Heat Transfer Lab Experiment Report

February 5, 2015 Heat Transfer lab

Batch 1 - Roll Nos:14 to 19 EVALUATION OF OVERALL HEAT TRANSFER COEFFICIENT AND THERMAL CONDUCTIVITY OF ALUMINIUM Chintaginjala Dileep Kumar, Fahd Bin Abdul Hasis, Gautam Kumar Jha, Ghosh Partho Shankar, Gowtham S, Himanshu Kumar B.Tech, Fourth semester Department of Aerospace Engineering Indian Institute of Space Science and Technology

ABSTRACT The experiment was conducted to determine the thermal conductivity of a sample by using an idealized one-dimensional conducting apparatus. The sample having unknown conductivity was placed in between two metal pieces having known conductivity. The temperature differences across each interface as well as across the sample were measured using thermocouples at regular intervals until steady state was achieved. Using the Fourier law of heat conduction, the required conductivity was calculated.

by applying the Fourier law of heat conduction.

THEORY When a temperature gradient exists in a body, experience shows us that there is an energy transfer from the hightemperature region to the low-temperature region. The energy thus transferred is by the process of conduction and the heat transfer rate per unit area is proportional to the normal temperature gradient.

INTRODUCTION Heat transfer can be defined as the process by which there is a transfer of energy from one location to other, provided a proper temperature gradient exists. This transfer can be caused by the various mechanisms like conduction, convection and radiation. Conduction can be defined as the transfer of energy from a higher temperature region to lower temperature region by means of the free electron movement and also the lattice vibration. The heat transfer rate for steady one dimensional transfer can be given as q = −KA(OT ), where k(W /mK) is called the thermal conductivity of the material. Thermal conductivity can be defined as the ability of the material to conduct heat and its reciprocal is called thermal resistivity. Materials having a higher k value will have higher heat transfer rate than the materials having a lower k value. Thus the materials of higher k values can be used as heat sink application and materials of lower k value can be used for the insulation applications. Generally, k value of the material depends on the temperature. Study of thermal conductivity finds an application in various fields like material science, insulation, high operating temperature conditions etc.. This report deals with the systematic way to find the thermal conductivity of the aluminum specimen and also the thermal contact resistance

q ∂T ∝ A ∂x When the proportionality constant is inserted, we get

q = −kA

∂T ∂x

where q is the heat transfer rate and ∂T ∂x is the temperature gradient in the direction of heat flow. The positive constant k is called the thermal conductivity of the material. The minus indicates that the flow is downhill. This equation is known as Fourier’s law of heat conduction. a) Determining the overall heat transfer coefficient The heated, intermediate and cooled sections are clamped tightly together, so that the end faces are in good thermal contact, and create a composite bar with a Aluminium section sandwiched between two Brass sections as shown below. For continuity, the steady heat flow through the successive 1

Figure 1. TEMPERATURE DISTRIBUTION ALONG THE LENGTH OF THE SPECIMEN

(=0.03m) Therefore,

pic_1.png

Thermocouples T3 and T6 are located 7.5 mm from the end face compared with a distance of 15mm between adjacent thermocouples (half the distance), therefore: In the case of heated section the temperature of the end face (hot face) will be lower than T3 and can be calculated as follows:

pic_2.png

Figure 2.

Q∆xint Aint (Thot f ace − Tcold f ace )

kint =

(T2 − T3 ) 2

Thot f ace = T3 −

TEMPERATURE DISTRIBUTION ACROSS THE INSULATOR

Similarly

pic_3.png

Tcold f ace = T6 + sections must be the same so Fourier’s Law can be applied to the three sections as follows:

c) Determining the thermal conductivity of an insulator Material such as paper and cork have very low values of thermal conductivity which means only a small amount of heat will pass through the material even though a high temperature difference may exist across its two faces. Such materials are known as insulators and are practically utilized in situations where it is required to reduce heat loss from a hot body to its surroundings. The heated and cooled sections are clamped tightly together with the cork disk in between to create a composite bar with the insulated disk of unknown thermal conductivity sandwiched between two brass sections. Because of the low value of k for an insulator the dimension must be small and only a small amount of heat(low power) must flow through the specimen to prevent a large temperature difference which will trip the thermostat.

kint ∆Tint Q khot ∆Thot kcold ∆Tcold = = = A ∆Xhot ∆Xint ∆Xcold From which it follows that: (T1 − T8 ) = (∆Thot + ∆Tint + ∆Tcold ) =

Q A



Xhot Xint Xcold + + khot kint kcold



or 1 Q = U(T1 − TB )where = A U



Xhot Xint Xcold + + khot kint kcold

where U is the overall heat transfer coefficient and sistance to heat flow R.

1 U

T6 − T7 2

 =R Q = kins is the re-

∆Tins where ∆Tins = (Thot f ace − Tcold f ace ) ∆xint

Therefore kins =

b) Determining the thermal conductivity of the metal The heated, intermediate and cooled sections are clamped tightly together so that the end faces are in full thermal contact, and create a composite bar with a metal specimen of unknown thermal conductivity sandwiched between two brass sections. int From Fourier’s law Q = kint Aint ∆T ∆xint where ∆Tint = (Thot f ace − Tcold f ace ) and ∆xint is the length of the intermediate specimen

q Ains (Thot

f ace − Tcold f ace )

In the case of heated section the temperature of end face (hot face) will be lower than T3 and can be calculated as follows Thot 2

f ace

= T3 −

(T2 − T3 ) 2

• Same procedure was repeated by changing voltage value (15V).

In the case of cooled section the temperature of end face (cold face) will be lower than T6 and can be calculated as follows Tcold

f ace

= T6 +

(T6 − T7 ) 2

OBSERVATIONS

APPARATUS

Table 1.

Aluminium heat conducting specimen A multi-section bar for the examination of linear conduction. Insulator material The aluminium specimen is sandwiched between two sections which are covered by insulator material and are clamped tightly. Electrical console Provides electrical power for heaters in the specimens and digital readout of the temperature at any of the selected points along the heat-conducting paths. Thermocouples Three thermocouples are located on each brass sections with a distance of 15mm between adjacent thermocouples. Heat Transfer Service device Thermocouples are connected to the Heat Transfer Service device which displays the temperature. Flow sensor Cold water is supplied in pipe which passes through a solenoid valve and a flow sensor. Flow sensor is used to measure the flow rate. Heat sink compound Heat sink compound is applied between the joints to ensure proper conduction between them.

OBSERVATION TABLE

Time (mins)

T1 oC

T2 oC

T3 oC

T6 oC

T7 oC

T8 oC

0

40.2

38.1

36.2

32.6

31.4

30.9

3

46.1

43.4

40.9

35.5

33.5

32.4

6

48.9

46.2

43.6

38

36.1

35.2

9

52.2

49.3

46.7

41.4

39.4

38.6

12

54.3

51.6

49

43.6

41.7

40.7

15

56.2

53.4

50.7

44.5

42.2

40.9

18

57.3

54.3

51.4

44.6

42

40.4

21

57.4

54.4

51.4

44.2

41.5

39.5

30

57.5

54.4

51.6

42.8

40.3

38.5

33

58

55.2

51.8

42.6

40.1

38.4

35

58.5

55.9

53.4

42.3

39.7

38.1

38

58.2

55.2

52

43.3

40.5

38.5

41

58.1

55

51.9

43.2

40.4

38.4

RESULTS Following results were observed after conducting the experiment.

PROCEDURE • All the thermostat cables were attached to the Heat Transfer service unit and to the temprature measuring points on the sample. • Aluminium specimen was fixed in between the heat source and sink. • Cold water supply for cooling the heat sink was turned on and auxiliary control was used to adjust the flow rate as required. • Voltage(10V for first case), current and water flow rate wasset. • Temperature values of thermocouples(T1,T2,T3,T6,T7,T8) are recorded for every 3 minutes. • . The graph of temperatures is plotted to find out the steady state temperatures. • Temperature values of hot and cold surfaces was found using steady state temperature values. • Thermal conductivity value of aluminium and overall heat transfer coefficient of composite bar was calculated.

1. The steady state assumption for specimen was verified and found to hold after a sufficient time interval. 2. Thermal conductivity of aluminium alloy was found to be 111.6019 W /m − K with a deviation of 38% from the actual value 180 W /m − K.

SOURCES OF ERROR 1. Heat loss through nylon insulation. 2. Fluctuations in the water flow rate. 3. There may have been fluctuations in the supply voltage 4. Convection losses may have caused faulty results. 5. Thermocouple error may have been present. 6. Thermal contact resistance between aluminium and brass section both on upper and lower side. 3

Figure 3.

TIME vs TEMPERATURE PLOT

Taking the observation at time t = 41 minutes (from 3 and 4): Thot f ace = 50.35 oC Tcold f ace = 44.6 oC Using 5: ∆T = (50.35 − 44.6) = 5.75 oC Now we finally use 6 to calculate the thermal conductivity of the intermediate material: kint =

10.5 × 0.03 W ( ) −4 4.909 × 10 × (50.35 − 44.6) m − K

Thus,

kint = 111.60 REFERENCES [1] http://en.wikipedia.org/wiki/ Thermal conduction [2] Yunus A. Cengel, Af- shin J. Ghajar, Heat and Mass transfer, McGraw Hill Education (India) Private Limited. [3] Heat Transfer Lab manual, Aerospace Engineering Department, IIST

W m−K

APPENDIX B: Error Analysis Individual errors in x,d,∆T,V and I would contribute in the net error while calculating the thermal conductivity k.

ω1 =

∂k Q˙ δx = × 0.0001 = 0.372 ∂x A∆T

APPENDIX A: Sample Calculations

Q˙ = V I W A = πr2 m2 T2 − T3 o Thot f ace = T3 − ( C) 2 T6 − T7 o Tcold f ace = T6 + ( C) 2 ∆Tint = (Thot f ace − Tcold f ace ) (oC) Qxint kint = Aint (Thot f ace − Tcold f ace )

ω2 =

˙ ∂k −8Qx δd = 3 × 0.0001 = −0.893 ∂d πd ∆T

ω3 =

˙ ∂k −Qx × 0.1 = −1.941 δ∆T = ∂∆T A∆T 2

(1) (2) (3) (4) (5)

ω4 =

∂k Ix δV = × 0.1 = 1.116 ∂V A∆T

ω5 =

∂k Vx δI = × 0.01 = 1.063 ∂I A∆T

(6)

V = 10V I = 1.05 A ωtotal =

q

(0.372)2 + (−0.893)2 + (−1.941)2 + (1.116)2 + (1.063)2

From 1: Q˙ = 10 × 1.05 = 10.5W Using 2:

A = π×

= 2.66

0.0252 2 m = 4.909 × 10−4 m2 4

%Error = 4

2.66 × 100 = 2.38% 111.60

APPENDIX C: Proposed Experiment Measurement of Thermal Contact Resistance In the analysis of the heat conduction through two bodies, we assume that the contact between them is perfect, and that no temperature drop appears across the interface. This is an ideal case that occurs when the surfaces are assumed to be entirely smooth without any irregularities. However, in practice there are always voids due to the irregularity formation and these voids are filled by air. Thus the interface offers resistance to heat transfer and such a resistance per unit area is called thermal contact resistance, Rc . Therefore, the heat transfer through two bodies pressed together is actually due to the point to point contact of each of the crusts of the irregularities, and also the heat transfer through the voids in the non contact areas which is actually a major contributor. Q = Qcontact + Qgap

Figure 4.

Qgap = hc A ∆Tinter f ace Thermal contact resistance,

Rc =

∆Tinter f ace Q/A

The value of thermal contact resistance depends on the surface roughness and the material properties as well as the temperature and pressure at the interface and the type of fluid trapped at the interface. Thus, to minimize this resistance, a thermally conducting liquid called a thermal grease such as silicon oil is applied on the surfaces before they are pressed together. Another method is to replace the air by better conducting gases such as helium or hydrogen (for specific applications that require minimum thermal resistance). It is because of the presence of thermal contact resistance that the thermocouples in our experiment are attached a little away from the interface. The air gaps cause a non-uniformity in the flow at the interface. But as we move away from the interface, the heat transfer becomes more uniform and can be considered as one-dimension heat flow. Various loses do occur other than due to contact resistance. There could be radiations loses, as well as convection loses if the bodies are not fixed tightly.

5

MEASURING THERMAL CONTACT RESISTANCE