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Calculations & Examples for selecting Water Heating Systems

Calculations & Examples

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Page 1: Calculations & Examples

Calculations & Examples for selecting

Water Heating Systems

Page 2: Calculations & Examples

Features of Heat-Pump Water Heating Systems

•  Energy efficient COP factors over 4.5 Can reach 6+ under ideal conditions

•  Can operate in low ambient conditions : between minus 7°C and + 43°C

•  Supplies constant hot water up to 55°C

– On demand, direct to taps

–  As supply to geyser, for storage

–  For under-floor or radiator panel heaters

Page 3: Calculations & Examples

C O P efficiency graph

Page 4: Calculations & Examples

Calculation Facts

Facts:

•  The energy required to raise one litre of water by one degree Centigrade is a calorie.

•  Water weighs 1kg / litre

•  1 kW heat equals 860 calories, a factor of 1.163

•  An “average guest in a 5 star hotel” uses 125 ~ 150 litres hot water per day

Page 5: Calculations & Examples

Hot Water useage

CIBSE:Chartered Institute of Building Service Engineers

ASHRAE : American Society Of Heating, Refrigeration & Air-conditioning

Institute of Plumbing

Suppliers of Water Heating products

Page 6: Calculations & Examples

Domestic Scenario # 1. Current situation: 4 adults, 2 bathrooms, 2 x 150 litre geysers of 3kW

Electricity Account 2007/08 = 21 563 kWhr = R 6 534.24 ( +- 33c/kWhr) •  Assume that 50% of electricity consumption is for hot water

generation

Then : 50% of 21 563kWhr ÷365 days = 29.54 kWhr/day

Therefore : 29.54kWhr ÷ 6kW elements = 4.92 hours

Hence each geysers work for an average of 4.92 hours per day.

•  Say 5hours / day to heat 150 litres of water, using 3kW power

Page 7: Calculations & Examples

Unit Selection & Savings •  Each geyser operates for 5 hours per day, consuming 3kW each.

By selecting an Airco ARSJ-72/300 we get: 7.2 kW heated water (additional 17% capacity) at 180 litres per hour ( 1.66 hours operation per day) using only 1.87 kW power (electricity saving of 69%)

Therefore :

29.54 kWhr/day less 1.87 kW used x 1.66 hours x 365 = 16 765 kWhrs @ 33c/kWhour = R 5 532.53

Saving of R 1 001.71 per annum at current rates.

Page 8: Calculations & Examples

Hotel # 2.

•  3 star hotel of 250 guests

•  “Make-up” or recovery time of 6 hours

•  Hot water required at 55°C

•  Supply water at 20°C supply

Page 9: Calculations & Examples

Calculations

An hotel with 250 guests @125 l/day = 30 000 litres consumption.

Hence: litres water x ▲T x factor 1.163 1 000 w x 6 hours recovery time

= 30 000 x (55 – 20) x 1.163 6 000

= 1 221 150 6 000

= 203 kW Energy Required to heat 30 000 litres to 55°C

litres water x Delta T x 1.163 factor 1 000 x hours (time limit for heating water)

30 000 x (55 – 20) x 1.163 1000 x 6

= 305 kW

Page 10: Calculations & Examples

Unit Selection

•  203 kW Energy required

No of units required = 3; therefore 203 kW ÷ 3 = 67 kW, however

•  30 000 litres of hot water to be produced in 6 hours = 5 000l/hr

5 000 l/hr ÷ 3 units = 1 666 litres per hour.

Airco ARSJ770/S-820 unit produces 77kW of constant hot water at 1 660litres / hour at 55°C

Page 11: Calculations & Examples

Selection & Savings

Model Airco ARSJ770/S-820 unit produces 77kW of constant hot water at 1 660litres / hour at 55°C

Total Energy output / unit = 77 kW x 3 = 231 kW

Total Power input / unit = 17.4 kW = 52.2 kW

Saving: 77.4% on electricity

C O P = 4.42

Page 12: Calculations & Examples

Clinic # 3. •  Existing: 1 x 4000 litre geysers with 72kW of electric elements

•  Make-up period, maximum of 4 hours.

•  Tariff @ R0.50 per kW/hr, plus max demand charge of R63.00 / kVa

Select Airco 2 x ARSJ380/S-820, each capable of 38.5kW @ 830l/hr. Total Heating : 77 kW @ 1 660l/hr at 60°C using 17.4 kW/hr

Energy Saving: (72kW – 8.7kW) x 2units = 54.6 kW/hr

Max demand saving: 54.6 x R0.50 x 12months = R 41 277.60

kVa demand = (72 kW – 8.7 kW) x R 63.00 x 2 units x 12 months = R 95 709.60 Total Savings: = R 136 987.20

Page 13: Calculations & Examples

Determining Back-up Electrical Capacity 22 room hotel x 1.5 people, 80% occupancy, x 125 l/person = 3 300 litres Allow 25% for losses & mixing = 4 125 litres Recovery period of 4 hours, therefore 2 x 2 000 litre tanks is adequate

Select: 2 x Airco ARSJ-200/S-540, power input/output of 4.35 / 20kW @ 430l/hr. In winter temperatures can drop to -2°C and COP reduces from 4.6 to 2.8

Revised COP

Page 14: Calculations & Examples

Back-up electrical element calcs.

•  Revised COP of 2.8 at -2°C, but Power input remains 4.35 kW

Therefore: 2.8 x 4.35 = 12 kW (revised unit capacity at -2°)

•  12 kW will raise 2 000 litres to 20.6°C in 4 hours and a further 12kW is required to raise temperature from 20° to 65°C in 4.5 hours

•  Electric back-up of 12 kW required in storage vessel, but only operational for 4,5 hours to maintain 2 000litres of water at 65°C

Page 15: Calculations & Examples

Electricity Tarriffs

•  Each municipality charges different rates – Fixed charge Demand charge or Consumption charge

Seasonal rates - Winter & Summer months

•  Some consumers get preferrential rates

If we can get a copy of : Electricity Tariff Schedule recent account We can calculate: Savings in Rands per month Payback period in months

Page 16: Calculations & Examples

Quality Approved & Certification

All units are manufactured to ISO standards and are internationally approved & certified