Wheatstone Bridge for a Resistance Temperature Detector with Self-Heating Error

Problem Statement

Consider measuring the temperature of a liquid using a resistance temperature detector (RTD) — an element whose resistance changes with temperature.

When current flows through the RTD, Joule heating raises the RTD's own temperature above that of the surrounding environment. This is known as "self-heating error." In this problem, we combine this thermal effect with circuit behavior to find the true liquid temperature.

A Wheatstone bridge circuit has four terminals $\mathrm{A, B, C, D}$ . The elements connected between terminals are:

Between $\mathrm{A}$ and $\mathrm{B}$ : RTD with resistance $R_1$
Between $\mathrm{B}$ and $\mathrm{C}$ : fixed resistor $R_2$
Between $\mathrm{C}$ and $\mathrm{D}$ : variable resistor $R_3$
Between $\mathrm{D}$ and $\mathrm{A}$ : fixed resistor $R_4$

A DC power supply of voltage $V_0$ is connected between terminals $\mathrm{A}$ and $\mathrm{C}$ (positive terminal at $\mathrm{A}$ ). A galvanometer is connected between terminals $\mathrm{B}$ and $\mathrm{D}$ . The internal resistance of the power supply and wire resistance are negligible. The fixed resistors satisfy $R_2 = R_4 = R_c$ .

The RTD $R_1$ has resistance as a function of its own temperature $T_s$ :

R_1(T_s) = R_0 \{1 + \alpha (T_s - T_0)\}

where $T_0$ is the reference temperature, $R_0$ is the resistance at reference temperature, and $\alpha$ is the temperature coefficient of resistance.

The Joule power dissipated by the RTD per unit time is $P$ . With the surrounding liquid temperature $T_f$ , Newton's law of cooling states that at thermal equilibrium (steady state):

P = K (T_s - T_f)

where $K$ is the heat dissipation constant representing how easily heat flows from the RTD to the liquid.

The RTD is immersed in a large amount of liquid, current flows through the circuit, and sufficient time passes to reach steady state. Then the variable resistor $R_3$ is adjusted until the galvanometer current reaches zero (bridge balance) at $R_3 = R_x$ .

Derive the true liquid temperature $T_f$ .

Constraints

DC supply voltage: $V_0 = 14 \, \mathrm{V}$
Fixed resistor value: $R_c = 120 \, \Omega$
Variable resistor at balance: $R_x = 90 \, \Omega$
Reference temperature: $T_0 = 300 \, \mathrm{K}$
RTD resistance at reference temperature: $R_0 = 100 \, \Omega$
Temperature coefficient of resistance: $\alpha = 5.0 \times 10^{-3} \, \mathrm{K^{-1}}$
Heat dissipation constant: $K = 8.0 \times 10^{-2} \, \mathrm{W/K}$

Input Format

Give the liquid temperature $T_f$ [ $\mathrm{K}$ ] as a positive integer (no units).

Problem Statement

Constraints

Input Format

Submit Answer