Time required for center temperature of cube?

Question

Determine the time required for center temperature of cube to reach 80C. the cube has volume of 125 cm³. Thermal conductivity of material is 0.4 W/(m C); density is 950 kg/m³, and specific heat is 3.4 kJ/(kg K). the initial temperature is 20C. The surrounding temperature is 90C. The cube is immersed in a fluid that results in negligible surface resistance to heat transfer.

Answer 1

 

Hello again, Nadeshiko -- looks like we want to heat 60C of the 70C dT ... a handy device

called a "time constant" suggests how long it takes to span about 2/3 of a remaining spread.

 

Two time constants would cover 60/70 ... time constants are long when we have a big pool

to drain with a thin, long pipe ... in heat exchange, small masses with little heat capacity

conducting through a large area will change temps quickly ...

 

We look at  m Cp L/ (k A) ~ 125m^3(1 kg/m^3)3.4J/kg-K(0.025m) divided by

(0.4J/s-m-K * 0.015m-m) clocks at 1700 seconds ... two time constants is 3400s ~ 1hr

Answer 2

 

An approximate answer can be obtained by assuming linear heat conduction from surface to core. The total heat transferred to the cube is

 

Q=mc ΔT.

 

We need the mass of the cube: m=950 kg/m³(125*10^-6m³)=0.1188 kg.

 

Let's assume that the average cube temperature is 85°C when the core has reached 80°C, so that ΔT=85-20=65 °C. Then

 

Q=0.1188 kg *3400 J/kg/K * 65 K = 26250 J

 

We find the time for the core to reach 80°C from the heat equation. Let's use as an approximation the 1-dimensional heat conduction equation,

 

Q/t = k A ΔT' /L, so t= Q L /(k A ΔT')

 

The volume of the cube is 125 cm³, so its side length is 5 cm and surface area is 6*25=150 cm²=0.015 m². Let's take for L the shortest surface-core distance, 2.5 cm=0.025 m. Let's assume the average surface temperature during the conduction process is (90+20)/2=55°C, while the average core temperature is 25°C (it increases exponentially). With ΔT'=55-25=30°C, we get

 

t= 26250 J * 0.025 /(0.4 W *0.015* 30) = 3646 s ≈ 1 hour

 

A better approximation would take the 3-dimensional nature of conduction within a cube into account.

Answer 3

I am on my way to bed, but I will give you a hint for setting up the problem.

 

This is a fairly typical application for the 4D heat equation (x,y,z,t) with non-zero boundary conditions.

 

For brevity, I will use the following shorthand for a partial derivative: partial of function u with respect to x is u_x.

 

Let us assume a cartesian coordinate system with the corner of the cube at the origin, and the cube extending into octant I. 

 

Assume there exists some (possibly non-trivial) function T(x,y,z,t) for all points and times inside the cube.

 

The sides of the cube are .05m on a side (cube root of 125 cm³)

 

T must solve the partial differential equation

 

T_t=a²(T_xx+T_yy+T_zz) where a is the thermal diffusivity of the material (you will have to calculate this from thermal conductivity, density, and specific heat provided (it is thermal conductivity / (specific heat capacity * density)))

 

Now, we know that the boundary of the cube is held at 90C, aka T(x,y,z,t) = 90 for all (x,y,z) on the surface of the cube, and for all t>0

 

We also know that T(x,y,z,0) = 20C for all (x,y,z) inside the cube as provided by our initial conditions

 

Applying a Fourier series method to the problem we see that 

 

T(x,y,z,t) = 90C + Σ_iΣ_jΣ_ka_i,j,kexp(-a²c_i,j,kt)sin(iπx/.05)sin(jπy/.05)sin(kπz/.05)

 

with c_i,j,k=(π/.05)²(i²+j²+k²)

 

and a_i,j,k=-(90-20)(2/π)³(1/ijk)(1-(-1)ⁱ)(1-(-1)^j)(1-(-1)^k)

 

Once you carry through all the numerical aspects, you can simply plug in .025 for x, y, and z and set the RHS of the equation to 80C, and pick your method of choice for solving / approximating t. 

 

 

-------

 

 

The following document provides a reasonable overview of heat and wave equations solved for various simple geometries in R² and R³.

 

http://ocw.mit.edu/courses/mathematics/18-303-linear-partial-differential-equations-fall-2006/lecture-notes/pde3d.pdf

 

Given that your major is food engineering, most things are not as convenient as the cube example and require approximate solutions rather than analytic ones. 

 

You may want to, for less contrived problems, look to find a good software package for performing finite element methods. If you are unfamiliar with the ideas of numerical methods (finite difference / element / volume as well as others) this paper will perhaps provide a bit of background on the idea http://www.f.kth.se/~jjalap/numme/FDheat.pdf

and this powerpoint for something shorter: http://auc.cz/summerschool2013/slides_NoveHrady_2013.pdf

 

for just a quick fix, sans the theory, the following matlab package should suffice http://www.mathworks.com/matlabcentral/fileexchange/40096-numerical-solutions-of-heat-equation though you will have to adapt it to your geometries

 

 

If that does not get you unstuck, I can help you tomorrow :)