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a. The resistance thermometer bridge circuit shown in FIGURE 1 has a designed maximum temperature of 200°C, ignoring the effects of connecting wire resistance. If the connecting loop is 250 m determine the smallest gauge (swg) of copper wire which must be used if the indicated maximum temperature is to be less than 202°C.

The temperature coefficient of resistance of the thermometer is 0.0042 °C–1 and the resistance of the thermometer is 112 Ω at 20°C. Assume the connecting leads are at 20°C.

b. Using the gauge of copper wire calculated in part (a), calculate the maximum indicated temperature using a 3-wire system (as shown in FIGURE 2) over a distance of 125 m.Relevant EquationsRt = R0(1 + at)

R1RT = R2RS

R = pL/A = Resistance = (Resistivity) x Length of wire / cross-sectional area

A = (pi x d2) / 4

d = squareroot(4A /pi)

R_t=R_20×(1+α×∆t)

α=0.0042

R20 = 112Ω

∆t=202-20=182

R_t=R_20×(1+α×∆t)

R_t=112×(1+0.0042×182)

R_t= 197.6128

As both resistors on figure 1 are 150Ω, they cancel each other out.

Find the resistivity of copper at 20°C

ρ=RA÷L

Where:

R = Resistance of copper at 20°C

A = area

L = length

From the properties of resistance wires table:

When diameter = 0.376mm, resistance = 0.155m-1

0.376mm diameter = 0.11103645mm2 for the area.

Converting resistance to per mm = 0.000155m-1

For length 1m = 1000mm.

Therefore, ρ=0.000155×0.111036450÷1000=1.72×10^(-8)

Resistivity of copper wire at 20°C = 1.72×10^(-8)Ωm-1

To find the area the equation would be A=(ρ×L)/R

A=(1.72×10^(-5) Ωmm×250000mm)/197.6128Ω

A = 0.02176mm2

Find diameter = 0.16645mm

Too small of a diameter.

Any help would be greatly appreciated. Ive been stuck for a week

The temperature coefficient of resistance of the thermometer is 0.0042 °C–1 and the resistance of the thermometer is 112 Ω at 20°C. Assume the connecting leads are at 20°C.

b. Using the gauge of copper wire calculated in part (a), calculate the maximum indicated temperature using a 3-wire system (as shown in FIGURE 2) over a distance of 125 m.Relevant EquationsRt = R0(1 + at)

R1RT = R2RS

R = pL/A = Resistance = (Resistivity) x Length of wire / cross-sectional area

A = (pi x d2) / 4

d = squareroot(4A /pi)

R_t=R_20×(1+α×∆t)

α=0.0042

R20 = 112Ω

∆t=202-20=182

R_t=R_20×(1+α×∆t)

R_t=112×(1+0.0042×182)

R_t= 197.6128

As both resistors on figure 1 are 150Ω, they cancel each other out.

Find the resistivity of copper at 20°C

ρ=RA÷L

Where:

R = Resistance of copper at 20°C

A = area

L = length

From the properties of resistance wires table:

When diameter = 0.376mm, resistance = 0.155m-1

0.376mm diameter = 0.11103645mm2 for the area.

Converting resistance to per mm = 0.000155m-1

For length 1m = 1000mm.

Therefore, ρ=0.000155×0.111036450÷1000=1.72×10^(-8)

Resistivity of copper wire at 20°C = 1.72×10^(-8)Ωm-1

To find the area the equation would be A=(ρ×L)/R

A=(1.72×10^(-5) Ωmm×250000mm)/197.6128Ω

A = 0.02176mm2

Find diameter = 0.16645mm

Too small of a diameter.

Any help would be greatly appreciated. Ive been stuck for a week

*Moderator note: Moved from a technical forum.*
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