This report made by koya university student of chemical engineering (shwan sarwan ).

1. Linear heat conduction
Name of student: Shwan Sarwan Sadiq
Group: A
Date of Exp. NOV 11th
2015
Submission date: NOV 18th
2015
Supervisor: Ms.Farah
Chemical Engineering Department
Heat transfer lab
3rd stage
(
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2. Table of content:
Aim of the experiment 3
Introduction 3
Apparatus 4
Performance of the
experiment
5
Data sheet 6
calculation 7&8
Graph(s) 9
Discussion 10
References 11

3. Aim of the experiment:
We are doing the linear heat conduction to verify the low
of the Fourier’s law and to find out how does the heat
changes linearly.
Introduction:
Each material particle has its own motion to transfer energy
it could be electrical, thermal or other types. According to the
ability of transformation and the mechanism of the energy
motion materials divided, if we take linear thermal change in
a tool and make several experiments each turn by changing
its specification, we will find that metals are good linear
conductors because of their outer shell electrons which
makes them very active and transfer the heat and other
types of energy. Also we will realize that the thermal energy
will transfer from the higher temperature side to the lower
side.

4. Apparatus:
1-display and control unit,
2-measuring object,
3-experimental set-up for radial heat conduction,
4-experimental set-up for linear heat conduction

5. Performance of the experiment:
1. Read the flowrate of the water and read the Tcold
2.Turn on the master switch, set the Q to 82W . and read the
Tcold .
3. Read the temperature where the 9 sensors are being
placed.
4. Read the Thot
5. Stabilize the flowrate of the water which the Tcold will be
the same but change the Q to 25W.
6. Read the 9 sensor temperature and the Tcold

6. Calculation paper:
For reading number 1
V =
21∗10−3
60
= 3.5 *10−4
m3
/s
m = 𝜌V = 998.3*3.5*10-4
= 0.349 kg/s
A =
𝛱
4
d2
=
𝛱
4
*0.025 2
= 4.908*10-4
m2
Cp of water at 23.5 is 4.1806 KJ/Kg.S by interpolation (3)
Q lost =mCp (T hot – T cold)= 0.349 * 4.1806 *(28-19) =13.13KJ
Q act = Q supply – Q lost = 82 – 13.13 = 68.87 KJ
K=
𝑄 ∆𝑋
𝐴 ∆𝑇
K 1 =
68.87∗1∗10−2
4.908∗10−4∗(120.8−112.6)
= 171.12
K2 = 154.19
K3 =98.12
K4 =126.41
K5 =133.63
K6 =91.71
K7 =129.92

7. Continue..
For reading number 2
V =
21∗10−3
60
= 3.5 *10−4
m3
/s
m = 𝜌V = 998.3*3.5*10-4
= 0.349 kg/s
A =
𝛱
4
d2
=
𝛱
4
*0.025 2
= 4.908*10-4
m2
Cp of water at 23.5 is 4.1806 KJ/Kg.S by interpolation (3)
Q lost =mCp (T hot – T cold)= 0.349 * 4.1806 *(26-19) =10.21KJ
Q act = Q supply – Q lost = 25 – 10.21 = 14.79 KJ
K=
𝑄 ∆𝑋
𝐴 ∆𝑇
K 1 =
14.79∗1∗10−2
4.908∗10−4∗(79.4−76.1)
= 91.31
K2 = 125.56
K3 =17.12
K4 =59
K5 =167.41
K6 =30.13
K7 =32.05

8. K8=88.63
Graph(s):
Temperature
K

9. Discussion:
From the charts and the calculations of the K and Q by using
temperatures and reference information we can’t get an
idea about the relation between K and temperature or K
and Q , but we could find the solutions . so it means that the
error is not in the calculation or dimensions , finding the
error in the technical experiment is more possible because
the temperature readings was not officially from the current
experiment but it was a given , also the reading of the Thot
and Tcold was not acceptable and unsure .

10. References:
1. http://www.chemistry.tutorvista.com/physical-
chemistry/specific-heat-capasity.html
2. http://www.gunt.de/static/s3684_1.php