Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

In an earlier contribution to this discussion,
Colin Wilson o.uk wrote:

Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

No - flow is flow, pressure is pressure.

You wouldn't try to equate amps to volts, would you - without knowing
something about the impedance of the circuit?
--
Cheers,
Roger
______
Email address maintained for newsgroup use only, and not regularly
monitored.. Messages sent to it may not be read for several weeks.
PLEASE REPLY TO NEWSGROUP!

"Roger Mills" wrote in message
...
In an earlier contribution to this discussion,
Colin Wilson o.uk
wrote:

Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

No - flow is flow, pressure is pressure.

Which are inversely proportional.

On 26 May, 01:40, Colin Wilson
o.uk wrote:
Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

The listed flow rate is the flow at which it can maintain a
temperature increase of IIRC 30 degrees. It will run hot water faster
if you want, but won't heat the water adequately. Often the flow is
restricted at the cold water input so that you cant exceed the 13L by
much, but that's only for convenience so that it's not fiddly to get
hot water out.

A

In message ews.net,
Doctor Drivel writes

"Roger Mills" wrote in message
...
In an earlier contribution to this discussion,
Colin Wilson o.uk
wrote:

Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

--
geoff

"raden" wrote in message
news
In message ews.net,
Doctor Drivel writes

"Roger Mills" wrote in message
...
In an earlier contribution to this discussion,
Colin Wilson o.uk
wrote:

Had mine tested the other day (Worcester Bosch contract) and the flow
rate on the DHW was put on the sheet as 13.1 L/min

Can this be equated to pressure (as in Bar), or are they sensitive to
back-pressure caused by something like a shower / steam bath
thingymajig possibly restricting flow, making it impossible to guess
at ?

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

Maxie, have you gone into the undertaking business? Do you have ways of
drumming up business?

The message
from raden contains these words:

In message ews.net,
Doctor Drivel writes

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

So Dribble is back. Shame they couldn't have kept him locked up for
longer but we must be thankful for small mercies. According to my
killfile log he has been absent from 2nd march to 25th May inclusive - 3
lunar months.

--
Roger Chapman

In article ,
Roger wrote:
So Dribble is back. Shame they couldn't have kept him locked up for
longer but we must be thankful for small mercies. According to my
killfile log he has been absent from 2nd march to 25th May inclusive - 3
lunar months.

That's a long rehab.

--
*The more I learn about women, the more I love my car

Dave Plowman London SW
To e-mail, change noise into sound.

On Fri, 01 Jun 2007 13:09:15 +0100, Dave Plowman (News) wrote:

In article ,
Roger wrote:
So Dribble is back. Shame they couldn't have kept him locked up for
longer but we must be thankful for small mercies. According to my
killfile log he has been absent from 2nd march to 25th May inclusive - 3
lunar months.

That's a long rehab.

Depends on the nature of the event(s) that put him offline.




--
Ed Sirett - Property maintainer and registered gas fitter.
The FAQ for uk.diy is at http://www.diyfaq.org.uk
Gas fitting FAQ http://www.makewrite.demon.co.uk/GasFitting.html
Sealed CH FAQ http://www.makewrite.demon.co.uk/SealedCH.html
Choosing a Boiler FAQ http://www.makewrite.demon.co.uk/BoilerChoice.html
Gas Fitting Standards Docs he http://www.makewrite.demon.co.uk/GasFittingStandards

"Roger" wrote in message
k...
The message
from raden contains these words:

In message ews.net,
Doctor Drivel writes

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

So

Please eff off you are a total idiot.

On 2007-06-01 22:16:32 +0100, "Doctor Drivel" said:


"Roger" wrote in message
k...
The message
from raden contains these words:

In message ews.net,
Doctor Drivel writes

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

So

Please eff off you are a total idiot.

Now look, you know what happened last time you started that. The next
three month stretch may not be as nice as the last one.

"Andy Hall" wrote in message
...
On 2007-06-01 22:16:32 +0100, "Doctor Drivel" said:


"Roger" wrote in message
k...
The message
from raden contains these words:

In message ews.net,
Doctor Drivel writes

No - flow is flow, pressure is pressure.

Which are inversely proportional.

Haven't you died yet ?

So

Please eff off you are a total idiot.

Now look, you know what happened last time you started that. The next
three month stretch may not be as nice as the last one.

It will be sailing in the Caribbean was nice. Avoid hurricanes though.

On 2007-05-26, 00:40 Z,
in t,
Colin Wilson wrote:
I had [my DHW (domestic hot water heater)] tested and the flow
rate on the DHW was put on the sheet as 13.1 L/min. Can this be
equated to pressure?

Colin: Let's take a simplistic example to simplify the problem.
Let's say you have a straight, horizontal, 4-m-long,
13-mm-inside-diameter pipe exiting near the top of the water tank.
Let's allow the water to flow freely out of the other end of the
pipe. The water exits the straight-cut, unrestricted pipe outlet
into the atmosphere, thus having pressure p2 = 0 Pa (gauge).
Because the pipe outlet is unrestricted, we know the flow rate is
q = 13.1 L/min = 0.0002183 m^3/s. Therefore, the exit velocity is
v2 = 1.6451 m/s. We'll assume the pipe connection to the tank has
a well-rounded fillet. The water velocity in the tank is
essentially v1 = 0. There is no pipe elevation change in this
example, so z2 = z1 = 0 m. The relative roughness of drawn copper
tubing of diameter D2 = 13 mm is e/D2 = (0.0015 mm)/(13 mm) =
0.0001154. Pipe length is L = 4 m. The density and absolute
viscosity of water at 50 deg C is rho = 988 kg/m^3 and mu =
0.000547 Pa*s, respectively. Therefore, the Reynolds number for
the pipe flow is Re = rho*v2*D2/mu = (988 kg/m^3)(1.6451
m/s)(0.013 m)/(0.000547 Pa*s) = 3.86e4. Therefore, the pipe flow
friction factor is f = 0.0226. Therefore, major head loss, due to
pipe flow friction, is hf = 0.5(v2^2)*f*L/D2 = 0.5(v2^2)(0.0226)(4
m)/(0.013 m) = 0.5(v2^2)(6.954). Minor head loss at the pipe inlet
is essentially zero; minor head loss at the unconstricted pipe
exit is zero. Bernoulli's equation is

p1/rho + 0.5*v1^2 + g*z1 = p2/rho + 0.5*v2^2 + g*z2 + hl,

where hl = major plus minor head losses = hf + 0. But we said v1 =
z1 = p2 = z2 = 0. Substituting the quantities stated above, and
simplifying, gives

p1 = 0.5(rho)(v2^2)(1 + 6.954)
= 0.5(7.954)(988 kg/m^3)(1.6451 m/s)^2 = 10 634 Pa (gauge),

where p1 = pressure in the tank. Please let me know if I worked
the problem wrong for the stated assumptions, or if the pipe
inside diameter is not 13 mm, or if the pipe has a different
length, or an elevation change.

If you give us the total cross-sectional area of an outlet device
restricting the pipe flow, then we could figure out the reduced
flow rate and corresponding pressure at the outlet, p2. The source
(tank) pressure, p1, remains essentially constant regardless of
any outlet restriction. (Conversion factor, 100 000 Pa/bar.)