EEEE3089 Seminar Lecture 4 examples
Temperature sensor examples RTD
RTD we use:
What is R if t = 180 degrees celcius? the device is a pt100, ppm
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R = 169.3000
Real devices are not usually perfectly linear so a data sheet might give you additional parameters, and the equation that defines the response.
clear variables
alpha = 3850*1e-6 ; %converting from ppm
R0 = 100 ; % this is what the 100 refers to
R = R0.*(1+alpha.*t); %ohms
display(R)
delta = 1.4999;
t2 = (0:800);
A = alpha+ ((alpha.*delta)/100);
B = ((-alpha.*delta)/(100*100));
R = R0.*(1+alpha.*t2);
R2 = R0.*(1+A.*t2+(B.*(t2.*t2)));
figure;plot(t2,R,t2,R2)
xlabel(‘Temperature (deg C)’);
ylabel(‘Resistance (Ohms)’);
legend(‘linear’,’real device’)
This example shows the linear and 2 paramter equation produce near identical values around the calibration temperature (zero degrees celcius) but the difference becomes noticable for large temperature changes. There are equations with even more parameters to allow better calibration over wider temperature ranges, though this is something we will not consider.
NTC Example
We have an NTC device with R0 of 10kOhm at T0 25 degrees celcius, and a beta parameter of 3435K. Let’s plot the resistance versus temperature for this device. Recall the equation for the response is:
T0 = 25+273;
T=(-20:100)+273 ;
Beta is in kelvin so temps in kelvin.
If you don’t convert the temperature and plot over a temperature range where T goes through zero you’ll have 1/0 in the exponent which should tell you somethig is wrong…
beta = 3435
beta = 3435
We can plot for degrees celcius – just remove the offset for = R0.*exp(beta.*((1./T)-(1/T0)));
plot(T,Rt);
xlabel(‘Temperature Kelvin’)
ylabel(‘Resitance Ohms’)
plot(T-273,Rt);
xlabel(‘Temperature Celcius’)
ylabel(‘Resitance Ohms’)
we can check this – we expect 10kOhms at 25 degrees Celcius, this looks correct from the graph.
Another NTC Example
Lets work out beta from the given values.
R0 is 10k at the calibration temperature of 25 degreees celcius, and Rt is measurfed to be 3838 ohms at 52 degrees celcius.
What is the beta value of this NTC device?
We need to do some rearranging of the equation.
B = log(3.838/10)/((1/T)-(1/T0))
T0 =273+25
Rt=3.838e3
; %25deg C in kelvin
; %52deg C in kelvin
; %3838 ohms
B = 3.4351e+03
units of this are in kelvin, we can tidy up the equation :
B = ((T0*T)/(T0-T))*log(Rt/R0)
B = 3.4351e+03
This form is a little neater and easier to input values.
As long as we know the calibration temperature and nominal resitance at the calibration temperature we can use this form.
if we don’t know these then we must make multiple measurements at known temperatures and then solve the resulting simultaneous equations.
Thermocouple Example
When we use the seebeck table to work out the sensitivity, we are assuming the response is linear, we get the voltage for a given temperatrure difference between the junctions.
close to the cold junction temperature the response is linear.
T = -20:420;
alpha = 40e-6; % this is essentially S for the thermocouple
emf_lin = c+(alpha.*(T));
plot(T(1:60),emf_lin(1:60));
xlabel(‘Temperature T’)
ylabel(‘V’)
axis([-20 50 -1e-3 1e-3])
Using the full equation gives the actual response of the device over a wide temperature range. Here close to the cold junction temperature the response is linear.
T = -20:420;
alpha = 40e-6; % this is essentially S for the thermocouple
beta = -0.1e-6;
emf = c+(alpha.*(T))+beta.*(T).*(T);
emf_lin = c+(alpha.*(T));
plot(T(1:60),emf_lin(1:60),T,emf);
xlabel(‘Temperature T’)
ylabel(‘V’)
legend(‘linear’,’response’)
axis([-20 50 -1e-3 1e-3])
Over a wide range the response is much different
plot(T,emf);
xlabel(‘Temperature T’)
ylabel(‘V’)
we can work out the neutral temperature and the inversion temperature (see notes)
Tn = (-alpha/(2*beta))
Tn = 200.0000
Ti = -alpha/beta
Ti = 400.0000
Pyrometer Example
We have the equation for the emitted radiation as:
Emissivity – materials and surface conidtion dependant constant = 5.670367*10^-8 W/m^2 /K^4
What is the thermal raditation (W/m^2) emitted by a plaster wall at 50 degrees celcius? Here = 0.89 (see table in notes) we plug in the values and see what we get.
T = 50+273; % in kelvin
J = 5.670367e-8*0.89*T^4 % in W/m^2 J = 549.3021
What is the temperature of aluminium foil, if the thermal radiation given off is measured to be 100W/m^2? (see table in notes)
rearrange the equation to get T
T = nthroot(100/(5.670367e-8*0.03),4)
T = 492.3987
T_c = T-273
T_c = 219.3987
% What is the temperature of Glass, if the thermal radiation given off is measured to be 5kW/m^2? (see table in notes)
T = nthroot(5000/(5.670367e-8*0.95),4)
T = 551.9615
T_c = T-273
T_c = 278.9615
For foil the emissivity was so low that the very low recorded thermal radiation corresponds to a high temperature. The glass value is ~50 times higher but is a similar temperature.
In the case of foil other background objects can easily make the reading inaccurate even if they are at a much lower temperature.
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