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Non-pressurised Cold Water Pipe Sizing
The method is similar to L.T.H.W. pipe sizing except that the pressure available is not from a pump but from the head available from the tank. The higher the tank is above the outlets the more head will be available to force the water through the outlets and overcome pipework resistances.
Head Available
The Head Available develops water pressure and this water pressure is used up in overcoming the frictional resistance of the pipe and in creating the velocity pressure for water flow at the outlet.
p1 - p2 = frictional resistance + velocity pressure or,
h1 - h2 = head loss in pipe due to friction + velocity head
where p = pressure (N/m2)h = head (m)
In practice, to avoid additional velocity pressure calculations, it is usual to calculate the available pressure by considering the difference in levels between the bottom of the storage tank and the height of the draw-off points.
The pressure losses in the system are frictional pipe losses and velocity pressure loss through sanitary fittings such as taps, cistern ball valves and shower heads.
Velocity head loss through fittings is as follows :-Pillar tap 1mShower head 1.5mBall valve 1m
Water Flow Rates
Cold water flow rates for sanitary appliances for small installations may be found from the table below.
Approximate hot or cold water demand Flow rate (l/s)
Basin (spray tap) 0.05Basin (tap) 0.15Bath (private) 0.30Bath (public) 0.60Flushing cistern 0.10Shower (nozzle) 0.15Shower (100mm rose) 0.40Sink (15mm tap) 0.20Sink (20mm tap) 0.30Wash fountain 0.40
In larger more complex buildings where many sanitary appliances are installed simultaneous demand should be considered from tables CIBSE Guide B (1986) B4.20 and B4.21
Pipe Size Procedure
1. Divide system into sections.
2. Calculate demand units if simultaneous demand is effective.
3. Estimate flow rates in each section.
4. Estimate pipe diameter.
5. Measure the pipe run for the section.
6. Calculate length of pipe equal to resistance of fittings.
7. Calculate effective pipe length.
8. Determine pressure loss due to friction for pipe from CIBSE guide tables.
9. Calculate pressure consumed by friction.
10. Calculate cumulative pressure consumed.
11. Calculate pressure available at the end of the section.Subtract pressure loss in section from static pressure available.
12. If pressure loss due to friction is less than pressure available,
Notes:- 1. Keep velocity below 2.0 m/s for noise reduction see Table 2.19 in CIBSE Guide G (2004) Public Health Engineering.2. An alternative method of pipe sizing is to use a nomogram.
This can be found in CIBSE Guide G (2004) Public Health Engineering Figure 2.21.
Example 1
Determine a suitable pipe size for the system shown below.DATAFittings include the following; exit from tank or large vessel, 3No. bends, 1No. gate valve, 1No. 15mm tap, Length of pipe run is 8 metres and copper pipe is to be used.The flow rate for a 15mm Sink Tap from above Table is 0.2 l/s.
The pressure available to force the water through the pipework and tap comes from the head of water above the tap.The formula below gives the relationship between pressure and head.
P = x g x hWhere;P = pressure (N/m2) = density (1000 kg/m3 for water)g = acceleration due to gravity (9.81 m/s2)h = head (m)
Therefore: P = 1000 x 9.81 x 2.0 = 19,620 N/m2
The resistance to flow is from the fittings and pipework.
Example 2
Determine suitable pipe sizes for the system shown below.The building is a three-storey Nursing Home.DATACopper pipe is to be used.Flow rates are to be obtained from above Table.
Answer:
From above Table the flow rate for a private bath is 0.3 l/s.The pipe sizes, flow rates and pressures are indicated on the drawing below.
HOT AND COLD WATER PIPE SIZING TABLE
1 2 3 4 5 6 7 8 9 10 11 12Ref Deman
dUnits if require
d
FlowRate
(l/s)
EstimatedPipe Dia.
(mm)
MeasuredPipe Run
(m)
Length ofPipe Equal to Resistance’s
(m)
EffectivePipe LengthCol . 5 + 6
(m)
Pipe Pressure Loss
(kPa/m)
PressureConsumedCol. 7 x 8
(kPa)
TotalPressure
Consumed(kPa)
PressureAvailable at End of
Section(Pa)
Final PipeSize
(mm)A 0.9 28 7.0 Factors for fittings:
1No.Exit large vessel = 0.41No.Gate Valve = 0.31 No. Bend = 1.01 No.28 x 28 x 22 tee = 0.2 ----------- Total 1.9T.E.L. = Total x le = 1.9 x 1.1 = 2.09 m
7 + 2.09 = 9.09 m
1.250 11.362 11.362 Static pressure =
4m x 9.810 = 39.24
Press. Available = 32.79 kPa= 39.24+4.91-11.362
28
B 0.6 28 3.0 1No.28 x 22 x 22 tee = 0.20with 28 x 22 reducer:A2 / A1 = x 0.0112 / x 0.0142. = 0.617 gives = 0.25 ----------- Total 0.45T.E.L. = Total x le = 0.45 x 1.0 = 0.45m
3.0 + 0.45 = 3.45 m
0.600 2.070 2.070 Static pressure = 3m x 9.810 = 29.43 kPa,Press. Available = 32.79 – (2.07-29.43) = 60.15 kPa
28
C 0.3 22 8.0 1No.Bend = 1.01No. Angle valve bath tap = 5.0 ----------- Total 6.0
8.0 + 4.2= 12.2 m
0.625 7,625 7,625 Static pressure = 2 m x 9.810 = 19.6 kPaPress. Available = 60.15 – (7,625-19.6) =
22
1 2 3 4 5 6 7 8 9 10 11 12T.E.L. = Total x le = 6.0 x 0.7 = 4.2 m
72.125 kPa
D 0.3 22 4.0 1No. Angle valve bath tap = 5.0T.E.L. = Total x le = 5.0 x 0.7 = 3.5 m
4.0 + 3.5 = 7.5 m
0.625 4,69 kPa 4.69 kPa Static pressure =
1m x 9.810 = 9.81
Press. Available = 32.79 – ( 4,69 + 9.81)= 18.29 kPa
22* add pump or raise the cistern
1m above or
use circular
tank with 2m height
E 0.3 22 3.0 1No. Angle valve bath tap = 5.0T.E.L. = Total x le = 5.0 x 0.7 = 3.5 m
3.0 + 3.5 = 6.5 m
625 4,063 4,063 Static pressure = 6m x 9810 = 58,860Press. Available = 58,860 – 14,683 = 44,177 Pa – 4,063 = 40,114 Pa
24
Re-calculate pipe ref. C for 15mm pipe
C 0.3 15 7.0 1No. 28 x 15 x 22 tee (already included)with 2 No.28 x 15 reducers:A2 / A1 = x 0.00752 / x 0.0142. = 0.287 gives = 0.47 2No. Reducers = 0.94 1No.Bend = 1.01No. Angle valve bath tap = 5.0 ----------- Total 6.94T.E.L. = Total x le = 6.94 x 0.5 = 3.47 m
7.0 + 3.47= 10.47 m
4000estimated
41,880 41,880 +14,683 = 56,563
Static pressure = 9 m x 9810 = 88,290Press. Available = 88,290 – 56,563 = 31,727 Pa
15
Example 3
Determine suitable pipe sizes for the system shown below.The building is a three-storey hotel.DATACopper pipe is to be used.Flow rates and simultaneous demand data are to be obtained from the CIBSE guide.
Hot and Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12
Ref DemandUnits if
required
FlowRate
(l/s)
EstimatedPipe Dia.
(mm)
Measured
Pipe Run
(m)
Length ofPipe Equal to Resistance’s
(m)
EffectivePipe
LengthCol . 5 +
6(m)
Pipe Pressure
Loss (Pa/m)
PressureConsumedCol. 7 x 8
(Pa)
TotalPressure
Consumed(Pa)
PressureAvailable at
End of Section
(Pa)
Final PipeSize
(mm)
Mains Water Pipe SizingPipe Sizing Procedure
1. Reference the pipe section.
2. Calculate flow rates from Table below.
3. Estimate flow rates in each section. Keep velocity below 2 m/s. See also CIBSE Guide G
(2003) part 2, Table 2.19.
4. Estimate pipe diameter from pipe sizing tables in CIBSE Guide C.
5. Measure the pipe run from drawings.
6. Calculate length of pipe equal to resistance of fittings. The Total equivalent length of a fitting = Equivalent
Length x Pressure Loss factor (Zeta). 7. Calculate effective pipe length.
8. Determine pressure loss due to friction from CIBSE Tables.
9. Calculate pressure consumed due to friction (Pa) = effective pipe length (m) x pressure loss due to friction (Pa/m)
10. Calculate total pressure consumed = Friction loss + Static pressure loss
11. Determine pressure at start of section.
12. Calculate pressure available at end of section = Pressure at start of section - Total pressure consumed If pressure available at end of section is less than the maximum allowable pressure drop then we can accept this pipe size.
13. Determine pressure required at end of section, this can be the minimum pressure that is required for terminal equipment.
14. If the pressure available at the end of the section is more than or equal to the pressure required at the end of the section then the pipe size is correct.
Water Flow Rates
Cold water flow rates for sanitary appliances for small installations may be found from the table below.
Approximate hot or cold water demand Flow rate (l/s)
Basin (spray tap) 0.05Basin (tap) 0.15Bath (private) 0.30Bath (public) 0.60Flushing cistern 0.10Shower (nozzle) 0.15Shower (100mm rose) 0.40Sink (15mm tap) 0.20Sink (20mm tap) 0.30Wash fountain 0.40
In larger more complex buildings where many sanitary appliances are installed simultaneous demand should be considered from tables CIBSE Guide B (1986) B4.20 and B4.21.
Notes:- An alternative method of pipe sizing is to use a nomogram.This can be found in CIBSE Guide G (2004) Public Health Engineering Figure 2.21.
Pipe Sizing table
Pressurised Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ref
Demand
Units if require
d
Flow
Rate
(l/s)
Estimated
Pipe Dia.
(mm)
Measured
Pipe Run
(m)
Length ofPipe
Equal to Fittings
Resistance’s(m)
Effective
Pipe LengthCol . 5
+ 6(m)
Pipe Pressure Loss (kPa/
m)
PressureConsumed due
to FrictionCol. 7 x
8(kPa)
Friction loss + Static
pressure loss = Total
PressureConsum
ed(kPa)
Pressure at Start
of Sectio
n
(kPa)
Pressure
Available at
End of Section
(kPa)
Pressure
Required at
End of Section
(Pa)
Final
PipeSize
(mm)
Pressurised Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Example 1
Calculate an appropriate pipe size for the system shown above.Use Copper Table X pipework for water at 10oC.. Answer The maximum allowable pressure drop along the length of pipe = 300,000 Pa – 250,000 Pa = 50,000 Pa
Pressurised Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ref
Demand
Units if require
d
Flow
Rate
(l/s)
Estimated
Pipe Dia.
(mm)
Measured
Pipe Run
(m)
Length ofPipe Equal to Fittings Resistanc
e’s(m)
Effective
Pipe LengthCol . 5
+ 6(m)
Pipe Pressure Loss (Pa/m)From CIBSE Tables
PressureConsumed due
to FrictionCol. 7 x
8(Pa)
Friction loss + Static
pressure loss = Total
PressureConsum
Pressure at Start
of Section
(Pa)
Pressure
Available at
End of Section
(Pa)
Pressure
Required at
End of Section
(Pa)
Final
PipeSize
(mm)
Pressurised Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12 13 14
ed(Pa)
A none 0.8 22 50 none 50 3500 175,000 175,000 300,000 125,000 250,000
Too smal
l
A none 0.8 28 50 none 50 1000 50,000 50,000 300,000 250,000 250,000
28
Pipe Sizing Procedure
1. Reference the pipe section - section A.
2. Calculate demand units or loading units from Tables in CIBSE guide (attached). – not required, see No.3 below.
3. Estimate flow rates in each section. Keep velocity below 2 m/s. - given
4. Estimate pipe diameter from pipe sizing tables in CIBSE Guide C. – 22mm (velocity too high at approx 2.4 m/s) or 28mm (velocity is 1.5 m/s).
5. Measure the pipe run from drawings. – 50m
6. Calculate length of pipe equal to resistance of fittings. – no fittings
7. Calculate effective pipe length. - 50m
8. Determine pressure loss due to friction from CIBSE Tables. See Table 4.18 in Guide C (CD version).
9. Calculate pressure consumed due to friction (Pa) = effective pipe length (m) x pressure loss due to friction (Pa/m). Column 7 x 8 in Pipe Sizing Table.
10. Calculate total pressure consumed = Friction loss + Static pressure loss. There are no vertical pipe sections and therefore no static pressure loss.
11. Determine pressure at start of section. Given in drawing as 300,000 Pa.
12. Calculate pressure available at end of section = Pressure at start of section - Total pressure consumed.
300,000 – 175,000 = 125,000 Pa. (22mm) …………..300,000 – 50,000 = 250,000 Pa (22mm).
If pressure available at end of section is less than the maximum allowable pressure drop then we can accept this pipe size.
13. Determine pressure required at end of section, this can be the minimum pressure that is required for terminal equipment. Given in drawing as 250,000 Pa.
14. If the pressure available at the end of the section is more than or equal to the pressure required at the end of the section then the pipe size is correct.
28mm pipe is correct, 22mm is too small since there is not enough pressure available at the end of the section and the water velocity is also too high.
Example 2
Calculate an appropriate pipe size for the system shown above.Use Copper Table X pipework for water at 10oC..
Answer The maximum allowable pressure drop along the length of pipe = 120,000 Pa – 90,000 Pa = 30,000 Pa
Pressurised Cold Water Pipe Sizing Table
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ref
Demand
Units if
required
Flow
Rate
(l/s)
Estimated
Pipe Dia.
(mm)
Measured
Pipe Run
(m)
Length ofPipe
Equal to Fittings
Resistance’s(m)
Effective
Pipe Lengt
hCol . 5
+ 6(m)
Pipe Pressu
re Loss (Pa/m)
From CIBSE Tables
Pressure
Consumed due
to FrictionCol. 7 x
8(Pa)
Friction loss + Static
pressure loss =
TotalPressur
eConsum
ed(Pa)
Pressure at Start
of Sectio
n
(Pa)
Pressure
Available at End of Sectio
n(Pa)
Pressure
Required at
End of Sectio
n
(Pa)
Final
Pipe
Size
(mm)
A none 0.5 22 14 1.6 15.6 1500 23,400 23,400 120,000
96,600 90,000 22
Pipe Sizing Procedure
1. Reference the pipe section - section A.
2. Calculate demand units or loading units from Tables in CIBSE guide (attached). – not required
3. Estimate flow rates in each section. Keep velocity below 2 m/s. - given
4. Estimate pipe diameter from pipe sizing tables in CIBSE Guide C. – 22mm (velocity is 1.5 m/s).
5. Measure the pipe run from drawings. – 14m
6. Calculate length of pipe equal to resistance of fittings. – 2 bends.
The Total equivalent length of a fitting = Equivalent Length x Pressure Loss factor (Zeta). See Pipe Sizing Heating Section - page 4 - pipe fitting losses.Copper pipe elbow (Zeta) = 1.0 x 2 bends = 2.0Determine equivalent length from CIBSE table C4.18, le = 0.8Total equivalent length of fittings = 0.8 x 2.0 = 1.6 metres.
7. Calculate effective pipe length. - 15.6m
8. Determine pressure loss due to friction from CIBSE Tables. See Table 4.18 in Guide C (CD version).
9. Calculate pressure consumed due to friction (Pa) = effective pipe length (m) x pressure loss due to friction (Pa/m). Column 7 x 8 in Pipe Sizing Table.
10. Calculate total pressure consumed = Friction loss + Static pressure loss. There are no vertical pipe sections and therefore no static pressure loss.
11. Determine pressure at start of section. Given in drawing as 120,000 Pa.
12. Calculate pressure available at end of section = Pressure at start of section - Total pressure consumed. 120,000 – 23,400 = 96,600 Pa.
If pressure available at end of section is less than the maximum allowable pressure drop then we can accept this pipe size.
13. Determine pressure required at end of section, this can be the minimum pressure that is required for terminal equipment. Given in drawing as 90,000Pa.
14. If the pressure available at the end of the section is more than or equal to the pressure required at the end of the section then the pipe size is correct.
22mm pipe is correct.
Pipe Sizing for Heating Systems
All pipe sizing in building services is based on the D'Arcy equation, where:
H = 4 . f . l . v2 / 2 . g . d
where H = head loss due to friction in a straight pipe (m)f = friction coefficientl = length of pipe (m)d = diameter (m)v = velocity of fluid (m/s)g = acceleration due to gravity (m/s2)
The object of pipe sizing is to obtain the smallest diameter of pipe without too high a water velocity or too high a pressure drop and therefore large pumps.
It is more convenient to use pipe sizing tables when sizing pipes rather than the D'Arcy equation.
This is because the water velocity and head loss (or pressure loss) are unknown at the time of pipe sizing, and the friction coefficient (f) varies with Reynolds number which in turn varies with velocity and diameter.
A FLOW of WATER in PIPES TABLE is provided in these notes.
The CIBSE guide provide pipe sizing tables in sections C4.11 to C4.45.
L.T.H.W. (Low Temperature Hot Water) PIPE SIZING
When sizing pipes for heating systems the water velocity should not exceed 1.0 m/s. (Except for large diameters - see CIBSE guide tables B1.17 & B1.18).
This reduces noise and wear.
The pressure drop should not exceed 300 Pascals per metre run of pipework to keep pumps down to a reasonable size.
This means that for every metre of pipework the resistance to water flow should be no more than 300 Pa which is about 30 mm head.
FLOW of WATER in PIPES TABLE
The table below shows pipe sizes for 15mm to 76mm diameter copper pipe.
To find a suitable pipe size for a heating circuit the flow rate is used to find an appropriate diameter.
If the flow rate of water is known then look down under any pipe diameter column to ascertain the corresponding pressure drop and velocity.
If the pressure drop and velocity exceed the criteria in the previous section then try the next pipe size up.
If the pressure drop and velocity are within the criteria then the pipe is sized correctly
EXAMPLE 1
Determine the smallest pipe which will carry 0.4 kg/s of water at 75oC using Copper, Table 'X'.
Consult the FLOW of WATER in PIPES TABLE (Copper pipe at 75oC).
The pipe diameters are written in bold type across the top of the table.
The pressure loss per unit length (Pa/m) and velocity (m/s) are written down the LHS of the table.
A red horizontal line is drawn across the table below 300 Pa/m.
This means that suitable pipe sizes will be found above this line.
The velocity follows a stepped line the lower blue line is the
1.0 m/s velocity line.
A 22 mm pipe will carry 0.4 kg/s but the pressure loss per unit length is below the red horizontal line and outside the table.
The pressure loss is in fact (790 Pa/m) and is too high since the maximum should be 300 Pa/m.
The velocity is also too high at about 1.3 m/s, the optimum being 1.0 m/s.
If this is the case then look at the next pipe size up, at 28 mm.
The flow rates closest to 0.4 are 0.394 and 0.414 kg/s.
0.4 kg/s is in between these two flow rates.
A 28 mm pipe will carry 0.4 kg/s with a pressure loss of about 230 Pa/m and a velocity of 0.7 m/s.
This meets the design criteria and therefore 28 mm would be a suitable pipe size.
EXAMPLE 2
Determine a suitable pipe size for L.T.H.W. copper pipe for a flow rate of 1.0 kg/s.
Answer:
A 42 mm pipe gives a flow rate of 1.0 kg/s with a pressure loss of 160 Pa/m and velocity of about 0.9 m/s.
EXAMPLE 3
Choose a pipe diameter for a heating system (L.T.H.W.) with a heat output of 32 kW.
Answer: First find the mass flow rate of water required.
A 35 mm pipe will give a flow rate of 0.76 kg/s with a pressure loss of 250 Pa/m and a velocity of 0.95 m/s.
EXAMPLE 4
Size the flow and return pipework to a 1.6 kW radiator.
A 15 mm pipe will give a flow rate of 0.038 kg/s with a pressure loss of 80 Pa/m and a velocity of about 0.25 m/s.
PIPE FITTING LOSSES
Pipework fittings such as bends, tees, reducers etc., cause pressure loss or resistance in a heating system. When making approximate calculations 10%, 15%, 20% or more may be added to the pressure loss in straight pipe runs.
For accurate calculations the fitting loss should be determined separately for each fitting.
The concept of equivalent length is used and is defined as the length of straight pipe which would give a friction pressure loss equivalent to one velocity head.
The D’Arcy equation is;
H = 4 . f . l . v2 / 2 . g . d
Where; H = head loss due to friction in a straight pipe (m)f = friction coefficientl = length of pipe (m)d = diameter (m)v = velocity of fluid (m/s)g = acceleration due to gravity (m/s2)
The D’Arcy equation can be rewritten for pressure instead of head.
Pressure drop (pl) = ( 4 . f . l ) / d x ( v2 . . g ) / 2 . g
Where; pl = Pressure loss in a pipe section (Pa)f = friction coefficient for pipe l = length of pipe (m)d = diameter of pipe (m)v = water velocity (m/s) = density of water (kg/m3)g = acceleration due to gravity (m/s2)
To simplify the above equation we get;
Pressure drop (pl) = ( 4 . f . l ) / d x ( v2 . ) / 2
or; Pressure drop (pl) = ( 4 . f . l ) / d x ( ½ . . v2 )
For the friction pressure loss to equal one velocity head;
Velocity pressure = ( ½ . . v2 )
Then 1.0 = ( 4 . f . l ) / d
The length (l) is now called equivalent length (le) and by rearranging the above formula we get;
1.0 x d = 4 . f . le
le = d / 4 .f
Values of equivalent length are given in the FLOW of WATER in PIPES TABLE for water at 75oC, see CIBSE guide C (2001) section 4 , Flow of Fluids in Pipes and Ducts, Tables 4.9 to 4.33 for various types of pipes.
These values should be corrected for each particular type of fitting.
The correction factors of Velocity pressure loss factors are called (Zeta) factors.
The resistance in a fitting is converted to equivalent straight lengths of pipe, e.g. a bend may have a resistance equivalent to 1.2 metres of straight pipe.
The TOTAL EQUIVALENT LENGTH OF A FITTING = Equivalent Length x Pressure Loss factor (Zeta).
TOTAL EQUIVALENT LENGTH OF A FITTING (m) = (le) x (Zeta factor).
See CIBSE guide C (2001) section 4.9 for more details of fittings zeta factors.
EXAMPLES OF zeta FACTORS
The following are some examples of pressure loss ( zeta) factors for pipe fittings:
PIPE SIZING TABLE
The pipe sizing table shown below is an aid to sizing more complicated circuits.
A description is given in the Table below.
EXAMPLE 5
Size the pipework for section (B) of a heating system shown below.
Section B is in red.
Also determine the pressure drop in section (B)
The total length of the section is 8 metres (includes flow and return)
Section B supplies hot water to Radiator No. 2 and No. 3.
The amount of heat to be transferred to radiators No. 2 & 3 is:
1.5 + 1.0 = 2.5 kW.
The downstream tees will be included in this section.
Index Circuit
The Index Circuit is the circuit with the highest resistance.
This only applies to systems where the circuits are divided.
The Index Circuit needs to be identified so that the pump can be sized.
Example 1
The system shown below is divided into two sub-circuits A & B.
A pipe sizing calculation would determine which of the two sub-circuits had the most resistance and therefore which was the Index Circuit.
The reason for finding the Index Circuit is to size the pump.
The pressure developed by the pump should be capable of overcoming the resistance in the Index Circuit.
If the pump pressure can overcome the resistance in the Index Circuit, then it can overcome the resistance in other circuits of lesser resistance.
If it was found that the Index Circuit was Circuit (B) in the above diagram then we would include the flow of water through radiators No. 3, 4 and 5.
If we examine Circuit (B) then the Index Circuit flows past Radiator No.3 and No.4 and through Radiator No.5.
This would be the circuit with the highest resistance.
If the pump is capable of forcing water through the pipework to Radiator No.5 then there will be enough pressure to force the water through Radiators No.3 & No.4 since they are closer to the pump.
This is the reason why only one radiator is included in the calculations for resistance in the Index Circuit.
Example 2
A heating system is shown below.
There are seven radiators and seven pipe sections.
The pipe sections under the radiators are from tee to tee.
The Index circuit is the one with the highest resistance.
This is normally the longest run to the radiator that is at the greatest distance from the pump. The first six radiators are not included in the Index circuit. The index circuit below includes all seven sections but only includes; the boiler and fittings around the boiler in section 1, tees in sections No.1 to No.6 and the last radiator and radiator valves in the section 7.
