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?
Which are inversely proportional.
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
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 isp1/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.)
HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.