used when the Biot number is less than 0.1.

A steel ball has a diameter of 0.3 meters at the Temperature of 273 K. It is suddenly plunged into a large water bath that has a temperature of 350 K. Using the Lumped-Sum model, estimate the time needed for the center temperature of the sphere to reach 349.9 K

you are given:

heat transfer coefficient= 10 W/m^2k

Thermal Conductivity=43 W/mk

Density: 7850 Kg/m^3

Heat capacity: 460 J/Kgk

The Lumped-Sum equation is given by:

$(\frac{T(t)-Tinf}{Ti-Tinf})=exp(\frac{-h*Asectional}{m*cp}*t)$

first, we calculate the cross sectional area

$Asectional=4*pi*r^2$

$=(4)(3.144)(0.15{)}^{1}2$

$=0.283$

the calculation for the volume is needed before the calculation of the mass

$Density=mass/volume$

volume of a sphere is given by:

$V=(\frac{4}{3}*pi*r^3)$

$V=0.0141m^3$

$Mass=Volume*Density$

$=(7850)*(0.0141)=110.685kg$

from there, we can calculate the time:

$1.298*10^{-}3=exp(\frac{(-10)(0.283)}{(110.685)(460)}*t)$

$1.298*10^{-}3=exp(-5.558*10-5*t)$

$t=119592.1314seconds*\frac{1hr}{3600s}$

$t=33.22hr$

you can now compare your results from the Lumped-Sum model vs. the result received Using the Heisler chart. the results from the two calculations show that there is not a significant difference in the results.

why so? what can you infer about the Biot number's relationship to heat transfer in the sphere?