5 replies to this topic

[bobackr]

Hi Everyone,

I want to know in design of equipment and in industry what exactly does pressure drop tell you. The 2 things I do know are:

1. In heat exchangers, an excess of pressure drop will decrease heat transfer.

2. In piping systems, if you are pumping from one vessel to another. Pump discharge is 26 psi and tank is at 14psi. If pressure drop in the system is greater than 12 psi, backflow will occur.

Is this what will occur in 2 or can other things occur. Please add any other helpful things pressure drop will tell you in industry.

Thanks

[Art Montemayor]

bobackr:

I’m going to assume you have completed at least a course in Fluid Mechanics, or are presently taking this course or something similar – like introductory Unit Operations & fluid flow. If I don’t assume these things, I fear my response would not be understood by you.

You say you want to know “what exactly does pressure drop tell you in design of equipment and in industry”. My response to this is:

1. When in the design phase, the pressure drop of a fluid indicates the relative amount of energy that must be imposed on that fluid to overcome the resistance to flow offered by the system. In other words, if you intend to pump a liquid from point “A” to point “B”, you’re going to have to create a driving force in order to create a positive flow in the direction you want the fluid to go. That means the pressure at point “A” must be greater than the pressure at point “B” in order for there to be flow. Initially, the pump builds up sufficient pressure at its discharge port in order to overcome the resistance to flow that the piping, pipe fittings, valves, changes in direction, entrances, and exits contribute. The ultimate result is that the pressure at point “A” is greater than the pressure at point “B”. Within certain parameters, the larger the pressure drop, the larger will be the flow rate of fluid through the same system. Parameters that affect the ultimate result are the fluid (Newtonian liquid, non-Newtonian liquid, subsonic gas, choked flow gas, supersonic gas, slurries, fluidized solids, etc.), fluid viscosity, Reynolds Number, friction factor, and length & diameter of pipe. This is all represented and correlated in the classical Darcy-Weisbach equation:

Pressure Drop = (Density) (FF) (L) (velocity)^2/(144) (D) (2g)
where,
Density = lb/ft3
FF = Friction Factor
L = pipe length,
velocity = ft/sec
D = pipe ID, ft

All practical formulas for the flow of fluids are derived from Daniel Bernoulli’s famous Theorem which bases itself on the conservation of energy, with modifications to account for losses due to friction.

2. In Industry, the existence of a pressure drop in fluid flow indicates the “trade-off”, or the price, one must pay for imposing the flow of the fluid. The pressure drop is an indication of the horsepower that must be inputted into the fluid (by a pump or compressor, depending on the fluid) and as such, the amount of expense or cost involved in effecting the flow.

Now, I want to address the two items you say you “know”:

1. Regardless of where you obtained the information, you are wrong in believing that “an excess of pressure drop will decrease heat transfer”. This statement is categorically wrong. In fact, within a certain range, the amount of heat transferred in a given heat exchanger will increase when there exists a larger pressure drop across the same exchanger. This is easily explained by the fact that with a larger pressure drop, there exists more flow resulting in more fluid turbulence within the heat exchanger. This is so because the fluid’s heat transfer film coefficient increases when subjected to increased turbulence. With increased pressure drop you have more energy consumed; however, the trade-off (there always is one) is that the heat transfer is increased.

2. When you cite fluid pressures, always state clearly what pressure base you are referring to: absolute pressure or gauge pressure. In your example this makes a big difference about what you are trying to describe. I will assume you mean “gauge pressure” (the local pressure as indicated by a pressure gauge). You state you have a supply tank that has a pressure of 14 psig (I assume you mean at the top of the liquid level – which is the conventional manner of describing it) and a pump is used to transfer the liquid to another tank. The pump’s discharge pressure is 26 psig and there exists a pressure drop in the system of 12 psi (note that pressure difference is neither absolute or gauge). Kindly refer to the attached simplified sketch I drew. You will note that you don’t state where your pressure gauges are located in order to define what you mean by the system’s pressure drop. You must do this in order to have an idea of what pressure drop you are referring to. Additionally, also note that it makes a big difference where your pressure gauges are located as well as what the supply liquid height is above the pump’s suction nozzle as well as the discharge tank’s liquid height above the pump’s suction nozzle. I can easily furnish you with a specific tank-pump-tank configuration where in it would be impossible for you to obtain backflow employing the same pressures you cite. Note that I can change the pump’s discharge pressure by changing the location where I introduce the liquid into the target tank – by simply diverting the flow from one place to another opening one discharge block valve and opening the other. I suggest you study the diagram and try locating the pressure gauges, together with the relative liquid heights at different locations and measure the pressure drops these would cause – as well as how you could have back flow.

Basically, I have the impression that you are confused with the existence of a pressure drop in a fluid system. I would remind you that the pressure drop is a result of fluid flow and the conditions under which the same fluid flow exists. It is analogous to voltage drop in electrical systems. The voltage drop (not the “volts”) is the driving force in accordance with Fourier’s Law.

Without a pressure drop (driving force), you cannot have fluid flow.

I hope I’ve addressed your concerns in detail and that my response helps you better understand the basic fluid mechanics involved.

Art Montemayor

[Art Montemayor]

bobackr:

For some unknown reason, I was unsuccessful in attaching the simplified sketch I made to my original response. Here's another attempt to do the same.

Art Montemayor

Attached Files

•  Basic_Liquid_Transfer_by_Pump.doc   26.5KB   145 downloads

[bobackr]

OK Thanks,
So in your diagram, the piping to the bottom of the tank would be the one to use since it has less pressure drop, correct? Would you know the pressure at the inlet to the second tank if you pump intothe bottom of it? Wouldn't the head in the tank cause extra pressure at the inlet of the tank?

Thanks Again

[Art Montemayor]

bobackr:

Yes, the piping circuit feeding the bottom of the target tank COULD have less pressure drop; but that will depend on the height of the liquid head existing in the target tank and the pressure drop suffered in using that circuit. Don’t overlook the fact that without stating the actual piping configuration and the actual dimensions, it is quite possible to have the circuit feeding the bottom section to be much longer (& with more change-of-direction fittings) than the circuit feeding the top. If that’s the case, the resulting friction (& subsequent pressure drop) might be more than that for the circuit feeding the top. Fluid mechanics is very dependent on specific identification of your system and its basic data. You cannot simply generalize in most cases. For example, you may have assumed that both tanks are located at the same grade level. It could be that the target tank is located at a lower elevation than the supply tank (a pump might still be used to ensure a higher flowrate of liquid).

Knowing the specific dimensions and flow basic data, you should be able to calculate the pressure of the fluid entering the bottom of the target tank. You similarly should be able to predict the pressure of the fluid entering the top of the target tank.

Of course, the static liquid head existing in the target tank will be a pressure (or “head”) that has to be overcome and, as a result, will add an additional pressure drop to the fluid flow. However, this is no different that the obstacle of extra head that the second circuit has to overcome in reaching the top of the tank, before discharging into it. In fact, under normal operating conditions the circuit feeding the top of the tank ALWAYS requires more pressure from a pump because the height of the fluid to overcome is always larger than the maximum height of the level in the tank. Nevertheless, there is a process safety reason for selecting the top-entry circuit in many chemical engineering applications. The safety reason is that the top entry nozzle does not permit the contents of the tank to back-flow (or “syphon”) towards the upstream direction when, for example, the pump suddenly shuts down or there is a pipe rupture or leak in the upstream location while the pump is shut down. Again, we note the existence of a trade-off.

If you are indeed a Chemical Engineering student, I strongly recommend you make an effort (or sacrifice) to obtain and study two excellent fluid flow classic works that form the basis for most of the fluid flow applications in industry:

1. Flow of Fluids Through Valves, Fittings, and Pipe; Crane Technical Paper No. 410; Crane Co.
2. Cameron Hydraulic Data; Ingersoll-Rand Pumps.

It is indeed unfortunate and a loss for engineering students that both the above works are not used as the basis of fluid flow courses at University level. The Crane manual is the basic “Dogma” of design engineers involved in fluid flow, while the Cameron book extends its efforts into pumping applications. Both should be mandatory engineering tools, in my opinion. Both can be obtained through the Internet for purchase.

I hope my efforts have not fallen short on explaining the answers to your queries and I hope that other, experienced Forum members also contribute their valued opinions to this thread.

Art Montemayor

[Ming Hooi]

Dear Art,

I refer to your initial response to boback's submission dated 25th March, as quoted below:

Regardless of where you obtained the information, you are wrong in believing that “an excess of pressure drop will decrease heat transfer”. This statement is categorically wrong. In fact, within a certain range, the amount of heat transferred in a given heat exchanger will increase when there exists a larger pressure drop across the same exchanger. This is easily explained by the fact that with a larger pressure drop, there exists more flow resulting in more fluid turbulence within the heat exchanger. This is so because the fluid’s heat transfer film coefficient increases when subjected to increased turbulence. With increased pressure drop you have more energy consumed; however, the trade-off (there always is one) is that the heat transfer is increased.

When I first read boback's post, my immediate thought was the same as the one you've stated above. But now I imagine that there are two sides to the coin. With larger pressure drop, both the below occur:
(a) they way you've explained, more turbulence allowing increased rate of heat transfer within the region where heat transfer takes place; and
( [i'm assuming for the same motor power], the flowrate would be decreased due to higher pressure drop. This decreases the chance of maintaining a large delta T at both the inlet and outlet of the heat exchanger (i.e. the region where heat transfer takes place). When delta T is smaller, the rate of heat transfer naturally decreases.

Without balancing between the above two and contradictive effects, we'd really do not know what actually happens to the heat transfer rate with increased pressure drop.