design of a refrigerator abdallah soliman masih ahmed ryan

Design of a Refrigerator

Department of Aerospace and Mechanical Engineering

Abdallah Soliman

Masih Ahmed

Ryan Seballos

EMAE 355 – Project 4

Professor

Dr. J.R. Kadambi

Teaching Assistants

Bo Tan

Henry Brown

December 4, 2015

Abstract:

The engineering team has been tasked with designing a refrigeration

system for a relocating restaurant. The design requirements were to keep the

refrigerated room at 20o F, shielded from the ambient air temperature is 80o F.

Using an environmentally friendly refrigerant - R 134a - the design team

completed the review with positive results. The final design outcome and

procedures are detailed in the following report. The total estimated cost is $

907.51, with a , , a , and a of 147 kJ/kgqL of 80.06 kJ/kgqH − 1 of 33.06win

.OP of 4.44646C

Table of Contents

Section 1: Introduction

Section 2: Methods

Section 3: Results

Section 4: Discussion

Section 5: Conclusions

Section 6: Acknowledgements

References

Academic Integrity Statement

Appendices

A1: R134a Tables

A2: P-h Diagrams

A3: Moody Diagram

Nomenclature

Symbol Description Units

COP Coefficient of Performance -

h Specific enthalpy kJ/kg

ṁ Mass flow rate kg/s

P Pressure MPa

q Specific heat kJ/kg

s Specific entropy kJ/kg-K

T Temperature ℃

w Specific work done by the compressor kJ/kg

ρ Density kg/m3

L Length m

Subscripts

Symbol Description

1 State between the evaporator and compressor

2 State between the compressor and condenser

3 State between the condenser and expansion valve

4 State between the expansion valve and evaporator

Carnot The maximum COP value that can, theoretically, be obtained

H Refers to a highest temperature, which is where the condenser transfer energy to

in Refers to specific work transferred into the system by the compressor

L Refers to a lowest temperature, which is where the evaporator transfer heat from

1. Introduction:

A very popular restaurant is planning to move into a new location. Based

upon the recommendation of the architect, they need a refrigeration system

with the load capacity of 1.25 tons (1 ton=12000Btu/hr) to keep the perishable

food storage room at 20o F. The ambient air temperature is 80o F. The team has

been selected to design an efficient refrigeration system for the restaurant.

Environmentally friendly refrigerant was to be used (e.g. R 134a).

2. Methods:

2.1.1 Assumptions

● The mass flow rate remains constant throughout the system ● Maximum density is used for Power loss calculations

2.1.2 Material Selection

Components:

After a final design space was picked, final components were picked based off the specifications required from each state.

Pipes:

Stainless steel is durable, long lasting, and most importantly, has a low roughness factor and will therefore lead to low pressure drops overall. For this reason, Stainless steel was selected as the material for the pipes.

2.1.3 Calculation Procedure

Before the calculations were carried out, a basic understanding of the refrigeration cycle chosen needs to be had or achieved. This design utilizes a vapor-compression refrigeration cycle. The physical components, as well as the defined and numbered states, are provided below:

Figure 2.1: Components and states of a vapor-compression refrigeration cycle

Now that the basic states and components are known or have been decided upon, knowledge

of the general vapor-compression cycle will be of great use in designing our refrigeration system. The

pressure-enthalpy diagram as well as temperature-entropy diagrams are provided below:

Figure 2.2: General pressure-specific enthalpy diagram with state numbers marked

First a temperature shall be chosen such that it is lower than the temperature of theT 1

refrigerated room (20 ℉ or -6.7 ℃). Next, a temperature shall be chosen such that it is higher thanT 3

the temperature of the ambient air that is exterior of the refrigerator (80 ℉ or 26.7 ℃). Using the

temperature for and the saturated vapor/liquid tables in Appendix A1, the specific enthalpy,T 1

specific entropy, absolute pressure, and density can be found under the saturated vapor columns that

are in line with the respective temperature for . Once this is done, using the temperature for T 1 T 3

and the saturated vapor/liquid tables in Appendix A1, the specific enthalpy, specific entropy, absolute

pressure, and density can be found under the saturated liquid columns that are in line with the

respective temperature for . Once this has been accomplished, the pressure for states 4 and 2 canT 3

also be found. Utilizing Figure 2.2, and . Now all the pressures of the four statesP 4 = P 1 P 2 = P 3

have been found. Also from this same figure one can see that . Utilizing Figure 2.3, we canh4 = h3

see that the compression process is isentropic and that as well as that for temperaturess2 = s1

. Using the now known values of entropy and pressure for state 2, the superheated vaporT 4 = T 1

tables can be used to determine the density, specific enthalpy, and temperature at state 2. Now all of

the property values needed for states 1 through 3 have been found. The entropy and density at state

4 do not need to be solved for as the values are not useful in the calculations.

Figure 2.3: General temperature-specific entropy diagram

Looking at Figure 2.2 again, a few values that need to be solved for are the specific q values

and the specific work inputted into the system by the compressor. These values are as follows:

qL = h1 − h4

qH = h3 − h2

win = h2 − h1

Now that the q values and specific work have numerically been found, a coefficient of

performance (COP) can be calculated as such:

OP C = h −h2 1

h −h1 4 = qLwin

To evaluate how efficient the design is, the COP of the Carnot cycle (or the theoretically

largest possible COP) can be calculated as such:

COP carnot =TL

T −TH L

The temperature values of and are the temperatures of the refrigerated room andT L TH

ambient air, respectively. Note that, realistically, the COP will never reach the Carnot cycle COP.

Mass Flow Rate Determination

To determine the required mass flow rate, the load capacity is first divided by qL and then by qH,

resulting in two, distinct mass flow rates.

mdot = qload capacity

The mass flow rate chosen must be greater than both of the resulting mass flow rates.

Power Loss Calculations

To calculate pressure losses and power required to overcome the losses in , the following equations were used:

Although the density is changing throughout the system, the highest density can be picked for a worst case scenario of power loss.

Power = Pressure *Volumetric flow rate

3. Results

Design Space

State Temperature (℃ )

Specific Enthalpy (kJ/kg)

Specific Entropy (kJ/kg-K)

Pressure (MPa) Density (kg/m3)

1 -20 386.66 1.7417 0.13268 0.02836

2 50 430.91 1.7417 1.0165 45.86

3 40 256.35 1.1903 1.0165 1146.5

4 -20 256.35 *** 0.13268 ***

30.31 qL = 1 kgkJ 74.56 qH = − 1 kg

kJ 4.25 win = 4 kgkJ OP .94486C = 2

[Design 1]





1 -15 389.72 1.7375 0.163975 8.28879

2 45 427.76 1.7375 0.88711 41.42

3 35 248.94 1.1668 0.88711 1167.25

4 -15 248.94 *** 0.163925 ***

40.78 qL = 1 kgkJ 78.82 qH = − 1 kg

kJ 8.04 win = 3 kgkJ OP .70084C = 3

[Design 2]





1 -10 392.75 1.7337 0.20052 10.0371

2 40 424.61 1.7337 0.77008 36.98

3 30 241.65 1.1432 0.77008 1187.2

4 -10 241.65 *** 0.20052 ***

51.1 qL = 1 kgkJ 82.96 qH = − 1 kg

kJ 1.86 win = 3 kgkJ OP .74262C = 4

[Design 3]





1 -12 391.55 1.7351 0.18516 9.30319

2 40 424.61 1.7351 0.81530 36.98

3 32 244.55 1.1527 0.81530 1179.3

4 -12 244.55 *** 0.18516 ***

47 qL = 1 kgkJ 80.06 qH = − 1 kg

kJ 3.06 win = 3 kgkJ OP .44646C = 4

[Design 4]

The fourth design space was selected.

Mass Flow Rate Determination

The provided load capacity is given in units of BTU/hr. To use it in the analysis, it must

first be converted to kJ/s.

2, 00 (0.00029307) 3.5168 1 0 hrBTU = s

kJ

Using the equation defined above, the necessary mass flow rate is first determined for

qL as follows:

391.55 244.55) 47.0 qL = h1 − h4 = ( − kgkJ = 1 kg

kJ

.0239 ṁ = qLload capacity = 147.0 kg

kJ3.5168 s

kJ

= 0 skg

The process is then repeated for qH:

244.55 24.61) 82.06 qH = h3 − h2 = ( − 4 kgkJ = − 1 kg

kJ

.01932 ṁ = QHload capacity = 182.06 kg

kJ3.5168 s

kJ

= 0 skg

The selected mass flow rate must be larger than qL, the the greater of the two values.

Applying a factor of safety of 1.5, the resulting mass flow rate is:

.5 .0239 .0359 ṁ = 1 * 0 skg = 0 s

kg

Power Loss Determination

Assuming all components are connected by straight pipes allows for the

assumption that there will be no power loss due to minor head loss. Although

this may seem far fetched, the fact of the matter is that this design will cool a

large industrial room sized refrigerator, and so a design wherein there are no

bends apart from inside components is possible. Again, because of the large

area, a small length and relatively large diameter can be used.

Using Moody-Diagram[Appendix] - [Use properties at state 3]

L=characteristic length = l = 2 meters

D= diameter = .25 meters

Density = 1179.3 kg/m3

Area = (D/2)2*pi = .049 m2

Volumetric flow rate = mdot/density = 42.33 m3/s

Velocity = Volumetric Flow rate/Area = 862.48 m/s

Viscosity @ 32 degrees = 196 * 10-6 Pa*s

Reynolds Number = 1.03*1010

Roughness of steel=[0.000015 m]

Relative roughness = Roughness/Diameter = .00006

Using Moody Diagram, at Reynolds number of 1.03*1010 and relative roughness

of .0006, the friction factor was found to be .017.

Pressure Drop = (.017)*(2m/.025m)*Density*Velocity2/2

Pressure Drop = 5.96529*106 Pa

Power = Pressure Drop * Volumetric Flow Rate = 2505.42 kW

Discussion:

The mass flow rate varies based on the design space selected. The larger

qL or qH value, the smaller the required mass flow rate will be. A smaller mass

flow rate is desirable as it will require less power to pump the refrigerant

through the system. The selected design has a relatively low mass flow rate and

therefore met this criteria. A factor of safety of 1.5 was applied to the mass flow

rate to ensure that the overall load capacity would remain under the required

12,000 BTU given in the design specifications.

The pressure drop and Power losses were calculated to be massive, but

these values were calculated from worst case scenarios and the high densities of

state 3. Additionally, the refrigerator needs to be reliable and withstand long

periods of time. In reality, the real pressure loss will be negligible compared to

this value. Furthermore, many of the components will contain their own pump,

and as such, a seperate pump will not be purchased.

4. Conclusions:

The Carnot Cycle should have a COP of 8. The team would choose design 3 as it has the highest COP,

but the temperature difference between the evaporator and the storage room as well as the

temperature difference between the condenser and the ambient air is very small. Heat can only be

transferred from the refrigerator to the evaporator if the evaporator is at a lower temperature.

Likewise, the condenser has to be at a higher temperature than the ambient air for heat transfer to

occur in the direction from the condenser to the ambient air. The gap in temperatures in design 3 are

close enough that the system runs the risk of not working if there’s a slight variation in temperature.

Therefore, to ensure that the refrigerator works the team picked the next efficient design with a

larger temperature gap. This particular design was design 4:





1 -12 391.55 1.7351 0.18516 9.30319

2 40 424.61 1.7351 0.81530 36.98

3 32 244.55 1.1527 0.81530 1179.3

4 -12 244.55 *** 0.18516 ***

47 kJ/kg qL = 1 80.06 kJ/kg qH = − 1 3.06win = 3 OP .44646C = 4

Using the above points for Specific Enthalpy, minimum specifications for each component were

determined.

Compressor Wattage = Enthalpy [State 2 - State 1]*mass flow rate = .971 kW

Evaporator Wattage = Enthalpy [State 4 - State 1]*mass flow rate = 5.49 kW

Condenser Wattage = Enthalpy [State 2 - State 3]*mass flow rate = 6.46 kW

Final BOM:

Part Description Source Unit Price Qty Total Price

Compressor Embraco FF8.5HBK1

Grainger $ 276.75 1 $ 276.75

Condenser True 950663

Condensing Unit

webstaurantstore $ 319.99 1 $ 319.99

Evaporator Sears Part #: 5303918284

SearsPartsDirect $186.30

1 $186.30

Expansion Valve

DELFIELD 3516273

PartsTown $ 124.47 1 $ 124.47

Total Price: $ 907.51

Section 6: Acknowledgements

Our project team would like to thank the following people for their assistance during this project:

o Dr. J. R. Kadambi – for the useful R134a tables, pressure-enthalpy diagrams, and suggestions pertaining to the design process.

References

[1] S. Turns, Thermal-fluid sciences . Cambridge: Cambridge University Press, 2006.

[2] '1997 ASHRAE Fundamentals Handbook', Building Services Engineering Research and Technology , vol. 2, no. 4, 1997.

Academic Integrity Statement

This report was written in accordance with the academic integrity policy described in the student handbook of Case Western Reserve University. The following signatures verify that each group member adhered to these standards.

Abdallah Soliman ________________________________ Date: ____________________

Masih Ahmed ________________________________ Date: ____________________

Ryan Seballos ________________________________ Date: ____________________

A1: R134a Tables

A2: P-h Diagrams

A3: Moody Diagram

ESSOM CO., LTD. 510/1 Soi Taksin 22/1 Taksin Rd. Bukkalo Thonburi Bangkok 10600 Thailand Tel. +66 2476 0034 Fax. +66 2476 1500 E-mail : [email protected], http://www.essom.com

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design of a refrigerator abdallah soliman masih ahmed ryan

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