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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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

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.

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

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

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

K 2 = 154.19

K3 =98.12

K4 =126.41

K5 =133.63

K6 =91.71

K7 =129.92

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

K 2 = 125.56

K3 =17.12

K4 =59

K5 =167.41

K6 =30.13

K7 =32.05

K8=88.63

Graph(s):

Temperature

K

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 .

References:

1. http://www.chemistry.tutorvista.com/physical-

chemistry/specific-heat-capasity.html

2. http://www.gunt.de/static/s3684_1.php