Distribution Systemsin Heating and Cooling

Distribution Systems

Equipment and Materials Necessary for Conveying the Heating and Cooling Media

• Hydronic Systems• Storage Tanks• Steam Systems• Air Distribution Systems

Pipe InsulationRequired Data:• Length of Pipe• Pipe Diameter• Temperature of Fluid• Hours of Operation• Days in Heating Season

• Boiler Efficiency• Fuel Cost• Insulation Levels

– Existing– Proposed

Savings = Length x ∆ Heat Loss x Hours x 1 x $/Unit .of Pipe Length Year Eff. Wh/Unit

Minimum Piping InsulationSystemHeating

Temperature Range (°C)

Insulation Thickness,(mm)

High Pressure Steam 150-230 37,5-87,5

Medium Pressure Steam 120-148 32,5-75

Low Pressure Steam up to 119 25-50

Condensate 88-105 25-75

Hot Water over 93 25-75

Hot Water up to 93 12,5-37,5

Cooling

Chilled Water 4,5-13 12,5-25

Refrigerant and Brine Below 0 25-37,5

Pipe Insulation Problem

Required Data• 87 m of pipe • 37,5 mm diameter • 82 °C water • 24 hours/day• 240 days/year

• 80% efficiency • 0,0287 $/kWh• No insulation• Add 25 mm of

insulation

Cost savings = pipe length x ∆ Heat Loss x Hours x 1 x $ .Length Year Eff. kWh

Heat Loss with Uninsulated PipeW/m of pipe

Temperature of Water in Pipe (°C)37,8 48,9 60,0 65,6 82,0

Heat Loss (W/m of pipe) with no Insulation12,5 18,9 31,2 43,5 45,4 67,120,0 23,6 35,4 53,0 65,2 80,325,0 29,3 47,2 63,3 80,3 100,231,3 35,0 55,8 75,6 96,4 121,037,5 40,6 64,3 88,4 112,5 141,850,0 48,2 75,6 106,3 132,3 160,762,5 56,7 88,8 125,7 154,5 189,075,0 66,1 104,0 149,3 181,4 222,1Pipe

Dia

met

er (m

m)

Heat Loss with InsulationW/m of pipe

Temperature of Water in Pipe (°C)37,8 48,9 60,0 65,6 82,0

Heat Loss (W/m of pipe) with 25 mm of Insulation12,5 2,8 5,7 8,5 11,3 12,320,0 2,8 5,7 8,5 11,8 15,125,0 3,3 6,1 8,5 11,8 15,131,3 3,3 6,1 9,0 12,3 16,137,5 3,8 7,6 9,5 14,2 19,050,0 4,7 10,4 10,9 16,1 21,362,5 6,6 10,4 14,2 21,3 23,675,0 9,5 13,2 19,0 24,6 28,4Pipe

Dia

met

er (m

m)

Tank Insulation Example• Surface temperature (oC)• Room temperature (oC)• Annual hours• Boiler efficiency

• Tank length (m)• Tank diameter (m)• U values (1/R value),

(W/m2 oC)• Energy cost ($/kWh)

⎟⎟⎠

⎞⎜⎜⎝

⎛+×=

42

2DLDA ππ )(

D

L

Area for heat loss:

Savings = (Uold – Unew) x A x ∆T x Hours x $Efficiency x kWh

Duct Insulation

• Data Required– Length of duct (m)– Size of duct (m)– Operating hours/day– Operating days/year

– Ambient air temperature (oC)– Air temperature in duct (oC)– System efficiency– Cost per kWh

Energy Savings =

Area x ∆ W x hours x days x 1 x $ .m2 day year eff. kWh

Energy savings =area x ∆ W/m2 x hours

• Data Required– Length of duct (m)– Area per length of duct (m2/m) – Ambient air temperature (oC)– Air temperature in duct (oC)– Hours of operation– Cost per kWh

Duct Insulation Example

• 30 m duct run• No insulation • 12 hours/day• 240 days/year• Natural gas @ 0,0273 $/kWh

• 1 m x 0.3 m duct size• Add 37,5 mm of insulation• 48.8°C air in duct• 21°C plenum temperature• 80% furnace efficiency

30 m

FurnaceEff.: 80%Fuel

(Natural Gas)

Air Duct (48.8 °C air temp)

21 °C

Duct Insulation Example (cont...)• 30 m duct run• No insulation • 12 hours/day• 240 days/year• Natural gas @ 0,0273 $/kWh

• 1 m x 0.3 m duct size• Add 37,5 mm of insulation• 48.8°C air in duct• 21°C plenum temperature• 80% furnace efficiency

Energy Savings = Area x ∆ W x hours x days x 1 x $ .

m2 day year eff. kWh

Find from the chart