Temperature Field in a Cooling Flange

Chemical reaction fluids can be cooled using glass flanges. The reaction fluid is passed through the flange and the air surrounding the flange then serves as the coolant. Engineers looking to optimize the cooling performance of such flanges can look to simulation for help.

Figure 1: Operating principle of the cooling flange.

First off, the physics to analyze in this case is heat transfer, involving both convection and conduction. Heat transfer occurs via convection with both interior and exterior surfaces of the flange, and via conduction through the glass. The tube is heated from the inside by the process fluid, conduction then diffuses the heat in the flange, which in turn heats the air, causing a convective flow due to buoyancy effects. The cooling performance of the flange therefore depends on both convection and conduction.

To simplify the simulation convection cooling is analyzed using the heat transfer coefficient, h. Since this describes the fluid flow influence and the convective fluxes, the flow field does not have to be computed.

This particular example uses external research data (Ref. 1) for the outer surface heat transfer coefficient, which was originally obtained through semi-empirical data for natural convection around a cylinder. For the tube-facing surface a coefficient for forced convection in a tube is used.

Model Definition

Figure 2 presents the modeled geometry.

Figure 2: Drawing of the cooling flange.

The glass flange consists of a 4 mm thick pipe and 4 mm thick and 10 mm tall flanges, with a connecting pipe that has a 3 mm thick wall and an inner diameter of 16 mm. Note that in this tutorial, a parametric study on the pipe inner diameter is performed while the other dimensions remain fixed.

During operation, the hot process fluid heats the inside of the tube. The flange conducts the heat and transfers it to the surrounding air. As the air is heated, buoyancy effects cause a convective flow.

The heat transfer within the flange is described by the stationary heat equation

where k is the thermal conductivity (W/(m·K)), and T is the temperature (K). On the flange’s exterior boundaries, which face the air and process fluid, the applicable boundary condition is

where n is the normal vector of the boundary, h, is the heat transfer coefficient (W/(m2·K)), and Tinf is the temperature of the surrounding medium (K). For this simulation, set Text to 298 K for the cooling air and to 363 K for the process fluid.

You can approximate the value for the heat transfer coefficient, h, on the process fluid side with a constant value of 15 W/(m2·K) because the fluid’s velocity is close to constant and the model assumes that its temperature decreases only slightly.

The h expression on the air side is more elaborate. Assume that the free-convection process around the flange is similar to that around a cylinder. The heat transfer coefficient for a cylinder is available in the literature (Ref. 1), and you can use the expression

where k is the thermal conductivity of air (26.2 mW/(m·K) at 298 K); L is the typical length, which in this case is the outer diameter of the flange (44 mm); and f(θ) is an empirical coefficient tabulated in Table 1 as a function of the incidence angle θ, which is shown in Figure 3. Finally, Gr is the Grashof number defined as

where αp is the coefficient of thermal expansion (1/K), which equals that for an ideal gas, g is the gravitational acceleration (9.81 m/s2), and ν is the kinematic viscosity
(18·10−6 m2/s). For the flange material, use silica glass.

Table 1: Empirical transfer coefficient vs. incidence angle.
θ (deg)	f(θ)
0	0.48
90	0.46
100	0.45
110	0.435
120	0.42
130	0.38
140	0.35
150	0.28
160	0.22
180	0.15

Figure 3: Definition of the incidence angle θ.

Results and Discussion

Figure 4 shows the flange surface temperature at steady state.

Figure 4: Stationary surface temperature of the flange.

As you can see in the surface temperature plot above to the left, the tube surface temperature is about 13 K higher than the flange shoulders. From the results, we can learn that the heat transfer from the outer surfaces of the flange is pretty efficient; there is a temperature difference of roughly 19 K between the outer flange surface and the air stream.

Figure 5 shows the heat transfer coefficient for the flange’s outer walls.

Figure 5: Heat transfer film coefficient, h, for the flange’s outer walls.

As you can see, the coefficient decreases significantly along the vertical position of the flange’s outer boundary.

Calculating the flange’s total cooling power by integrating the normal total heat flux over the outer surfaces gives a value of 0.51 W which correspond to 1.02 W for the full geometry.

The surface temperature plot shown on Figure 4 further gives light to an inefficiency in the current flange design. When it enters the flange the process fluid is 363 K, while the inner surface of the pipe is only 32 K lower. Increasing the tube diameter would improve the heat transfer here, which is performed in a second stage of the modeling by varying the pipe’s diameter, but keeping all other factors constant. By plotting in Figure 6 the global cooling power as a function of the inner pipe radius, you can now analyze how altering the radius impacts the performance of the cooling flange.

Figure 6: Total cooling power versus inner pipe radius.

Reference

1. B. Sundén, Kompendium i värmeöverföring [Notes on Heat Transfer], Sec. 10-3, Dept. of Heat and Power Engineering, Lund Inst. of Technology, 2003 (in Swedish).

Heat_Transfer_Module/Thermal_Processing/cooling_flange

Modeling Instructions

From the menu, choose .

New

In the window, click

Model Wizard

In the window, click

In the tree, select .

Click .

Click

In the tree, select .

Click

Global Definitions

Parameters 1

Define parameters for relevant temperatures, material properties, and geometric dimensions. The parameters related to the geometry dimensions make it easier to perform parametric studies where you let some of these dimensions vary.

In the window, under click .

In the window for , locate the section.

In the table, enter the following settings:


Name	Expression	Value	Description
k	26.2[mW/(m*K)]	0.0262 W/(m·K)	Thermal conductivity, air
T_inner	363[K]	363 K	Temperature, process fluid
Hh	15[W/(m^2*K)]	15 W/(m²·K)	Heat transfer coefficient
nu	18e-6[m^2/s]	1.8E-5 m²/s	Kinematic viscosity
r_inner	8[mm]	0.008 m	Inner pipe radius
l1	12[mm]	0.012 m	Pipe length excluding flanges
t1	3[mm]	0.003 m	Pipe thickness
t2	4[mm]	0.004 m	Pipe thickness, flange section
hf	10[mm]	0.01 m	Flange height
wf	4[mm]	0.004 m	Flange width
D	2*(r_inner+t2+hf)	0.044 m	Outer flange diameter

Next, create an interpolation function defined by the data in Table 1.

Interpolation 1 (int1)

In the toolbar, click

and choose .

In the window for , locate the section.

In the text field, type f.

In the table, enter the following settings:


t	f(t)
0	0.48
90	0.46
100	0.45
110	0.435
120	0.42
130	0.38
140	0.35
150	0.28
160	0.22
180	0.15

Locate the section. In the table, enter the following settings:


Argument	Unit
t	deg

In the table, enter the following settings:


Function	Unit
f	1

Click

Geometry 1

In a first geometry modeling stage, create a 2D geometry by following the steps below.

Work Plane 1 (wp1)

In the toolbar, click

In the window for , click

Work Plane 1 (wp1)>Plane Geometry

In the window, click .

Work Plane 1 (wp1)>Rectangle 1 (r1)

In the toolbar, click

In the window for , locate the section.

In the text field, type 2*wf.

In the text field, type t2.

Locate the section. In the text field, type r_inner.

In the toolbar, click

Click the

button in the toolbar.

Work Plane 1 (wp1)>Quadratic Bézier 1 (qb1)

In the toolbar, click

and choose .

In the window for , locate the section.

In row , set to r_inner+t2.

In row , set to wf/2, and to r_inner+t2.

In row , set to wf/2, and to r_inner+t2+wf/2.

Work Plane 1 (wp1)>Line Segment 1 (ls1)

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Locate the section. From the list, choose .

Locate the section. In the text field, type wf/2.

Locate the section. In the text field, type 3*wf/2.

Locate the section. In the text field, type r_inner+t2+wf/2.

Work Plane 1 (wp1)>Quadratic Bézier 2 (qb2)

In the toolbar, click

and choose .

In the window for , locate the section.

In row , set to 3*wf/2, and to r_inner+t2+wf/2.

In row , set to 3*wf/2, and to r_inner+t2.

In row , set to 2*wf, and to r_inner+t2.

Work Plane 1 (wp1)>Line Segment 2 (ls2)

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Locate the section. From the list, choose .

Locate the section. In the text field, type r_inner+t2.

Locate the section. In the text field, type 2*wf.

In the text field, type r_inner+t2.

Drag and drop below .

Work Plane 1 (wp1)>Convert to Solid 1 (csol1)

In the toolbar, click

and choose .

Select the objects , , , and only.

In the toolbar, click

Click the

button in the toolbar.

Work Plane 1 (wp1)>Rectangle 2 (r2)

In the toolbar, click

In the window for , locate the section.

In the text field, type wf.

In the text field, type hf-wf.

Locate the section. In the text field, type wf/2.

In the text field, type r_inner+t2+wf/2.

In the toolbar, click

Click the

button in the toolbar.

Work Plane 1 (wp1)>Circle 1 (c1)

In the toolbar, click

In the window for , locate the section.

In the text field, type wf/2.

In the text field, type 180.

Locate the section. In the text field, type wf.

In the text field, type r_inner+t2+hf-wf/2.

In the toolbar, click

Click the

button in the toolbar.

Work Plane 1 (wp1)>Array 1 (arr1)

In the toolbar, click

and choose .

Click in the window and then press Ctrl+A to select all objects.

In the window for , locate the section.

From the list, choose .

In the text field, type 3.

Locate the section. In the text field, type 2*wf.

In the toolbar, click

Click the

button in the toolbar.

Work Plane 1 (wp1)>Polygon 1 (pol1)

In the toolbar, click

In the window for , locate the section.

From the list, choose .

Locate the section. From the list, choose .

In the text field, type 0 0 -l1 -l1 -l1 -l1 -2*l1/3.

In the text field, type r_inner+t2 r_inner r_inner r_inner r_inner+t1 r_inner+t1 r_inner+t1.

Work Plane 1 (wp1)>Cubic Bézier 1 (cb1)

In the toolbar, click

and choose .

In the window for , locate the section.

In row , set to -2*l1/3, and to r_inner+t2.

In row , set to -l1/3, and to r_inner+t2.

In row , set to r_inner+t2.

Work Plane 1 (wp1)>Convert to Solid 2 (csol2)

In the toolbar, click

and choose .

Select the objects and only.

In the toolbar, click

Click the

button in the toolbar.

In the window, right-click and choose .

You obtain the following 2D geometry.

Next, revolve the embedded 2D geometry to create the 3D model geometry.

Revolve 1 (rev1)

In the toolbar, click

In the window for , locate the section.

Click the button.

In the text field, type 180.

Locate the section. Find the subsection. In the text field, type 1.

In the text field, type 0.

In the toolbar, click

Click the

button in the toolbar.

Definitions

Variables 1

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Select Boundaries 3, 7, 9, 10, 13, 19, 23, 25, 29, 32, 35, 38–42, 45, 51, 55, 57, 61, 64, 67, 70–74, 77, 83, 87, 89, 93, 96, 99, and 102–106 only.

To do this, first copy the list of boundaries from this text, then use the button in the section, to paste these numbers.

For use when specifying the boundary condition for the flange’s outer surface, create a selection.

Click

In the dialog box, type Outer Boundaries in the text field.

Click .

In the window for , locate the section.

In the table, enter the following settings:


Name	Expression	Unit	Description
alphap	1/ampr1.T_amb		Coefficient of thermal expansion
Gr	g_constalphap(T-ampr1.T_amb)*D^3/nu^2		Grashof number
theta	atan(y/z)+90[deg]	rad	Incidence angle
Hc	kf(theta)Gr^0.25/D		Heat transfer film coefficient

Add Material

In the toolbar, click

to open the window.

Go to the window.

In the tree, select .

Click in the window toolbar.

In the toolbar, click

to close the window.

Definitions

Ambient Properties 1 (ampr1)

In the toolbar, click

and choose .

In the window for , locate the section.

In the Tamb text field, type 298[K].

Heat Transfer in Solids (ht)

The next steps set the temperature shape function to in order to reduce computational costs without significant deterioration in accuracy.

In the window, under click .

In the window for , click to expand the section.

From the list, choose .

Initial Values 1

In the window, under click .

In the window for , locate the section.

In the T text field, type 323.15[K].

Symmetry 1

In the toolbar, click

and choose .

Click the

button in the toolbar.

Select Boundaries 2, 8, 12, 15, 21, 22, 24, 27, 33, 34, 44, 47, 53, 54, 56, 59, 65, 66, 76, 79, 85, 86, 88, 91, 97, and 98 only.

Heat Flux 1

In the toolbar, click

and choose .

Select Boundaries 4, 6, 16, 18, 48, 50, 80, and 82 only.

In the window for , locate the section.

From the list, choose .

In the h text field, type Hh.

In the Text text field, type T_inner.

Click the

button in the toolbar to go back to the original view.

Heat Flux 2

In the toolbar, click

and choose .

In the window for , locate the section.

From the list, choose .

Locate the section. From the list, choose .

In the h text field, type Hc.

From the Text list, choose .

Mesh 1

Mapped 1

In the toolbar, click

and choose .

Click the

button in the toolbar. Do this twice to see the boundary mesh.

Select Boundaries 8, 21, 22, 33, 53, 54, 65, 85, 86, and 97 only.

Size 1

In the toolbar, click and choose .

In the window for , click

Free Triangular 1

In the toolbar, click

and choose .

Select Boundaries 34, 66, and 98 only.

In the window for , click

Click the

button in the toolbar.

Swept 1

In the toolbar, click

Distribution 1

Right-click and choose .

In the window for , locate the section.

In the text field, type 30.

Click

Study 1

In the toolbar, click

Results

Temperature (ht)

The first default plot group shows the temperature field on the surface. To get a more intuitive view with gravity along the vertical, rotate the geometry in the window. You can preserve a view for a plot by creating a feature node as follows:

Click the

button in the toolbar.

In the dialog box, in the tree, select the check box for the node .

Click .

View 3D 3

In the window, under right-click and choose .

Use the toolbox to get a satisfying view.

In the window for , locate the section.

Select the check box.

Next, apply the view to the temperature plot.

Temperature (ht)

In the window, under click .

In the window for , locate the section.

From the list, choose .

In the toolbar, click

Compare with Figure 4.

Delete the second plot group to make a new surface plot of the heat transfer film coefficient.

Isothermal Contours (ht)

In the window, under right-click and choose .

Click to confirm.

Heat Transfer Film Coefficient

In the toolbar, click

and choose .

In the window for , type Heat Transfer Film Coefficient in the text field.

Locate the section. From the list, choose .

Surface 1

In the toolbar, click

In the window for , click in the upper-right corner of the section. From the menu, choose .

In the toolbar, click

Compare this plot with that in Figure 5.

Outgoing Heat Flux

In the toolbar, click

and choose .

In the window for , type Outgoing Heat Flux in the text field.

Locate the section. From the list, choose .

Click in the upper-right corner of the section. From the menu, choose .

Click

Table

Go to the window.

The integrated value, approximately 0.51 W, appears in the tab below the window. Taking both flange halves into account, the total cooling power of the flange is thus roughly 1 W.

Finally, extend the model by performing a parametric sweep over the inner pipe radius. Before adding a separate study for this purpose, define a variable for the total cooling power.

Definitions

Integration, Outer

In the toolbar, click