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E NERGY SAVINGS : Central AC vs. Convective River Cooling By Drew Anderson & Joshua Cluff.

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Presentation on theme: "E NERGY SAVINGS : Central AC vs. Convective River Cooling By Drew Anderson & Joshua Cluff."— Presentation transcript:

1 E NERGY SAVINGS : Central AC vs. Convective River Cooling By Drew Anderson & Joshua Cluff

2 H OME AIR CONDITIONING ❖ The average home spends nearly $400.00 for three months of air conditioning for the Summer months. ❖ Methods to cool homes have been attempted in all areas. ❖ What if you could use a nearby river/stream/lake to cool air to return it into your home? ❖ Is it possible and affordable? What would the conditions have to be and how much money could you save?

3 T HE SETUP A home near a river pumps internal air from the home into a pipe that travels into a river to be cooled convectively. ❖ Temp Air from home: 32 C ❖ Temp Water: 17 C ❖ Temp Air out of pipe: 17 C ❖ River Velocity: 2 m/s ❖ Air Velocity: 4.56 m/s Assumptions: ❖ No pressure losses in pipe. ❖ Infinite heat sink ❖ Convective coefficient of water is ~1500 W/m-K

4 Home Design Assumptions Home Criteria of Comparison: ❖ Utah based home ❖ US average cost of electricity: 11.65 ¢/kw-hr ❖ 935 Operating hrs/yr ❖ 2000 ft 2 home ❖ 3-ton Central Air Unit ❖ Furnace Fan only operates at: ~ 600 Watts ❖ Central Air operates at: ~ 4200 Watts

5 Problem Solution ❖ Air Calculations: ❖ Water Calculations: ❖ Solve for L by subbing R Tot into above equation. ❖ Power consumption: ❖ Determine costs, and optimize if possible to maximize cost vs. realism

6 Cost Calculation based on Aluminum Piping 1m/s river

7 Cost Calculation based on Aluminum Piping 2m/s river

8 R ELATIONSHIPS AND RESULTS  Changing the material of the pipe makes little difference  Slow moving river verse faster river is significant  Internal air speed of pipe is the major factor; slower is better  Compare 2 inch Dia volume flow rate to 6 inch Dia of current duct of current systems makes cost a 1900% decrease in cost for new system

9  From the data it becomes obvious that there are various obstacles to overcome in order to make this method of cooling effective. The first and foremost obstacle is creating a scenario that reaches the necessary amount of cooling needed for a 2000 ft 2 home. Major obstacles to this are being able to create a large enough heat flux in the pipe and air flowing through the pipe. Issues arise with the large volumetric flow rate needed to cool a home being too large to reach a feasible pipe length. The ratio between pipe diameter and length can be maximized but additional issues arise when the diameter of the pipe is so small it would require significant pump power to maintain the needed flow rate. If possible to reduce the air flow in the duct, perhaps in creating a scenario in which the furnace fan runs more often but at lower velocities, it would be significantly more reasonable. Conclusions

10 Appendix  Acknowledgments:  Cost Estimates to use Central Air:  https://answers.yahoo.com/question/index?qid=20070125151027AAbwG1q https://answers.yahoo.com/question/index?qid=20070125151027AAbwG1q  US Electricity Rates: 11.65 cents/kw-hr  http://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a http://www.eia.gov/electricity/monthly/epm_table_grapher.cfm?t=epmt_5_6_a  HVAC Calculator:  http://www.hvacopcost.com/ http://www.hvacopcost.com/  Costs to Operate various appliances:  http://www.glendalewaterandpower.com/rates/appliance_operating_costs.aspx http://www.glendalewaterandpower.com/rates/appliance_operating_costs.aspx

11 Properties:  Air @ 26.5 C  rho = 1.1613 kg/m^3  cp = 1007 J/kg-K  k =.0263 W/m-K  Pr =.707  mu = 184.4*10^-7 N-s/m^2  nu = 15.86*10^-6  alpha = 22.5*10^-6  Water @ 17 C  cp = 4184 J/kg-K  mu = 1076*10^-6 N-s/m^2  k =.5982 W/m-K  Pr = 7.53 Appendix


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