gani Posted November 24, 2013 Report Share Posted November 24, 2013 my main broke at my house and i'm switching from quest 3/4 pipe to 3/4 pex pipe. I'm using the crimp connections and they turn my 3/4 pex joints/connectors to roughly a 1/2 inch. I have 5 sinks, 3 toilets, 2 bathtubs and 2 spouts outside also ice maker, washing machine, and dishwasher are all branched from my main water pipe T. Will I have enough pressure to run 2 showers, dishwasher, washing machine and a sink or 2 with 3/4 pex? Quote Link to comment Share on other sites More sharing options...
Scottpat Posted November 24, 2013 Report Share Posted November 24, 2013 my main broke at my house and i'm switching from quest 3/4 pipe to 3/4 pex pipe. I'm using the crimp connections and they turn my 3/4 pex joints/connectors to roughly a 1/2 inch. I have 5 sinks, 3 toilets, 2 bathtubs and 2 spouts outside also ice maker, washing machine, and dishwasher are all branched from my main water pipe T. Will I have enough pressure to run 2 showers, dishwasher, washing machine and a sink or 2 with 3/4 pex? Going to PEX I would seriously consider 1". You also must take into consideration how high the water needs to go to reach all of the fixtures. The higher it must travel the more pressure and volume you need. Pressure is not really a concern if you had good pressure before, a concern is volume! Quote Link to comment Share on other sites More sharing options...
Jim Katen Posted November 25, 2013 Report Share Posted November 25, 2013 Upsize to 1" and use full bore fittings. Quote Link to comment Share on other sites More sharing options...
John Dirks Jr Posted November 25, 2013 Report Share Posted November 25, 2013 Like Scott said, volume is the important factor, not pressure. It's why both he and Jim recommend the 1" line. Quote Link to comment Share on other sites More sharing options...
rjbrown2 Posted November 25, 2013 Report Share Posted November 25, 2013 When you do the math for pressure drop in a circular pipe in the kind of situation you find in residential plumbing, you note that the drop is directly proportional to the volume flow rate (gallons/minute) but inversely proportional to the fourth power of the diameter of the pipe. This is so overwhelming that whenever there is the slightest question about sufficient supply based on pressure, go bigger even if costs a little more. Saying it in other words, for the same pressure drop you will be able to deliver a lot more water to the point that something else will be the limiting factor. Quote Link to comment Share on other sites More sharing options...
Marc Posted November 25, 2013 Report Share Posted November 25, 2013 When you do the math for pressure drop in a circular pipe in the kind of situation you find in residential plumbing, you note that the drop is directly proportional to the volume flow rate (gallons/minute) but inversely proportional to the fourth power of the diameter of the pipe. This is so overwhelming that whenever there is the slightest question about sufficient supply based on pressure, go bigger even if costs a little more. Saying it in other words, for the same pressure drop you will be able to deliver a lot more water to the point that something else will be the limiting factor. More like the square of the radius of the pipe which is to say the doubling the diameter reduces pressure drop by a factor of 4, which is the square of 2. Marc Quote Link to comment Share on other sites More sharing options...
rjbrown2 Posted November 26, 2013 Report Share Posted November 26, 2013 Marc: I was putting into words the classic Hagen-Poiseuille (perhaps one of your relates?) equation for laminar flow of an incompressible Newtonian fluid in a long constant cross section pipe. To wit, the pressure drop is expressed as: Delta P = (128*Mu*L*Q)/(Pi* D^4) or in terms of R Delta P = (8*Mu*L*Q)/(Pi*R^4) Where Delta P = pressure loss, L = pipe length, Q = volume flow rate, Mu = dynamic viscosity, D = diameter, R = radius, and Pi is our constant 3.14159...etc Quote Link to comment Share on other sites More sharing options...
Marc Posted November 26, 2013 Report Share Posted November 26, 2013 Marc: I was putting into words the classic Hagen-Poiseuille (perhaps one of your relates?) equation for laminar flow of an incompressible Newtonian fluid in a long constant cross section pipe. To wit, the pressure drop is expressed as: Delta P = (128*Mu*L*Q)/(Pi* D^4) or in terms of R Delta P = (8*Mu*L*Q)/(Pi*R^4) Where Delta P = pressure loss, L = pipe length, Q = volume flow rate, Mu = dynamic viscosity, D = diameter, R = radius, and Pi is our constant 3.14159...etc Ok. But isn't that equation intended for higher viscosities than water? Water at 20C has a viscosity of only 0.001 Marc Quote Link to comment Share on other sites More sharing options...
rjbrown2 Posted November 26, 2013 Report Share Posted November 26, 2013 Marc: Poiseuille conducted his original experiments in the 1840s using distilled water and eventually extended his work to other fluids. The H-P equation is very much applicable to water. I think I know the root of our difference here (as if anyone cares now except you and me) The form of H-P changes when Velocity is included in the Delta P equation instead of Q Delta P = (32*Mu*L*V)/D^2 The reason I discussed the form that I did was because the variables that were relevant in the discussion were: pressure and flow rate Quote Link to comment Share on other sites More sharing options...
Jim Katen Posted November 26, 2013 Report Share Posted November 26, 2013 A slightly bigger pipe gives a whole lot more flow. Quote Link to comment Share on other sites More sharing options...
Marc Posted November 26, 2013 Report Share Posted November 26, 2013 Marc: Poiseuille conducted his original experiments in the 1840s using distilled water and eventually extended his work to other fluids. The H-P equation is very much applicable to water. I think I know the root of our difference here (as if anyone cares now except you and me) The form of H-P changes when Velocity is included in the Delta P equation instead of Q Delta P = (32*Mu*L*V)/D^2 The reason I discussed the form that I did was because the variables that were relevant in the discussion were: pressure and flow rate Alright. Thanks. I was looking at it from strictly as a function of circular area vs radius but I see the other factors in that now. Marc Quote Link to comment Share on other sites More sharing options...
Ben H Posted November 27, 2013 Report Share Posted November 27, 2013 A slightly bigger pipe gives a whole lot more flow. Best answer. Quote Link to comment Share on other sites More sharing options...
rjbrown2 Posted November 29, 2013 Report Share Posted November 29, 2013 That's where I started, but a gave just a little more than that in words, because the question revolved around the relative importance of pressure vs. flow. It only got mathematical when I was questioned on it. ?It can scarcely be denied that the supreme goal of all theory is to make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience.? Albert Einstein Happy T Day! Quote Link to comment Share on other sites More sharing options...
Marc Posted November 29, 2013 Report Share Posted November 29, 2013 That's where I started, but a gave just a little more than that in words, because the question revolved around the relative importance of pressure vs. flow. It only got mathematical when I was questioned on it. ?It can scarcely be denied that the supreme goal of all theory is to make the irreducible basic elements as simple and as few as possible without having to surrender the adequate representation of a single datum of experience.? Albert Einstein Happy T Day! Not to worry. All's fine. Marc Quote Link to comment Share on other sites More sharing options...
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