Thermal Conductivity of Liquid & Gases

Introduction and theory

Thermal conductivity In physics, thermal conductivity, (k), is the property of a material that indicates its ability to conduct heat basis of conduction heat transfer is Fourier Law. This law involves the idea that the heat flux. Thermal conductivity, k, a property of proportional to the temperature gradient in any direction materials that is temperature dependent, is the constant of proportionality. Heat always moves from warmer objects to cooler objects. The composition of a material affects its conduction rate. If a copper rod and an iron rod are joined together end to end, and the ends placed in heat sources, the heat will conduct through the copper more quickly than the iron because copper has a K value of 92 (W/m k), whereas, iron has a K value of 11(W/m k). It should be noted that heat can also be transferred by Thermal radiation and/or convection, and often more than one of these processes occur in a particular situation. The law of heat conduction, also known as Fourier law, states that the rate, heat transfer through a material is proportional to the negative gradient in the temperature and to the area at right angles, to that gradient, through which the heat is flowing:

q =-KA( dT/dr) A : across sectional area is given by

A=2ΠrL r : radial coordinate L:length of the cylinder q r = ( 2 Π L k) (T1-T2) ln(R2/R1) k=

ln(R2/R1) qr / (T1-T2)(2ΠL)

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Apparatus and procedures Apparatus Thermal conductivity of liquids and gas unit (model HE 156) Consists two coaxial concentric cylindrical plugs with a thin radial clearance in between the clearance is made extremely small which is 0.3 mm to reduce the natural heat convection the heat sourced from the center of the coaxial concentric cylindrical plugs made of copper

figure (1)

figure (2)

procedures



Thermal conductivity of air

1) set power regulators anti clock 2) connected water supply and electric supply on 3) set the temperature at 100 ◦C 4) on the heater switch and power regulators to about 20 w, 30 w , 40 w 5) record the power and T1 , T2



Thermal conductivity of acetone

Repeat all the previous steps in the previous section

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Results and discussion Results: To determination of thermal conductivity of air :

• R1(m ) 0.016 65

k= R2(m ) 0.016 95

ln(R2/R1) qr

/

(T1-T2)(2ΠL)

L(m) 0.1

T2(◦C)

∆T(k )

q lost (W)

(W)

41.2

25.7

15.5

2.4

15.1

31

55.3

26

29.3

4.6

26.4

42.3

67.1

26.5

40.6

6.2

36.1

Q

T1

(W)

(◦C)

17.5

qr

ln(R2/R 1) 0.017857 62 0.017857 62 0.017857 62

2×π×L (m) 0.628 0.628 0.628

K (

W/m

k) 0.0277018 72 0.0256212 42 0.0252839 57 Kavg = 0.026

TABLE (1) To determination of thermal conductivity of acetone :

•

k=

ln(R2/R1) qr

R1(m )

R2(m )

L(m)

0.016 65

0.016 95

0.1

Q (W)

T1 (◦C)

/

(T1-T2)(2ΠL)

T2(◦C)

∆T(k )

q lost (w)

qr (W)

14.6

36.6

26

10.6

1.5

13.1

19.8

41.7

26.1

15.6

2.4

17.4

26.5

45

26.3

18.7

3

23.5

TABLE (2)

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ln(R2/R 1) 0.017857 62 0.017857 62 0.017857 62

2×π×L (m)

K (w/ (m*k))

0.0351422 29 0.0317167 0.628 38 0.0357346 0.628 99 Kavg= 0.03419 0.628

Discussion:.



The calculated value of thermal conductivity for Air and Acetone is not equal

the tabulated value,. Due to the system did not reach completely to steady-state.



Air is a good insulator, which can be seen from the value of tabulated thermal

conductivity or from the experimental data.

Conclusion and Recommendations Conclusion

 The thermal conductivity has a low value for gases.  The thermal conductivity for liquid between gases and solids.  Heat losses must be noticed due to using of non-perfect conductor.  The thermal conductivity determine using Fourier Law

Recommendations

We recommend carrying out the test on different types of liquids and gases

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References

1-

Dr.Rami Jumah 'Heat transfer laboratory' “laboratory manual July

2009”

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Appendices Calculation: 

For Air ∆T = T1 – T2

•

= 41.2 - 25.7 = 15.5

o

C

From figure 2 from manual heat transfer :

q lost vs ∆T:

Q = 17.5 w

q •

lost

= 2.4 w

q r = Q – q lost = 17.5 – 2.4 = 15.1 w

.

R1 = 0.01665 m •

R2 = 0.01695 m

ln (R2 /R1) ln (0.01695/0.01665) =0.01786

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•

2×Π×L = 2 × Π ×0.1

m

= 0.628

•

k=

ln(R2/R1) qr

/

(T1-T2)(2ΠL)

= ln(0.01695/0.01665)( 15.1)/ (2 × Π ×0.1)( 15.5)



•

= 0.0277 W/m K For acetone : ∆T = T1 – T2 = 36.6 - 26 = 10.6

o

C

From figure 2 from manual heat transfer : q

lost

vs ∆T:

Q = 14.6 W

q •

lost

= 1.5 W

q r = Q – q lost = 14.6 – 1.5 = 13.1 w.

R1 = 0.01665 m •

R2 = 0.01695 m

ln (R2 /R1) ln (0.01695/0.01665) =0.01786

•

2×Π×L = 2 × Π ×0.1 7

= 0.628 m

•

k=

ln(R2/R1)( qr) / (T1-T2)(2ΠL)

= ln(0.01695/0.01665)(13.1)/ (2 × Π ×0.1)( 10.6) = 0.035 W/m k

Equipment specification figure (2) Inner cylinder plug Outer diameter (mm) length (mm) material

: 33.3 : 100 : copper

outer cylinder plug inner diameter (mm) length (mm) material

: 33.9 : 100 : copper

cylinder water jacket material

: stainless steel

outer radius of the inner cylinder R 1 ( m )

: 0.01665

inner radius of the outer cylinder R 2 ( m ) length of the cylinder L( m )

: 0.01695 : 0.10

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