GEOTHERMAL GREENHOUSE GEOTHERMAL GREENHOUSE DESIGN *DESIGN *

John W. LundJohn W. LundNational Renewable Energy LabNational Renewable Energy Lab

Golden, ColoradoGolden, Colorado

* Based on work by K. Rafferty and Toni Boyd

Greenhouses Greenhouses –– Introduction IIntroduction I

•• Optimize growthOptimize growth•• Reduce operating costs (up to 35%)Reduce operating costs (up to 35%)•• Allows operation in colder climatesAllows operation in colder climates

Geothermally heated Greenhouses:Geothermally heated Greenhouses:

•• Allows operation in colder climatesAllows operation in colder climates•• Uses relatively lowUses relatively low--grade heat (25grade heat (25oo to to

9090ooC)C)•• Better humidity control to prevent Better humidity control to prevent

condensation and reduce problems related condensation and reduce problems related to disease controlto disease control

Greenhouses Greenhouses –– Introduction IIIntroduction II

•• 23,300 TJ/yr (6,500 GWh/yr) or 0.8 million TOE 23,300 TJ/yr (6,500 GWh/yr) or 0.8 million TOE savedsaved

•• WorldWorld--wide ~ 1200 ha (2010)wide ~ 1200 ha (2010)

WorldWorld--wide Utilization Iwide Utilization I

•• WorldWorld--wide ~ 1200 ha (2010)wide ~ 1200 ha (2010)•• Russia: 2,500 ha of agricultural land heated? Russia: 2,500 ha of agricultural land heated?

–– covered and uncoveredcovered and uncovered

•• Hungary: 120 ha of greenhousesHungary: 120 ha of greenhouses•• Japan: 23 ha greenhouses Japan: 23 ha greenhouses

–– some as tourist attractionssome as tourist attractions

Greenhouses Greenhouses –– Introduction IIIIntroduction III

•• Iceland: over 20 ha of greenhousesIceland: over 20 ha of greenhouses–– 10.5 of soil warming10.5 of soil warming–– along with a horticulture college at Hvergerdialong with a horticulture college at Hvergerdi

WorldWorld--wide Utilization IIwide Utilization II

–– along with a horticulture college at Hvergerdialong with a horticulture college at Hvergerdi

•• USA: 40 ha of greenhousesUSA: 40 ha of greenhouses–– Mainly cut flowersMainly cut flowers

•• Italy: 22 ha at Mt. Amiata plus drying operationItaly: 22 ha at Mt. Amiata plus drying operation•• Tunisia: 110 haTunisia: 110 ha•• Turkey, Russia, Hungary, China, and ItalyTurkey, Russia, Hungary, China, and Italy

20

30

40

50

60

70

75

DRIER

GREENHOUSES

Mt. Amiata, Italy

AUG SEP OCT NOV DEC JAN FEB MAR APR MAY JUN JULY

10

•15 MWe – 184oC

•22 ha houses

•86oC – 2000t/hr

•Flowers/plants

•Feed/vegetables

•650 employees

32 50 68 86 104

75

100

125

Temperature 0F

LETTUCE TOMATO

0 5 10 15 20 25 3530 400

25

50

Temperature 0C

CUCUMBER

General Design Criteria & Equipment IGeneral Design Criteria & Equipment I

•• 0.2 to 0.4 ha for 0.2 to 0.4 ha for individualindividual buildingsbuildings

•• Rectangular size pf approx. 36 by 110 mRectangular size pf approx. 36 by 110 m–– 10 by 30 to 50 m for plastic film types10 by 30 to 50 m for plastic film types

Most efficient and economical greenhouse Most efficient and economical greenhouse shape and size:shape and size:

–– 10 by 30 to 50 m for plastic film types10 by 30 to 50 m for plastic film types

•• 6 L/s of 606 L/s of 60ooC to 80C to 80ooC Fluid required for peak C Fluid required for peak heatingheating

•• Load factor of 0.4 to 0.5 (0.48 in 2010)Load factor of 0.4 to 0.5 (0.48 in 2010)–– ~0.45 average~0.45 average

•• Storage tank for peak demandStorage tank for peak demand

•• Major commercial operation approx. 6 ha Major commercial operation approx. 6 ha –– 15 greenhouses15 greenhouses

General Design Criteria & Equipment IIGeneral Design Criteria & Equipment II

•• Glass Glass

–– most most expensiveexpensive and least energy efficientand least energy efficient

••

Construction MaterialConstruction Material

•• Plastic FilmPlastic Film

–– Frequent replacementFrequent replacement

•• Fiberglass or similar rigid plasticFiberglass or similar rigid plastic

•• Combination of plastic film and fiberglassCombination of plastic film and fiberglass

Steel or aluminum Steel or aluminum frameframe (wood frames in the past)(wood frames in the past)

General Design Criteria & Equipment IIIGeneral Design Criteria & Equipment III

•• Arch or peaked roofArch or peaked roof•• Double or single layer Double or single layer

of plastic filmof plastic film

ShapeShape

of plastic filmof plastic film

CostCost

•• US$ 50 to 100 per sq. US$ 50 to 100 per sq. metermeter

Mt. Amiata, Italy

glass houses

Burgett’s greenhouses, NM

High snow load design

Masson Greenhouse

New Mexico

6.4 ha

Potted flowers and tropical plants

Heating Pipe Network

Production Wells

Fan Coil

Thermostat T

BypassGW - in

GW - out

A

Typical Large Greenhouse LayoutTypical Large Greenhouse Layout

Storage Tank

Return WellCiculating Pump

T

300 m 300 m

110 m

3 m

36 m

B

Heating System Design IHeating System Design I

In most applications, a heat exchanger is In most applications, a heat exchanger is required to separate the actual heating required to separate the actual heating equipment from the geothermal fluid equipment from the geothermal fluid –– due to due to scaling and corrosion problemsscaling and corrosion problems

•• Plate heat exchangersPlate heat exchangers–– 33oo to 6to 6ooC temperature lossC temperature loss

•• ShellShell--andand--tube heat exchangers tube heat exchangers –– 88oo to 11to 11ooC temperature lossC temperature loss

•• Homemade configurationsHomemade configurations–– 1111oo to 22to 22oo C temperature lossC temperature loss

650 C600 C

ProductionWell

HeatExchanger

FinnedCoil

FloorTubes

Heat Exchanger Heat Exchanger SchematicSchematic

500 C450 C

ToDisposal

CirculatingPump

Exchanger

Hot brine Hot water

Shell and Tube Heat Shell and Tube Heat ExchangerExchanger

Cold brine Cold water

Geothermal out

Treated water in

Plate Heat Plate Heat ExchangerExchanger

Geothermal in

Treated water out

Heating Requirement IHeating Requirement I

•• Heat loss composed of two componentsHeat loss composed of two components–– Transmission loss through the walls and roof, Transmission loss through the walls and roof,

Need to determine peak heating requirements Need to determine peak heating requirements for the structurefor the structure

–– Transmission loss through the walls and roof, Transmission loss through the walls and roof, andand

–– Infiltration and ventilation losses due to the Infiltration and ventilation losses due to the heating of cold outside airheating of cold outside air

Heating Requirement IIHeating Requirement II

Transmission loss:Transmission loss:Trans. Heat loss = (surface area of structure) Trans. Heat loss = (surface area of structure)

X (inside X (inside –– outside temperature) outside temperature)

X (heat loss factor (UX (heat loss factor (U--factor))factor))

= SA1 X = SA1 X ∆∆TD1 X HLF1 TD1 X HLF1

+ SA2 X + SA2 X ∆∆TD2 X HLF2 TD2 X HLF2

+ …….+ …….

(if constructed of more than one material)(if constructed of more than one material)

See Table 1 for HLF at 24 km/hr outside wind speed, and See Table 1 for HLF at 24 km/hr outside wind speed, and

Table 2 for wind speed correction factorsTable 2 for wind speed correction factors

Heating Requirement IIIHeating Requirement IIIInfiltration loss:Infiltration loss:

Analyzed based on air changes/time (from leakage)Analyzed based on air changes/time (from leakage)

Infil. Heat loss = (air changes/hr) Infil. Heat loss = (air changes/hr)

x (volume of greenhouse) x (volume of greenhouse)

x (inside x (inside –– outside temperature)outside temperature)x (inside x (inside –– outside temperature)outside temperature)

= AC/H x V x = AC/H x V x ∆∆TDTD

see Table 3 for AC/Hsee Table 3 for AC/H

if V is in mif V is in m33 and and ∆∆TD in TD in ooC, thenC, then

x 0.088 to get kg calx 0.088 to get kg cal--hrhr

x 0.102 to get kWx 0.102 to get kW

X 1.21 to get kJ/hrX 1.21 to get kJ/hr

Heating Requirement IVHeating Requirement IVTotal greenhouse heating requirement Total greenhouse heating requirement (peak/hr)(peak/hr)

Total heating requirementsTotal heating requirements

= trans. Heat loss + infil. Heat= trans. Heat loss + infil. Heat= trans. Heat loss + infil. Heat= trans. Heat loss + infil. Heat

Annual heating requirementsAnnual heating requirements

= total x load factor x time= total x load factor x time

(load factor approx. = 0.4 to 0.5)(load factor approx. = 0.4 to 0.5)

Heating Systems ConsiderationsHeating Systems Considerations

Must provide for ventilation Must provide for ventilation –– ceiling ceiling openingsopenings

Also, may need to restrict light transmission Also, may need to restrict light transmission Also, may need to restrict light transmission Also, may need to restrict light transmission (glass) or increase it (plastic film) + (glass) or increase it (plastic film) + lightinglighting

Heating at night less for plantsHeating at night less for plants

Heating SystemsHeating Systems

Accomplished by:Accomplished by:

•• Circulation of air over finnedCirculation of air over finned--coil heat exchanger s coil heat exchangers carrying hot watercarrying hot water

•• Often used with perforated plastic tubes running th e Often used with perforated plastic tubes running th e length of the greenhouse to maintain uniform heat length of the greenhouse to maintain uniform heat length of the greenhouse to maintain uniform heat length of the greenhouse to maintain uniform heat distributiondistribution

•• HotHot--water circulating pipes or ducts located in (o r water circulating pipes or ducts located in (or on) the flooron) the floor

•• Finned tube units located along the walls and under Finned tube units located along the walls and under benches, orbenches, or

•• Bare pipes running along the walls (Iceland: ROT Bare pipes running along the walls (Iceland: ROT -- 2 2 m of pipe for 1 mm of pipe for 1 m 22 of floor area) = 0.5 ft per sq. ft.of floor area) = 0.5 ft per sq. ft.

•• Combination of the above Combination of the above -- typicaltypical

Heating System DesignHeating System Design

Usually installed along the walls (long Usually installed along the walls (long dimension), and one third can be installed dimension), and one third can be installed across the floor to improve heat distributionacross the floor to improve heat distribution

27

19 C

18

17

20 2021 21

222323

20222427

Temperature profiles in a greenhouse heated with ra diation piping along Temperature profiles in a greenhouse heated with ra diation piping along the side wallsthe side walls

Heating System Design IIHeating System Design IIFinned pipe heating system:Finned pipe heating system:

Warm Air Hot Water

Constructed of steel or copper with steel or Constructed of steel or copper with steel or aluminum fins attached to the outside aluminum fins attached to the outside ––generally based on 93generally based on 93ooC average water C average water temperature and 18temperature and 18ooC entering air C entering air temperature temperature –– heating expressed in W/mheating expressed in W/m

AirWarm Water

Hot Water

Warm Water

Floor Heating SystemFloor Heating System

Basis:

Roof/floor area = 1.3/11 change/hour8km/h wind

200

300

400

Inside-outside temperature differencein degrees C

100

10 20 30 40 500

Greenhouse Greenhouse DesignDesign CalculationsCalculations II1. Greenhouse 1. Greenhouse heatheat lossloss

2.5 m

36 m

Inside design temp. = 20Inside design temp. = 20ooCC

Outside design temp. = 5Outside design temp. = 5ooCC

Design wind velocity = 40 km/hDesign wind velocity = 40 km/h

Double poly roof: U = 14.9 kJ/mDouble poly roof: U = 14.9 kJ/m22hrhrooCC

13 m Fiberglass walls: U = 21.6 kJ/mFiberglass walls: U = 21.6 kJ/m22hrhrooCC

End area of arc = 0.15 wEnd area of arc = 0.15 w 22 (approx. ¼ circle)(approx. ¼ circle)

Arc length = 1.12 wArc length = 1.12 w

Area of double poly roof: 1.12 wL = 1.12 x 13 x36 = 524 mArea of double poly roof: 1.12 wL = 1.12 x 13 x36 = 524 m2 2

Area of fiberglass: 2(2.5 x 36 + 2.5 x 13 + 0.15 x 13Area of fiberglass: 2(2.5 x 36 + 2.5 x 13 + 0.15 x 1322) = 296 m) = 296 m22

Greenhouse Design Greenhouse Design CalculationsCalculations IIII1a. Transmission loss (see Tables 1 and 2 for U1a. Transmission loss (see Tables 1 and 2 for U--Values)Values)

Double poly: Double poly:

A x A x ∆∆T X U = 524 mT X U = 524 m22 x (20x (20--5)5)ooC x 14.9 kJ/mC x 14.9 kJ/m22hrhrooCC

= 117,114 kJ/hr= 117,114 kJ/hr

Fiberglass: Fiberglass: Fiberglass: Fiberglass:

A x A x ∆∆T X U = T X U = 296 x (20296 x (20--5) x 21.6 5) x 21.6

= 95, 904 kJ/hr= 95, 904 kJ/hr

Total = 213, 018 kJ/hrTotal = 213, 018 kJ/hr

Table 1. Table 1. Glazing material Glazing material UU--valuesvalues (based on 24 (based on 24 km/hr wind speed)km/hr wind speed)

MaterialMaterial kJ/mkJ/m22 hr hr ooCC W/mW/m22 ooCC

GlassGlass 22.522.5 6.26.2

FiberglassFiberglass 20.420.4 5.75.7

Single PolySingle Poly 23.523.5 6.56.5

Double PolyDouble Poly 14.314.3 4.04.0

Table 2. Table 2. UU--values at values at VariousVarious Wind Velocities Wind Velocities (kJ/m(kJ/m22 hr hr ooC)C)

Wind VelocityWind Velocity

00 88 1616 3232 4040 48 km/hr48 km/hr

MaterialMaterial 00 2.242.24 4.474.47 8.948.94 11.211.2 13.4 m/s13.4 m/sMaterialMaterial 00 2.242.24 4.474.47 8.948.94 11.211.2 13.4 m/s13.4 m/s

GlassGlass 15.615.6 19.419.4 21.221.2 23.323.3 23.723.7 24.124.1

FiberglassFiberglass 14.214.2 17.717.7 19.419.4 21.121.1 21.621.6 22.022.0

Single PolySingle Poly 16.516.5 20.420.4 22.322.3 24.324.3 24.724.7 25.125.1

Double PolyDouble Poly 10.910.9 12.912.9 13.813.8 14.614.6 14.914.9 15.015.0

Greenhouse Design Greenhouse Design CalculationsCalculations IIIIII1b. Infiltration loss (1b. Infiltration loss (seesee Table 3 for air change data)Table 3 for air change data)

Volume of greenhouse: Volume of greenhouse:

36 x 13 x 2.5 + 0.15 x 1336 x 13 x 2.5 + 0.15 x 1322 x 36x 36

=2083 m=2083 m33

Heat loss Heat loss Heat loss Heat loss

AC/H x V AC/H x V x x ∆∆T X 1.21 kJ/mT X 1.21 kJ/m 3o3oC = C =

= 1.50/hr x 2083 m= 1.50/hr x 2083 m 33 x (20x (20--5)5)ooC x 1.21 kJ/mC x 1.21 kJ/m 3o3oCC

= 56,710 kJ/hr= 56,710 kJ/hr

1c. Total 1c. Total heatingheating lossloss

Total (q) Total (q) = 213,018 + 56,710 + 269, 728 kJ/hr ~= 213,018 + 56,710 + 269, 728 kJ/hr ~ 270,000 kJ/hr270,000 kJ/hr

Table 3. Table 3. Air Change Data for Various Glazing Air Change Data for Various Glazing MaterialsMaterials

MaterialMaterial Air changes/hrAir changes/hr

Single glassSingle glass 2.5 to 3.52.5 to 3.5

Double glassDouble glass 1.0 to 1.51.0 to 1.5

FiberglassFiberglass 2.0 to 3.02.0 to 3.0FiberglassFiberglass 2.0 to 3.02.0 to 3.0

Single polySingle poly 0.5 to 1.00.5 to 1.0

Double polyDouble poly 0.0 to 0.50.0 to 0.5

Single poly / w low fiberglass sidesSingle poly / w low fiberglass sides 1.0 to 1.51.0 to 1.5

Double poly / w low fiberglass sidesDouble poly / w low fiberglass sides 0.5 to 1.00.5 to 1.0

Single poly / w high fiberglass sidesSingle poly / w high fiberglass sides 1.5 to 2.01.5 to 2.0

Double poly / w high fiberglass sidesDouble poly / w high fiberglass sides 1.0 to 1.51.0 to 1.5

Greenhouse Design Calculations IVGreenhouse Design Calculations IV1d. Geothermal 1d. Geothermal flowflow

Assuming hot water is available thru a heat exchanger (plate) as Assuming hot water is available thru a heat exchanger (plate) as shown (loss thru HE = 5shown (loss thru HE = 5ooC) and geothermal water is available at C) and geothermal water is available at 6565ooCC

60oC65oC

Flow required = Q = q/(15,040 Flow required = Q = q/(15,040 ∆∆T) T)

= 270,000 kJ/hr [15,040 (60= 270,000 kJ/hr [15,040 (60--50)50)ooC]C]

= 1.80 l/s = 108 l/min = 155 m= 1.80 l/s = 108 l/min = 155 m33/day/day

Geothermal Greenhouse

50oC55oC

Greenhouse Design Calculations VGreenhouse Design Calculations V2. Finned tube heating pipe design I2. Finned tube heating pipe design I

a.a. Find average water temperature (at 1.80 l/s)Find average water temperature (at 1.80 l/s)

∆∆T = q/15,040 Q = 270,000 / (15,040 x 1.80) = 10T = q/15,040 Q = 270,000 / (15,040 x 1.80) = 10ooCC

(which (which wewe already knew as: (60already knew as: (60--50)50)ooCC

AWT = TAWT = Tss –– ((∆∆T/2) = 60 T/2) = 60 –– 10/2 = 5510/2 = 55ooCCAWT = TAWT = Tss –– ((∆∆T/2) = 60 T/2) = 60 –– 10/2 = 5510/2 = 55 CC

(60(60oo in and 50in and 50oo out)out)

Greenhouse Design Greenhouse Design CalculationsCalculations VV2. Finned tube heating pipe design II2. Finned tube heating pipe design II

b. Length of finned tube pipeb. Length of finned tube pipe

Assume: 108 fins/m (Table 4) @ 3,873 kJ/hr/m capacity @ 93Assume: 108 fins/m (Table 4) @ 3,873 kJ/hr/m capacity @ 93ooCC

Correction factor for 55Correction factor for 55ooC (Table 5) = 0.3927C (Table 5) = 0.3927

Actual capacity = 3,873 x 0.3927 = 1,521 kJ/hr/mActual capacity = 3,873 x 0.3927 = 1,521 kJ/hr/m

Total length required = 270,000 kJ/hr) / (1,521 kJ/hr/m)Total length required = 270,000 kJ/hr) / (1,521 kJ/hr/m)

= 178 m= 178 m

Number of lengths in greenhouse needed Number of lengths in greenhouse needed assuming 35 m long inside = 178 / 35 ~ 5 assuming 35 m long inside = 178 / 35 ~ 5 lengths lengths –– of which approx. 1/3 placed of which approx. 1/3 placed near center of greenhousenear center of greenhouse2.5 m

13 m

Greenhouseend area

Pipes

Table 4. Table 4. Hot Water Ratings of Hot Water Ratings of HeatingHeating ElementsElements

Base Heating Base Heating ElementElement RowsRows kJ/hr/mkJ/hr/m

W/mW/m(hot water ratings)(hot water ratings)

108 fins/m108 fins/m 11 38733873 10761076

(average water (average water temperaturetemperature 9393ooC (200C (200ooF) F) –– Base CaseBase Case

(33 fins / ft)(33 fins / ft) 22 67096709 18641864

33 87148714 24212421

131 fins/m131 fins/m 11 42194219 11721172

(40 fins/ft)(40 fins/ft) 22 69506950 19321932

Table 5.Table 5.DeDe--Rating Rating FactorsFactors for Heating for Heating Elements (<93Elements (<93ooC)C)

Average Water (Average Water (ooC)C) FactorFactor

8888 0.900.90

8282 0.800.80

7171 0.620.62

6060 0.470.47

4949 0.300.30

3838 0.170.17

Greenhouse Design Calculations VIGreenhouse Design Calculations VI3. Standard unit heaters (see Tables 6 and 7)3. Standard unit heaters (see Tables 6 and 7)

Assume 2 unit heaters Assume 2 unit heaters –– one at each end with a capacity of each one at each end with a capacity of each

= 270,000 / 2 = 135,000 kJ/hr= 270,000 / 2 = 135,000 kJ/hr

Correction factor for entering water temperature (EWT = 60Correction factor for entering water temperature (EWT = 60ooC), and C), and entering air temperature (EAT = 15entering air temperature (EAT = 15ooC) C)

= 0.571 (Table 7)= 0.571 (Table 7)= 0.571 (Table 7)= 0.571 (Table 7)

•• Therefore: q (needed) = 135,000/0.571 = 236,000 kJ/hrTherefore: q (needed) = 135,000/0.571 = 236,000 kJ/hr

the proper selection would then be an the proper selection would then be an “E” unit“E” unit from Table 6 @ from Table 6 @ 236,000 kJ/h236,000 kJ/h

b.b. Flow required: 2 x 1.39 l/s = 2.78 l/s from plate HE (Table 6)Flow required: 2 x 1.39 l/s = 2.78 l/s from plate HE (Table 6)

this would require a decreased DT of:this would require a decreased DT of:

DT = 270,000/(15,040 x 2.78) = 6.5DT = 270,000/(15,040 x 2.78) = 6.5ooC instead of 10C instead of 10ooCC

therefore the design is more than adequatetherefore the design is more than adequate

Table 6.Table 6.Hot Hot waterwater unit heater ratings (Modine) unit heater ratings (Modine) (93(93ooC EWT and 15.6C EWT and 15.6ooC EAT)C EAT)

ModelModel kJ/hrkJ/hr L/sL/s mm33/min /min (air)(air)

Final air Final air temperature temperature

((ooC)C)

hp hp -- WW

AA 95,00095,000 0.570.57 5050 4343 1/6 1/6 –– 125125AA 95,00095,000 0.570.57 5050 4343 1/6 1/6 –– 125125

BB 140,000140,000 0.840.84 9292 3838 1/3 1/3 –– 250250

CC 146,000146,000 0.880.88 8282 4242 1/3 1/3 –– 250250

DD 209,000209,000 1.261.26 129129 3939 1/2 1/2 -- 375375

EE 236,000236,000 1.391.39 130130 4242 1/2 1/2 -- 375375

FF 288,000288,000 1.701.70 145145 4242 1/2 1/2 -- 375375

Table 8.Table 8.Maximum Maximum RecommendedRecommended Mean Water Mean Water Temperature (Temperature (ooC)C)

BurialBurial Steel PipeSteel Pipe Polybutylene TubePolybutylene Tube

Depth (mm)Depth (mm) k = 6k = 6 k = 9k = 9 k = 6k = 6 k = 9 k = 9

2525 4444 4141 5151 4444

5050 4747 4343 5555 49495050 4747 4343 5555 4949

7575 5050 4646 5959 5353

100100 5252 4747 6262 5555

125125 5353 4949 6464 5757

150150 5757 5252 6969 6161

k = 6 Btu in/hr ftk = 6 Btu in/hr ft 22 ooF = 0.864 W/m F = 0.864 W/m ooCC

k = 9 Btu in/hr ftk = 9 Btu in/hr ft 22 ooF = 1.30 W/m F = 1.30 W/m ooCC

Greenhouse Design Calculations VIIGreenhouse Design Calculations VII4. Fan coil 4. Fan coil unitsunits

•• Larger units with larger fan motorsLarger units with larger fan motors

•• More expensiveMore expensive

•• However, greater However, greater ∆∆T from the waterT from the water

•• Can approx. cool water down to 8 to 14Can approx. cool water down to 8 to 14ooC of space temperatureC of space temperature

(Our design = 20(Our design = 20ooC) or C) or ∆∆T = 60T = 60ooC C –– (20(20oo + 8+ 8oo) = 32) = 32oo

∆∆T = 32T = 32ooC could be achievedC could be achieved

Using 2.78 l/s Using 2.78 l/s –– the total heat that can be supplied is:the total heat that can be supplied is:

q = 15,040 x 2.78 l/s x 32q = 15,040 x 2.78 l/s x 32ooC = 1,338,000 kJ/hrC = 1,338,000 kJ/hr

or 1,338,000 / 270,000 = 5, or or 1,338,000 / 270,000 = 5, or 5 greenhouses5 greenhouses could be could be supplied with the same flowsupplied with the same flow

Greenhouse Design Calculations VIIIGreenhouse Design Calculations VIII5. Soil heating I5. Soil heating I

a.a. Calculate required floor temperature to meet the heat loadCalculate required floor temperature to meet the heat load

Floor area (less 10%) = 13 x 36 x 0.90 = 421 mFloor area (less 10%) = 13 x 36 x 0.90 = 421 m22

Heat load / floor area = 270,000 / 421 = 641 kJ/hr/mHeat load / floor area = 270,000 / 421 = 641 kJ/hr/m22

b.b. Calculate inside surface temperature (IST) Calculate inside surface temperature (IST) b.b. Calculate inside surface temperature (IST) Calculate inside surface temperature (IST)

double poly: double poly:

IST = 20IST = 20ooC C –– (0.0291 x 14.9 x 20) = 11.3 (0.0291 x 14.9 x 20) = 11.3 ooCC

single fiberglass: single fiberglass:

IST = 20IST = 20oo –– (0.0291 x 20.4 x 20) = 8.1(0.0291 x 20.4 x 20) = 8.1ooCC

average temperature of unheated surface average temperature of unheated surface

(AUST) = (490 x 11.3 + 301 x 8.1)/(524+296) = 9.7(AUST) = (490 x 11.3 + 301 x 8.1)/(524+296) = 9.7ooCC

Greenhouse Greenhouse DesignDesign Calculations IXCalculations IX5. Soil heating 5. Soil heating IIII

c.c. Determine floor surface temperature (TDetermine floor surface temperature (Tpp) (see Appendix 1)) (see Appendix 1)

641 kJ/hr m641 kJ/hr m22 = 1.70 {[(1.8 T= 1.70 {[(1.8 Tpp + 492)/100]+ 492)/100]44

–– [(1.8 x 9.7 + 492)/100][(1.8 x 9.7 + 492)/100]44 + 7.87(T+ 7.87(Tpp –– 20)20)1.321.32

TTpp = 32.6= 32.6ooC (by trial and error)C (by trial and error)

The maximum recommended for comfort ~ 30The maximum recommended for comfort ~ 30ooC, thus 32.6 is C, thus 32.6 is high, however, can cut back by having the floor system provide high, however, can cut back by having the floor system provide only some of the total heating load. At 30only some of the total heating load. At 30ooC only 524 kJ/hr mC only 524 kJ/hr m22 is is provided, or (524/641) 100 = provided, or (524/641) 100 = 82% of load provided82% of load provided

Greenhouse Greenhouse DesignDesign Calculations XCalculations X5. Soil 5. Soil heatingheating IIIIII

d.d. Determine the depth and spacing of tubesDetermine the depth and spacing of tubes

assume: 100 mm depthassume: 100 mm depth

soil k = 1.30 W/msoil k = 1.30 W/mooC (Table 8)C (Table 8)

∆∆T of water = 10T of water = 10ooCC∆∆T of water = 10T of water = 10ooCC

AWT = 60 AWT = 60 –– 10/2 = 5510/2 = 55ooC C

(Table 8 recommends 55 max (Table 8 recommends 55 max –– OK)OK)

Greenhouse Greenhouse DesignDesign Calculations XICalculations XI5. Soil heating 5. Soil heating IVIV

d.1 using a surface temperature of 30d.1 using a surface temperature of 30ooC (82% of load satisfied) C (82% of load satisfied) and, AWT = 55and, AWT = 55ooC, the tubeC, the tube--toto--surface surface ∆∆T = (55T = (55--30) = 2530) = 25ooCC

Using Figure 8 (Heat output for radiant floor system) for 19 mm Using Figure 8 (Heat output for radiant floor system) for 19 mm (ID) tubing(ID) tubing(ID) tubing(ID) tubing

tube output = 9.00 kJ/hr m tube output = 9.00 kJ/hr m ooCC

for 25 mm (1 inch) diameter tubing for 25 mm (1 inch) diameter tubing –– use ratio of outside surface use ratio of outside surface areaarea

= 9.00 x (33.0/26.7) = 11.1 kJ/hr m = 9.00 x (33.0/26.7) = 11.1 kJ/hr m ooCC

(where 25 mm ID has 33.0 mm OD and 19 mm ID has 26.7 mm (where 25 mm ID has 33.0 mm OD and 19 mm ID has 26.7 mm OD)OD)

Table 7.Table 7.Unit heater Unit heater correctioncorrection factors (Modine) factors (Modine) (EWT = entering water temperature, EAT = (EWT = entering water temperature, EAT = entering air temperatureentering air temperature

EWT (EWT (ooC)C) EAT (EAT (ooC)C)

4.44.4 15.615.6 26.726.7 37.837.8

2727 0.2930.293 0.1430.143 00 00

3838 0.4390.439 0.2860.286 0.1400.140 0.0690.069

4949 0.5850.585 0.4290.429 0.2790.279 0.1370.137

6060 0.7310.731 0.5710.571 0.4190.419 0.2730.273

7171 0.8780.878 0.7140.714 0.5590.559 0.4100.410

8282 1.0241.024 0.8570.857 0.6990.699 0.5470.547

9393 1.1701.170 1.0001.000 0.8330.833 0.6840.684

Steel(1.30 w/m C)

PB(1.30 w/m C)

100

150

Steel(0.864 w/m C)PB

(0.864 w/m C)

PB =polybutylene

0 5 10 15 200

50

Tube Output in kJ/hr m C

Greenhouse Greenhouse DesignDesign Calculations XIICalculations XII5. Soil heating V5. Soil heating V

d.2 thus, the heat output per lineal meter is (19 mm ID):d.2 thus, the heat output per lineal meter is (19 mm ID):

= 9.00 (55 = 9.00 (55 –– 30) = 30) = 225 kJ/hr m225 kJ/hr m

And, the tube spacing using 524 kJ/ hr mAnd, the tube spacing using 524 kJ/ hr m22

= 524/225 = 2.33 m/m= 524/225 = 2.33 m/m22 or 1/2.33 = 0.429 m on centeror 1/2.33 = 0.429 m on center

d.3 correct for heat loss downward and at edges by a safety factor d.3 correct for heat loss downward and at edges by a safety factor of 10%of 10%

design spacing = 0.429 x 0.90 = design spacing = 0.429 x 0.90 = 0.386 m on center0.386 m on center

and a total of: (2.33/0.90) x 36 x 13 x 0.90 = and a total of: (2.33/0.90) x 36 x 13 x 0.90 = 1090 m required1090 m required

Greenhouse Design Calculations XIIIGreenhouse Design Calculations XIII5. Soil heating VI5. Soil heating VI

d.4 using the recommend fluid velocity of ~ 1 m/sd.4 using the recommend fluid velocity of ~ 1 m/s

the volume carried in the 19 mm ID tubingthe volume carried in the 19 mm ID tubing

V = p (0.0192/2)V = p (0.0192/2)22 x 1.0 m/s = 0.000284 mx 1.0 m/s = 0.000284 m33/s /s

= = 0.284 l/s0.284 l/s= = 0.284 l/s0.284 l/s

the flow needed for the flow needed for ∆∆T = (60T = (60--50) = 1050) = 10ooC @ 82% load:C @ 82% load:

Q = (270,000 x 0.82) / (15,040 x 10) = Q = (270,000 x 0.82) / (15,040 x 10) = 1.47 l/s1.47 l/s

therefore, the number of circuits required are:therefore, the number of circuits required are:

1.47 / 0.284 = 5.17 1.47 / 0.284 = 5.17 -- 6 for the total flow6 for the total flow

each: 1090/6 = each: 1090/6 = 182 m long182 m long

Greenhouse Design Calculations IXXGreenhouse Design Calculations IXX5. Soil heating 5. Soil heating VIIVII

d.4 cont. d.4 cont.

since since ∆∆T = T = 1010ooCC > 8> 8ooC recommended, therefore use:C recommended, therefore use:

double serpentine piping layout double serpentine piping layout

182 m @ 12 m wide x 35 m long floor layout ~ 16 182 m @ 12 m wide x 35 m long floor layout ~ 16 182 m @ 12 m wide x 35 m long floor layout ~ 16 182 m @ 12 m wide x 35 m long floor layout ~ 16 runs across the floorruns across the floor

[revised spacing: 35 m / (16 x 6) = 0.365 m][revised spacing: 35 m / (16 x 6) = 0.365 m]

note: remaining (100 note: remaining (100 –– 82) = 18% of heat load must 82) = 18% of heat load must be met by above ground heating system (unit be met by above ground heating system (unit heaters or finned tube)heaters or finned tube)

Greenhouse Greenhouse DesignDesign Calculations XXCalculations XX5. Soil heating 5. Soil heating VIIIVIII

d.5 Layout (pland.5 Layout (plan view of floor)view of floor)

12 m + 4 morecircuits

60oCsupply

50oCsupply

0.365 m spacing

5.84 m per circuit

circuits

Greenhouse Design Greenhouse Design CalculationsCalculations XXIXXI

6. Bare pipe system I6. Bare pipe system I

assume: 60assume: 60ooC supplied to greenhouse from HEC supplied to greenhouse from HE

∆∆T = 10T = 10ooCC

2020ooC air temperatureC air temperature

a.a. Flow required = 270,000 / (15,040 x 10) = 1.80 l/sFlow required = 270,000 / (15,040 x 10) = 1.80 l/sa.a. Flow required = 270,000 / (15,040 x 10) = 1.80 l/sFlow required = 270,000 / (15,040 x 10) = 1.80 l/s

b.b. Average water temp.: AWT = TAverage water temp.: AWT = Tss –– ∆∆T/2 = 60 T/2 = 60 –– 10/2 10/2 5555ooCC

Greenhouse Greenhouse DesignDesign Calculations XXIICalculations XXII6. Bare pipe 6. Bare pipe systemsystem IIII

c. Calculate the heat output/meter of 19 mm ID tubing c. Calculate the heat output/meter of 19 mm ID tubing based on AWT based on AWT

TTaveave = 255.6 + (AWT = T= 255.6 + (AWT = Tairair)/2 = 255.6 + (55 + 20)/2 = )/2 = 255.6 + (55 + 20)/2 = 293.1293.1

DT = AWT DT = AWT –– (T(Tairair = 3) = 55 = 3) = 55 –– (20+3) = 32(20+3) = 32ooCC

TT11 = 255.6 + AWT = 255.6 + 55 = 310.6= 255.6 + AWT = 255.6 + 55 = 310.6ooCC

TT33 = (AUST + T= (AUST + Tairair)/2 = 9.7 + 20)/2 = 15)/2 = 9.7 + 20)/2 = 15ooCC

TT22 = 255.6 + T3 = 255.6 + 15 = 270.6= 255.6 + T3 = 255.6 + 15 = 270.6ooCC

Greenhouse Design Greenhouse Design CalculationsCalculations XXIIIXXIII

6. Bare pipe 6. Bare pipe systemsystem IIIIII

area of 26.7 mm OD pipe (19 mm ID) = [p(0.0267)] x 1 area of 26.7 mm OD pipe (19 mm ID) = [p(0.0267)] x 1 m = 0.0838 m2/mm = 0.0838 m2/m

q/l = {[4.422 x (1/OD)q/l = {[4.422 x (1/OD)0.2 0.2 x (1 / (1.8 x Tx (1 / (1.8 x Taveave + 32))+ 32))0.1810.181 x x ((∆∆T)T)1.2661.266] + (15.7 x 10] + (15.7 x 10--1010) x [(1.8 x T) x [(1.8 x T11 = 32)= 32)4 4 ––((∆∆T)T) ] + (15.7 x 10] + (15.7 x 10 ) x [(1.8 x T) x [(1.8 x T11 = 32)= 32) ––(1.8T(1.8T22 +32)+32)44)]} x 11.345 x A)]} x 11.345 x A

q/l = {[4.422 x (1/26.7)q/l = {[4.422 x (1/26.7)0.20.2 x (1/(1.8 x 293.1 + 32))x (1/(1.8 x 293.1 + 32))0.1810.181 x x (32)1.266] + (15.7 x 10(32)1.266] + (15.7 x 10--1010) x [(1.8 x) x [(1.8 x310.6 + 32)310.6 + 32)44 –– (1.8 x 270.6 + 32)(1.8 x 270.6 + 32)44]} x ]} x 11.345 x 0.083811.345 x 0.0838

= = 129.6 kJ/hr m129.6 kJ/hr m

Greenhouse Design Calculations XXIVGreenhouse Design Calculations XXIV6. Bare pipe system IV6. Bare pipe system IV

d.d. Thus, the total length (L) required to meet the design load is:Thus, the total length (L) required to meet the design load is:

L = 270,000 / 129.6 = L = 270,000 / 129.6 = 2083 m2083 m of 19 mm tubing, andof 19 mm tubing, and

2083/35 = 2083/35 = 60 times60 times the length of the housethe length of the house

which is generally placed under benches and along the wallswhich is generally placed under benches and along the walls

e.e. If 50 mm (2 inch) ID pipe (63.5 mm OD) is used instead, then If 50 mm (2 inch) ID pipe (63.5 mm OD) is used instead, then approximately approximately 1000 m1000 m, or half of the above length needed, or half of the above length needed

NOTE:NOTE: which is approximately 2 times the floor area of 468 m2which is approximately 2 times the floor area of 468 m2

(the Icelandic ROT (the Icelandic ROT –– Sverrir Thorhallsson)Sverrir Thorhallsson)

SummarySummary

•• The greenhouse heating systems usually The greenhouse heating systems usually consist of two components. The actual system consist of two components. The actual system chosen depends on the type of crops and work chosen depends on the type of crops and work space needed.space needed.

•• Artificial lighting is now becoming more Artificial lighting is now becoming more •• Artificial lighting is now becoming more Artificial lighting is now becoming more important important –– even during the daytime.even during the daytime.

•• Labor can be a major problem.Labor can be a major problem.

•• Marketing is important Marketing is important –– no market no market –– no sales.no sales.

•• COCO22 can enhance growth can enhance growth –– from the from the geothermal waters. geothermal waters.