Thermocouple Response Time Calculation for Gas Stream Temperature Measurement
This problem involves determining the time it takes for a thermocouple junction to reach a specific temperature when exposed to a different environment. This is a classic transient heat transfer problem that can often be solved using the lumped system analysis, provided the internal resistance to heat conduction is small compared to the resistance to convection.
Lumped System Analysis Principle
The lumped system analysis assumes that the temperature within the object (the thermocouple bead in this case) is uniform at any given time. This is valid when the Biot number ($Bi$) is less than 0.1. The transient heat transfer is governed by Newton's law of cooling:
$$ \rho V c \frac{dT}{dt} = -h A (T - T_\infty) $$Where:
- $\rho$ = Density of the thermocouple material ($\frac{kg}{m^3}$)
- $V$ = Volume of the thermocouple bead ($m^3$)
- $c$ = Specific heat of the thermocouple material ($\frac{J}{kg \cdot K}$)
- $T$ = Temperature of the thermocouple bead at time $t$ ($^\circ C$)
- $t$ = Time ($s$)
- $h$ = Convective heat transfer coefficient ($\frac{W}{m^2 \cdot K}$)
- $A$ = Surface area of the thermocouple bead ($m^2$)
- $T_\infty$ = Temperature of the surrounding fluid (gas stream) ($^\circ C$)
The solution to this first-order ordinary differential equation is:
$$ \frac{T(t) - T_\infty}{T_i - T_\infty} = e^{-\left(\frac{hA}{\rho V c}\right)t} $$The term $\tau = \frac{\rho V c}{hA}$ is known as the thermal time constant.
$$ \frac{T(t) - T_\infty}{T_i - T_\infty} = e^{-t/\tau} $$Rearranging to solve for time $t$:
$$ t = -\tau \ln\left(\frac{T(t) - T_\infty}{T_i - T_\infty}\right) $$Calculating Necessary Parameters
First, let's list the given values and calculate the required geometric parameters for the spherical thermocouple junction.
- Diameter, $D = 0.706~\text{mm} = 0.706 \times 10^{-3}~\text{m}$
- Radius, $r = D/2 = 0.353 \times 10^{-3}~\text{m}$
- Convective heat transfer coefficient, $h = 400~\frac{W}{m^2 \cdot K}$
- Thermal conductivity, $k = 20~\frac{W}{m \cdot K}$
- Specific heat, $c = 400~\frac{J}{kg \cdot K}$
- Density, $\rho = 8500~\frac{kg}{m^3}$
- Initial temperature, $T_i = 30^\circ C$
- Gas stream temperature, $T_\infty = 300^\circ C$
- Target temperature, $T(t) = 298^\circ C$
Geometric Properties of the Sphere
For a sphere:
- Surface Area, $A = \pi D^2$
- Volume, $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{\pi D^3}{6}$
The ratio $\frac{A}{V}$ is crucial for calculating the time constant:
$$ \frac{A}{V} = \frac{\pi D^2}{\frac{\pi D^3}{6}} = \frac{6}{D} $$ $$ \frac{A}{V} = \frac{6}{0.706 \times 10^{-3}~\text{m}} \approx 8498.58~\text{m}^{-1} $$Calculating the Thermal Time Constant ($\tau$)
Now, we can calculate the thermal time constant $\tau$:
$$ \tau = \frac{\rho V c}{hA} = \frac{\rho c}{h \left(\frac{A}{V}\right)} $$ $$ \tau = \frac{(8500~\frac{kg}{m^3}) (400~\frac{J}{kg \cdot K})}{(400~\frac{W}{m^2 \cdot K}) \left(\frac{6}{0.706 \times 10^{-3}~\text{m}}\right)} $$ $$ \tau = \frac{8500 \times 400}{400 \times \frac{6}{0.706 \times 10^{-3}}} = \frac{8500 \times 0.706 \times 10^{-3}}{6} $$ $$ \tau = \frac{5990 \times 10^{-3}}{6} = \frac{5.99}{6} \approx 0.9983~\text{seconds} $$Calculating the Time Taken
We need to find the time $t$ when the thermocouple temperature reaches $298^\circ C$. Plugging the values into the rearranged formula:
$$ t = -\tau \ln\left(\frac{T(t) - T_\infty}{T_i - T_\infty}\right) $$Substitute the temperatures:
$$ \frac{T(t) - T_\infty}{T_i - T_\infty} = \frac{298^\circ C - 300^\circ C}{30^\circ C - 300^\circ C} = \frac{-2}{-270} = \frac{2}{270} = \frac{1}{135} $$Now calculate the time $t$:
$$ t = -(0.9983~\text{s}) \ln\left(\frac{1}{135}\right) $$Using the property $\ln(1/x) = -\ln(x)$:
$$ t = (0.9983~\text{s}) \ln(135) $$Calculate the natural logarithm of 135:
$$ \ln(135) \approx 4.9053 $$Finally, calculate the time:
$$ t \approx (0.9983~\text{s}) \times 4.9053 \approx 4.900~\text{seconds} $$Conclusion
The time taken for the thermocouple bead to reach $298^\circ C$ is approximately 4.9 seconds. This result aligns with option 4.
