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Vikki [24]
3 years ago
8

Thermal conductivity of AISI 316 Stainless Steel at 90ºC is 14.54 W/m K. Convert this value to IP system.

Engineering
1 answer:
abruzzese [7]3 years ago
7 0

Answer:

the value of conductivity in IP is 8.406\dfrac{Btu}{ft.hr.F}

Explanation:

Given that

Thermal conductivity K=14.54 W/m.K

This above given conductivity is in SI unit.

     SI unit                                           IP unit              Conversion factor

    m                                                      ft                      0.3048

   W                                                       Btu/hr               0.293          

 

The unit of conductivity in IP is Btu./ft.hr.F.

Now convert into IP divided by 1.73 factor.

0.57\dfrac{Btu}{ft.hr.F}=1 \dfrac{W}{m.K}

So

0.57\times 14.54\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}

8.406\dfrac{Btu}{ft.hr.F}=14.54 \dfrac{W}{m.K}

So the value of conductivity in IP is 8.406\dfrac{Btu}{ft.hr.F}

 

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Answer:

b)Poly crystalline and amorphous materials with small diameter

Explanation:

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Fibers poly crystalline and amorphous materials with small diameter.

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For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

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