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Eg2002 Heat Exchanger

In: Science

Submitted By thomas92
Words 3933
Pages 16
EG2002 Process Engineering
Continuous Assessment Report

Heat Exchange Laboratory
By
Thomas A. Lindie
51011245

School of Engineering
University of Aberdeen
Kings College
2011-12

Table of Contents

Contents

Table of Contents I List of Figures II List of Tables II Symbols and Abbreviations III 1. Introduction 1 1.1 Background 1 1.2 Aims and Objectives 1 1.3 Structure of the Report 2 2. Background Theory 3 3. Experimental Methodology 6 4. Results 8 4.1 Tables of Co-Current and Counter-Current flow taken from Result Table 8 4.1.1 Table of Co-Current Flow at Steady State 8 4.1.2 Table of Counter-Current Flow at Steady State 8 4.2 Log Mean Temperature Different (LMTD) Calculations 9 4.3 Calculating the Duty of the HEX and the Efficiency 10 4.4 Graphs of Results for Co-Current and Counter-Current Flow 12 5. Discussion and Analysis 13 5.1 Log Mean Temperature Different (LMTD) Calculation Analysis 13 5.2 Efficiency of the Heat Exchanger 14 5.3 Errors in Laboratory 14 6. Conclusions and Recommendations 15 Bibliography 16

List of Figures

Figure 1: 3D View of Shell and Tube heat exchanger taken from http://www.secshellandtube.com/ 3 Figure 2: Shell and Tube heat exchanger flow pattern taken from http://www.cheresources.com/content/articles/heat-transfer/specifying-a-liquid-liquid-heat-exchanger 3 Figure 3: Screenshot taken from co-current experiment on Armfield Programme 7 Figure 4:Graph of Co-Current Flow 12 Figure 5: Graph of Counter-Current Flow 13

List of Tables

Table 1: Table of sample number 74 for co-current flow at steady state 8
Table 2: Table of sample number 53 for counter-current flow at steady state...........................................9

Symbols and Abbreviations

Symbol | Description | Units | Q | Heat Transfer | W | Qh | Duty of Hot Stream | W | Qc | Duty of Cold Stream | W | Cp | Specific Heat Capacity | kJ/kg K | gh | Mass Flow Rate of Hot Stream | kg/s | gc | Mass Flow Rate of Cold Stream | kg/s | ∆T | Tout-Tin | K | Ƞ | Efficiency of Heat Exchanger | % | U | 1/Resistance (Overall Heat Transfer Coefficient) | | A | Area | m^2 | ∆Tlm | Driving Force (LMTD – Log Mean Temperature Difference) | K | h | Film Heat Transfer Coefficient | W/m^2 K | k | Thermal Conductivity | | r | Radius | m | θ1 | See Background Theory for Description | DegC | θ2 | See Background Theory for Description | DegC |

1. Introduction
A heat exchanger is regarded as a very important piece of equipment in the world of Engineering and it is used specifically for very efficient heat transfer from one medium to another, to put it in simpler terms, when a solid, liquid or gas needs to be heated up or cooled then a heat exchanger is used. Heat exchangers are used in a number of areas such as refrigeration, power plants, internal combustion engines and a number of other areas. They are one of the most common unit operations and they come in three types: * Shell and Tube * Concentric Tube * Plate
The ‘Shell and Tube’ heat exchanger is the type that we will focus on in this experiment.
Our main area of research in this laboratory is investigating the effect of counter-current and co-current flow on the performance of the ‘Shell and Tube’ heat exchanger. 1.1 Background
Carrying on from the previous information of the heat exchanger (HEX), the two fluids present in the HEX are separated by a physical barrier which is usually a tube wall or a metal plate. (Heat Exchangers) Heat exchangers are very efficient and for this to be true, they are designed to maximize the surface area of the tube walls or metal plate between the two fluids and this then minimizes the resistance to fluid flow through the heat exchanger.
As stated above, from the three types of heat exchangers available, we will be focusing on the ‘Shell and Tube’ heat exchanger for our experiment. This heat exchanger is a common type of heat exchanger for use in oil refineries and large chemical processes and it is mainly suited for high-pressure uses. As can be determined from the name, the ‘Shell and Tube’ heat exchanger consists of a series of tubes known as a tube bundle encased in a shell where one fluid that will be heated or cooled flows through the tubes and the second fluid flows over the tubes to transfer the heat between the two fluids.
1.2 Aims and Objectives
The main aim of the heat exchanger laboratory was: * To investigate the effect of counter-current and co-current flow on the performance of the ‘Shell and Tube’ heat exchanger.
From this we would be calculating the Log Mean Temperature Difference (LMTD) of the heat exchanger in both counter-current and co-current, as well as calculating the Duty of the Shell and Tube heat exchanger and also to calculate the efficiency.
We will also compare our results with the results achieved using the Armfield software.
Also, the laboratory can be used to familiarise ourselves with the heat exchanger setup and apparatus used, along with the Armfield programme used on the computer to set the values of T1,T2,T3 and T4 as well as the flow rate values of the cold water and hot water and record our results in a table and in a graphical form.
1.3 Structure of the Report
This heat exchanger laboratory report will have 6 main parts which are; introduction, background theory, experimental methodology, results, discussion and analysis and the conclusion.
Some of the sections consist of subsections so I will provide a little more detail below: 1. Introduction – There are 3 subsections within the introduction, 1.1 Background, 1.2 Aims and Objectives and 1.3 Structure of the Report. 2. Background Theory – This section has been mainly used to state and describe a number of equations that are used to calculate values relating to the heat exchanger. 3. Experimental Methodology – This section of the report gives a detailed description of the experiment including the step by step process and a list of the apparatus used. 4. Results – This is where I have stated the results gained from the experiments undertook in the laboratory. 5. Discussion and Analysis – This part of the report will include the reasons and theory as to how the results were obtained. 6. Conclusion – I have stated my conclusions in relation to my aims and objectives in Section 1.2 of the report. *

2. Background Theory
As stated previously, a ‘Shell and Tube’ heat exchanger is one of three heat exchanger designs and it can be regarded as the most common type of heat exchanger. In our experiment the Tube Inlet will be warm water and this will travel through the tube bundles and the Shell Inlet will be the cold water that will travel over the tube bundles in the shell encasing. |
Figure [ 1 ]: 3D View of Shell and Tube heat exchanger taken from http://www.secshellandtube.com/
The picture above in Figure 1 shows the 3D view of a ‘Shell and Tube’ heat exchanger and the one used in our laboratory was relatively similar to this. From the left hand side of the picture you can distinguish where the tube bundles are and the shell has also be noted on the diagram. |
Figure 2: Shell and Tube heat exchanger flow pattern taken from http://www.cheresources.com/content/articles/heat-transfer/specifying-a-liquid-liquid-heat-exchanger
The above picture in Figure 2 shows a more detailed image of a ‘Shell and Tube’ heat exchanger and it also shows the flow pattern of the heat exchanger. There are 4 main values that I would like to draw attention to and these are:
Tube Outlet can be denoted by T1 ,
Tube Inlet can be denoted by T2 ,
Shell Inlet can be denoted by T3 , and
Shell Outlet can be denoted by T4 .
From the aims and objectives given in Section 1.2, one of the values we have to calculate is the duty of the heat exchanger and this is the amount of heat that we want to remove or add using the heat exchanger.
The Duty of can be given by the equation:
Q= Cp g Tout-Tin
For the hot stream the equation is:
Qh= Cp gh ∆T [1]

And for the cold stream the equation is:
Qc= Cp gc ∆T [2]
Where,
Qh = Duty of Heat Exchanger
Cp = Specific Heat Capacity gh = Mass Flow Rate of Hot Stream gc = Mass Flow Rate of Cold Stream
∆T = Tout-Tin

Now that we have the two duty equations, we can put them into the equation for the efficiency of the heat exchanger which is given by:

Ƞ= QcQh [3]

Another one of our aims was to calculate the overall heat transfer value by using the equation:

Q=U A ∆Tlm [4]
Where,
Q = Heat Transfer
U = 1/Resistance = (Overall Heat Transfer Co-efficient)
A = Area
∆Tlm = Driving Force (Log Mean Temperature Difference = LMTD)

To obtain the value of U we need the equation:
1U=1ho+1hos+roln⁡(rori)ks+rori1hi+rori1his

Where, h = Film Heat Transfer Co-efficient r = Radius k = Thermal Conductivity
The final value that we have to calculate in the experiment is the driving force which is given as the Log Mean Temperature Difference (LMTD), using the equation:

LMTD= ∆Tlm= (θ1-θ2)ln⁡(θ1θ2)

Where, θ1 = Difference between Inlet Temperatures for Co-current Flow θ2 = Difference between Outlet Temperatures for Co-Current Flow.
However, the value of θ1 and θ2 vary when looking at a counter-current flow and they vary because the inlet and outlet temperatures of the second fluid are reversed as the second fluid flows from the opposite end of the first fluid.
Therefore, for counter-current flow, θ1 would be the difference between the inlet temperature of fluid 1 and the outlet temperature of fluid 2. θ2 would then be the difference between the outlet temperature of fluid 1 and the inlet temperature of fluid 2.
3. Experimental Methodology
Before we commenced with the laboratory we were given a talk by the demonstrator briefly about the background to the experiment. We then split into a group of 7 people and chose one of the three heat exchangers that were available and the heat exchanger that we were going to be carrying out our experiment using was the ‘shell and tube’ heat exchanger.
To begin with we had a brief look at the setup we were faced with before beginning and we established that the heat exchanger was hooked up to a computer which would record our results and plot a graph from these. So, clicking on the Armfield icon on the computer started the necessary software to carry out the experiment and we then selected the ‘counter-current’ option. There were a number of slides to read providing us with some useful information before carrying out the experiment.
Clicking the ‘diagram’ icon on the Armfield programme brought up a diagram of the heat exchanger with a number of buttons that we were able to click and adjust. There were a number of adjustments to be made and these were: * Set the heater to automatic control with the set-point at 50°C. * Set the flow rate of the cold water to manual and 50% capacity. * Set the flow rate of the hot water to manual and 50% of capacity.
After adjusting these values to the ones stated, we were then able to switch the power on button on the diagram and click GO to start recording our results.
Next to the ‘diagram’ button on the Armfield programme is a ‘graph’ button which we clicked to bring up a set of axis containing no data points. Secondly, in the same window we clicked the ‘y = mx + c’ button to bring up a box where we selected the inlet and outlet temperatures which were T1,T2,T3 and T4 to go on the y-axis and then selected the two flow rates to go on the x-axis.
Following this, we then observed the behaviour of the inlet and outlet temperatures as well as the two flow rates for around 50-70 readings until steady-state had been achieved.
Saving our results to an excel spreadsheet, we then saved on a memory stick and distributed to all the members in the group for processing.
As we had just completed the experiment for the counter-current operation, we then repeated the experiment for the co-current operation following the exact same process and exact same recording of results. Before beginning the co-current experiment, we asked the demonstrator if she would change the water for us so we could start a fresh and have no discrepancies in our results.
Below is a screenshot of the setup that we were faced with on the Armfield Programme and the buttons and values that were required to be changed can be seen.

Figure 3: Screenshot taken from co-current experiment on Armfield Programme.

4. Results
Below are the results for the heat exchanger laboratory and they are split into four subsections and these are; * 4.1 Tables of Co-current and Counter-Current flow taken from result tables. * 4.2 Log Mean Temperature Difference (LMTD) Calculations * 4.3 Calculating the Duty of The Hex and The Efficiency * 4.4 Graphs of Results for Co-Current and Counter-Current Flow 4.1 Tables of Co-Current and Counter-Current flow taken from Result Table
Below are the results for co-current and counter-current flow that has been taken from the results table on excel.
Instead of putting a whole table of samples in this report, I have chosen the numbered sample where the results started to reach a steady-state.
4.1.1 Table of Co-Current Flow at Steady State
As stated previously, I have chosen a steady-state value for the co-current flow and it can be seen in the table below. The sample number that I chose was 74. Sample Number | Temp T1 (°C) | Temp T2 (°C) | Temp T3 (°C) | Temp T4 (°C) | Cph (kJ/kgK) | Cpc (kJ/kgK) | Hot Fluid Avg Temp (°C) | Density Hot Fluid (kg/mᶟ) | 74 | 47.1 | 50.7 | 10.7 | 18.1 | 4.180 | 4.186 | 48.9 | 988.5 |

Cold Fluid Avg Temp (°C) | Density Cold Fluid (kg/mᶟ) | T hot (°C) | T cold (°C) | Hot Mass Flow Rate qmh (kg/s) | Cold Mass Flow Rate qmc (kg/s) | LMTD (K) | Overall Heat Transfer Coefficient U | 14.416992 | 999.2 | 3.6 | 7.3 | 0.075 | 0.035 | 34.19 | 1764.82 |
Table 1: Table of sample number 74 for co-current flow at steady state.

4.1.2 Table of Counter-Current Flow at Steady State
Following on from Subsection 4.1.1, below is the table for sample number 53 for the counter-current flow. Sample Number | Temp T1 (°C) | Temp T2 (°C) | Temp T3 (°C) | Temp T4 (°C) | Cph (kJ/kgK) | Cpc (kJ/kgK) | Hot Fluid Avg Temp (°C) | Density Hot Fluid (kg/mᶟ) | 53 | 51.4 | 47.8 | 12.8 | 20.1 | 4.180 | 4.184 | 49.5828 | 988.2 |

Cold Fluid Avg Temp (°C) | Density Cold Fluid (kg/mᶟ) | T hot (°C) | T cold (°C) | Hot Mass Flow Rate qmh (kg/s) | Cold Mass Flow Rate qmc (kg/s) | LMTD (K) | Overall Heat Transfer Coefficient U | 16.4302 | 998.9 | 3.6 | 7.3 | 0.079 | 0.037 | 33.12 | 1884.39 |
Table 2: Table of sample number 53 for counter-current flow at steady state.

4.2 Log Mean Temperature Different (LMTD) Calculations
As stated in Section 1.2 Aims and Objectives, we are required to calculate the Log Mean Temperature Difference (LMTD) for the heat exchanger in both co-current and counter-current flow.
Below is the first calculation for the LMTD calculation for co-current flow.
The equation is:
LMTD=∆Tlm=θ1-θ2lnθ1θ2
Where, θ1 = T1 (hot inlet) – T3 (cold inlet) θ2 = T2 (hot outlet) – T4 (cold outlet)
LMTD=47.1-10.7-(50.7-18.1)ln(47.1-10.7)(50.7-18.1)
LMTD= 3.80.11026
LMTD=34.465 K

Secondly, here is my LMTD calculation for counter-current flow;

Where, θ1 = T1 (hot inlet) – T4 (cold outlet) θ2 = T2 (hot outlet) – T3 (cold inlet)
LMTD=51.4-20.1-(47.8-12.8)ln(51.4-20.1)(47.8-12.8)
LMTD=-3.7-0.1117
LMTD=33.12 K

4.3 Calculating the Duty of the HEX and the Efficiency
We are also required to calculate the duty of the heat exchanger and then calculate the efficiency of the heat exchanger.
Due to the fact that we have done the experiment in both co-current and counter-current flows, there will be four calculations for the duty.
The duty equation for the hot stream is;
Qh=Cp . gh (Tout-Tin)
Where,
Cp = Specific Heat Capacity of Hot Stream (kJ/kg K) gh = Hot Stream Mass Flow Rate (kg/s)
The first calculation for the hot stream is the duty of hot stream in co-current flow;
Qh=4180 0.075 47.1-50.7
Qh= -1128.6 W
The second calculation for the hot stream is the duty of hot stream in counter-current flow;
Qh=4180 0.079 47.8-51.4
Qh=-1188.792 W
Following on from calculating the duty of the heat exchanger for the hot stream in both co-current and counter-current flow, I then calculated the value of the heat transfer to cold fluid in both co-current and counter-current flow;
Qc=Cp gc (Tout-Tin)
For co-current;
Qc=(4186) (0.035) (18.1-10.7)
Qc=1084.17 W
For counter-current;
Qc=(4184) (0.037) (20.1-12.8)
Qc=1130.1 W
Now that we have the two values of Qc and Qh, we can use these to calculate our efficiency of the heat exchanger in both co-current flow and counter-current flow;
The efficiency of the heat exchanger is defined by the following equation;
Ƞ= QcQh
To calculate the efficiency of the heat exchanger in co-current flow, we use the values;
Qh=1128.6 W
Qc=1084.17 W
Note that, we can ignore the negative in the Qh value as this notes the direction of the Heat Transfer only and does not need to be considered for this calculation.
Ƞ= 1084.171128.6
Ƞ=0.96=96%
Next, we calculate the value of the heat exchanger in counter-current flow using the values;
Qh=1118.792 W
Qc=1130.1 W
Ƞ= 1130.11188.792
Ƞ=0.95=95%

4.4 Graphs of Results for Co-Current and Counter-Current Flow
Below are the two graphs of the results taken from the excel spreadsheet generated by the Armfield Programme.
The first graph is that of the Co-Current flow;

Figure 4: Graph of Co-Current Flow.
The second graph is that of the Counter-Current flow;

Figure 4: Graph of Counter-Current Flow.

5. Discussion and Analysis
5.1 Log Mean Temperature Different (LMTD) Calculation Analysis
Throughout the discussion and analysis I will be closely relating to my results gained in section 4 above and I will also be referring to the results that the Armfield Programme produced and drawing some comparisons.
Firstly, I will look at the Log Mean Temperature Difference (LMTD) calculations obtained from the two chosen sample numbers that were at a steady state flow;
The value obtained for the LMTD of co-current flow using sample number 74 was;
LMTD=34.465 K
And the value obtained for the LMTD of counter-current flow using sample number 53 was;
LMTD=33.12 K
Now, from the results that the Armfield Programme obtained, the value obtained for the LMTD of co-current flow was;
LMTD=33.19 K
And the value obtained for the LMTD of counter-current flow was;
LMTD=33.12 K
Comparing these values, it can be seen that the value I obtained for the co-current flow using sample number 74 which was at steady-state flow was slightly different from the value that the Armfield Programme calculated.
However, it can then be seen that the value obtained for the LMTD of counter-current flow using sample number 53 which was at steady-state flow was exactly the same as the value that the Armfield Programme calculated.
5.2 Efficiency of the Heat Exchanger
Secondly, I will look at the Efficiency of the Heat Exchanger.
It was calculated that the efficiency value of the heat exchanger in co-current flow was calculated as 96% and the efficiency value of the heat exchanger in counter-current flow was calculated as 95%. In theory, the counter-current flow should be more efficient than co-current flow but due to some slight errors and problems in the experiment, this is not the case for my calculations.
5.3 Errors in Laboratory
There are a number errors throughout the laboratory and these are; * Rounding error – Throughout the calculation some of the numbers have roughly 6-9 numbers after the decimal point and there is always the choice of whether to round these numbers to 1 or 2 decimal points instead of keeping the full value. This creates a slight error in the final answer and could be a reason why the value calculated and the value from the software is slightly different.

* The first experiment we done was the counter-current flow on a ‘Shell and Tube’ heat exchanger and from our results, it appears we may have stopped the experiment too early. If we had let the experiment run on for roundabout 20 more samples, the end results would have looked a lot closer to steady-state flow than the ones that can be seen in our graph.

* As with every piece of equipment there is always going to be a slight error in calibration and this may slightly effect the values obtained for the temperatures and other values in the results tables.

6. Conclusions and Recommendations
There are a number of conclusions that can be drawn from this experiment: 1. From the experiment it was calculated that a co-current flow is more efficient on a ‘Shell and Tube’ Heat Exchanger and a counter-current flow is less efficient, however, in reality, this is not the case, a counter-current flow should be more efficient on a ‘Shell and Tube’ Heat Exchanger than a co-current flow but as mentioned in the Discussion and Analysis, a number of errors prevented this.

2. Using sample number 74 which was at a steady-state flow for co-current, I conclude that the value I calculated was slightly higher than the value obtained from the Armfield Programme. Secondly, using sample number 53 which was at a steady-state flow for counter-current, I conclude that the value I calculated was the exact same as the value obtained from the Armfield Programme.

3. From the Duty of the Heat Exchanger, it can be seen that the duty was larger for the hot stream in a counter-current flow than in a co-current flow.

4. The heat transfer to cold fluid is also larger for the cold stream in a counter-current flow than in a co-current flow.

1.

Bibliography
Heat Exchangers. (n.d.). Retrieved December 6th, 2011, from http://lorien.ncl.ac.uk/ming/Webnotes/Main/Where/heat.html

--------------------------------------------
[ 1 ]. http://lorien.ncl.ac.uk/ming/Webnotes/Main/Where/heat.html

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