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Koya University Faculty of Engineering

Chemical Engineering Department

Laboratory of Heat transfer Experiment Number Two Radial heat conduction

Instructor: Dr.Barham & Mrs.Anfal

Name of Student: Aree Salah Tahir

Experiment Contacted on: 12/ Nov/2014

Report Submitted on: 24/Nov/2014

Group:A

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List of content:

Objectives……………............................................3

Introduction..………………………………… ….4

Background Theory ...……..…………...…..…….5

Procedure ………………..…………..……...……6

Equipment and components used………………...7

Table of Reading……………….…………......….8

Calculation…………...............................9-10-11-12

Discussion …..………………………….…...13-14

References ……………………...…………..…16

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Objectives:

It allows us to obtain experimentally the coefficient of thermal

conductivity of some unknown materials ,and in this way to

understand the factors and parameters that affect the rate of heat

transfer.

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Introduction:

Heat transfer is a common phenomenon encountered in many areas in daily life. It

is an important subject in natural sciences and even more so in engineering and the

field of environmental physics. Knowledge about heat transfer is essential for the

various possible ways to cut down on energy use. For example, to economize on

home-heating one has to optimize insulation, i.e. minimize heat transfer. Improving

cooling processes on the other hand implies maximizing heat transfer to the coolant.

Engineering problems that involve heat transfer are generally concerned with

optimization of heat transfer processes. For instance, in the development of novel

materials that cannot sustain extreme temperatures heat transport towards the

material is to be minimized, whereas the transport away from this material is to be

maximized. Research on heat transfer focusses on electronic equipment, energy

systems, fire and combustion systems, gas turbines, as well as manufacturing and

materials processing. There is even a Journal of Heat Transfer whose emphasis is

on energy transfer in applied thermodynamic processes in all fields of (mechanical)

engineering and related industries. Therefore it is important to understand the basic

concepts of heat transfer and the way it is mathematically described. You should be

already familiar with the different mechanisms that result in heat transfer from the

lecture. The most important heat transfer mechanisms are:

• Conduction; on a microscopic or atomic scale kinetic energy is transferred

through collisions between the atoms so that on a macroscopic scale thermal

energy is transferred.

• Convection; if the carrier of thermal energy is mobile, the thermal energy may

be transferred through mass transfer.

• Radiation; thermal energy may be transformed into electromagnetic energy,

emitted, absorbed and with that it is transformed into thermal energy again

{1}

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Background Theory:

Conduction: is the molecular transfer of heat in solid, liquid, and gaseous media

under the influence of a temperature difference.

Cylindrical and spherical systems often experience temperature gradients in the

radial direction only and may therefore treat as one dimensional. A common

example is the hollow cylinder, whose inner and outer surfaces are exposed to fluids

at different temperatures, as shown in Figure 1.

Figure (1)

For radial conduction, the electrical heating element is bonded to the center part of

a circular brass plate (heat source). The cooling water flows through the edge of the

plate that acts as a heat sink for heat discharge. The other surfaces of the plate are

well insulated to simulate radial heat conduction from the plate center to its edge

when the heating element is switched on. The brass plate has a radius, r plate = 50

mm and thickness, t = 4 mm. Thermocouples are embedded in the circular plate, at

r = 0, 10, 20, 30, 40, and 50 mm. A simple mimic diagram for heat conduction

along an well-insulated cylindrical rod is shown as below:

Figure (2)

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Procedure:

1. Set up the unit as per sect. 2.3 and adjust the cooling water flow rat (a

cooling water flow rate of approx. 11th is required to dissipate a heater

power of 90 watts at a temperature difference of 90 c)

2. Connect up the power and data cables appropriately.

3. Switch on the unit and adjust the desired temperature drop via the

power setting on the control and display unit.

4. When the thermal conduction process has reached a steady state

condition, I.e. the temperature at the individual measuring points are

stable and no longer changing, note the measurement results at the

individual measuring points and the electrical power supplied to the

heater..

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Equipment and components used:

1-display and control unit,

2-measuring object,

3-experimental set-up for radial heat conduction,

4-experimental set-up for linear heat conduction {2}

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Table of Reading:

Temperature of barrel is 21.3 c

Q=40watt →Tout=31.8 c

Q=70watt → Tout=38 c

Q T1 T2 T3 T4 T5 T6 40 0 50 43.5 X 33.9 33.5 70 0 70 59.3 X 43.9 42.9

Q T1 T2 T3 T4 T5 T6 40 0 50 43.5 38.7 33.9 33.5 70 0 70 59.3 51.6 43.9 42.9

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Calculation:

At Q = 40 watt

For No 1:

Length of (Brass) =40mm = 0.04m

Q= 40 watt T2= 50 𝑐0 T3= 43.5 𝑐0 ∆T=T2 − T1 =6.5 𝑐0

r=10mm = 0.01m

Q=K(2πL∗∆T

ln𝑟𝑜𝑢𝑡𝑟𝑖𝑛

) → K=𝑞∗𝑙𝑛

𝑟𝑜𝑢𝑡𝑟𝑖𝑛

2𝜋𝐿∗𝛥𝑇

K =40(0.693)

2π∗0.04∗(6.5) = 16.98

W

m.C°

The same way for others……….

NO. R (m)

T

(𝒄𝟎)

ΔT

(𝒄𝟎) Ln

𝒓𝒐𝒖𝒕

𝒓𝒊𝒏 K

(𝑾𝒎. 𝒄𝟎⁄ )

1 _ _ _ _ _

2 0.01 50 _ _ _

3 0.02 43.5 6.5 0.693 16.98

4 0.03 38.7 4.8 0.4055 13.45

5 0.04 33.9 4.8 0.2877 9.54

6 0.05 33.5 0.4 0.2231 88.81

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At Q = 70 watt

For No 1:

Length of (Brass) =40mm = 0.04m

Q= 70 watt T2= 70 𝑐0 T3= 59.3 𝑐0 ∆T=T2 − T1 =10.7 𝑐0

r=10mm = 0.01m

Q=K(2πL∗∆T

ln𝑟𝑜𝑢𝑡𝑟𝑖𝑛

) → K=𝑞∗𝑙𝑛

𝑟𝑜𝑢𝑡𝑟𝑖𝑛

2𝜋𝐿∗𝛥𝑇

K =70(0.693)

2π∗0.04∗(10.7) = 18.05

W

m.C°

The same way for others……….

NO. R

(m)

T

(𝒄𝟎)

ΔT

(𝒄𝟎) Ln

𝒓𝒐𝒖𝒕

𝒓𝒊𝒏 K

(𝑾𝒎.𝒄𝟎⁄ )

1 _ _ _ _ _

2 0.01 70 _ _ _

3 0.02 59.3 10.7 0.693 18.05

4 0.03 51.6 7.7 0.4055 14.68

5 0.04 43.9 7.7 0.2877 10.412

6 0.05 42.9 1 0.2231 62.17

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For Q=40watt

Volume=200ml & time=267.48 s

Volume flow rate= V/T → Volume flow rate=200/267.48=0.74

Volume flow rate=0.74 *10-6 m3 /s

Mass flow rate=0.74 *10-3 kg /s

Q=m*cp* ΔT→ m*cp*(Tout-Tin)

Cp=4.182

Qlose=(0.74 *10-3)* 4.182*10.5

Qlose=32 J/s =32watt

Qactual=Qsupply - Qlose

Qactual=8 watt

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For Q=70watt

Volume=200ml & time=267.48 s

Volume flow rate= V/T → Volume flow rate=200/267.48=0.74

Volume flow rate=0.74 *10-6 m3 /s

Mass flow rate=0.74 *10-3 kg /s

Q=m*cp* ΔT→ m*cp*(Tout-Tin)

Cp=4.182

Qlose=(0.74 *10-3)* 4.182*16.7

Qlose=51.6 J/s =51.6watt

Qactual=Qsupply - Qlose

Qactual=18.4 watt

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Discussion:

In this experiment we see that the relation between ( ∆T) & (Distance) is linear

with decrease ∆T decrease Distance

50

43.5

38.7

33.9 33.5

0

10

20

30

40

50

60

0 0.01 0.02 0.03 0.04 0.05 0.06

Tem

per

atu

re(c

)

Distance(m)

Plot between Temperature and DistanceAt Q=40

70

59.3

51.6

43.9 42.9

0

10

20

30

40

50

60

70

80

0 0.01 0.02 0.03 0.04 0.05 0.06

Tem

per

atu

re(c

)

Distance(m)

Plot between Temperature and DistanceAt Q=70

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Question:

Why did we count Qlose ?

Answer:

we count Qlose because it is importance to determin Qactual

and we can determin Qactual by this equation :

Qactual=Qsupply - Qlose

…………………………………………………………………………………

We find the cp and density in this tabe by temperature

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

For Q=40

Taverag = (Tin+Tout)/2

(21.3+31.8)/2 = 26.55 at this temperature(26.55)

cp=4.18 density=996.82

………….

For Q=70

cp=4.179

density=995.95

………….

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1-Why we neglect the first rate of temp. ?

Because of the distance is zero and read temp.as minus.

2-Here are the factors that affect the rate of conduction

A) Temperature difference

B) Cross-sectional area

C) Length (distance heat must travel)

D) Time

3-In contact point will make error because when we join the peace of

material will make a space that cause to heat losses.

4-

To measure the temperature distribution for steady-state conduction of

energy through a composite plane wall and determine the Overall Heat

Transfer Coefficient for the flow of heat through a combination of

different

materials in series

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By increasing area and ΔT thermal conductivity decrease:

But by increasing ΔX, thermal conductivity increase

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Why we neglect the first rate of temp. ?

Because of the distance is zero and read temp.as minus.

7-

What is the advantage of cooling water?

To make the difference between the temp.

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References:

1- www.laserspectroscopy.ucc.ie/PY3011/Manual_HeatCo

nd.pdf

2- http://www.gunt.de/static/WL372

3- http://www.engineeringtoolbox.com/water-thermal-

properties-d_162.html