MATH WIZARDS

Visual Analysis

Industrial Thermics for Mathematica

What Can You Do with Industrial Thermics?

Industrial Thermics for Mathematica is a software pack that allows you to perform analytical analysis in all fields of heat transfer. It provides you with innovative solution techniques for thermal calculations; you can create, change, and optimize parametric mathematical models quickly and intuitively.

What Do You Gain?

Great Results

More than 100 examples are explained in detail in the manual and included in on-line notebooks.

Advanced Know-How

The comprehensive manual includes every theory you need to know about heat transfer.

Rapid Modeling:

No longer must you wait for manual calculations; you get fast, accurate solutions from an advanced, easy-to-use tool.

Is Industrial Thermics Designed for You ?

Industrial Thermics was specially designed for the engineer, but it is also a "must" for the technical student; Industrial Thermics provides all that you need to understand heat transfer

Topics covered

Design of Heat Exchangers

The Overall Heat Transfer Coefficient 
Design and Prediction of the Performance of Heat Exchanger
The Log Mean Temperature Difference – Finding The Heat Flow	
First Law Analysis of a Heat Exchanger – Finding the Mass Flow Rate	
Parallel-Flow and Counterflow Heat Exchangers	
Cross-Flow Heat Exchangers
Vessel Heat Exchangers	
Other Types of Heat Exchangers	
Compact Heat Exchangers

Steady Conduction Heat Transfer

Walls, Hollow Cylinders and Hollow Spheres with one or more Layers	
An Alternative Conduction Analysis
Walls, Hollow Cylinders and Hollow Spheres with Inner Heat Sources
Heat Transfer from Extended Surfaces
Rectangular Plates	
Conduction Shape Factors

Transient Conduction

Lumped-Heat-Capacity Systems
Semi-Infinite Solids	
Heat Flow in Walls, Cylinders and Spheres
Inner and Moving Heat Sources

Convection Heat Transfer

Calculation of Nusselt Numbers and Convection Heat Transfer Coefficients of Plates, Cylinders, Noncircular Cylinders, Spheres, Tube Banks, Circular Tubes Noncirculare Tubes, Concentric Tube Annulus, Fluid Beds, Enclosures

Radiation Heat Transfer

Gray Body Radiation	
Radiation Exchange between Surfaces
Radiation Shape Factors
Shape Factor Algebra
Radiation Shape Factor Relations in Enclosures
Heat Exchange between Blackbodies
Heat Exchange between Non-Blackbodies
Radiation Shields

Copy and Paste Examples

Design of Heat Exchangers

Multipass, Double-Pipe, and Compact Heat Exchangers
Fouling Factors
Optimization of Heat Exchangers
Car Radiator

Steady Conduction Heat Transfer

Optimization of Building Insulation
Heat Loss of Steel Tubes to Ambient Air
Heat Transfer from Fins of Different Shapes
Heat Transfer from Water Pipes to the Ground

Transient Conduction

Temperature Distribution in Fire Resistant Walls
Quenching of Steel Plates
Cooling of a Moving Steel Belt in a Oil Bath
Temperature Distribution in Insulated Building Walls
Cooling of Hot Bowls
Electrical Heating Wires
Temperature Distribution in Plates due to Heating theirs Surface with a Moving Laser

Convection Heat Transfer

Cooling of Electrical Components
Design of Air Conditioners
Power Optimization of Silicon Chips
Drying Processes in Fluid Beds

Radiation Heat Transfer

Radiation Shape Factor Algebra
Temperature Distribution in an Electrical Oven
Heat Gain of Cryogenic Fluids with Radiation shields

....and much more

Selected Examples

Steady-State Conduction

Extended Surfaces

An aluminium fin  (lambda=200 W/(mK))  that is 3mm thick and 7.5cm long protrudes from a wall. The base temperature is maintained at 300°C, the ambient air temperature is 50°C and the convection heat transfer coefficient is  h=10 W/(m^2K)) . Find the heat loss from the fin per unit depth when rectangular or triangular fins are used.

Solution

The temperature distribution in the rectangular fin is calculated by the function CoolFin

T=CoolFin[lambda,h,a,L,b,{T0,TF},x]

The substitution of the numerical parameters in the expression for the temperature distribution yields

T=T/.{lambda->200,a->0.003,L->0.075,T0->300, 
TF->50,h->10}

Now we can plot the temperature distribution against the x-coordinate in the rectangular fin

Plot[%,{x,0,0.075},AxesLabel->{"x","T"},
PlotRange->{Automatic,{200,300}}]

The heat flow transferred from a rectangular fin to the ambient fluid is calculated by the function CoolFin

q=CoolFin[lambda,h,a,L,1,{T0,TF},x,Result->HeatFlow]

The substitution of the numerical parameters in the expression for the heat flow yields

q=q/.{lambda->200,a->0.003,L->0.075,T0->300,TF->50,
h->10}//N

The temperature distribution in the triangle fin is calculated by the function CoolFin

T=CoolFin[lambda,h,a,L,b,{T0,TF},x,Geometry->Triangle]

The substitution of the numerical parameters in the expression for the temperature distribution yields

T=T/.{lambda->200,a->0.003,L->0.075,T0->300,
 TF->50,h->10}

Now we can plot the temperature distribution against the x-coordinate in the triangular fin

Plot[%,{x,0,0.075},AxesLabel->{"x","T"},
PlotRange->{Automatic,{200,300}}]

The heat flow transferred from the triangular fin to the ambient fluid is calculated by the function CoolFin

q=CoolFin[lambda,h,a,L,1,{T0,TF},x,
Result->HeatFlow, Geometry->Triangle]

The substitution of the numerical parameters in the expression for the heat flow yields

q=q/.{lambda->200,a->0.003,L->0.075,T0->300,
TF->50,h->10}//N

Transient Conduction

Quenching of a Steel plate

In a manufacturing process a 64mm thick steel plate (rho=7800 kg/m^3, cp=510 J/(Kg K), lambda=18 W/(mK))  initially at 85°C, is quenched by submersion in an oil bath kept at 5°C with a convection heat transfer coefficient of 1000  W/(m^2K) on both sides of the steel plate. The plate is very long so we can neglect the heat transfer from the small sides of the plate. Plot the temperature distribution in the plate that depends on the coordinate and the time. Consider a time 2min into the cooling process. What is the surface temperature and the mid-plane temperature of the plate? What is the average temperature? What is the total amount of heat lost within the period of 2min? Plot the temperature distribution in the plate after 2min. What is the required time to reduce the temperature in the mid-plane of the plate to 10°C?

Solution

The initial temperature, the density, the specific heat, the thermal conductivity and the thickness of the plate are

T0=85; rho=7800; cp=510; lambda=18; L=0.064;

The temperature of the oil bath and the convection heat transfer coefficient are

TF=5; h=1000;

The cooling time is

Time=2*60;

The temperature distribution is determined by the function TransientFluid1D

T=TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},x,t]

Now we can plot the temperature distribution in the plate that depends on the x-coordinate and the time

Plot3D[T,{x,-L/2,L/2},{t,0,500},AxesLabel->{"x","t","T"},
ViewPoint->{1.870,1.957,2.030}]

The surface temperature of the plate is calculated by the function TransientFluid1D

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},L/2,Time]

The temperature in the mid-plane of the plate is calculated by the function TransientFluid1D, the x-coordinate is 0

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},0,Time]

The average temperature of the plate is calculated by the function TransientFluid1D. For the average temperature, the option name Result is set to Average.

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},x,Time,
Result->Average]

The total amount of heat lost   within the 2min period is calculated by the function TransientFluid1D. For getting the total amount of heat lost, the option name Result is set to Energy.

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},x,Time,
Result->Energy]

The temperature distribution after 2min is determined by the function TransientFluid1D

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},x,Time]

Now we can plot the temperature distribution in the plate that depends on the x-coordinate

Plot[%,{x,-L/2,L/2},AxesLabel->{"x","T"}]

The temperature in the mid-plane of the plate is determined by the function TransientFluid1D

TransientFluid1D[{rho,cp,lambda},L,h,{T0,TF},0,t]

Now we can calculate the very moment when the temperature in the mid-plane of the plate is 10°C

FindRoot[%==10,{t,200}]

Convection Heat Transfer

Array of Silicon Chips

An array of 5 silicon chips, each of length L=5mm  on a side, is isolated on one surface and cooled on the opposite surface by air with TFluid=290K . When in use, the same electrical power is dissipated in each chip, maintaining a uniform surface heat flux. Determine the position where the maximum surface temperature is reached. Plot the temperature distribution at this position that depends on the fluid speed in the range of 1 m/s  to 45 m/s  and the power per chip of 0 to 2W. Determine for a fluid speed of 30 m/s  the maximum allowable power per chip so that the temperature may not exceed 80°C.

Solution

Normally, properties should always be evaluated at the film temperature. Because we don’t know the plate temperature, we take the properties at the free-stream conditions for an initial calculation.

TF=290;A=0.005^2;L=0.005;Vmax=80;

prop=AirProperties[TF]

The convection heat transfer coefficient in the range of   and   is calculated by the function LocalAlphaPlate

h=LocalAlphaPlate[prop,20,5*L,HeatKind->HeatFlux]

Now the surface temperature of the copper can be evaluated

T=TF+Q/(h*A)

The following graph shows the temperature at the position

pl=Plot3D[T/.x->5*L,{w,1,45},{Q,0,2},
DisplayFunction->Identity];

Show[pl,AxesLabel->{"Fluid Speed","Power","Temperatur"},
PlotLabel->"Maximum Temperature",
DisplayFunction->$DisplayFunction]

Now we will determine the maximum allowable power per chip so that the temperature may not exceed 80°C when the fluid speed is 30 m/s. We will perform the calculation in two steps. In the first step we will determine the maximum allowable power per chip for the thermophysical properties of the air. In the second step we will determine the film temperature and repeat the calculation with the now-known film temperature.

· First Step

We convert the temperature with the function TemperatureConversion

v=30;

Tmax=TemperatureConversion [80, C -> K]

Then we substitute the maximum length and the fluid speed

T==%/.{x->5*0.005,w->v}

Now we solve the equation for the power

Solve[%,Q]

The maximum allowable power is

· Second step

First we determine the maximum heat flux

q=Q/A/.%[[1]]

With the approximation for the maximum heat flux we calculate a better film temperature and repeat all calculations. The film temperature is calculated by the function LocalAlphaPlate

hl=LocalAlphaPlate[prop,v,5*L,
HeatKind->HeatFlux,Message->Off]/.{x->5*L,w->v}

Tcor=(2*q/(3*hl)+2*290)/2

The thermophysical properties of the air are determined by the function AirProperties

propcor=AirProperties[Tcor]

The corrected convection heat transfer coefficient is calculated by the function LocalAlphaPlate

hcor=LocalAlphaPlate[propcor,45,5*0.005,
HeatKind->HeatFlux,Message->Off]

We substitute the maximum length and the fluid speed

Tcor=TF+Q/(hcor*A)/.{x->5*L,w->v}

Now we solve the equation for the power

Tcor==Tmax

Solve[%,Q]

The corrected maximum allowable power is

Radiation Heat Transfer

Heat Transfer between two Parallel Disks

Two parallel disks of radius 0.2m are centered along the same normal. Plot the radiation shape factor that depends on the radius of the lower disk and the distance of the two disks.

Solution

The shape number can be taken from Table 4.2.1. The radiation shape factor is determined by the function RadiationShapeFactor

RadiationShapeFactor[7,{R1,R2,DR}]

The substitution of the numerical parameters in the expression for the radiation shape factor yields

RSF=%/.{R2->0.2}

Now we can plot the temperature against the distance

Plot3D[%,{DR,0.001,1},{R1,0.01,1},
AxesLabel->{"DR","R1",
"Shape factor"},PlotRange->{0,1},
ViewPoint->{2.008,1.781,2.060}]

Clear[L1]

Heat Exchanger

Compact Heat Exchanger

A heat exchanger with one shell pass and two tube passes is used to cool hot air to the temperature of the ambient air. The inlet temperature T1in  of the hot air is 60°C. Inside the shell tube, there is a bundle of small tubes. Each bundle consists of 150 staggered tubes. The longitudinal pitch measured between the tube centers is 8mm and the transverse pitch measured between the tube centers is 3mm. The outer diameter of such a tube is 4mm and the inner diameter is 3.2mm. The inlet temperature T2in  of the cold air is 30°C. The mass flow rates of the two air streams are 0.083 . The inlet area of the cold air stream is 0.02 m^2  . The thermal resistance of the tubes are negligible. Find the outlet temperatures of the two air streams.

Solution

To determine the outlet temperature, the heat exchanger is divided up into six coupled single pass cross-flow heat exchangers (see Figure 5.7.2 and numbers 1 to 6 in Figure 5.7.1). In each of this single cross-flow heat exchangers one fluid is mixed and the other is unmixed. So we can use a cross-flow heat exchanger with one row of tubes as a theoretical model.

The hot air stream has a temperature T1, and the cold air stream has a temperature T2. The first number of the indices is the number of each heat exchanger and the second number of the indices is the direction of the flow, where 1 means inlet and 2 means outlet. The inlet temperatures of the air streams are

T1in=60; T2in=30;

The mass flow rates of the air streams are

m1=0.083; m2=m1;

The number of the tubes is

n=150;

The longitudinal pitch and transverse pitch are

s1=0.008; s2=0.003;

The inner and outer diameter of one small tube are

di=0.0032; da=0.004;

The length of one small tube is

length=1.8;

The inlet area of the cold air is

A2=0.02;

The thermophysical properties of the hot air at a mean temperature of 50°C are determined by the function AirProperties

prop1 = AirProperties[50+273]

The thermophysical properties of the cold air at a mean temperature of 40°C are determined by the function AirProperties

prop2 = AirProperties[40+273]

The effective flow area of a small tube is

A1=n*di^2/4*Pi

Now the velocities of the two air streams can be calculated

v1=m1/(A1*prop1[[1]])//N

v2=m2/(A2*prop2[[1]])

The convection heat transfer coefficient of the cold air stream is determined by the function AlphaTube

h1=AlphaTube[prop1,v1,length,di]

The convection heat transfer coefficient of the hot air stream is determined by the function AlphaBank

h2=AlphaBank[prop2,v2,s1,s2,da,Aligned->False]

k=CompositeHeatTransition1D[{Infinity},{di,da,},
length,{h1,h2},{T1,T2}, Geometry->Tube,kValue->True]

The specific heat capacity rates are

C1=.083*prop1[[2]]

C2=.083*prop2[[2]]

The effectiveness of each cross-flow heat exchanger is determined by the function CrossedHeatExchanger

e=CrossedHeatExchanger[N[k*Pi*da*0.3*n],{C1,C2},
{T1in,T2in,T1out,T2out},RowNumber->1]

The equations are entered as follows

· for the coupling of the first five cross-flow heat exchangers

tab1=Table[T2[i+1,1]-(1-e)*T2[i,1]-e*T1[i,1]==0,{i,5}]

· for the coupling of the last cross-flow heat exchanger

tab2={-(1-e)*T2[6,1]-e*T1[6,1]+T2[6,2]==0}

· for the known temperatures

tab3={T2[1,1]==30,T1[6,1]==60}

· for the application of the energy balances

tab4={C1*( T1[3, 1] - T1[6, 1]) +
 C2*(-T2[6, 1] + T2[6, 2]) == 0, 
      C1*( T1[2, 1] - T1[3, 1]) +
 C2*(-T2[3, 1] + T2[4, 1]) == 0, 
      C1*( T1[1, 1] - T1[2, 1]) +
 C2*(-T2[2, 1] + T2[3, 1]) == 0, 
      C1*(-T1[1, 1] + T1[4, 1]) +
 C2*(-T2[1, 1] + T2[2, 1]) == 0, 
      C1*(-T1[4, 1] + T1[5, 1]) +
 C2*(-T2[4, 1] + T2[5, 1]) == 0, 
      C1*(-T1[5, 1] + T1[5, 2]) +
 C2*(-T2[5, 1] + T2[6, 1]) == 0}

Now the lists of equations are joined

eq=Join[tab1,tab2,tab3,tab4]

Then all the variables of the system are defined

var1=Table[T1[i,1],{i,1,6}]

var2=Table[T2[i,1],{i,1,6}]

var=Join[var1,var2,{T1[5,2],T2[6,2]}]

Now all the solutions of this linear system of equations are calculated

sol=Solve[eq,var]

The outlet temperature of the hot air and the outlet temperature of the cold air is extracted

{T1[5,2],T2[6,2]}/.sol[[1]]

The effectiveness of the heat exchanger is extracted

(T2[6,2]-30)/(60-30) /.sol[[1]]

Finally, the heat flow of the heat exchanger is extracted

C1*(60-T1[5,2])/.sol[[1]]

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