# Difference between a fan and a pump?

Not sure if this is in the right topic, but does anyone know the difference between a fan and a pump? I'm working on something where I need to distinguish a fan from a pump. I need to do so without limiting the pump to any specific type. Basically I need to allow for a piston & cylinder type, screw type, impeller, diaphragm type and any other type of pump that isn't a fan. I've looked up definitions and I'm having trouble finding ways to differentiate the two. My ideas are to include so other structure such as a housing or shroud such that pressure or suction is created, and maybe focus on what is being moved. The problems I see here are liquids and gases are both fluids, and also, not all types of pumps have a shroud or real housing. Anyways, here are the definitions I found so far. Anybody got any ideas or better definitions???

Main Entry: pump
1: a device that raises, transfers, or compresses fluids or that attenuates gases especially by suction or pressure or both
2: a mechanical device that moves fluid or gas by pressure or suction

Main Entry: air pump
1: a pump for exhausting air from a closed space or for compressing air or forcing it through other apparatus

Main Entry: fan
1: an instrument for producing a current of air: as a : a device for cooling the person that is usually shaped like a segment of a circle and is composed of material (as feathers or paper) mounted on thin rods or slats moving about a pivot so that the device may be closed compactly when not in use b : a device that consists of a series of vanes radiating from a hub rotated on its axle by a motor c slang : an airplane propeller
2: a device for creating a current of air by movement of a surface or surfaces

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A fan is a compressor and a compressor is a device which moves fluids in a gaseous state.

A pump is device which is used to move a liquid.

Both devices work in similar manners in that one design can be used to move a liquid or a vapor---a screw type device for instance is used in automobiles as a blower and but used on farms as irrigation pumps. The device design itself does not differentiate one type of system from another; however, the state of the fluid being moved is used to demarkate between the two catagories.

Most thermodynamics text books will have some form of definition if you want a more 'engineering' definition.

Hope this helped a little.
Good luck.

If a pump is a device to move liquid, then what exactly is an air pump?

clbarnes said:
If a pump is a device to move liquid, then what exactly is an air pump?
A misnomer. A compressor.

russ_watters
Mentor
I tend to disagree. Pumps are generally designed to create pressure and fans to create flow. But depending on the application, those roles can switch. Ie, in large HVAC systems, air handlers are used more like pumps than fans - their controls regulate pressure and not flow. According to the dictionary, the word "fan" only applies to gases and the word "pump" can apply to a gas or liquid. So I guess that makes fans a subset of pumps.

Besides, at the pressure most fans work at, air is essentially incompressible, meaning there is no functional difference between a gas and a liquid.

A lot of this is just a matter of convention though....

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russ_watters said:
I tend to disagree. Pumps are generally designed to create pressure and fans to create flow. But depending on the application, those roles can switch. Ie, in large HVAC systems, air handlers are used more like pumps than fans - their controls regulate pressure and not flow. According to the dictionary, the word "fan" only applies to gases and the word "pump" can apply to a gas or liquid. So I guess that makes fans a subset of pumps.

Besides, at the pressure most fans work at, air is essentially incompressible, meaning there is no functional difference between a gas and a liquid.

A lot of this is just a matter of convention though....
Well I guess we'll disagree here then because I've worked in many places where the purpose of the 'pump' was to create 'flow' as in the feed pump for a steam generator. Yeah, the pump had to boost the pressure to the operating pressure of the S/G but more importantly, the pump had to match the volumetric flow rate of the S/G. Also for automtive applications, fuel pumps are rated less on pressure and more on flow meaning for an FI application as long as pressure stays about 40psi(or more) the regulator will will keep pressure at 40psi but the pump also needs to maintain a flow rate to keep the old engine running.

Meh, like you said we're just arguing semantics now.

russ_watters
Mentor
Yeah, it is a little bit of semantics, but I'd like to add one more little thing: while depending on the application, a pump may be sized/selected based more on flow than pressure, all pumps (that I know of) are pressure generators, whereas not all fans are. Bladed fans have virtually no ability to generate pressure, and the ability to generate pressure is why centrifugal fans are used in HVAC. The ability to generate pressure is also why most pumps are centrifugal.

Well, two things: something I've found in my job is that pressure is very often neglected, even by engineers. In a recent job I did some troubleshooting on, a condensate return pump 10 feet below the top of a low-pressure boiler was sized for 80 feet of head by a professional engineer, when it really only needed about 25 - he saw 30 gpm and selected the pump. Unfortunately, with no backpressure, the pump ended up putting out too much flow and overamped its own motor.

russ_watters said:
Yeah, it is a little bit of semantics, but I'd like to add one more little thing: while depending on the application, a pump may be sized/selected based more on flow than pressure, all pumps (that I know of) are pressure generators, whereas not all fans are. Bladed fans have virtually no ability to generate pressure, and the ability to generate pressure is why centrifugal fans are used in HVAC. The ability to generate pressure is also why most pumps are centrifugal.

Well, two things: something I've found in my job is that pressure is very often neglected, even by engineers. In a recent job I did some troubleshooting on, a condensate return pump 10 feet below the top of a low-pressure boiler was sized for 80 feet of head by a professional engineer, when it really only needed about 25 - he saw 30 gpm and selected the pump. Unfortunately, with no backpressure, the pump ended up putting out too much flow and overamped its own motor.
All fans are pressure generators too. The motive force for the flow through a fan or blower is a pressure difference. The magnitude of pressure generated by a fan is less than that produced by a centrifugal pump moving water or a positive displacement compressor; however, fans still produce a pressure differential.

I mean air moving at 30m/s only has a DP of:

$$v^2=\frac{2\Delta P}{\rho}$$

thus:

$$\Delta P=\frac{v^2\rho}{2}=\frac{(30m/s)^2(1.169)}{2}=0.526kPa$$

or for those cracy Americans:

To make a wind of 67 MPH one only needs .07psi of DP.

Not a whole lot of DP is needed to move air but a DP is needed none the less. A pressure differential was generated.

I agree that various terms used inadvertently for various fluid moving devices is purely semantics. All the fluid moving devices develop or don't develop pressure depending upon the connected system characteristic. If the discharge is open to atmosphere, as the system resistance is hypothetically zero, no pressure is developed and this indicates runout condition of a pump or a fan. The maximum pressure developed will be at shut off condition where the system offers the maximum resistance that the fan or pump can handle. It's not true to say that any device should be selected based on either flow or pressure. The selection should always be at the point where the system resistance curve meets the performance curve of a fluid moving device.

All centrifugal fluid moving devices develop pressure at the expense of centrifugal force. For same specific speed, the pressure developing capacity of a centrifugal fluid moving device depends upon the geometry of the device. For example, axial flow fans have higher capacity than the radial fans. Backward curved radial fans develop higher pressure that that of forward curved fans(with same specific speed).

Either a pump or a fan doesn't develop pressure if it is not required and this is purely a system characteristic.

FredGarvin
You beat me to it Quark.

Pumps and fans do the same thing. They impart energy to a fluid. They do not impart pressure. The pressure is a resultant of the the resistance to the flow created by the pump/fan by the system it is connected to.

I would have to say that the nomenclature depends heavily on the usage not on a rigid definition seperating the two components. I think we're also fighting a long time period of slang usage and personal preferences that have made their way into regular usage over time.

Thanks for the help guys.

Clausius2
Gold Member
faust9 said:
I mean air moving at 30m/s only has a DP of:

$$v^2=\frac{2\Delta P}{\rho}$$

thus:

$$\Delta P=\frac{v^2\rho}{2}=\frac{(30m/s)^2(1.169)}{2}=0.526kPa$$
This makes no sense. Where does it come from? Can you elaborate why you equal such pressure differential to kinetic energy? I will be glad to hear it.

Depending on how are designed, fans and pumps can create an STATIC pressure differential (which is usually very small), or merely they move fluids.

I think it was discussed here: https://www.physicsforums.com/showthread.php?t=77131&highlight=stagnation+enthalpy+pump

Clausius2 From there said:
Engineers usually talks that way: pumps causes an increasing in kinetic energy and not a pressure increasing. Such pressure increasing is due to the hydraulic circuit downstream them.

In fact, pumps and compressors cause an increasing of stagnation enthalpy. In liquids, the increasing of stagnation enthalpy approximately is equal to an increasing of stagnation pressure. Therefore, you can obtain an increasing of stagnation pressure no matter static pressure remains constant, by varying kinetic energy. (stagnation pressure: ). You can check it in the next way: take a pump attached upstream to some pipe through which flows water. Leave downstream outlet pipe being discharging to an atmosphere of uniform pressure. Well, you should realise the static pressure remains constant from upstream to downstream pipelines. But there have been some impulsion or change in kinetic energy. What happens if there is some hydraulic circuit downstream the pipe?. Hydraulic actuators usually provide pressure constraints to the circuit. In Hydraulic Science, pressure is imposed by the circuit and actuators, and not by the pump. One have to choose a pump accordingly with the maximum pressure developed in the circuit, which might be provided by external loads acting upon actuators (pistons).

This is not completely possible with compressors. Gases flowing at moderate Mach numbers (I am not including Pneumatic applications), have pressure and kinematic fields coupled. So you cannot govern pressure and speed by separate. Also, the example I have given you above couldn't be possible with gas flow, because it could be a backflow due to a compressible behavior. Therefore, compressors usually change velocity and static pressure.

Clausius2 said:
This makes no sense. Where does it come from? Can you elaborate why you equal such pressure differential to kinetic energy? I will be glad to hear it.
It comes from Bernoulli's equation. The equation is used in some form or another ($v=k\sqrt{\Delta P}$) for Pitot tubes, venturis, orifices and flow rates of gases across low pressure compressors like the air pump used to feed additional O2 into the exhaust stream---at least in the part of the auto industry I work in.

Clausius2
Gold Member
faust9 said:
It comes from Bernoulli's equation. The equation is used in some form or another ($v=k\sqrt{\Delta P}$) for Pitot tubes, venturis, orifices and flow rates of gases across low pressure compressors like the air pump used to feed additional O2 into the exhaust stream---at least in the part of the auto industry I work in.
Bernoulli equation CANNOT be employed along an stream which passes through a fan, pump, compressor, turbine or any device which extracts/adds mechanical energy to the fluid. At least it CANNOT be employed in a reference frame attached to ground, but it CAN be emplyed in a reference frame rotating with the machine rotor. In this last case you should have employed flow velocities RELATIVE to the fan blade, which you haven't done.
Although it will give you a figure, that figure has a 100% of error. Think of it, the proper jump of pressures should be based also in the power of the turbomachine, which is not shown in your formula.

faust9 said:
thus:

$$\Delta P=\frac{v^2\rho}{2}=\frac{(30m/s)^2(1.169)}{2}=0.526kPa$$.
I agree with Clausius2 here. The Delta P is referring to pressure at a point; the dynamic pressure. It is the difference between static and stagnation pressure. Expanding its use to "across" something, specifically a pump would not be appropriate as it is written.

Crane 410 does not deal with flow "across" pumps, neither does my Engineering Fluid Mechanics textbook. I have looked through much of my mechanics trade school literature and tried to find simplified uses of bernouilli's equation in reference to low pressure pumps. No luck. Perhaps the reference that faust9 used makes some "grandiose" assumption that is good for specific applications.This is what the engineering world tends to do. Good thread on definitions though !

Clausius2
Gold Member
pete worthington said:
I agree with Clausius2 here. The Delta P is referring to pressure at a point; the dynamic pressure. It is the difference between static and stagnation pressure. Expanding its use to "across" something, specifically a pump would not be appropriate as it is written.

Crane 410 does not deal with flow "across" pumps, neither does my Engineering Fluid Mechanics textbook. I have looked through much of my mechanics trade school literature and tried to find simplified uses of bernouilli's equation in reference to low pressure pumps. No luck. Perhaps the reference that faust9 used makes some "grandiose" assumption that is good for specific applications.This is what the engineering world tends to do. Good thread on definitions though !
We should clear it up forever and ever. Bernoulli cannot be used along streamlines which are provided with additional energy, as anyone passing through a turbomachine. This statement is true for any inertial reference frame. However, we can write the equation for energy conservation. Neglecting heat flux through turbomachine walls, which it is almost certain at $$Re\cdot Pr>>>1$$ (being Re=Reynolds number and Pr=Prandtl number), then one obtains:

$$W=\dot m (h_{o2}-h_{o1})$$

that is, the power communicated to fluid is equal to the mass flow multiplied by the difference of stagnation enthalpies. Those who have studied something about Turbomachines will know this equation, which belongs to the "energy world" is linked to another so-called Euler Turbomachines Equation which belongs to the "mechanical world" in the design of blades. The equation exposed above is not any Bernoulli equation. For liquids, this difference of enthalpies are almost negligible in most systems (as in an absorption chiller), but it is not zero. In most power plant systems the pump has the mission of recirculating the flow. For instance, in an absorption refrigeration machine the pump must recirculate the flow from the absorber (which is at low pressure) to the generator (which is at high pressure). As I underlined yet before in my self-quotation, pressure is a magnitude externally imposed in hydraulics systems. Pumps do not create main pressure differences, but they works against such pressure differences moving fluid from low pressure zones to high pressure zones. However, when designing a pump, an special blade shape could give you a desired pressure jump. This pressure jump which belongs to blade design, is usually small enough and pressure is ultimately accomodated to the one which reigns in the exhausting atmosphere. Although the work done by moving fluid in a power plant is usually neglected, the pressure jump cannot be neglected too.

As fans are concerned, a fan usually only moves the fluid. But a fan conveniently arranged with some cowling around it, can provoke a jump in static pressure due to the proper blade design. It can be proved via Bernoulli. How? Well, as it can be demonstrated, Bernoulli equation can be employed in non inertial frames rotating at an angular speed $$\omega$$ with the turbomachine rotor. Think of it, from this point of view the fluid only has to undergo with centrifugal forces caused by the rotating motion. Thus the sum of static pressure, dynamic pressure based on the velocity $$w$$ relative to blade and centrifugal pressure is constant along any streamline:

$$\frac{P}{\rho}+\frac{w^2}{2}+\frac{\omega^2r^2}{2}=const$$

for a constant radius $$r=const.$$ the sum of relative dynamic pressure and static pressure is constant. In particular, a fan increases the absolute velocity flow and decreases the relative to blade velocity flow (draw a simple kinematic triangle behind the blade), and therefore increases static pressure. Such increasing of pressure can be modulated for instance being zero. If we put a fan into a room, there will be an slight jump of static pressure which will make the wake to be curved inwards and continuosly stretched as the flow continues downstream. Eventually, the pressure of any transverse section of the wake is almost equal to ambient pressure at moderately large distances downstream the blades. Therefore, no net jump of static pressures are produced.

Hope this may be helped about this stuff, which has been widely discussed in this forum. I will be glad to hear the opinions of those who don't agree with me.

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faust9,

You are just showing the ability, of a pump/fan, to develop pressure at the expense of the velocity. The condition where velocity head is totally converted to pressure head is an assumption with same elevation, no frictional losses in the pump/fan and no mechanical losses. This case can be replicated at shutoff condition for very small amount of time.

What Classius states is the Bernoulli's equivalent of First Law of TD aka Steady Flow Energy Equation and this is the best thing to analyze total energy of a moving fluid.