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Thermal Systems Design

MEM 440 Design Project Group 1: Alexandria Ruggeri & Yong Peng Zhou

9/1/2012

Table of Contents

Executive Summary ...................................................................................................................................... 3

System Definition and Problem Definition.................................................................................................. 4

Results .......................................................................................................................................................... 6

Discussion and Conclusion ........................................................................................................................... 9

References .................................................................................................................................................. 10

Appendix A ................................................................................................................................................. 11

2

Executive Summary

In order to design a good system for heat recovery from exhaust air, optimization on cost and performance is the most important. To maximize the performance and reduce the heat cost and annual cost on the recovery system, heat exchangers are the main part for the optimization. The use of the Newton‐Raphson method is the most effective way to find the minimum cost of the recovery system. There are four constraint equations and six unknowns for the given system. MATLAB is used as the main method to solve them. The total cost of the system including initial cost, electrical heating cost, pump power cost, and the component’s cost is $13251 for a year. The initial cost is $7240, which is half of the total cost. The initial cost mainly comes from the design of the heat exchanger. For instance, fin spacing, number of rows of tubes, length of tubes, etc. The most important finding is to use different methods to determine the optimization of the thermal system.

3

System Definition and Problem Definition

For this design project, students were given the challenge of optimizing an ethylene glycol runaround system for heat recovery from exhaust air. A physical illustration of the system to be optimized can be seen below in Figure 1.

Figure 1. Heat‐recovery system to be optimized

To optimize this system means to optimize, or set up parameters to obtain the best possible solution, the heating cost. The savings to be maximized can be found as the maximum difference between the the reduced heating costs and the annual costs. The optimum system for a set of heat exchangers with a pump for the ethylene glycol and an electric heater should consist of the most favorable combination of the following variables: length of the tubes in the coil, height (or number of tubes high) of the coil, number of rows of tubes deep (parallel to the path of the airflow), fin spacing, and glycol flow rate.

These variables are dependent upon each other as well as a few other unknowns to the system.

Some data to include in the problem was given. The outdoor‐air and exhaust‐air flow rates

should both be 3.0 m3/s, the coil circuiting is made up of vertical headers that feed horizontal tube circuits in parallel, which can be seen in Figure 2.

4

Figure 2. Schematic of the heating system

The flow of glycol through the U bends can be considered counter to the air flow, and pure counterflow can be assumed for the coils. The copper tubes have an OD of 16mm and wall thickness of 1 mm. Tube spacing (both horizontal and vertical) is 41mm, and the average outdoor temperature is 5⁰C for 250 days of 24 hour operation. The life of the system is 10 years with 0% interest rate. Electricity cost is $0.16 per kWh, and both the motor and pump efficiencies are 75%. The initial costs of the pump and motor are to be assumed as $400 each, while the interconnecting piping is assumed constant at $150.

5

Results

A diagram of all of the components can be seen in Figure 2. As stated earlier, the students were to minimize the cost of running this system using constraint equations. In these constraint equations, it is best to minimize the variables: length of the tubes in the coil (L), height (or number of tubes high) of the coil (NR), number of rows of tubes deep parallel to the path of the airflow (W), fin spacing (NF), and glycol flow rate (Veg). Some other variables seen in the constraint equations are Ta,I, which is the temperature of the air right before the electric heating occurs, and Qeg, which is the volumetric flow rate of the ethylene glycol.

With all of this being said, the objective function is to minimize cost. There are a few cost equations that are to be considered, and the total of all of these will be the function to minimize. The cost equations to be considered are as follows:

Initial Cost:

0.26

18

0.024

500

1

Electrical Heating Cost:

0.75

3 1.77 24

,

10 250 24 0.16

Pump Power Cost:

0.75

10 250 24 0.16

Initial Cost of Pump, Motor, & Piping:

400

400

150

This would make the total cost:

In order to minimize this equation, we must give it some constraints. First, we can substitute Qeg with the given area of 0.000154m2 times the velocity of the glycol, so that C3 becomes

0.75

0.000154

10 250 24 0.16

6

Then, we have an equation for the pressure differential in terms of W, L, and Veg, so that we can substitute the following equation into the C3 equation, but also use it as a constraint.

5.2 0.15

0.0875

1

0.3

.

This leads to a new C3 equation of

0.75

0.000154

0.3

10 250 24 0.16

.

5.2

0.15

0.0875

1

In order to optimize a system, there must be more unknowns, n, than constraint equations, m.

Since the number of constraints must be less than the unknowns, we have a perfect amount to continue with our analysis, since there are 4 equations with 6 unknowns and m>n. To solve for an optimum solution when there are multiple constraints and unknowns, it is best to use a processing software, such as Matlab. This was done by the students. In this case, it was also ideal to use the Newton‐Raphson method, which uses successive substitution until a value converges to have very little error. The entirety of the code can be seen in the Appendix, but a snip of it below in Figure 3 shows the equations used.

Figure 3. Snip of Matlab code containing the equations used

As seen in the Appendix, the code was set up with initial guesses for the unknowns were given as well as all of the constraint equations. Then, a while loop was set up to reiterate the function of converging using error found in the matrices that were set up. The Newton‐Raphson method was set up, and then at the end of the code, the total for each cost, as well as the overall cost can be seen.

An optimized solution was found using this method. There were a few issues along the way, but

the results were found, nonetheless. As for the results, the individual costs as well as total costs can be seen below in Table 1.

7

Table 1. Cost of the System

Initial Cost

$7,240.20

Electic Heating Cost

$3,845.00

Pump Power Cost

$1,215.90

Cost of Pump, Motor, & Piping $950.00

Total Cost

$13,251.10

The initially guessed values were changed during the iterations. The optimum values for each of these can be seen below in Table 2.

Table 2. Optimized Values for Unknowns

Unknown

NR

W

L

NF

Veg

Tai

Initial Guess

5

1

0.5

15

10

23.9

8

Optimum Value

5

1

0.5

15

35.69

24

Discussion and Conclusion

By using the Newton‐Raphson method, the students were able to solve for an optimized system

for the given heating system. There were a few complications when using Matlab. The main issue was that there was an issue with the number of iterations that were performed. This was solved by simply changing the initial values so that it would converge to the optimum solution. There were no assumptions in values that needed to be made to simplify the number of unknowns and constraints.

Overall, the students learned about optimization, the Newton‐Raphson method, Matlab coding,

and thermal systems. This project was a well‐rounded assignment that students were able to gain much experience from.

9

References

Janna, W.S. (2009). Design of Fluid Thermal Systems (3rd ed.). Thomson Learning.

Stoecker, W.F. (1989). Design of Thermal Systems (3rd ed.). Mc‐Graw‐Hill.

1 0

Appendix A: Matlab Script

% Solve non-linear algebraic equations using Newton-Raphson method

% Problem 6.15 (Stoecker Text)

% The problem formulation gives the following system of equations:

%

f(1)= 0.26*(18+0.024)*(x(4)+500)*x(2)*(x(3)*(x(1)+1)); %%initial cost

%

f(2)= 0.75*3*1.77*1005.7*(24-x(7))*10*250*24*0.16; %%electrical heating cost % f(3)= 0.75*x(6)*0.000154*x(5)*10*250*24*0.16; %%pump power cost

%

f(4)= 400+400+150; %%cost of pump, motor, and piping

%

f(5)= f(1)+f(2)+f(3)+f(4);

%

f(1)=

0.26*(18+0.024)*(x(4)+500)*x(2)*(x(3)*(x(1)+1))+0.75*3*1.77*1005.7*(24x(7))*10*250*24*0.16+0.75*x(6)*0.000154*x(5)*10*250*24*0.16+400+400+150;

% This Program starts with initial guesses and continuously updates the

% solution

% set initial guess clear all; clc; %initial guess x(1)=5; x(2)=1; x(3)=0.5; x(4)=15; x(5)=10; x(6)=23.9999;

% Here x(1) is NR, x(2) is W, x(3) is L, x(4) is NF, x(5) is Veg, x(6) is

% Qeg and x(7) is Tai disp(sprintf('Initial Condition:\n')); disp(sprintf('NR W

L

NF

Veg

Tai')); disp(sprintf('------------------------------------------------------')); disp(sprintf('%0.1f

%0.1f

%0.1f

%0.1f

%0.1f

%0.1f\n',x(1),x(2),x(3),x(4),x(5),x(6)));

% initialize the function array f = zeros(length(x),1);

% initialize the derivative array df = zeros(length(x),length(x));

% initialize the 'error' error=1E8; % initialize iteration number i = 0; maxi = 30; disp(sprintf('Solution:\n')); 1 1

disp(sprintf('Iteration No.

NR

W

L

NF

Veg

Tai')); disp(sprintf('-------------------------------------------------------------------')); % do the iteration while error > 1E-5 i = i + 1;

% calculate the f values f(1)= 0.26*(18+0.024)*(x(4)+500)*x(2)*(x(3)*(x(1)+1)); %%initial cost f(2)= 0.75*3*1.77*1005.7*(24-x(6))*10*250*24*0.16; %%electrical heating cost f(3)= 0.75*5.2*(0.15*x(2)*x(3)+0.0875*(x(2)1)+0.3)*(x(5)^1.75)*0.000154*x(5)*10*250*24*0.16; %%pump power cost f(4)= 400+400+150; %%cost of pump, motor, and piping

%

f(5)= 5.2*(0.15*x(2)*x(3)+0.0875*(x(2)-1)+0.3)*x(5); %%constraint equation for differential pressure

% calculate the derivative values

% df(1,1) stores the value of derivative df1/dx1 df(1,1) = (0.26*(18+0.024)*(x(4)+500)*x(2)*x(3)); df(1,2) = 0.26*0.024*(x(4)+500)*x(3)*(x(4)+1); df(1,3) = 0.26*(18+0.024)*(x(4)+500)*x(2)*(x(1)+1); df(1,4) = 0.024*0.26*x(2)*x(3)*(x(1)+1); df(1,5) = 0; df(1,6) = 0; df(2,1) = 0; df(2,2) = 0; df(2,3) = 0; df(2,4) = 0; df(2,5) = 0; df(2,6) = 0.75*3*1.77*1005.7*10*250*24*0.16; df(3,1) = 0; df(3,2) = 0; df(3,3) = 0; df(3,4) = 0; df(3,5) = x(6)*0.75*10*250*24*0.16*0.000154; df(3,6) = x(5)*0.75*10*250*24*0.16*0.000154; df(4,1) = 0; df(4,2) = 0; df(4,3) = 0; df(4,4) = 0; df(4,5) = 0; df(4,6) = 0;

%

df(5,1) = 0;

%

df(5,2) = 0.52*x(5)*(0.15*x(3)+0.0875);

%

df(5,3) = 0.52*x(5)*(0.15*x(2));

%

df(5,4) = 0;

%

df(5,5) = 5.2*(0.15*x(2)*x(3)+0.0875*(x(2)-1)+0.3);

%

df(5,6) = 0;

%

df(5,7) = 0; df; f; y=-df/f'; 1 2

%x(1)=x(1)+y';

%x(2)=x(2)+y';

%x(3)=x(3)+y';

%x(4)=x(4)+y';

%x(5)=x(5)+y';

%x(6)=x(6)+y';

%x(7)=x(7)+y';

error=sqrt(y(1)*y(1)+y(2)*y(2)+y(3)*y(3)+y(4)*y(4)+y(5)*y(5)+y(6)*y(6)); error(i)=sqrt(f(1)*f(1)+f(2)*f(2)+f(3)*f(3)+f(4)*f(4)+f(5)*f(5)); % set the 'A' marix

A=[

df(1,1) df(2,1) df(3,1) df(4,1) x=[

df(1,3) df(2,3) df(3,3) df(4,3) df(1,4) df(2,4) df(3,4) df(4,4) df(1,5) df(2,5) df(3,5) df(4,5) df(1,6); df(2,6); df(3,6); df(4,6)]; x(1); x(2); x(3); x(4); x(5); x(6)]; f=[

df(1,2) df(2,2) df(3,2) df(4,2) f(1); f(2); f(3); f(4)]; % calculate dx xc=linsolve(A,f); % correct the solution (x) x=x-xc; disp(sprintf('\t%d \t\t\t%0.3f\t%0.3f \t%0.3f

\t%0.3f\t%0.3f\t%0.3f\t%0.3f',i,x(1),x(2),x(3),x(4),x(5),x(6)));

% calculate the relative error error=max(abs(xc/x)); if (i > maxi) error = 0; s=sprintf('****Did not converge within %3.0f iterations.****',maxi); disp(s) end

% continue loop end f(1)

1 3

f(2) f(3) f(4) f(1)+f(2)+f(3)+f(4) 1 4

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...Motivation I cannot trace back the moment when I first heard this statement that the more energy from sunlight strikes the earth in an hour than all of the energy consumed on the planet in one year but I wondered at that time, why are we having energy problems on earth. This was the start of fascination to look keenly into the problem of making alternate energy sources more practical. This fascination about renewable energy made me think more and more with open mind about the available large amount of energy which is gift of the nature to us. If we have made our life worse by not taking care of the nature then to use nature positively is the only way for us to save our planet. But I always wondered why we could not still use these sources for all our needs. But I knew if I had to go into the root of this thing, my aspiration should be coupled with the strong technical base and other necessary dependent aspects. In my third year I took course on “Renewable Energy” that gave me knowledge about different kinds of renewable energy systems. The course structure covered major part of it - Solar Energy, Wind Energy and Bio-Energy etc. And it was more like the base for my future endeavor. Fortunately having centre of Excellence in Energy in our college (Indian Institute of Technology Rajasthan) gave me several opportunities to get exposure of Energy field. In the 5th semester, The Centre of Excellence in Energy at IIT Rajasthan conducted a Renewable Energy Consortium in 2010 in......

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...to compare LiquefinTM with the competing process, specially the reputed most efficient ones: C3/MR, C3/MR followed by nitrogen cycle, dual mixed refrigerant process with spiral wound exchangers. To compare properly one process to another, it must be done with the same gas, with the same site conditions, with the same gas turbines and with the same cooling medium temperature (air or water). That done, to compare processes like for like is still not that easy. For instance, it seems fair to take the same efficiencies for the compressors, however, axial and centrifugal compressors do have different efficiencies. Similarly, equal basis leads to have the same temperature approach for the air-cooler (or water coolers), however between mixed refrigerant and propane, the heat exchange area will be much lower for mixed refrigerant if the approach is kept identical. The end flash vapour quantity has also a big influence on the process efficiency, but each process has a different fuel gas consumption. Those added factors may lead to wide differences. Axens has calculated the effect of all those parameters on efficiency. The equipment characteristics play also an important role in the comparison: the limitations of axial compressors, of centrifugal compressors (Mach number) and possibly spiral-wound exchangers maximum size do have to be taken into account. Even the gas turbines or alternative drivers chosen can be well adapted to one process, but not to the other. Another important......

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