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Chapter 18: Problem 42

The thermo emf of a thermo-couple is found to depend on temperature \(T\) (in degree celsius) as \(E=4 T-\frac{T^{2}}{200}\), where \(T\) is the temperature of the hot junction. The neutral and inversion temperatures of the thermocouple are (in degree celsius) (a) 100,200 (b) 200,400 (c) 300,600 (d) 400,800

Short Answer

Expert verified
The neutral and inversion temperatures are 400掳C and 800掳C, respectively.

Step by step solution

01

Understand Thermocouples

Thermocouples generate an electromotive force (emf) across two different metal junctions. The emf varies with the temperature of the junction.
02

Discuss Key Temperatures

The neutral temperature is where the emf is maximum, which happens when the derivative of the emf with respect to temperature equals zero. The inversion temperature is where the thermo emf equals zero again after reaching its peak.
03

Find the Neutral Temperature

Given the equation for emf: \( E = 4T - \frac{T^2}{200} \). To find the temperature for maximum emf, differentiate with respect to \( T \) and set the derivative equal to zero: \( \frac{dE}{dT} = 4 - \frac{T}{100} = 0 \). Solving this gives \( T = 400 \).
04

Solve for the Inversion Temperature

Set the emf equation to zero: \( 4T - \frac{T^2}{200} = 0 \). Rearrange to form a quadratic equation: \( T(4 - \frac{T}{200}) = 0 \). This gives \( T = 0 \) (irrelevant because it's not a hot junction) or \( 4 = \frac{T}{200} \), which solves to \( T = 800 \).
05

Conclusion on Correct Option

With the neutral temperature at 400掳C and the inversion temperature at 800掳C, we identify the correct answer as option (d): 400 and 800.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermo emf
Thermo emf, or thermal electromotive force, is the voltage generated by a thermocouple due to a temperature difference across its junctions. It encapsulates a principal function of thermocouples, which are devices used extensively for temperature measurement. When two dissimilar metals are joined at one end and heated, they produce an emf that can be measured. This phenomenon is based on the Seebeck effect. Understanding thermo emf is crucial because it forms the basis of how a thermocouple functions.

The thermal emf depends on several factors:
  • The types of metals used in the thermocouple.
  • The temperature difference between the hot and cold junctions.
  • The characteristics of the temperature change, like linearity.
The exercise you have involves finding specific temperatures related to thermo emf expressed by the function:\[ E = 4T - \frac{T^2}{200} \]This formula gives us the emf in terms of the temperature at the hot junction.
Neutral temperature
The concept of neutral temperature in thermocouples refers to the temperature at which the thermo emf reaches its maximum value. This is a critical point because it indicates optimal conditions for the thermocouple's operation. To find the neutral temperature, we differentiate the given emf expression with respect to temperature and set the result equal to zero.

The process is as follows:
  • Given the emf equation: \( E = 4T - \frac{T^2}{200} \).
  • Differentiating with respect to \( T \) gives: \( \frac{dE}{dT} = 4 - \frac{T}{100} \).
  • Setting the derivative equal to zero provides \( 4 - \frac{T}{100} = 0 \).
  • Solving this equation gives the neutral temperature as \( T = 400 \) degrees Celsius.
At this temperature, the thermocouple will exhibit the maximum thermo emf, which implies the greatest sensitivity to changes in temperature, maximizing its effectiveness.
Inversion temperature
Inversion temperature in the context of a thermocouple is where the thermo emf first reaches zero again after achieving its maximum value. This signifies a point beyond which the generated emf reverses sign, leading to a decrease in sensitivity.

To find the inversion temperature in the given exercise, follow these steps:
  • Use the emf equation: \( E = 4T - \frac{T^2}{200} \).
  • Setting this equation to zero determines where emf ceases: \( 4T - \frac{T^2}{200} = 0 \).
  • Factorizing gives \( T(4 - \frac{T}{200}) = 0 \).
  • Solving, we find \( T = 0 \) (not relevant as the hot junction is not zero) or \( 4 = \frac{T}{200} \).
  • Solve \( \frac{T}{200} = 4 \) to find that \( T = 800 \) degrees Celsius.
At 800掳C, the emf resets to zero, representing the inversion temperature. Recognizing and calculating this point is crucial for understanding where and how the thermocouple's sensitivity shifts or diminishes.

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Most popular questions from this chapter

In cosmic rays \(0.15\) protons \(\mathrm{cm}^{-2} \mathrm{sec}^{-1}\) are entering the earth's atmosphere. If the radius of the earth is \(6400 \mathrm{~km}\), the current received by the earth in the form of cosmic rays is nearly. (a) \(0.12 \mathrm{~A}\) (b) \(1.2 \mathrm{~A}\) (c) \(12 \mathrm{~A}\) (d) \(120 \mathrm{~A}\)

There is a current of \(0.21 \mathrm{~A}\) in a copper wire whose area of cross- section is \(10^{-6} \mathrm{~m}^{2}\). If the number of free electrons per \(\mathrm{m}^{3}\) is \(8.4 \times 10^{28}\), then find the drift velocity, \(\left(e=1.6 \times 10^{-19} \mathrm{C}\right.\) ) (a) \(2 \times 10^{-5} \mathrm{~ms}^{-1}\) (b) \(1.56 \times 10^{-5} \mathrm{~ms}^{-1}\) (c) \(1 \times 10^{-5} \mathrm{~ms}^{-1}\) (d) \(0.64 \times 10^{-5} \mathrm{~ms}^{-1}\)

One junction of a certain thermocouple is at a fixed temperature \(T_{r}\) and the other junction is at temperature \(T\). The thermoelectric force for this is expressed by $$ E=K\left(T-T_{r}\right)\left[T_{0}+\frac{1}{2}\left(T^{2}+T_{r}^{2}\right)\right] $$ At temperature \(T=T_{0} / 2\), the thermoelectric power is (a) \(\frac{1}{2} K T_{0}\) (b) \(\frac{3}{2} K T_{0}\) (c) \(\frac{1}{2} K T_{0}^{2}\) (d) \(\frac{1}{2} K\left(T_{0}-T_{r}\right)^{2}\)

The earth's surface has a negative surface charge density of \(10^{-9} \mathrm{C} / \mathrm{m}^{2}\). The potential difference of \(400 \mathrm{kV}\) between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only \(1800 \mathrm{~A}\) over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth's surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe. Radius of earth \(\left.=6.37 \times 10^{6} \mathrm{~m}\right)\) (a) \(273 \mathrm{~s}\) (b) \(263 \mathrm{~s}\) (c) \(283 \mathrm{~s}\) (d) \(205 \mathrm{~s}\)

An electric lamp is marked \(60 \mathrm{~W}, 230 \mathrm{~V}\). The cost of kilowatt hour of power is \(1.25 .\) The cost of using this lamp 8 hour a day for 30 days is (a) \(楼 10\) (b) ? 16 (c) ? 18 (d) ? 20
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