Joseph Cooper: Project Lead
Kylie Rhoades, Clara Echavarria, Jonathon Locke, Alex Gee

• Background
• Problem Statement (Input on EER table)
• Customer Needs
• Functional Decomposition
• Specifications/Constraints/Given
• Concept Experiment
• Concept Development (Input on alternate designs)
• Concept Decision
• Heat Exchanger Calculations (Input on inlet temperatures)
• Initial Visual Representation of Unit Design

• RIT has a goal of becoming carbon neutral by 2030 and a continuous vision of campus expansion.
• RIT will soon be the home of a brand new ice arena as well as the current home of Ritter Arena.
• Mission: Design a method to extract the cooling energy from a volume of ice (generated from an ice rink) effectively and efficiently.
• On a game day at an ice rink, there are approximately 5 Zamboni
“dumps”, summing up to 500 ft 3 (14.15 m 3 )
• On a typical day of operation, 100 ft 3 (2.83 m 3 ) is discarded.
• According to a density test, this will weigh approximately 2000 kg per load or 10,400 kg on a game day (per 5 loads)

• Create a testing unit to which will demonstrate the feasibility of obtaining a cooling capacity from waste ice. This small scale proof-of-concept will be in the form of an air cooling unit.
• This testing unit is to be comparable (ideally found much better) to cooling efficiencies of a typical water or evaporative cooled condensing unit with a COP of 3.8
• http://www.centerpointenergy.com

Equal to a
COP of 3.8

• Purpose:
• Suspicion of creating an air gap around a pipe is thought of in theory
• Run test to find if we are able to have a vertical heat exchanger pipe in the ice box, and observe ice behavior during melting in this case.
• After about 35 minutes:

Pros
Auto Ice Settling
Known Ice-Heat exchanger SA
Cons
Crush Piping
Pump Required
Closed Loop Allows for Possible Coolant Piping Cost ($$)
No Filter Necessary Maintenance of Coolant Additives
Cleaning of Tank Around Pipes

Pros
Open system without refrigerant.
Auto-settling of ice.
No pipes needed.
Cons
Unknown ice behavior during melting.
Recirculation of the same water (will not get the full cooling effect).
Pump required.
Need screen for pumping loop.
High enough flow rate?

Pros
No pump  Less power in.
No pipes needed.
Water in is about 55 0 F – consistent input temp.
Cons
Waste water.
Need to source the water.
Enough pressure from water to outweigh line losses?
More testing/less theory.

Rank Proposed Mitigation Risk
Is there a high enough flow rate to avoid cavitation?
2
Unknown amount of exposure between working fluid and ice causing a low cooling rate.
3
Based on pump flow rate, start with enough water to eliminate this risk until we understand how the system behavior.
Different spray patterns over top of ice to ensure even melting. If this is not helping, revert to heat exchange
Debris entering pump loop causing pump/system failure.
Water leakage into component area.
3
1
Implement a screen to filter out any unwanted debris.
Keep elements elevated from base of their compartment and/or seal them off.
2
3
Likelihood Scale
1 Low Risk Likelihood
Moderate Likelihood
High Likelihood

• Concept 2 includes benefits from both 1 and 3.
• Can be fitted with a heat exchanger (Concept 1) if needed for appropriate cooling.
• Heat exchanger will require:
• Design
• Lead Time
• Budget/Cost

Background:
• Initial calculations are done with copper tubing
• Future plans are to use a finned radiator
• Coolant has been chosen as water
• Air is to be moved evenly by 2 DC fans with flow rates required by radiator
• Pump to be sized based on radiators and associated head losses

Cross Flow = Air
Tube Flow = Water

• Water Inlet Temperature = 0°C
• Qwater = 1 gpm
• Air Inlet Temperature = ~30°C
• Air Flow Rate = 105.9 CFM or 3 m 3 /min
• ½” Copper Tubing

• Assume:
• Pure Ice at 0 o C
• 5 gallon tank
• 3.5 gallons of ice
• 1.5 gallons of H
2
O
• 300,000 J/kg latent heat of ice
• 917 kg/m 3 density of pure ice
• 736 kg/m 3 experimental density of Zamboni shaved ice
• 2773 BTU storage in Zamboni Ice
• 3992 BTU/hr Cooling Load of Heat Exchanger
• 45 Minutes of Run Time

Total Cooling Load= 1.08 KW or 3692 BTU/hr
Required length of ½” diameter tubing= 96 ft
Tubing Layout:
• 15” of straight tube
• 1.5” diameter elbows
• 1” gap between tubes
Tubing section (HeightxWidthXDepth)=16.5”x3.5”x.5”
Total Size (HeightxWidthXDepth)= 16.5”x19.375”x8”