Design of a Refrigerator. Department of Aerospace and Mechanical Engineering. Abdallah Soliman. Masih Ahmed. Ryan Seballos. EMAE 355 Project 4

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1 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

2 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 20 o F, shielded from the ambient air temperature is 80 o 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 $ , with a q L of 147 kj/, q H of kj/, a of 33.06, and a C OP of w in

3 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

4 Nomenclature Symbol Description Units COP Coefficient of Performance - h Specific enthalpy kj/ ṁ Mass flow rate /s P Pressure MPa q Specific heat kj/ s Specific entropy kj/-k T Temperature w Specific work done by the compressor kj/ 3 ρ Density /m 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 H in L The maximum COP value that can, theoretically, be obtained Refers to a highest temperature, which is where the condenser transfer energy to Refers to specific work transferred into the system by the compressor Refers to a lowest temperature, which is where the evaporator transfer heat from

5 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 20 o F. The ambient air temperature is 80 o 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).

6 2. Methods: Assumptions The mass flow rate remains constant throughout the system Maximum density is used for Power loss calculations 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 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.

7 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 the T 1 refrigerated room (20 or -6.7 ). Next, a temperature shall be chosen such that it is higher than T 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 T 1. Once this is done, using the temperature for 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 can T 3

also be found. Utilizing Figure 2.2, P 4 = P 1 and P 2 = P 3. Now all the pressures of the four states have been found. Also from this same figure one can see that. Utilizing Figure 2.3, we can h 4 = h 3 see that the compression process is isentropic and that s 2 = s 1 as well as that for temperatures.

8 also be found. Utilizing Figure 2.2, P 4 = P 1 and P 2 = P 3. Now all the pressures of the four states have been found. Also from this same figure one can see that. Utilizing Figure 2.3, we can h 4 = h 3 see that the compression process is isentropic and that s 2 = s 1 as well as that for temperatures. Using the now known values of entropy and pressure for state 2, the superheated vapor T 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: q L = h 1 h 4 q H = h 3 h 2 w in = h 2 h 1

9 Now that the q values and specific work have numerically been found, a coefficient of performance (COP) can be calculated as such: C OP h1 h = 4 h 2 h = q L 1 w in 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 = T L TH TL The temperature values of T L and T H are the temperatures of the refrigerated room and 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 q L and then by q H, resulting in two, distinct mass flow rates. m dot = load capacity q 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.

10 Power = Pressure *Volumetric flow rate 3. Results Design Space State Temperature ( ) Specific Enthalpy (kj/) Specific Entropy (kj/-k) Pressure (MPa) Density (/m 3 ) *** *** q L = [Design 1] kj q H = kj w in = kj C OP = State Temperature ( ) Specific Enthalpy (kj/) Specific Entropy (kj/-k) Pressure (MPa) Density (/m 3 ) *** *** q L = [Design 2] kj q H = kj w in = kj C OP = State Temperature ( ) Specific Enthalpy (kj/) Specific Entropy (kj/-k) Pressure (MPa) Density (/m 3 )

11 *** *** q L = [Design 3] kj q H = kj w in = kj C OP = State Temperature ( ) Specific Enthalpy (kj/) Specific Entropy (kj/-k) Pressure (MPa) Density (/m 3 ) *** *** q L = 147 [Design 4] kj q H = kj w in = kj C OP = 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. BTU 12, 000 hr ( ) = kj s Using the equation defined above, the necessary mass flow rate is first determined for q L as follows: q L = h 1 h 4 = ( ) kj = kj ṁ = kj s load capacity q = L kj = s

The process is then repeated for q H : q H = h 3 h 2 = ( 244.55 424.61) kj = 182.06 kj ṁ = kj 3.5168 s load capacity Q = H 182.06 kj = 0.

12 The process is then repeated for q H : q H = h 3 h 2 = ( ) kj = kj ṁ = kj s load capacity Q = H kj = s The selected mass flow rate must be larger than q L, the the greater of the two values. Applying a factor of safety of 1.5, the resulting mass flow rate is: Power Loss Determination ṁ = 1.5 * s = 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. s Using Moody-Diagram[Appendix] - [Use properties at state 3] L=characteristic length = l = 2 meters D= diameter =.25 meters

13 Density = /m 3 Area = (D/2) 2 *pi =.049 m 2 Volumetric flow rate = m dot /density = m 3 /s Velocity = Volumetric Flow rate/area = m/s 32 degrees = 196 * 10-6 Pa*s Reynolds Number = 1.03*10 10 Roughness of steel=[ m] Relative roughness = Roughness/Diameter = Using Moody Diagram, at Reynolds number of 1.03*10 10 and relative roughness of.0006, the friction factor was found to be.017. Pressure Drop = (.017)*(2m/.025m)*Density*Velocity 2 /2 Pressure Drop = *10 6 Pa Power = Pressure Drop * Volumetric Flow Rate = kw Discussion: The mass flow rate varies based on the design space selected. The larger q L or q H 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

14 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: State Temperature ( ) Specific Enthalpy (kj/) Specific Entropy (kj/-k) Pressure (MPa) Density (/m 3 ) *** ***

15 q L = 147 kj/ q H = kj/ w in = C OP = 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 $ $ Condenser True webstaurantstore $ $ Condensing Unit Evaporator Sears Part #: SearsPartsDirect $ $ Expansion Valve DELFIELD PartsTown $ $ Total Price: $

16 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.

17 References [1] S. Turns, Thermal-fluid sciences. Cambridge: Cambridge University Press, [2] '1997 ASHRAE Fundamentals Handbook', Building Services Engineering Research and Technology, vol. 2, no. 4, 1997.

18 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:

19 A1: R134a Tables

24 A2: P-h Diagrams

29 A3: Moody Diagram

30 MOODY DIAGRAM Friction factors for any type and size of pipe. (From Pipe Friction Manual, 3rd ed., Hydraulic Institute, New York, 1961) ESSOM CO., LTD. 510/1 Soi Taksin 22/1 Taksin Rd. Bukkalo Thonburi Bangkok Thailand Tel Fax essom@essom.com,

Refrigeration. 05/04/2011 T.Al-Shemmeri 1

Refrigeration. 05/04/2011 T.Al-Shemmeri 1 Refrigeration is a process of controlled removal of heat from a substance to keep it at a temperature below the ambient condition, often below the freezing point of water (0 O C) 05/04/0 T.Al-Shemmeri