Jan 21st 2016, 02:25 AM #1 Senior Member   Join Date: Jun 2014 Posts: 306 head loss can anyone explain what does the statement ' The head loss represents the additional heights that the fluid needs to be raised by a pump in order to overcome the frictional losses in pipe ' on the right of the notes ? Attached Thumbnails
 Jan 21st 2016, 06:57 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,354 If you understand that "head" is a measure of pressure, then the loss in pressure as the fluid flows through a pipe can be referred to as "head loss." For example 10 psi is equivalent to 23 feet of water head (i.e. the pressure 23 below the surface of the body of water = 12 psi). If the pressure of fluid at the entrance of a horizontal pipe is, say, 100 psi, and at the exit of the pipe is 90 psi, then there has been a 10 psi drop in pressure along the length of the pipe. So we could say that the pressure loss is 23 feet of head. Stated another way - if the entrance of the pipe is raised 23 feet, we would expect that the pressure at the exit would be 90 psi + 10 feet of head = 90 psi+ 10 psi= 100 psi. Consequently the head loss of 23 feet can be made up for by raising the entrance of the pipe by 23 feet.
Jan 21st 2016, 08:17 AM   #3
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 Originally Posted by ChipB If you understand that "head" is a measure of pressure, then the loss in pressure as the fluid flows through a pipe can be referred to as "head loss." For example 10 psi is equivalent to 23 feet of water head (i.e. the pressure 23 below the surface of the body of water = 12 psi). If the pressure of fluid at the entrance of a horizontal pipe is, say, 100 psi, and at the exit of the pipe is 90 psi, then there has been a 10 psi drop in pressure along the length of the pipe. So we could say that the pressure loss is 23 feet of head. Stated another way - if the entrance of the pipe is raised 23 feet, we would expect that the pressure at the exit would be 90 psi + 10 feet of head = 90 psi+ 10 psi= 100 psi. Consequently the head loss of 23 feet can be made up for by raising the entrance of the pipe by 23 feet.
do you mean that for the entrance point to achieve 90Pa from 100Pa , it has to drop 10Pa . To drop 10Pa , as the pressure is given by the formula = h(rho)g , the height of entrance point need to be raised so that h would decrease to the point where pressure = pressure at exit point (90Pa) ?

 Jan 21st 2016, 12:30 PM #4 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 The friction within the pipe causes a drop in pressure. The drop in pressure caused by the friction is equivalent to the pressure loss that would (theoretically) be seen in a perfectly friction free pipe if the original head was lowered by 23 feet. __________________ You have GOT to Laugh !
Jan 21st 2016, 04:16 PM   #5
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 Originally Posted by MBW The friction within the pipe causes a drop in pressure. The drop in pressure caused by the friction is equivalent to the pressure loss that would (theoretically) be seen in a perfectly friction free pipe if the original head was lowered by 23 feet.
If the original head was lowered by 23 feet , the pressure will increase , right ? Sincep P= hρg , h decrease . Why you said it pressure will decrease ? Or do you mean when the entrance head is raised by 23 feet , the entrance head and exit head will by different by 10Pa ?

Last edited by ling233; Jan 21st 2016 at 04:22 PM.