GSO002 Problem 8

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 $\mathrm{A, B, C, D}$A,B,C,D. The elements connected between terminals are:

• Between $\mathrm{A}$A and $\mathrm{B}$B: RTD with resistance $R_1$R1​
• Between $\mathrm{B}$B and $\mathrm{C}$C: fixed resistor $R_2$R2​
• Between $\mathrm{C}$C and $\mathrm{D}$D: variable resistor $R_3$R3​
• Between $\mathrm{D}$D and $\mathrm{A}$A: fixed resistor $R_4$R4​

A DC power supply of voltage $V_0$V0​ is connected between terminals $\mathrm{A}$A and $\mathrm{C}$C (positive terminal at $\mathrm{A}$A). A galvanometer is connected between terminals $\mathrm{B}$B and $\mathrm{D}$D. The internal resistance of the power supply and wire resistance are negligible. The fixed resistors satisfy $R_2 = R_4 = R_c$R2​=R4​=Rc​.

The RTD $R_1$R1​ has resistance as a function of its own temperature $T_s$Ts​:

$$R_1(T_s) = R_0 \{1 + \alpha (T_s - T_0)\}$$R1​(Ts​)=R0​{1+α(Ts​−T0​)}

where $T_0$T0​ is the reference temperature, $R_0$R0​ is the resistance at reference temperature, and $\alpha$α is the temperature coefficient of resistance.

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

$$P = K (T_s - T_f)$$P=K(Ts​−Tf​)

where $K$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 $R_3$R3​ is adjusted until the galvanometer current reaches zero (bridge balance) at $R_3 = R_x$R3​=Rx​.

Derive the true liquid temperature $T_f$Tf​.

Constraints

• DC supply voltage: $V_0 = 14 \, \mathrm{V}$V0​=14V
• Fixed resistor value: $R_c = 120 \, \Omega$Rc​=120Ω
• Variable resistor at balance: $R_x = 90 \, \Omega$Rx​=90Ω
• Reference temperature: $T_0 = 300 \, \mathrm{K}$T0​=300K
• RTD resistance at reference temperature: $R_0 = 100 \, \Omega$R0​=100Ω
• Temperature coefficient of resistance: $\alpha = 5.0 \times 10^{-3} \, \mathrm{K^{-1}}$α=5.0×10−3K−1
• Heat dissipation constant: $K = 8.0 \times 10^{-2} \, \mathrm{W/K}$K=8.0×10−2W/K

Input Format

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