Requirement of temperature scale - thermodynamics, Mechanical Engineering

Assignment Help:

Requirement of Temperature Scale:

The temperature scale on which temperature of the system can be read is required to assign the numerical values to the thermal state of the system. It requires the selection of basic unit and reference state.

Establish a correlation between Centigrade and Fahrenheit temperature scales.
Sol:
Let the temperature't' be linear function of property x. (x can be length, resistance volume, pressure etc.) Then by using equation of Line;

t = A.x + B                                                                                                                         ...(i)

At Ice Point for Centigrade scale t = 0°, then

0 = A.xi  +B                                                                                                                         ...(ii)

At steam point for centigrade scale t = 100°, then

100 =A.x S + B                                                                                                                      ...(iii)

From equation (iii) and (ii), we get

a = 100/(xs - xi ) and b = -100xi/(xs  - xi) Finally equation becomes in centigrade scale is;

t0 C = 100x/(xs  - xi ) -100xi/(xs  - xi)

t0 C = [(x - xi )/ (xs  - xi )]100                                                                                           ...(iv)

Likewise if Fahrenheit scale is used, then

At Ice Point for Fahrenheit scale t = 32°, then

32 = A.xi  + B                                                                                                                        ...(v)

At steam point for Fahrenheit scale t = 212°, then

212 =A.xS + B                                                                                                                       ...(vi)

From equation (v) and (vi), we get

a = 180/(xs  - xi ) and b = 32 - 180xi/(xs  - xi) Finally general equation becomes in Fahrenheit scale is;

t0 F = 180x/(xs  - xi ) + 32 - 180xi/(xs  - xi)

t0 F = [(x - xi )/ (xs  - xi )]180 + 32                                                                                ...(vii)

Likewise if Rankine scale is used, then

At Ice Point for Rankine scale t = 491.67°, then

491.67 = A.xi  + B                                                                                                                    ...(viii)

At steam point Rankine scale t = 671.67°, then

671.67 = A.xS + B                                                                                                                      ...(ix)

From equation (viii) and (ix), we get

a = 180/(xs  - xi ) and b = 491.67 - 180xi/(xs  - xi) Finally equation becomes in Rankine scale is;

t0 R = 180×/(xs  - xi ) + 491.67 - 180xi/(xs  - xi)

t0 R = [(x - xi )/ (xs  - xi )] 180 + 491.67                                                                        ...(x)

Likewise if Kelvin scale is used, then

At Ice Point for Kelvin scale t = 273.15°, then

273.15 = A.xi  + B                                                                                                                      ...(xi)

At steam point Kelvin scale t = 373.15°, then

373.15 = A.xS + B                                                                                                                     ...(xii)

From equation (xi) and (xii), we get

a = 100/(xs  - xi ) and b = 273.15 - 100xi/(xs  - xi) Finally equation becomes in Kelvin scale is;

t0 K = 100x/(xs  - xi ) + 273.15 - 100xi/(xs  - xi)

t0 K = [(x - xi )/ (xs  - xi )] 100 + 273.15                                                                    ...(xiii)

Now compare between above four scales:

(x - xi )/ (xs  - xi ) = C/100                                                                                           ...(A)

= (F-32)/180                                                                                                                ...(B)

= (R-491.67)/180                                                                                                         ...(C)

= (K - 273.15)/100                                                                                                     ...(D)

Now joining all 4 values we get following relation

K = C + 273.15

C = 5/9[F - 32]

= 5/9[R - 491.67] F = R - 459.67

= 1.8C + 32

 


Related Discussions:- Requirement of temperature scale - thermodynamics

Heat loss in acylindrical pipe, Ask question #Minimum 1. Research the topic...

Ask question #Minimum 1. Research the topic of heat loss in a cylindrical pipe. Explain which laws of physics are used to discuss heat loss in a pipe, and briefly explain how the f

Describe inversion and mechanism, Describe Inversion and Mechanism. Illustr...

Describe Inversion and Mechanism. Illustrate in detail all inversions of 4-Bar chain Mechanism. (a) Explain all the types of constrained motion with examples. (b) Explain in

Equilibrium of body on the rough inclined plane, Equilibrium of body on the...

Equilibrium of body on the rough inclined plane: If the inclination is less than angle of friction, the body remains in equilibrium without any external force. If body is to b

Reduce system to a single force - mechanics, Reduce system to a single forc...

Reduce system to a single force: Q: A system of parallel forces is acting on rigid bar as shown in the figure given below. Reduce this system to  a single force

Cantilever beam stress, Dear Engineer! I have an aluminum tube(6061) fixed...

Dear Engineer! I have an aluminum tube(6061) fixed at one end (Cantilever Beam). The length is 180” and the load on the other end is 150 lbs. Question: What size of tubes I s

Tempering and stabilization, Tempering and Stabilization: Tempering o...

Tempering and Stabilization: Tempering of tool steel in heat treatment is again a significant step. The quenching of such steels reasons the existence of untempered martensit

Compute the force in the rod, Compute the force in the rod: A rod ABC ...

Compute the force in the rod: A rod ABC rotating at 20 rpm about a vertical axis through A, supports a ball of mass 10 kg at its lower end. It is fixed in position by the rod

Cad, nc & cnc machines

nc & cnc machines

Relationship between two specific heat, Relationship between two specific h...

Relationship between two specific heat ? Sol:   dQ = dU + dW; for a perfect gas dQ at constant pressure dU at Constant volume; = mC v dT = mC v (T 2  - T 1 ) dW at

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd