The thermo emf of a therocouple , one junction of which is kept at `0^(@)C` , is given by `varepsilon=at+bt^(2)` where a and b constants of the thermo

Question

The thermo emf of a therocouple , one junction of which is kept at `0^(@)C` , is given by `varepsilon=at+bt^(2)` where a and b constants of the thermo

The thermo emf of a therocouple , one junction of which is kept at `0^(@)C` , is given by `varepsilon=at+bt^(2)` where a and b constants of the thermocouple . Calculate the neutral temperature and peltier and Thomson coefficients.

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

The emf `varepsilon` of a Cu-Fe thermocouple varies with the temperature `theta` of the hot junction (cold junction at `0^(@)C` ), as `varepsilon(mu V

2 Answers 7 Views

The thermo emf of a thermocouple is given by `E=alpha(theta-theta_(g))+beta(theta^(2)-theta_(0)^(2))` Determine the neutral temperature and inversion

2 Answers 5 Views

With the cold junction at `0^(@)C` the neutral temperature for a thermo-couple is obtained at `270^(@)C`. The cold junction temperature is now lowered

2 Answers 4 Views

Thermo e.m.f. E of a Cu-Fe thermocouple varies with the temperature `theta` of the hot junction ( cold juction `0^(@)C`) as `E(muV)=12theta-0.02 theta

2 Answers 6 Views

The cold junction of a thermocouple is kept at `10^(@)`C. Calculate the temperature at which thermo emf would be maximum. Given that thermo emf change

2 Answers 6 Views

Near room temperature , the thermo emf of a thermo couple is `40 muV` per degree. A galvanometer of `25 Omega` and capable of detecting current of the

2 Answers 4 Views

The thermo emf of a copper-constantan thermocouple with one junction at `0^(@)C` and the other at `t^(@)C` is given by the relation `varepsilon=at=(1)

2 Answers 5 Views

The emf in a lead-iron thermocouple , one junction of which is at `t^(@)C` is given by `varepsilon=178t-2.4t^(2)( "in "muV)` where t is the temperatur

2 Answers 5 Views

For a certain thermocouple , `varepsilon=alphatheta+(1)/(2)betatheta^(@),where theta.^(@)C` is the temperature of the hot junction and `0^(@)C` is the

2 Answers 4 Views

A thermocouple with one junction at ice-point , has emf given by the relation, `varepsilon =atheta+(1)/(2)btheta^(@)` , where `theta^(@)C` is the temp

2 Answers 5 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - `varepsilon=alphatheta+(1)/(2)betatheta^(2) therefore (dvarepsilon)/(d theta)=alpha+betatheta`
At `theta=theta_(n), (dvarepsilon)/(d theta)=0 " or "alpha+betatheta_(n)=0`
`therefore theta_(n)=-(alpha)/(beta)=-(10)/(-1//20)=200^(@)C`
At `theta=theta_(i), varepsilon=0 " or "alphatheta_(i)+(1)/(2)betatheta_(i)^(2)=0 " or "alpha+(1)/(2)betatheta_(i)=0`
`therefore theta_(i)=-(2alpha)/(beta)=2xx200=400^(@) C`.