Question

Over a limited temperature range, the relation between electrical resistance \(R\) and temperature \(T\) for a resistance temperature detector is $$ R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right] $$ where \(R_{0}\) is the resistance, in ohms \((\Omega)\), measured at reference temperature \(T_{0}\) (in \(\left.{ }^{\circ} \mathrm{C}\right)\) and \(\alpha\) is a material constant with units of \(\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The following data are obtained for a particular resistance thermometer: \begin{tabular}{lcrr} \hline & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \multicolumn{2}{c}{\(R(\Omega)\)} \\ \hline Test \(1\left(\mathrm{~T}_{0}\right)\) & 0 & \(\left(\mathrm{R}_{0}\right)\) & \(51.39\) \\ Test 2 & 91 & \(51.72\) \\ \hline \end{tabular} What temperature would correspond to a resistance of \(51.47 \Omega\) on this thermometer?

Short answer

The temperature corresponding to a resistance of 51.47Ω is around 23.22°C.

Step-by-step solution

Understanding the given formula

The relation between resistance, R, and temperature, T, is given by: \[ R = R_0 \big[ 1 + \alpha(T - T_0) \big] \] where \( R_0 \) is the resistance at reference temperature \( T_0 \), and \( \alpha \) is the temperature coefficient of resistance.

Identify the given values

From the problem, the following resistance and temperature values are given: 1. \( T_0 = 0^{\circ}\mathrm{C} \) and \( R_0 = 51.39 \Omega \)2. \( T = 91^{\circ}\mathrm{C} \) and \( R = 51.72 \Omega \).

Calculate the material constant \alpha

Use the formula to solve for \( \alpha \): \[ R = R_0 \big[ 1 + \alpha(T - T_0) \big] \] Substitute the known values from Test 2: \[ 51.72 = 51.39 \big[ 1 + \alpha(91 - 0) \big] \]Rearrange to solve for \( \alpha \): \[ 51.72 = 51.39 \big[ 1 + 91 \alpha \big] \]\[ 1 + 91 \alpha = \frac{51.72}{51.39} \]\[ 91 \alpha = \frac{51.72}{51.39} - 1 \]\[ \alpha = \frac{\frac{51.72}{51.39} - 1}{91} \]Calculate \( \alpha \).

Substitute values for T

The next step is to find the temperature (T) when the resistance (R) is given as \( 51.47 \Omega \). Use the formula: \[ R = R_0 \big[ 1 + \alpha(T - T_0) \big] \]Substitute known values: \[ 51.47 = 51.39 \big[ 1 + \alpha(T - 0) \big] \]Knowing the value of \( \alpha \) from the previous step, solve for T.

Solve for temperature T

Rearrange the formula to solve for T: \[ 51.47 = 51.39 \big[ 1 + \alpha T \big] \]\[ 1 + \alpha T = \frac{51.47}{51.39} \]\[ \alpha T = \frac{51.47}{51.39} - 1 \]\[ T = \frac{\frac{51.47}{51.39} - 1}{\alpha} \]. Substitute the value of \( \alpha \) to obtain T.

Key concepts

Electrical Resistance

Electrical resistance is a measure of the opposition to the flow of electric current through a material. It is denoted by the symbol \(R\) and measured in ohms (\(\Omega\)). Resistance determines how much current will flow through a component for a given voltage according to Ohm's Law: \(V = IR\), where \(V\) is voltage, \(I\) is current, and \(R\) is resistance.

In a resistance temperature detector (RTD), the resistance changes with temperature. This property makes RTDs suitable for measuring temperature precisely. Essentially, as temperature increases, the resistance typically increases.

When analyzing an RTD, understanding how resistance correlates with temperature is crucial. The linear relationship provided explains this, where the resistance at any given temperature can be found if you know the resistance at a reference temperature and the material's temperature coefficient.

Temperature Coefficient

The temperature coefficient is denoted by the symbol \(\alpha\) and has units of \(\left(\degree C \right)^{-1}\). It represents the rate at which the resistance changes per degree of temperature change. For most metals used in RTDs, \(\alpha\) is positive, indicating that resistance increases with temperature.

In calculating \(\alpha\), you utilize the formula:
\[ R = R_0 \left[ 1 + \alpha (T - T_0) \right] \],
Here, \(R\) is the resistance at temperature \(T\), \(R_0\) is the resistance at a reference temperature \(T_0\), and \(\alpha\) is the temperature coefficient.

For example, given:
- \(T_0 = 0\degree C\)
- \(R_0 = 51.39 \Omega\)
- \(T = 91\degree C\)
- \(R = 51.72 \Omega\),
We derive \(\alpha\) using:
\[ 51.72 = 51.39 \left[ 1 + 91\alpha \right] \]
Simplifying this, you will calculate the precise value of \(\alpha\), demonstrating how the resistance of the material increases per degree Celsius change from the reference temperature.

Calculation of Temperature

Once you've determined the temperature coefficient \(\alpha\), you can find the temperature corresponding to any given resistance. Using the formula:
\[ R = R_0 \left[ 1 + \alpha (T - T_0) \right] \],
and knowing \(R\), \(R_0\), and \(\alpha\), you can solve for \(T\).

Given:
- \(R_0 = 51.39 \Omega\)
- \(R = 51.47 \Omega\)
- Found \(\alpha\) (from previous steps),
substitute these into the equation and solve for T:
\[ 51.47 = 51.39 \left[ 1 + \alpha T \right] \],
Simplifying, you get:
\[ T = \frac{\left(\frac{51.47}{51.39} - 1\right)}{\alpha} \]
This calculation will determine the temperature corresponding to the resistance of \(51.47 \Omega\) using RTD properties.

In summary, by understanding electrical resistance, the temperature coefficient, and the corresponding calculations, you can effectively use an RTD to measure temperatures accurately.