What is in this article?:
- Design Tips on Dielectric Waveguide
- Plating and preparing the raw substrate
- Attenuation of plated-dielectric waveguide & joining waveguide and fitting flanges
- Rf heating effect and the effect of solar heat
- Heat transfer and temperature stability
- Lightweight waveguide components, antennas, and feed systems
Rf heating effect and the effect of solar heat
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To determine the theoretical temperature rise of the waveguide core, the following formula can be used:
Q = heat flowing along the waveguide and discharge in BTU/hr;
C = conductivity in BTU/hr/ft2/°F/in.;
A = cross section of the waveguide in in.2; and
l = length of the waveguide in feet.
t2 = guide temperature with power flowing
t1 = cold guide temperature
To measure heating effects along a length of copper-plated dielectric waveguide, test equipment can be set up as shown in Fig. 5.
Typical test conditions might be:
Rf input power, 320 W cw;
Frequency, 3.5 GHz;
Heat sensor, 1, 12 in. from input flange;
Heat sensors 2, 3, 4 and 5 located along the waveguide as shown;
Heat sensor 6 set to sample the ambient temperature;
Heat sensors to be calibrated against a bulb thermometer prior to test.
A typical set of measured test results are shown in Fig. 6.
Effect of solar heat
Where the waveguide is located outdoors, the effects of solar heat must also be considered. The formula for heat transfer Q is:
k = constant depending on the material
A = area under the sun in ft2
∆X = average distance the heat travels from the core of the waveguide to one side.
The solar heat felt on the antenna surface depends on the sun angle and the antenna axis with respect to the sun. If the antenna is rotating, the solar heat felt will vary from maximum to minimum. How to calculate the average effect of the sun is illustrated by an example: For the antenna in our test, the maximum solar heat felt on its surface was 270 BTU/hr/ft2. If rotating, the average heat is: (1/2)(270 BTU/hr/ft2)(sin 45°) = 95.4 BTU/hr/ft2.
A black surface will absorb 0.9 (95.4) = 85.9 BTU/hr/ft2.
A medium surface will absorb 0.7 (95.4) = 66.8 BTU/hr/ft2.
A light surface will absorb 0.4 (95.4) = 38.2 BTU/hr/ft2.