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Piping systems for Chillers equipments

Dr. Ali Hammoud

EDU -2015

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The pipe size in any pipe systems depends on:

1. Pipe material & interior surface roughness.2. Water velocity limitations.3. Pipe diameter from pipe flow chart. 4. Pipe length+ Equivalent length (return pipe). 5. Water Piping Diversity “DF” to prevent over

design condition & cost .For large systems DF is between 0.75 up to 0.85 of the total gpm.

Pipe sizing

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The materials most commonly used in piping in HVAC systems are the following:1. Steel - black and galvanized.2. Wrought iron – black and galvanized.3. Copper – soft and hard.The table below illustrates the recommended materials for various services.

Pipe Material

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Pipe Absolute roughness ( practical values)

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Rela

tive

roug

hnes

s

Roug

hnes

s

Flow rate in gpm / pipe size

Design velocity

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Water piping diversity procedure:

A diversity factor of 0.75-0.85 was multiplied by the total gpm through each main pipe . The Pipe flow chart presented below was used to determine the pipe diameter taking into consideration the given boundaries.The principle of diversity evaluation and calculation of the reduced water quantity was used since the system water quantity actually required during normal operation is less than the total quantity required for the peak design conditions.

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Closed system

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The velocities recommended for water piping depend mainly on two conditions:

Is based on the service for which the pipe is to be used.

Water velocity limitation ASHRE

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Piping systems -2 for Chillers & Radiators

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Applications

• Direct return piping layout

• Reveres return piping layout

• Primary-secondary pumping layout

• Example problems “ Pipe sizing “

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Direct return piping layout

Figure 1

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Reverse return piping layout

Figure 2

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a) In the pressure distribution of a direct return system (in Figure 1), the available pressure drop for a load circuit is greatest near the system pump and decreases the farther away the load is from the source-pump. Care must be taken by the designer to size and select the control valves to ensure adequate flow distribution and proper close-off. b) In the pressure distribution of a reverse return system (in Figure 2), the pressure drop for a load circuit is uniform (if the load pressure drops are similar), even as the distance is increased from the source pump. A key reason for the reverse return design is to assist the two-way control valve with a more uniform pressure drop. Selection of control valves must ensure adequate flow and proper close-off, but this is not as critical as the direct return design.

Direct return & reverse return system

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Riser diagram

Chiller Piping Systems

ChillerPipe Systems

Two pipe Systems Three pipe Systems Four pipe Systems

Heavy duty equipment

Two-pipe systems

Consists of one supply and one return pipe for either chilled or hot water supply . The two-pipe system is limited during changeover season.

Boiler - Off

Chiller - Off

Two-pipe systems

Three-Pipe System

It has three pipes to each terminal unit. These pipes are a cold water supply, a hot water supply and a common return. The return pipe has a mixture of chilled & hot water during changeover operation,. These systems are rarely used today because they consume excess energy.

Four-pipe systems have a cold water supply, cold water return, hot water supply, and hot water return. The terminal unit usually has two independent secondary water coils: one served by hot water, the other by cold water.

Four-Pipe System

4 pipe systems

Comparison

The four-pipe system has the following advantages:

1- Responding quickly to load changes.

2- Operates with the summer-winter changeover.

3- Efficiency is greater and operating cost is lower, thoughinitial cost is generally higher.

4-The system can be designed with no interconnection of thehot and cold water secondary circuits.

Two & Three way -valve

Open & Closed Systems

• A “Open” hydronic system is one that has only one interface with air

• An “Closed” hydronic system is one that ,the water flow is not exposed to the atmosphere at any point. But some times contains an open expansion tank, but water area exposed is insignificant.

LOAD

LOAD

LOAD

SOURCE

EXPANSION

BUILDINGHEIGHT

BalancingValve

3-Way Control Valve

Isolating Valve

LOAD

Closed Systems

System Resistance Curve

CAPACITY

Friction Losses

“Closed” System Resistance Curve

Open Systems

Static Suction Head

CONDENSER

Static Head

HEA

D

System Head Curve

CAPACITY

Friction Losses

Total Static Head

“Open” System Resistance Curve

Chiller Equations

Summer 2004

Chiller Pump discharge gpm& Pipe sizing

)65.5(,8.10

arg./..

)(500..

CaboutFTincoilFanacrosswaterofdropetemperaturtheisTT

edischpumporflowratetheisgpmhrBTUinloadcoolingtheisLC

TTgpmLC

outin

outin

−=∆

−

−××=

We could also using the basic equation to determine the flow rate in the evaporator (chiller flow): :

In the condenser, the heat transferred to the condenser water includes the heat from the Evaporator, plus the heat of compression. For most practical comfort air-conditioning applications, a value of 14,400 Btu/h may be used as the total heat transferred to the condenser water.

Bth/hr)(ΔTQ(gpm)500q W ××=

ΔTTons28.8gpm

ΔTQ(gpm)50040041×

=

××=

What chilled water flow will a 100 -ton chiller handle for a ΔT=12°F rise in water temperature and a Δ= 8°F drop in tower water ? First, determine the flow rate in the evaporator (chiller flow):

Example 1:

gpmΔT gpm ΔTTons ;

1200/500 24Tons 24 100 Tons 24

gpm 200ΔT 12 F

× ×= =

× ×= = =

360F8

00128.8gpm =×

=

ΔTTons28.8gpm ×

=The flow in the condenser:

Suppose you want to calculate the required flow rate in gpm for a chiller Fan coil , assuming that, the required C.L. is 1 Tons-ref and ∆T =10.8 º F, (5-6 ºC )Solution:

ChartFromissizepipeingcorrespondThe "2/1

gpm22.2RateflowTheF.810gpm500120001

==

××=×

Black steel pipe is recommended

22.2F10.8

24Tons1ΔT

24Tonsgpm

;24

ΔTgpmTons

=×

=×

=

×=

Example

Or

Closed system

Chiller Pump selectionCare should be taken when selecting chiller pump :

1- The required flow rate & head

2- Up feed or down feed , If Up feed system is used ,the shut off head of the pump must be greater than the required head to push water up to highest level, However if down feed system is used remember to include gravity assist flow effect.

Example

Suppose you want to calculate the required flow rate for a chiller centrifugal pump sizing the main pipe & ,assuming that, the C.L. is 180 Tons and ∆T = 6 ºC = 10.8 º F. The D.F=75-80% is included in the calculation of the total Cooling Load.Solution:

??sizepipengcorespondiThegpm400dischargePumpThe

F.810gpm50012000180

==

××=×

Typical Fan coil arrangement

Fan Coil

The Fan coil inlet cold water temperature is about 7 ºCthe outlet 13 ºC. Whereas the hot water temperature inlet is about 40 ºC.

Practically, the temperature change across the Fan coil varies from 5 -7 º C in the summer. Whereas ,in Winter the hot water temperature difference is about 11ºC.

Maximum water operating pressure is about 3 Bars.

C’ont Fan Coil

The Fan coil air flow rate in CFM ranges from 200,300, 400, 600, 800 and1200.

The Fan coil gpm is about = 2.4 gpm/ton. Based on temperature difference between in & out

10 ºF, which is about 5.5 ºC.

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Chiller Piping Arrangements

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In general, we see multiple chiller designs applied in parallel instead of series because this permits adding chillers in the future due to single or multiple building additions, such as campus-type site plans: The designer must allow space in the chiller plant for the additions and the distribution mains must be sized for the future flows (see Figure ).Early chiller applications employed three-way valves on unit terminals to permit part-load terminal control by reducing the coil flow to bypass flow around the coil. The result is that the chilled water pumping power is a constant and cannot be reduced with load. As the size of the system increases, this means that the distribution system is pumping a constant volume and is not energy efficient.The use of the two-way valve gives the opportunity to reduce the distribution pumping with the load but, as discussed before , must have provision to reduce the potential valve differential by staging parallel pumps or by reducing the distribution pump's capacity with a variable speed pumping control system. The chillers are piped in parallel in a primary- production loop (see Figure ) with a common bridge to hydraulically decouple the chiller pumps from the distribution pumping, and determine from recirculation at part load if chillers should be shut down. The location of the common bridge determines how the chillers will be loaded or unloaded.

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47(Figure a). (Figure b).

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Primary –Secondary pumping system

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Controlling water temperature in a primary-secondary pumping arrangement with a two way valve is another approach to improving valve coil control, as shown figure (3-a ) .This permits constant flow at all times at a variable temperature in the coil circuit, at the design velocity, to maximize coil heat transfer. A common pipe (a-b )is connected to both the primary and secondary circuits with no pressure drop. This common-pipe is usually located in a bridge between the supply and return mains of the primary. The common pipe is selected with "no-pressure drop" to either the secondary or primary circuits.

Primary-secondary pumping

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Figure (3-a)

The primary-secondary concept allows the distribution pumping of the source supply from a central pumping facility. (see Figure 4-a) or distributing the pumping to remote buildings or zones of a large facility (see Figure 4-b).

The primary-secondary concept allows continual flow through the source and still permits two-way valve control in the loads. There is flexibility in dedicating a pump to a chiller or boiler or manifolding the pumps. This might simplify the need for having backup pumps for every system. Pumps, manifolds, accessories and associated pumping control may be assembled to match installation constraints, or can be factory prepackaged as an assembly for a designated mounting location. The designer work to determine the best arrangement .

Figure (4-a)

Figure (4-b)

ASME Dr.Hammoud BAU-2012 54

Cooling SystemsApplication

Chillers system applicationPrimary/Secondary System

Primary Pumps

Secondary Pumps

Common Pipe

Typical load with two way valve

ASME Dr.Hammoud BAU-2012

Primary/Secondary System at Design

56.0 °F

56.0 °F

56.0 °F

44.0 °F

44.0 °F

44.0 °F

Primary Pumps

1000 GPM Each

3000 GPM @ 56.0 °F

Secondary Pumps

3000 GPM @ 44.0 °F

Typical Coil

No flow

44.0 °F

56.0 °F

500 ton chillers1000 GPM Each56.0-44.0°F

100% system load

Primary/Secondary System at Part Load

53.0 °F

53.0 °F

53.0 °F

44.0 °F

44.0 °F

44.0 °F

Primary Pumps

1000 GPM Each

3000 GPM @ 53.0 °F

Secondary Pumps

2250 GPM @ 44.0 °F

Typical Coil

44.0 °F

56.0 °F

750 GPM @ 44.0 °F

2250 GPM @ 56.0 °F

75% System Load

Primary/Secondary System

53.0 °F

53.0 °F

44.0 °F

44.0 °F

Primary Pumps

1000 GPM Each

2000 GPM @ 53.0 °F

Secondary Pumps

1500 GPM @ 44.0 °F

Typical Coil

44.0 °F

56.0 °F

500 GPM @ 44.0 °F

1500 GPM @ 56.0 °F

OFF 50% System Load

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Parallel Chiller arrangement

Dr. Ali Hammoud

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Suppose we have a chiller-system consists of three chillers, as shown in Figure (A) below. Each chiller is fed by a constant-speed chiller pump (with a check-valve) that Operates only when the chiller is on. Each chiller pump is sized to achieve a ∆T of 16°F (8.8 °C), and the chiller is controlled by a discharge thermostat at 42°F(5.5 °C). The chiller inlet temperature is 58 °F (14.3 °C ), With the common bridge located between the load and the production sections (see Figure ) .Now the system need to operate at part-load (1200 gpm, 800 ton load), Calculate the required load in chiller 2 & 3 and the corresponding flow rate in the main pipe and return pipe when chiller 1 is off.Solution:Due to the hydraulics inherent in the design, chillers 2 and 3 will load proportionally and chiller 1will shut down. Chillers 2 and 3 receive the same temperature water from the return

i hi h i l l t d f ll

Example

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Fig.A

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The chillers load in proportion to the ratio of their flow rates to the total load flow. Chiller 2 loads to 320 tons and chiller 3 loads to 480 tons (each is 64% of their full load output because the load is 800/ 1250= 64 % of the combined chiller flow rate). Because the chiller pumps are constant speed, chiller 2 delivers 750 gpm and chiller 3 delivers 1125gpm, for a total of 1875 gpm. Because the load is calling for 1200gpm, 675 gpm must flow in the common-bridge. The calculation is shown below:Mixing equation of water occurs at the return tee: 675 gpm at 42F° + 1200 gpm at 58F° = 1875gpm at 52.24°(675× 42F°) + (1200 × 58F°) / 1875 = 52.24F° (return water temperature)The load on chiller 2 is:Tons = gpm × ∆T /24 =750 × (52.24°- 42°) / 24 = 320 tonsFor on chiller 3 is:Tons = 1125 × (52.24°- 42°) /24 = 480 tons

65Dr. Ali Hammoud

Fig.B

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Determine the pipe sizing and the total pressure drop for the zone A-B of the system shown below. Assuming the following:

1. Heating coil rated 38 gpm & ∆P=3ft wg.

2. One gate valve -2in screwed type.

3. One Control valve -1.5 in rated 38 gpm & ∆P=8.2ft wg.

4. One Balance valve -2 in rated 40gpm & ∆P=2ft wg.

5. Two tees ( branch flow )

6. Four Elbows (screwed type).

7. Total pipe length L= 300 ft.

Use the K values methods

Class exercise 1

Closed system

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Closed system

Closed loop

Sc-40

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From the pipe flow chart closed loop & for flow of 40 gpm the corresponding pipe diameter is 2”. The pressure drop per 100 ft is 3.1 ft/100 ft which is less the limited 4 ft/100ft [boundary] . However if 1.5 in pipe is selected the pressure drop will be 8.5 ft/100ft this value is outside the acceptable boundary. Then 2 in pipe is selected. The corresponding flow velocity is 4 ft/s.

The pressure drop due to pipe friction

∆P = 3.1 ft/100ft×300 ft = 9.3 ft.

Adding the aging effect (15 %) we get:

∆P =9.3×1.15 =10.7 ft.

Since V= 4 ft/s , V2/2g=0.227 ft.

Solution:

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The K values were selected fromthe corresponding charts. Thepressure for each fitting iscalculated from ∆ P=K(V2/2g)The K value for 2 in gate valve = 0.17 ft

The K value for 2 in 90 regular elbow =1 ft

The K value for 2 in Tee branch = 1.4 ft

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1. Heating coil rated 38 gpm ∆P=3ft wg. (given)

2. One gate valve -2in screwed ∆P=0.17(0.277)=0.04ft wg.

3. One Control valve-1.5 in rated 38 gpm & ∆P=8.2ft wg (given)

4. One Balance valve -2 in rated 40gpm & ∆P=2ft wg. (given)

5. Two tees ( branch flow ) ∆P=2 ×1.7(0.277)=0.64ft wg

6. Four Elbows (screwed type) ∆P=4×1(0.277)=0.91ft wg

The total pressure drop:

∆P= Coil+ pipe friction+ balance valve +control valve + gate valve+ tee branches + elbows=

= 3+10.7+2 +8.2+0.04+0.64+0.91 =25.49 ft

Pressure drop due to fittings & valves

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Figure below shows an open re-circulating pipe system where a centrifugal pump is used to circulate 250 gpm of water from the cooling tower to the condenser. During the pump’s operation, it is assumed that the water level in the cooling tower tank keeps a constant level. Calculate the total head of the pump and pump shaft power Assuming that: The pump efficiency η = 60 % , The pipe material is galvanized steel pipe .All the necessary dimensions and fittings are shown in the corresponding figure.

Class exercise

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Ø 6 "

Cooling tower

Condenser

30 m

2 mPump

4 m

3 mP

3 m Valve

Srainer

5 m

S

3 m

250 gpmQ= 57 m /hØ 4"

3

2 m

3 m

6 m

1

E

4 m

25 m

4 m

Pressure dropThrough strainer 0.27 bar.

Tower nozzles pressure required: 0.6 Bar

Including ( entr. & Exit losses)Through CondenserPressure Drop

Pressure drop

0.88 bar

3 m

ValveCheck

5”

Open system

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Determine the total pressure drop of the system & the pump’s power in Kw. Assuming the following:

1. The flow Q = 250 gpm .Suction diameter D=5” & Discharge diameter D= 4 “.

2. Tower nozzles pressure required ∆P=0.6 bar.

3. Pressure drop through Condenser including (entry& exit losses) ∆P=0.88 bar

4. All elbows are flanged long radius

5. Elevation difference in the cooling tower is 4 m (13 ft)

6. Pump efficiency 60%

Use the “Le” equivalent technique

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Solution:The total head of the pump is given by:

The head loss between 1 to S is hL1-S :

The total energy loss due to the pipe friction, fittings:

Where hl is the head loss per 100 ft, which can be determined from the pipeflowchart for pipe as follows:

For Q = 250 gpm & D= 5 “,the head loss per 100 ft is = 2 ft/100 ft.

The total effective length:

L is the length of straight pipe L1-S= 2+25+5 = 32 m = 106.5 ft

From equivalent length tables, we select the equivalent length for fittings.

2 (5”)long radius (90°) elbows = 2x 8.2 = 16.4 ft.

The equivalent length of full open gate valve D= 5” is 6 ft,

The pressure drop due to the presence of 5”strainer = 80 ft from table 2

then Le = 16.4 + 6 + 80 = 102.4 ft

The head loss between 1-S hL1-S is = 4 ft add the aging effect 1.15 x4≈5 ftft089.49.0822/100.Lhh effective1L =×==

ft.9082.4021106.5Leffective =+=

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Open loop

Sc-40

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The head loss between P to E is hLP-E :

Discharge pipe:

The length of straight pipe L= 4+3+2+3+3+3+30+6+3+4 = 61 m (203.3 ft)

From the pipe flow chart for pipe we get:

For Q = 250 gpm & D = 4 “ the pressure drop due to pipe friction =3 ft/100 ft.

The corresponding flow velocity is V= 5.7ft/s (1.71 m/s)

The equivalent length in ft:

8 long radius (90°) elbows = 8 × 6.7 =53.6 ft

2 Gate valves = 2× 4.5 = 9 ft

Check valve swing type = 40 ft

The total effective length in ft:

The head loss between P-E hLP-E is ≈ 9.177 ft. add the aging effect 10.5 ft

ftLeffective 9.3054096.533.203 =+++=ft9.177(305.9)ft/100ft3.Lhh effective1L ===

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ASME Dr.Hammoud BAU-2012 80

End of the lecture