Domestic Water-supply - Example

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Water Flowing in Pipes -real systems (2)

These pages explain how to choose thecorrect sizes of pipe when plumbing ahouse, and why it matters. This sectionincludes a practical worked example.The theory is explored in part 1 .

In this section I show how to calculate the flow-rate ina real domestic water-supply system by using acouple of design tools that link flow-rate to theavailable head - the pressure that makes the watermove. The worked example starts here . If all you needis the graph that links flow rate to the rate-of-pressure-drop for standard pipe sizes, it's here .

But before I design a water supply system for a realhouse, I should explain how the apparentlyimpossible calculation problems involving turbulentflow can be quickly and easily tackled in practice.

How much pressure is needed?

If the water has to move at a couple of metres persecond, or thereabouts, how much pressure isneeded?

It's a simple question, but unfortunately there is nosimple answer. It depends on what pipes are fitted,and how long they are. Each case must beindividually calculated. But don't despair - thecalculation is very easy.

The main thing to remember about pressure is this:

Pressure supplies the energy to push the water alongthe pipe. Each bit of pipe resists the flow. Energy islost as the water moves along the pipe, so the

What makes waterflow?

How fast will it flow?

Real pipes in realhouses

How much pressure isneeded?

What size pipe do youneed?

How fast will thewater flow?

Darcy-Weisbach

lost as the water moves along the pipe, so thepressure falls too. There's a pressure differencebetween the ends of the pipe.

The longer the pipe, the more energy is lost, and thegreater the pressure drop. The rate of pressure drop(that is, the pressure drop per metre of pipe) dependson the pipe diameter and the speed of flow, as youwould expect.

The design goal is to choose the pipe sizes that willgive the flow rates you want. Each length of pipe wllhave a pressure drop along its length. So the aim is tochoose pipes that will drop just enough of theavailable pressure (from the header tank, or from themains water supply in the street) to give the requiredflow rates. This means checking the pressure dropalong each pipe.

Pressure difference calculation

One way to find the pressure difference between theends of a pipe is to use the Darcy-Weisbach equationI mentioned in part 1. This predicts how muchpressure would be needed to push the water along apipe at a particular speed. The formula looks like this:

Here, the pressure difference P needed to achieve aflow velocity v depends on the length L and diameterD of the pipe as well as the density of the fluid (ρ -about 1,000 kg/m3 for cold water). It also depends onf, a fiddle factor - sorry, "friction factor", which isincluded to account for the effects of the Reynoldsnumber. This graph (and its equation) shows therelationship between Re and f.

The equation includes √f on both sides, and looksimpossible to solve. In fact, it was quitestraightforward. The trick is to begin by guessing avalue for f (say, 0.01), putting this value (and Re) inthe right-hand side, working out the value of the left-hand side, and hence finding f. This new value for f iscloser to the actual value than the initial guess, so youplug it back into the right-hand side and do thecalculation again . After a couple of iterations theanswer is usually close enough to be useful. (By theway, the friction factor used by American engineers isfor some reason four times bigger than this. But then,most things in America are bigger than they are inEngland.)

The graph appears to show that the "friction factor"decreases as the Reynolds number goes up. Morespeed giving less friction? Hardly likely, is it? In fact,that's not what the graph is saying. The "frictionfactor" is purely a measure of how the pipe affects theflow, and as the water becomes more turbulent thepipe itself plays a smaller part in events.

Example - a kitchen sink

Theory is all very well, but let's see someactual numbers. The kitchen sink is fed by15mm pipe. How much pressure will it take toget hot water (at about 60oC, say) moving outof the tap at 2 metres/second, and is this headachievable?

1. CalculateReynoldsnumberfromwaterspeed,pipesize,

size,density,&viscosity.

2. Lookupfrictionfactorfonthegraph.

3. CalculatepressuredropfromDarcy-Weisbachequation.

Start by calculating the Reynolds number:

Re = Speed x Diameter x (Density /Viscosity)

We know that the speed is to be 2 m/sec, andthe internal diameter of 15mm pipe is13.6mm. From Table 1, (ρ/μ) for water at 60o isabout 3.1 x106. Then the Reynolds number inthis case is:

Re = 2 x 13.6 x10-3 x 3.1 x106 = 84,000

near enough. From the graph above, this Rehas a "friction factor" f of about 0.019. So inthe pressure-difference equation

we know f (0.019) and v (2 m/s) and ρ (992.1)and D (13.6 mm). For now, assume that thelength L is just 1.0 metre. Then the pressuredifference (per metre) needed to get the waterflowing is:

P = 0.019 x 22 x (992.1 / 2) x (1.0 / 13.6x10-3) = 2,800 N/m2

This means that each metre length of the15mm pipe must have a pressure differenceof 2,800 N/sq.m. between its ends to push

The log-loggraph

1 bar  = 100,000  N/sq.m

1 lb/sq.in  = 7,000  N/sq.m

1 foot of water  = 3,000  N/sq.m

7 m. of water(the minimumwater pressureguaranteed in theUK)  = 69,000  N/sq.m

1 Pascal  = 1 N/sq.m

of 2,800 N/sq.m. between its ends to pushwater though it at 2 m/sec. If the pipe is 10mlong, the total pressure difference betweenthe ends of the pipe (that is, the headrequired) would be 28,000 N/sq.m. Or, to put itanother way, the water will flow at 2metres/second if the head happens to beexactly 28,000 N/sq.m.

If you're more comfortable with pressure expressedas the head in feet, the conversion factor is:

A head of 1 foot ofwater  ≈  3,000 N/sq.m.

So 28,000 N/sq.m. is about the same as a head of 9 feet(or 3m) of water. But if the head is not exactly 9 feet -and in practice, Sod's Law says it won't be - the waterwill flow at a different speed! More on this later.

Pressure difference from a graph - the basicdesign tool

The equations are useful if you ever need to calculateaccurately, but in practice it's easier to check from agraph that what you plan to do will work.

This graph shows pressure drop per metre for a givenflow rate and pipe size. You'll find something similarin the relevant British Standard. It was constructedfrom the pressure-drop equation and covers waterspeeds from 2.0 m/sec (at the top) down to 0.2 m/sec,and is valid for all normal temperatures. It's sayingthat the pressure drop along a length of pipe is(nearly) proportional to the square of the flow ratein the pipe.

The graph tells you nearly all you need to know. Use itlike this:

1. Decidetheflowrateyouneed(sink:0.3litres/sec;bath:0.5litres/sec,say).

2. Chooseapipesizethatwillcarrythisflow

flowatlessthan2m/sec.

3. Usethegraphtofindtherateofpressuredrop,permetreofpiperun.

Thistellsyoutheheadyouwillneed.

Example - a bathroom sink

A bathroom sink is fed with 15mm pipe andneeds a flow rate of 0.3 litres/sec.

From the graph, this means the water speedwill be 2 metres per second and the headrequired to achieve this flow rate will be 4,000N/sq.m. (or 1.3 feet height of water) per metreof pipe. So if the sink is fed from a tank 13 feetabove it, the pipe run could have an(equivalent) length of 10 metres. If the pipe isshorter, the water will flow faster.

Example - 22mm pipe connected to the water main

Suppose that the stop-tap offers a 22mmconnection, and that the water pressure hereis 2 bar. Assume a horizontal straight 22mmpipe is connected to the stop-tap. What will

pipe is connected to the stop-tap. What willthe flow rate be if the pipe is 10m long? Whathappens if it's 100m long?

From the graph, 10m of 22mm pipe carrying0.7 litres/sec ( = 2 m/sec water speed) has apressure drop of

P = 10 x 2,500 N/sq.m = 25,000 N/sq.m

If a pressure of nearly ten times this (and 2 bar= 200,000 N/sq.m) is applied, the graph can'tpredict what would happen. I would guessthat the flow rate would exceed 2 litres/secand the noise level would be scary. This is nota good idea!

However, with 100m of pipe, the 200,000N/sq.m mains pressure works out at a moremodest 2,000 N/sq.m per metre. The graphsays this delivers about 0.6 litres/sec (36litres/minute) at a water speed of somethingunder 2 m/sec. It would work fine.

Example - a fountain

Suppose that the 10m length of 22mm pipeconnected to the stop-tap points verticallyupwards. The 2 bar pressure at the stop-tapwill presumably cause water to squirt out ofthe top. How high will it go?

The weight of water in the vertical pipe exertsa pressure downwards, towards the stop-tap,of

Pressure = Length x density x g(N/sq.m)Pressure = 10 (m) x 1,000 (kg/cu.m) x9.8 (m/sec/sec) ≈ 100,000 (N/sq.m)

This pressure acts downwards, opposing the200,000 N/sq.m upwards pressure at the stop-tap. The net upwards pressure is reduced to100,000 N/sq.m. Over the 10m length, there isnow 10,000 N/sq.m per metre. This is off thegraph, as it represents a water speed of wellover 2 metres/sec. It might give a flow rate ofabout 1.5 litres/sec.

The cross-sectional area of 22mm pipe is 320sq.mm., so 1.5 litres in 22mm pipe occupies a

sq.mm., so 1.5 litres in 22mm pipe occupies alength of

(1.5/1,000) / (320 x 10-6) = 4.7 metres

which means that when the water leaves thetop of the pipe it is moving at 4.7 metres/sec.How high will it go? The equation I learnt atschool relates speed and distance for a bodymoving under gravity like this

v2 = u2 - 2 g s

where u and v are the initial and finalvelocity, s is distance, and g is 9.8 m/sec/sec asusual. Here u = 4.7 m/sec and v = 0 (becausethe water stops rising, pauses, then begins tofall) so

4.72 = 2 x 9.8 x s   ...   s = 4.72 / (2 x 9.8)= 1.1 metres ( ≈ 3.5 feet).

So at the end of a 10m vertical pipe - that is, atrooftop height, 30 feet in the air - mains waterpressure would still produce a fountain aboutas high as a child! No wonder watercompanies' pipes leak.

What size pipe do you need?

How do you go about choosing the correct sizes for allthe different pipes in the house?

Here's a simplified sketch of the hot- and cold-watersupply system in a two-storey house. The cold-waterheader tank in the loft feeds a bath on the first floor,and the kitchen sink on the ground floor. It also feedsthe hot water pipes via the cylinder.

The first step is to sketch the layout and choose thepipe sizes such that the water flows fast enough to fillthe bath and the sink in a sensible time.

Then calculate what will actually happen, and decidewhether anything needs to be changed.

So, here:

To fill a 10-litre kitchen sink in half a minute, theflow rate of the pipe feeding it must be close to 0.3litres/second, and 15mm pipe can probably handlethis.

The flow rate for a bath should be higher, but as asingle 22mm pipe can comfortably deliver morethan 0.5 litres/second, two 22mm pipes (hot andcold) will be more than adequate.

This house doesn't have a shower. Showers use about10 litres per minute - that is, about 0.17 litres/second -so 15mm pipes would be quite big enough if theowner ever decided to install one. A five-minuteshower only uses about 30 litres of hot water. That'swhy it's cheaper to shower than to have a bath. It'scheaper still when you share with a friend,apparently.

Cold water pipes

The design starts with the cold feeds. The kitchen sinkneeds 0.3 litres/sec, and according to Table 4 a 15mmpipe will only deliver 0.22 litres/sec at a water speedof 1.5 metres/sec. The choice is, to pay more and use

of 1.5 metres/sec. The choice is, to pay more and use22mm pipe, or to fit 15mm pipe and put up with asmall amount of extra noise. Which would you go for?

A cautious person might ask, how much more noise?A mountain stream, or Niagara Falls?

That's easy to answer. Increasing the flow rate by 30%means that the water flows 30% faster - 2.2 metres/secinstead of 1.5 m/s. The noise level would roughlydouble. That shouldn't be a big problem.

The kitchen sink cold feed can therefore be 15mm, atleast up to the junction with the bath cold feed. Thepipe from here to the bottom of the cylinder servestwo purposes, though. Someone might be running abath whilst someone else is downstairs washing up.What then?

Suppose that the bath cold tap and the kitchen sinkcold tap are both running at once, with 0.3 litres/secgoing to the sink downstairs and (say) 0.5 litres/secgoing into the bath. The total flow-rate would be 0.8litres/sec, and 15mm pipe would complain at that.Will 22mm pipe do, or should it be 28mm? You mightask how likely is it that both taps would be on at thesame time, and if it did happen, would anyone mindtoo much if the cold flow slackened off for a fewseconds? Probably not (unless they were having ashower!) 22mm pipe should be adequate.

Finally, there's the pipe from the header tank to thebottom of the cylinder. This one is more importantthan it looks - it not only carries cold water to the tapsbut also refills the cylinder as hot water is taken fromthe top. Water flows through this pipe to every tap inthe house. It would be sensible to make it 28mm,which can carry over a litre per second.

Hot water pipes

The hot-water pipes are easy to size, because thethinking has already been done for the cold pipes. Thekitchen sink will be fed in 15mm from the tee underthe bath, and then in 22mm from the top of thecylinder.

The vent pipe leading from the cylinder to above theheader tank should also be 22mm (as local authority

planning laws usually require). This pipe is only thereas a safety measure - if something goes wrong, andthe water in the cylinder boils, it can siphon up safelyinto the tank instead of bursting the cylinder andruining all the carpets.

What's the actual flow rate?

It's all very well calculating pipe sizes by assuming aflow rate, but what will actually happen in a realhouse in practice? How fast will the water flow out ofthe kitchen sink tap? How long will it really take to fillthe bath?

It is possible to predict how a real system will behave.In this section I show how to calculate what willhappen in the two-storey house design described earlier.Each step is explained in some detail in order to makeit easier to adapt the calculation to the differentproblem you may be trying to solve.

Pipes often go round corners

The pressure driving the water along the pipes is thehead. For the bath, this is 3 metres (say), and for thekitchen sink on the floor below it's 5 metres (say). Thispressure is opposed by the friction losses in the pipes,which can be thought of as the pressure-difference-per-metre needed to push the water along at the flowrate you want. The log-log graph can be used to find theflow rate in a pipe run when the head is known.

There is one small difficulty. Real pipe goes roundcorners, and through tees, and valves, and otherfittings. Each fitting creates its own bit of turbulenceand absorbs some energy. How can this be taken intoaccount?

Quite easily, as it happens. In just the same way that alength of straight pipe needs a pressure difference topush water through it, so does an elbow, or a valve.The pressure difference required across a 15mmelbow to move water though it at, say, 0.2 litres/seccan be measured. Whatever this number is, it must bethe same as the pressure difference required to movewater through some length of straight 15mm pipe atthe same speed. In fact, this equivalent length is about

Fittings:equivalentlength

the same speed. In fact, this equivalent length is about0.4 metres for a 15mm elbow. So the pressure drop inthe elbow can be included by pretending that the15mm pipe is really straight, but 0.4 metres longerthan it actually is. The "equivalent lengths" of somecommon fittings are listed below.

Table 5: The equivalent lengths (inmetres) of some standard fittings

Pipesize

Elbow Tee:through

Tee:into

branch

Tee:from

branch

15mm 0.4 0.05 0.7 0.6

22mm 0.6 0.09 1.1 1.0

28mm 0.9 0.12 1.6 1.4

One common fitting that doesn't appear in the table isthe shower head. Its function is to take the stream ofwater flowing in a 15mm pipe and split it into manylittle streams, each about 1mm in diameter. Thisprocess takes a lot of energy. In terms of equivalentpipe length, a shower head might represent as muchas 10-20 metres of 15mm, or even more, and this has aserious impact on flow-rate. That's why many peopleopt for a pumped shower, or one run directly frommains pressure via a combi boiler.

Example - the kitchen sink feed from the bathroom

The 15mm pipes run under the bathroom floor, thendown to the ground floor, along to the sink, then upagain to the taps.

There are 5 elbows (right-angle bends) in each pipe.According to Table 5, each elbow causes the samepressure drop as 0.4 metres of 15mm pipe. So theelbows represent 5 x 0.4m = 2m of pipe. The pipesthemselves are about 7m long, so the total equivalentlength of each one is 7m + 2m = 9m of pipe.

Then from the log-log graph, to achieve a water flowrate of 0.3 litres/second, the head would have to be9m x 4,000 N/sq.m = 36,000 N/sq.m.

Pipes are different sizes, too

Suppose someone turns on the cold tap at the kitchensink. What will happen?

Water will begin to flow out of the header tank, downthe 28mm pipe to the cylinder, along the 22mm pipeto the bath, then down the 15mm pipe to the sink.How fast it flows depends on the head and theopposing frictional pressure drop. The head is knownto be 5m, but the opposing frictional loss must becalculated.

The problem is that each different pipe size offers adifferent resistance to the same flow rate. What'sneeded is some way of expressing these differentresistances in some common unit, so that they can bejust added together.

A clue comes from the log-log graph. The lines are(nearly) parallel. This means that the rate of pressuredrop (RPD) for 22mm pipe (say) is always somefraction of that for 15mm pipe, at the same flow rate.

At 0.05 litres/sec, 15mm pipe has a RPD of about 150N/sq.m/m, whilst for 22mm RPD is just 20 N/sq.m/m -about seven times smaller.

(RPD for 15mm pipe) / (RPD for 22mm pipe)  =  7/ 1

And at 0.2 litres/sec the figures are 1900 and 270 -again, a ratio of about 7 to 1. So to get the same flowrate, 15mm pipe needs seven times the pressuredifference that 22mm needs!

1m of 15mm pipebehaves like 7m of

22mm pipe.

These figures aren't exact, but they're near enough tobe useful in the real world.

1m of 22mm pipebehaves like (1/7)m -

0.13m - of 15mm pipe.

The idea can be extended to the other pipe sizes. Thetable below shows the length of each standard size

What is the flowrate out of thekitchen coldtap?

table below shows the length of each standard sizepipe that is equivalent to a 1 metre length of 15mmpipe. It says, for example, that just 3.5cm of 28mmpipe has the same pressure drop as 1m of 15mm pipe.

Table 6: The lengths (in metres) ofstandard pipe sizes equivalent to 1m of

15mm

10mm

15mm

22mm

28mm

35mm

42mm

54mm

7 1.0 0.13 0.035 0.012 0.0047 0.0013

Flow rate calculations

So, back at the sink...

The question was, how fast will water come out of thecold tap at the kitchen sink?

1. Workouttheequivalentlengthofthe15mmsection.

2. Workouttheequivalentlengthsofthe22mmandthe28mmsections.

3. Convertthe22mmand28mmlengthstotheir

theirequivalent15mmlength.

4. Addallthelengthsof15mmequivalenttogether.

5. Workoutthetotalpressuredrop(fromhead,ρ,g).

6. Findtheaveragerateofpressuredrop(dividebypipelength).

7. Lookupthecorrespondingflowrateonthelog-loggraph.

The 15mm section runs from the kitchen tap itself upto the tee with the bath tap. It is about 7m long withfive elbows, so it has an equivalent length of

[15mm actual]  =  7.0m (the pipe) + (5 x 0.4m) (the

... 0.35litres/sec!

[15mm actual]  =  7.0m (the pipe) + (5 x 0.4m) (theelbows)  =  9.0m.

The 22mm section includes two tees, and the pipeitself. If the 22mm pipe is (say) 3.5m long, thisrepresents an equivalent length of

[22mm actual]  =  3.5m (the pipe) + (0.09m + 1.1m)(the tees: 1 in, 1 through)  =  4.7m.[Convert 22mm actual --> 15mm equivalent]  = 4.7m x 0.13  =  0.6m.

The 28mm pipe is 6m long, with two elbows, giving anequivalent length of

[28mm actual]  =  6.0m (the pipe) + (2 x 0.9m) (theelbows)  =  7.8m.[Convert 28mm actual --> 15mm equivalent]  = 7.8m x 0.035  =  0.3m.

So the total equivalent length of 15mm pipe is:

9.0m (15mm) + 0.6m (22mm) + 0.3m (28mm) =9.9m.

Now, the head is 5m, and we know that:

Pressure  =  Length x Density x g

so putting in numbers for density and g, the pressureat the kitchen tap will be:

5 [m.of water] x 1,000 [kg/m3] x 9.8 [m/sec2]  = 49,000 N/sq.m

This pressure drop is shared out along the pipe run -that is, along the 9.9m equivalent length of 15mm -which means the average rate of pressure drop is

49,000 / 9.9 = 5,000 N/sq.m per metre

more or less. From the log-log graph, 15mm pipe with aRPD of 5,000 N/sq.m per metre has a flow rate ofabout 0.35 litres/second. This is what will come out ofthe tap, and more by luck than by skilful design, it'sclose to the 0.3 litres/second that it should be.

But is this figure true? Cross-check the result byworking backwards. Breaking it down, the answersays that the 9m of real 15mm pipe accounts for (9 x5,000) = 45,000 of the 49,000 N/sq.m of availablepressure, the 22mm length takes (0.6 x 5,000) = 3,000N/sq.m, and the 28mm needs (0.3 x 5,000) = 1,500N/sq.m. This adds up to 49,500, which is close enoughto the expected figure of 49,000. This is supposed to beengineering, not physics.

What is the flowrate out of thebath cold tap?

Then the flow rate in the actual 4.6m of 22mm pipe atits RPD of (3,000 / 4.6m) = 652 N/sq.m per metre is,from the log-log graph, about 0.35 litres/second. And forthe actual 7.8m of 28mm at its RPD of (1,500 / 7.8) =192 N/sq.m per metre, the flow is once again 0.35litres/second. Each pipe is carrying the same flowrate, as it should do. So the kitchen sink tap really willdeliver 0.35 litres/second.

What if the pipes are too noisy?

In a different design - perhaps one with with fatterpipes, or fewer elbows, or a larger head - thecalculation might have predicted a much higher flowrate. In that case you would expect the pipes to benoisy when the water is running. To make themquieter, the water has to be slowed down, and this isactually very easy to do. Any competent plumberinstalling a system will have included valves atstrategic points, so that sections of the system can beisolated - when, for example, you need to change a tapwasher.

All you have to do is find the right valve and turn itdown a bit. The extra resistance this adds will reducethe flow rate to a more sensible value. Halving theflow rate would reduce the noise by a factor of four.

Running a bath

This calculation is a bit more complicated, because itinvolves both the hot and cold water pipes in the two-storey house sketched above. The approach is exactlythe same: find the equivalent lengths, convert themto the same size pipe, add them up, find the pressuredrop per metre, look up the corresponding flowrate.

Cold feed only: Think about the cold water first. The22mm pipe from the tap is 3.5m long and includes twotees. It has an apparent length of:

[22mm actual]  =  3.5m + (1.1m + 1.0m)  =  5.6m.

Similarly, the apparent length of the 28mm pipe is:

[28mm actual]  =  6.0m + (2 x 0.9m)  =  7.8m.

Since there is no 15mm pipe involved in the runs tothe bath, it seems silly to convert these lengths to

... 0.9 litres/sec!

the bath, it seems silly to convert these lengths totheir equivalent 15mm lengths, then add themtogether, then convert them back again to 22mm.Instead, I'll simply convert the 28mm length to itsequivalent 22mm value, using the figures in Table 6:

[28mm actual --> 22mm equivalent]  =  7.8m x(0.035 / 0.13)  =  2.1m.

Then the total equivalent length of 22mm is:

5.6m + 2.1m = 7.7m.

The head is 3m, which corresponds to a pressure of:

3 [m.of water] x 1,000 [kg/m3] x 9.8 [m/sec2]  = 29,400 N/sq.m

So the average rate of pressure drop is:

29,400 / 7.7 = 3,800 N/sq.m per metre

which according to the log-log graph means a (rathernoisy) flow rate of close to 0.9 litres/second for acold bath, rather than the 0.5 litres/second one mighthave hoped for. Still, things will change when the hottap is running too.

Hot feed only: Now for the hot water. The hot pipe isall 22mm, which makes it slightly easier. The pipe runto the top of the cylinder is (let's say) 6m long, andincludes two tees and three elbows. So:

[Hot: 22mm actual]  =  6m + (1.1m + 1.0m + [3 x0.6m])  =  9.9m.

However, the hot water leaving the cylinder isreplaced by cold water flowing from the header tank.The cylinder itself is only a kind of fitting, and it toohas resistance, just like an elbow. The resistance ofthe whole circuit must be calculated.

The 22mm run is only 1m or so, plus a tee and anelbow. The cylinder's resistance is equivalent to about1.6m of 22mm pipe. Adding these up gives:

[Cold: 22mm actual]  =  1m + (1.0m + 0.6m +1.6m)  =  4.2m.

Finally, there's the 28mm pipe from the header tank.I've already calculated that this is 7.8m (actual) and2.1m (22mm equivalent), so the total equivalentlength of 22mm pipe in this circuit is:

9.9m + 4.2m + 2.1m = 16.2m.

The head is still 3m, or 29,400 N/sq.m, so the average

The head is still 3m, or 29,400 N/sq.m, so the averagerate of pressure drop is

29,400 / 16.2 = 1,800 N/sq.m per metre

which the log-log graph says represents close to 0.6litres/second for a hot bath - pretty much what itshould be. The flow rate from the hot tap is less thanfrom the cold tap because of the resistance of all theextra pipe this water has to flow through.

Both hot and cold: Most people turn on both tapswhen they are running a bath. What happens then?It's a more difficult problem, because now the 28mmpipe from the header tank is carrying cold water bothto the bath and to the bottom of the cylinder. A higherflow rate means a greater resistance. How muchgreater? That depends on the flow rate it's carrying,and that in turn depends on its resistance!

Breaking this circle demands a little algebra, sincethere are now two unknown (and inter-dependent)quantities: the flow rates from each of the bath taps. Idon't know yet what they are, so I'll call the flow rateout of the hot tap H litres/sec, and that from the coldtap C litres/sec.

Now, the hot water circuit runs from the tee (with the28mm pipe) up through the cylinder, down and alongto the hot tap. It has an equivalent length of (9.9m +4.2m) = 14.1m. This pipe run is carrying H litres/sec.

Similarly, the effective length of the cold watercircuit, from the cold tap to the same junction, is 5.6m.This pipe run is carrying C litres/sec.

And the 28mm pipe, with an effective length of 7.8m(or 2.1m of 22mm equivalent), has to carry (H + C)litres/sec.

I know that H and C must be less than 0.6 and 0.9litres/sec respectively, because those are the flowrates with only one tap open. The flow rates with bothtaps open must be smaller, because the hot and coldflows share space in the 28mm pipe, and it will offergreater resistance to the flow, so (for now) guess thatH = 0.5 litres/sec. From the graph, this implies a Rate ofPressure Drop (RPD) of 1,400 N/sq.m per metre.

The effective length of the pipes carrying just hot

The straight-linelog-log graph couldalso be written as apower law.

For 15mm pipe, itwould be:   RPD =35,000 x FR1.83

The effective length of the pipes carrying just hotwater is 14.1m. The total pressure drop along thesepipes would then be (14.1 x 1,400) = 19,700 N/sq.m.The head is 29,400 N/sq.m, so the pressure differencebetween the water surface in the header tank and thejunction of the hot and cold circuits - at the tee nearthe bottom of the cylinder - would be (29,400 N/sq.m -19,700 N/sq.m) = 9,700 N/sq.m. I'll come back to thisfigure in a moment.

But the same pressure of 19,700 N/sq.m that drives thehot water flow is driving the cold water flow too. Theeffective length of the pipes carrying just cold water is5.6m, so the RPD for the cold-water pipes is (19,700 /5.6) = 3,500 N/sq.m per metre, and the graph says thatthis implies a flow rate of about C = 0.82 litres/sec.

My original guess was that H was 0.5 litres/sec, andthis guess resulted in a predicted value for C of 0.82litres/sec. In other words, C is 1.64 times bigger thanH. But this ratio depends only on the pipe layout. It'sindependent of the actual values of H and C.Whatever the real figures are, this ratio will stay thesame.

If my original guess that H was 0.5 litres/sec had beencorrect, then the combined flow in the 28mm pipewould have been (0.5 + 0.82) = 1.32 litres/sec. The graphsays that the RPD of 28mm pipe carrying 1.32litres/sec is about 2,700 N/sq.m per metre, so the totalpressure drop along its effective length of 7.8m is(2,700 x 7.8) = 21,000 N/sq.m.

But I have already calculated that if H really had been0.5 litres/sec, the pressure drop along the 28mm pipewould have been 9,700 N/sq.m - only half as much.The original guess was plainly wrong! So how can theproblem be solved?

A Useful Approximation

The relationship between the two quantities ofinterest - flow rate and pressure drop - is extremelycomplex, but fortunately it can be approximated by arather simple formula:

Rate of Pressure Drop (RPD)  =  A x (Flow Rate)2

+ B

35,000 x FR1.83

For 22mm pipe, itwould be:   RPD =5,000 x FR1.85

+ B

- where A and B are constants that depend only onpipe size. I give values for A and B in the table below.

Table 7: Values of constants A and B in the UsefulApproximation

Pipesize 10mm 15mm 22mm 28mm 35mm 42mm 54mm

A 400,000 44,000 5,300 1,400 450 160 40

B 100 70 40 30 18 20 14

The approximation is accurate when the pipe is carrying aflow rate of between 30% and 100% of its maximumcapacity.

Running the bath

Here is a simplified diagram showing only the piperuns to the hot and cold bathtaps. Water is flowingfrom both taps.

The cold water feed pipe is 5.6 metres long andcarrying C litres/second. The hot water pipe is 14.1metres long and carries H litres/sec. The commonfeed, carrying cold water to the tap and also into thebottom of the cylinder - that is, (C + H) litres/sec - is7.8 metres long, from the header tank to the tee.

Now, from the Useful Approximation, the totalpressure difference between the ends of a pipe is

Pressure drop in pipe = Length x [A x (FlowRate)2 + B]

The hot and cold pipes are both fed from the commonpipe, and both end in open taps. The pressuredifference between the common point and each tapmust be the same. So by applying the formula, doing abit of algebra, and discarding terms that are too smallto matter, we get a relationship between the flowrates that just depends on pipe lengths:

This is really just a more formal way of expressing theidea that the hot and cold flow rates will always bearthe same ratio to each other. But we also know that,for the whole system:

Head = (Pressure drop in common pipe) +(Pressure drop in hot [or cold] pipe)

and this, with a bit of algebra, can be made to yield anexpression for the actual hot or cold flow rate interms of numbers we already know! To make theequation as general as possible, I have used thesymbols Lc and Lh to stand for the lengths of the coldand hot pipe runs respectively, and L28 to mean thelength of the common 28mm pipe. For the cold flowrate, C, the equation is:

This equation looks forbiddingly complex, but findinga value for C is simply a matter of substituting knownnumbers for all the variables and calculating theanswer. The head is 29,400 N/sq.m., Lc is 5.6m., and Lhis 14.1m. It's important that all the lengths beexpressed in the same units, so L28 is 2.1m (of 22mmequivalent) rather than the actual figure of 7.8m.Finally, from Table 7, the constants for 22mm pipe areA = 5,300 and B = 40.

The answer I got was C2 = 0.49 litres/second, so C = 0.7litres/second. And since (C / H)2 = (14.1 / 5.6), Icalculate that H = 0.44 litres/second.

The answer can be checked by working out theindividual pressure drops using the UsefulApproximation. In the hot and cold pipes:

Cold pipe: Pressure drop = 5.6m x [5,300 x (0.7)2

+ 40] = 14,800 N/sq.m

Hot pipe: Pressure drop = 14.1m x [5,300 x(0.44)2 + 40] = 15,000 N/sq.m

which is near enough the same, as it should be, and inthe common pipe,

Common pipe: Pressure drop = 2.1m x [5,300 x(0.7 + 0.44)2 + 40] = 14,500 N/sq.m

making a total of about 29,500 N/sq.m. The actualhead is 29,400 N/sq.m. I think the conclusion is thatthe sums really do add up. The method works.

If you want another look at the theoreticalbackground to all this, you'll find it here in Part 1.

Copyright © John Hearfield 2007, 2012