Thermocouple time to indicate temp

In summary, the homework statement asks for a solution to a problem where the user does not know how to start, and needs help with figuring out what equations to use. The user also needs to know some information about the junction material, such as its density and its geometry. They can then use that information to find the mass of the junction material.
  • #1
williamcarter
153
4

Homework Statement


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Homework Equations


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The Attempt at a Solution


I do not know exactly how to start this problem.
I would really appreciate it if you could give me some hints.
I know that e^ - t/tau=T(t)-Twall/Ti-Twall
where tau=m*cp/hA
where h=convective heat tr.coeff
and T(t) is desired temp that we need to get to.
The exercise says that T(t)=99%Tinitial
We also know that tau=mcp/UA=mcp/hA
 
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  • #2
You can begin by calculating the things that are easy to find given the Relevant Equations in your second image. For example, you should be able to determine the time constant of the system from the given information.

Note that the problem doesn't say that T(t) is 99% Tinitial. It says that 99% of the initial temperature difference is indicated. So try to formulate the system decay equation in terms of ΔT rather than Ti and T.

Another thing to keep in mind as you go, and as a check on your result, is that an engineering rule of thumb for exponential decay is that most of the excitement is all over after five time constants :wink:
 
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  • #3
gneill said:
You can begin by calculating the things that are easy to find given the Relevant Equations in your second image. For example, you should be able to determine the time constant of the system from the given information.

Note that the problem doesn't say that T(t) is 99% Tinitial. It says that 99% of the initial temperature difference is indicated. So try to formulate the system decay equation in terms of ΔT rather than Ti and T.

Another thing to keep in mind as you go, and as a check on your result, is that an engineering rule of thumb for exponential decay is that most of the excitement is all over after five time constants :wink:

Thank you for your answer!

tau=m_dot*cp*delta T
however m_dot is not given.
I know m_dot=ro*u*A=vol flowrate*density
delta T also unknow
 
  • #4
Your second image has a formula for ##\tau## that doesn't use the flow rate. Rather it uses the given heat transfer coefficient between the gas and junction.
 
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  • #5
gneill said:
Your second image has a formula for ##\tau## that doesn't use the flow rate. Rather it uses the given heat transfer coefficient between the gas and junction.
Thank you for your reply,much appreciated.
the formula is tau=M*cp/h*A
I have cp,h,and A cand get from diameter.Don't know about the M
 
  • #6
williamcarter said:
Don't know about the M
What other information do you have about the junction material?
 
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  • #7
gneill said:
What other information do you have about the junction material?

Thank you for your reply.
This is all the info I have
Exercise:http://imgur.com/a/apwob
Formulas:http://imgur.com/a/nQGCp

I can get m_dot or u(velocity) from Dittus Boelter.
However I don't know Mew(Viscosity)
 
  • #8
Do you not have the density of the junction material? You also know its radius...
 
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  • #9
gneill said:
Do you not have the density of the junction material? You also know its radius...
Thank you for you answer.
By having D, I can get A=pi*D^2/4
and Density I have it as 8500kg/m^3
A*density*velocity=m_dot
I do not have the velocity, I can get it from Dittus-Boelter , however I do not know the viscosity Mew
 
  • #10
The only thing you really need to know about the gas is the heat transfer coefficient. The M in the time constant formula is the mass of the junction material.
 
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  • #11
gneill said:
The only thing you really need to know about the gas is the heat transfer coefficient. The M in the time constant formula is the mass of the junction material.
Thank you for reply,they do not provide me with M of junction material, how to find it?
 
  • #12
williamcarter said:
Thank you for reply,they do not provide me with M of junction material, how to find it?
They give you its material density. You also know its geometry. How does one find the mass of an object given its density?
 
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  • #13
gneill said:
They give you its material density. You also know its geometry. How does one find the mass of an object given its density?
ro=mass/volume
=>m=ro*V
 
  • #14
williamcarter said:
ro=mass/volume
=>m=ro*V
So what's the volume of the junction? What's its density? Hence, what's its mass?
 
  • #15
so m=density*Vol
density=8500kg/m^3;
Vol=vol sphere=4/3*pi*r^3=4/3*pi*(10^-3/2)^3=5.23*10^-10 m^3
=>m=8500*5.23^10^-10
m=4.45*10^-6 kg

Asphere=4*pi*r^2
Now tau=m*cp/h*A
tau=4.45*10^ - 6 *320/210*4*pi*(10^-3/2)^2
tau=2.21s
 
  • #16
williamcarter said:
so m=density*Vol
density=8500;
Vol=vol sphere=4/3*pi*r^3=4/3*pi*(10^-3/2)^3=5.23*10^-10 m^3
=>m=8500*5.23^10^-10
m=4.45*10^-6 kg

Asphere=4*pi*r^2
Now tau=m*cp/h*A
tau=4.45*10^ - 6 *320/210*4*pi*(10^-3/2)^2
tau=2.21

Okay, so ##\tau## is about 2.2 seconds.

Now you need to work on the exponential decay formula.
 
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  • #17
gneill said:
Okay, so ##/tau## is about 2.2 seconds.

Now you need to work on the exponential decay formula.

t(time)= - tau *ln(T(t)-Twall/Ti-Twall)
I need to know Tinitial,Twall and he said that T(t) is 99% of initial difference
 
  • #18
Start with the formula as given in your Relevant Equations image. You should be able to place ΔT's into the temperature difference specifications.
 
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  • #19
gneill said:
Start with the formula as given in your Relevant Equations image. You should be able to place ΔT's into the temperature difference specifications.

T(t) is 99% of initial temp difference
=>T(t)=99%*(Ti-Tw)

but e^ -t/tau=T(t)-Twall/Ti-Twall
now this becomes e^-t/tau = 99%*(Ti-Tw)-Twall/Ti-Twall

However I do not know Twall.
I am stuck here
 
  • #20
williamcarter said:
T(t) is 99% of initial temp difference
=>T(t)=99%*(Ti-Tw)

but e^ -t/tau=T(t)-Twall/Ti-Twall
now this becomes e^-t/tau = 99%*(Ti-Tw)-Twall/Ti-Twall

However I do not know Twall.
I am stuck here
The temperature portion of the given expression is:
$$\frac{T(t) - T_{\infty}}{T_i - T_{\infty}}$$
But ##T_i - T_{\infty}## is the initial ##\Delta T##, right? So what might ##T(t) - T_{\infty}## represent in terms of ##\Delta T##?
 
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  • #21
gneill said:
The temperature portion of the given expression is:
$$\frac{T(t) - T_{\infty}}{T_i - T_{\infty}}$$
But ##T_i - T_{\infty}## is the initial ##\Delta T##, right? So what might ##T(t) - T_{\infty}## represent in terms of ##\Delta T##?
99%deltaT?
Then time t= -tau*ln(99/100)
t=0.022s
 
  • #22
williamcarter said:
99%deltaT?
Then time t= -tau*ln(99/100)
t=0.022s
Nope. 99% of the ##\Delta T## is "used up". What remains?
 
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  • #23
gneill said:
Nope. 99% of the ##\Delta T## is "used up". What remains?
Ow yes,you are right, my bad ,I read that up wrong.
It will be 0.01 or 1%
t= - tau* ln(1/100)
t= -2.21* ln(1/100)
t=10.17s
 
  • #24
Good. Compare this with the "engineer's rule of thumb" approximation.
 
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  • #25
time constant tau =2.21
t=10.17s

Applying rule of thumb multiplying tau by 5
gives
5*tau=11.05 seconds, close to our t(that is 10.17s)
 
  • #26
Yup.
 
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  • #27
gneill said:
Yup.
Thank you very much for your professionalism and time, much appreciated!
 
  • #28
You're very welcome.
 

1. What is a thermocouple and how does it measure temperature?

A thermocouple is a type of sensor that measures temperature by utilizing the Seebeck effect, which is the conversion of temperature differences into electrical voltage. It consists of two dissimilar metal wires joined together at one end, called the hot junction, and connected to a voltmeter at the other end, called the cold junction. When there is a difference in temperature between the hot and cold junctions, a small voltage is produced, which can be used to determine the temperature.

2. How long does it take for a thermocouple to indicate the temperature?

The time it takes for a thermocouple to indicate the temperature depends on several factors such as the size and type of the thermocouple, the temperature range being measured, and the accuracy required. In general, most thermocouples can provide a temperature reading within a few seconds to a few minutes.

3. Can a thermocouple measure extreme temperatures?

Yes, a thermocouple can measure a wide range of temperatures, from -200°C to 1750°C. However, the type of thermocouple used will determine the maximum and minimum temperatures that can be accurately measured. Some thermocouples are designed specifically for high-temperature applications, while others are better suited for low-temperature measurements.

4. How accurate is a thermocouple in measuring temperature?

The accuracy of a thermocouple depends on several factors such as the type of thermocouple, the temperature range being measured, and the calibration method used. Generally, most thermocouples have an accuracy of around ±1°C to ±2°C, but this can vary depending on the conditions and specifications of the thermocouple.

5. Do thermocouples require any special handling or maintenance?

Thermocouples are relatively simple and durable devices, so they do not require any special handling or maintenance. However, they should be periodically checked and calibrated to ensure accurate temperature readings. Additionally, if the thermocouple is exposed to extreme temperatures or harsh environments, it may need to be replaced more frequently.

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