Pump & Compressor Sizing

Pump Sizing (Incompressible Flow)

Purpose:

In this exercise we will obtain enough information that we can get a reasonable idea of the cost of the pumps well need to build the process. As you know from the CAP COST program we need to determine the:

Type of Pump

Once the type of pump is determined the following information is also required:

Shaft Power

Pressure at Suction

Materials of Construction

Because CAPCOST does not require it this document intentionally does not address the important subject of pump Net Positive Suction Head (NPSH). The reader is encouraged to refer to the Turton for some background on this subject.

Pump Sizing and Selection:

The type of pump best suited for the application is usually selected based on the required flow rate and the differential pressure (or differential head). The heuristics in Turton (and in the rules of thumb section) provide some guidelines on making the appropriate selection.

The proper choice of materials of construction will be covered in a later exercise.

The Pressure at the suction (or "suction pressure" as it often called) is usually easy to determine from the pressure profile you've already done. It will be accurate enough for the level of detail our estimate requires. However, you should understand the assumptions we're making when we make that sweeping statement so that you can apply your engineering judgement to situations that invalidate those assumptions. The assumptions are:

- The frictional pressure drop of the suction piping is negligible as compared to the frictional pressure drop of the discharge piping. This is because the suction piping will be oversized, it will be as short as possible, it will have as few fittings (elbows, valves etc.) as possible and there won't be any equipment (ie. Exchangers) or control valves in it. All of these reasons are for the purpose of maximizing the allowable NPSH (Net Positive Suction Head). See Turton pg 306 for an elaboration on NPSH.

- What little frictional pressure drop there is will be offset by the elevation change of the suction line.

The shaft power however is going to dictate the overall size and cost of the pump, and it's going to require us to do some calculations. But first, what is shaft power? It's a measure of the energy the motor must put into the pump to increase a certain volumetric flow by a certain pressure. Here's what you need to know to calculate it

Shaft Power (also known as Brake Horsepower or BHP) = Hydraulic Power / ( Hydraulic Efficiency * Motor Efficiency )

(We'll use 50 % efficiency for our pumps (hydraulic & motor efficiency combined), which is reasonable of most 100 US gpm or less pumps.)

Hydraulic Power (Horsepower) = Flow rate (US gpm) * Pressure Rise across pump (psi) / 1714 (conversion factor)

Or

Power = volumetric flow rate * pressure differential (use consistent units)

Determining the volumetric flow rate

Since we have prepared a simulation of the process we have a good idea of the flows which will be encountered in the process. It's advisable to increase the flow rates in the process by 20% to account for any future expansion that might be required and to ensure small errors in our calculations don't impact the ability of the process to operate when it's built. We call this the "rated flow" of the pump.

As you shall see in examples below, if there are multiple destinations for the fluid (i.e. 3 m3/hr to destination A, 7 m3/hr to destination B, and 6 m3/hr to destination C) then the pump volumetric flow should be the sum of these flows.

All that's left to figure out now is the pressure rise across the pump.

Determining the required pump pressure rise:

This is where well spend most of our time doing the pump sizing calculation and its where most people have the greatest difficulty. However if we approach the problem systematically we'll find the calculations are straightforward.

If you keep the following points in mind the math is simple.

Since pumps are "differential" devices treat the suction side of the pump independently of the discharge side.

If the liquid is moving upwards (against gravity) well need more differential pressure from the pump.

If the liquid is moving downwards (with gravity) well need less differential pressure from the pump.

The friction of the piping (also control valves and equipment i.e. heat exchangers) always causes us to need more pressure from the pump.

Lets look at a diagram that illustrates these points.

Example #1

Pump Suction Side

Its common (but not always i.e. in the case of wells, sumps or pits) for the liquid on the pump suction to be above the centreline of the pump. This height difference is called suction static head (if the liquid level was below the pump its usually called static lift). In this example, since the liquid level is above the pump suction we have gravity helping us out, so the required differential pressure of the pump is reduced by the amount of static head we have (make sure this is clear in your mind before you continue). Also, when a fluid is flowing in a pipe the frictional resistance causes the pressure to drop along the length of piping.

Therefore, in this case, we have the effects of gravity and friction working against each other. To calculate the pressure at the pump suction we only need to know the pressure in the tank, add the static head (since the fluid is flowing downwards), and subtract the frictional losses.

The Suction Pressure of the pump is:

Suction Pressure = Source Pressure (i.e. pressure in drum) + static head (if the liquid level is above the pump, subtract if its static lift) - piping line losses

As mentioned in the rules of thumb section, for our purposes, you can assume the suction pressure is the source pressure (ignore the static head and piping line losses).

There are a couple of other notable features here. First when we look at what happens when flow rates increase or decrease you should understand that the static head does not change with flow rate (make sure this is clear in you mind before you continue its a common mistake).

Pressures due to Static Head do not change when the flow rate changes

However the frictional losses do change with flow rate. In reality, this is just the definition of an item that is static or frictional. From now on youll need to apply your intuition to think about whether the flow rate will cause a change in pressure differential to all the various items you would find in a pipe system (elbows, filters, valves, etc). If you're not sure there's an excellent book commonly referred to as the Crane Manual (entitled Crane technical manual #410) that can help you.

A Second notable feature is that the suction side static head is measured from the pump centre line to a point that is usually the bottom of the tank. Why not up to the normal liquid level (50% up the tank) in the tank, as most texts will indicate?

First, its always best to be conservative, think about the equation and look for the conditions that give you the worst case (highest required differential pump pressure). If the lower liquid level would require a larger pump (and it does in this case) then use this value. Second, although during normal operation the liquid level spend a majority of the time at the 50% tank level there may be times when the tank will need to be pumped out, or possibly a process upset would cause the liquid level to drop to the bottom of the tank. These are all possible operating conditions which the pump must satisfy for proper operation. At the same time, its important that the pump doesnt run dry. As a general rule pumps should never be allowed to operate without liquid inside to provide cooling and lubrication. Running a pump dry can often lead to a seal failure or bearing problem. We nearly always ensure a pump has liquid in it by providing some sort of control system which maintains a liquid level in the tank and also an interlock that shuts the pump down when the liquid level drops below a predetermined value. This minimum allowable liquid level would be estimated by us at this time and then finalized when we have selected the pump and understand its NPSH limitations (for more details on NPSH refer to Turton).

Pump Discharge Side

The final destination of a pump can be above, below or at the same level of the pump. If the destination is below the pump you should be weary of the fact that the liquid may syphon through your piping system when the pump is shut off (this can provide unexpected and possibly undesirable effects). In this case, we are pumping the liquid up to the top of a vertical vessel so there is gravity that well need to be working against. Also, there is frictional pressure drop in the piping. Therefore you can see that the pressure at the discharge of the pump is:

Pump Discharge Pressure = Destination Pressure (psig) + static head (psi) + frictional losses (psi)

Note the sign change of the static head in this equation between the pump suction and discharge. This throws a lot of first time pump designers off. If its not clear why the sign has changed go back and study the diagram and ensure you understand the reasoning.

Normally the pump system will have frictional losses which include piping, control valves, heat exchangers etc. The pressure drop of these types of components varies with flow rate. Occasionally you will encounter filters in the pump discharge. How do you think the pressure drop of a filter varies?

Solution:

Filters are like pipes in that their pressure drop will vary with flow rate (to the square of the volumetric flow rate see the rules of thumb section). However filters start off clean when theyre first installed and even though the flow may never change they gradually increase in pressure drop over time as they become dirty. A properly designed pump system will accommodate the fact that the filters pressure will vary from 2 to 15 psi differential (or whatever the particular filter is designed for) simply from the fact that they clog over time. As the designer you need to determine which of these two values to use when looking for worst case scenarios. For example you may to assume 15 psi differential to calculate the pump discharge pressure, on the other hand may need to assume 2 psi differential when calculating the worst case control valve pressure drop (more on the control valve in the next example).

That's the end of example number 1. We should now have an understanding of how to calculate the differential pressure of a pump based on static head and frictional losses.

Example #2

The following diagram illustrates a more difficult example of pump sizing problem. In this problem there are two destinations for the fluid and therefore apparently two different pump discharge pressures. A pump with two discharge pressures is impossible and therefore the key to sizing a pump for this situation is to understand which components pressure drops can vary and which ones are fixed.

The suction side is the same situation as discussed in the example above so we don't need to go into any detail here.

The discharge piping has two control valves in it but notice that the flow splits. What might not be so obvious is the fact that if one column operates at 5 bar(g), and the other 9 bar(g) there's a large difference in discharge pressure the pump is required to produce for either of the columns. However, there is only one pump and so it must be providing enough pressure to meet the process with the greater pressure requirements.

Does this mean we feed the 5 bar(g) column with 9 bar(g) fluid? No, it means that the control valve going to this distillation column has more pressure drop (about 4 bar more) than the control valve for the 9 bar(g) process.

To understand this conclusion you can only do so by realizing what pressures in the system are allowed to 'float' and what pressures are fixed. In this case we rather arbitrarily declared the pressure in the destinations were fixed. There will always be a pressure in the system that is fixed, but it may not be as obvious as in this case. Here are some guidelines to help decide which pressures have precedence over others (in order of greatest precedence).

1. If a tank or pipe is open to atmosphere then it will always be at atmospheric pressure.

2. A pressure control system will maintain a pressure at the point at which the pressure is measured very close to the 'set point' of the controller.

3. Occasionally situations will arise where the vapour pressure of the liquid inside a vessel will fix the operating pressure of the vessel and become a fixed pressure. For instance a tank half filled with water and containing nothing but water vapour in the space above the liquid will have a pressure equivalent to the saturated conditions of the water. For example, a tank with water at 100 deg C will have a pressure of 1 atmosphere inside it. Other gases, usually non-condensables, in the vapour space will invalidate this statement.

4. Every other point in a process has a pressure which floats based on the frictional and static losses in the piping.

To find a pressure on the discharge of a pump you'll need to look forward into the process until you find a fixed pressure and then work backwards accounting for frictional and static losses until you come back to the pump discharge.

Control Valves

Determining the pressure at a point in the piping becomes more complicated when control valves are present. To explain why we need to understand what a control valve does and how it works.

You can think of a control valve as being a long piece of pipe. A pipe which you can vary the length of at any time.

A control valve can be thought of as a long piece of pipe with a variable length

The pressure drop through a valve changes as the flow varies, just like a piece of pipe. Valves could be sized in terms of 'equivalent pipe length.' (i.e. 50 ft of equivalent piping pressure drop) But since the valve is adjustable we would have to say the valve has an equivalent pipe length of 5 to 50 ft. It's inconvenient to use this sizing system since we would also have to state the diameter of the pipe and it's friction factor etc. so instead we have a term we call Cv.

Cv = flow * sqrt(fluid specific gravity / pressure drop)

The normal units of this empirical equation are:

Flow in US gal / min, and

Pressure Drop in psi (sorry metric).

Every control valve will have a minimum and a maximum Cv which it can be operated with. The minimum Cv is when the valve is nearly closed so there is minimal flow and maximum pressure drop across the valve. Conversely when the valve is at it's maximum Cv there is maximum flow and minimum pressure drop across the valve. When you open the tap on your sink you are increasing the Cv of the valve.

For any given Cv there's an infinite number of flows possible, given that the pressure drop is allowed vary, just like a pipe. Said another way, if the pressure drop across the valve is allowed to increase the flow will increase and vice versa (just like a piece of pipe). However, In the case of a tap on your sink you can turn the handle, this varies the Cv of the valve and the flow changes, but the differential pressure (water supply pressure - room pressure) stays relatively constant. On the other hand, if we vary the valves' Cv we can maintain a constant flow across it even though the pressure drop has changed (as you might if someone flushed the toilet and the supply pressure dropped). This last sentence is how we solve the problem we presented above of having one pump with different destination pressure requirements.

Back to Pump Sizing

In summary, to solve these types of pump systems, you first need to determine the required pump discharge pressure for both processes (treat them separately at two independent cases). Then choose the higher discharge pressure of the two (the worst case). Next, go back and put the necessary pressure drop across the other control valve to satisfy the fact that the two control valves will have the same inlet pressure.

The pump differential pressure is:

Differential Pressure = Discharge Pressure - Suction Pressure

The Discharge Pressure of the pump is:

Discharge Pressure = Destination Pressure (i.e. column operating pressure at its feed point) + static head (elevation change) + piping line losses + control valve losses + heat exchanger losses + anything else that could cause pressure drop

If, as in above, there is more than one destination, you need to evaluate both destinations and choose the one with the highest discharge pressure for the pump sizing.

By this time we should have an understanding of how to calculate pump discharge pressures for most situations. All that remains is the mechanics of calculating static head and frictional losses for piping and components.

Calculating differential pressure from static head

You can convert static head (elevation change) to pressure using rho*g*head, or:

pressure (psi) = elevation change (ft) * 0.43352 * specific gravity of fluid relative to water

Static head is relatively easy to determine given a decent equipment elevation view, and only slightly more complicated given an equipment list with vessel sizes.

Calculating frictional loss of components

The next item to calculate is the frictional losses and especially the piping pressure drop. In industry a rough measurement of the line length based on the plot plan or piping layout drawings is often utilized. To speed up our project, for our purposes we're going to make an assumption about the pressure drop of the piping. For all pipes inside the process unit, assume there is 15 psi pressure drop.

A Tip And Trick Every Engineer Should Know About Pressure Drop

The frictional pressure drop of piping and fittings and most everything is based upon the square of the velocity or flow rate (everything else, like density or viscosity, being equal).

So, if you know the pressure drop of an item is 5 psi at 10 gal/ min then the pressure drop at 15 gal/min will be close to 5 * (15/10)^2 = 11.25 psi.

For any items that might be in the discharge of the pump you can apply the following rules of thumb.

PRIVATEControl ValvesTo make the calculations easier we're going to assume a minimum of 10 psi pressure drop for every control valve.

Minimum Flow BypassesA minimum flow bypass is often installed to prevent control valve(s) which could completely close from damaging the pump (vibration, overheating etc. due to not having enough fluid flowing through the pump).

Each pump is slightly different in terms of it's minimum flow requirements but for our purposes we shall assume that all centrifugal pumps need at least 15% of the pumps 'rated' flow passing through the minimum flow bypass.

Minimum flow bypasses are usually a piece of pipe with an orifice plate installed that connects the pump discharge to the tank on it's suction. The orifice plate provides the required differential pressure between the pump discharge pressure and the tank pressure.

Flow MetersAssume 3 psi for every flow meter.

Heat ExchangersAs a guideline, assume 10 psi pressure drop across an exchanger at rated flow. Use a heat exchanger sizing program if you need more accurate values.

FiltersFilters usually get changed when the differential pressure exceeds 10 to 12 psi. Ultimately, it's the mechanical strength of the filter media that determines the maximum differential pressure allowed (at some diff. press the media will collapse).

Because of the relationship between pressure drop (differential pressure) and volumetric flow we can draw a graph which we call the "system curve" that relates the flow pressure drop through the piping to any flow rate.

If there was a control valve in the system we could draw a series of different curves each with a different valve opening (Cv setting).

This curve would be much more useful if we had a similar relationship for the pump plotted with it. Where the two curves intersect it would determine the actual operating flow and pressure. Refer to Turton Section 12.2 for an elaboration on this subject.

Example #3

The pump is a reflux pump as in example number 2. The simulation indicates the fluid has a specific gravity of 1.0, the suction supply pressure is 1.5 bar(g), there is 50 ft of static head to destination #1 with a required flow of 20 m3/hr and 80 ft of static head to destination #2 with a required flow of 50 m3/hr. What is the required input to CAPCOST?

Solution:

1. Determine the flow rates

We'll make the assumption this is a centrifugal pump first (then check that assumption using the heuristics). Given that it is a centrifugal pump and there are control valves that might shut off we should provide a minimum flow bypass.

A sketch of the proposed system:

The rated flow rate is:

Rated flow to Destination A = 20 m3/hr * 120% = 24 m3/hr

Rated flow to Destination B = 50 m3/hr * 120% = 60 m3/hr

Pump Rated flow = (20 m3/hr * 120% + 50 m3/hr * 120%) / (1 - 15%) = 98.8 m3/hrIf the destinations are operating at their normal flow rates then the resulting flow through the minimum flow bypass must be:

Min flow Bypass Normal flow = 98.3 m3/hr - (20 m3/hr + 50 m3/hr) = 28.3 m3/hr

2. Calculate the Suction Pressure

Since the simulation data indicates the vessel is operating at 1.5 bar(g) then we'll assume the static head and frictional losses offset each other and the pump suction pressure is 1.5 bar(g).

3. Calculate the Pump Discharge Pressure

Since this pump has 3 destinations we'll work out each one independantly

Destination 1:

Discharge Pressure = destination pressure (5 bar(g) ) + Static Head (50 ft * 0.43352 * spec grav 1.0 = 21.6 psi = 1.49 bar) + Piping Line Losses (15 psi = 1.03 bar) + Control Valve Losses (minimum 10 psi = 0.69 bar)

Min Discharge Pressure to satisfy Destination 1= 8.2 bar(g)

Destination 2:

Discharge Pressure = destination pressure (9 bar(g) ) + Static Head (80 ft * 0.43352 * spec grav 1.0 =34.68 psi = 2.4 bar) + Piping Line Losses (15 psi = 1.03 bar) + Control Valve Losses (minimum 10 psi = 0.69 bar)

Min Discharge Pressure to satisfy Destination 1= 13.1 bar(g)

Destination 3 (Minimum Flow Bypass):

The minimum flow bypass has a restriction orifice in it that acts like a very long piece of pipe. It's treated differently from the other destination calculations since the pressure drop of the orifice can be almost any value. Its pressure drop is unknown at this point in the design, but can instead be calculated later. For now what we need to know is that the minimum flow bypass will never dictate the pump discharge pressures.

The worst case discharge pressure (the highest value) is 13.1 bar(g).

Knowing this we can now calculate the required control valve pressure drop on the destination 1 line and the pressure drop of the minimum flow bypass orifice.

Destination 1 control valve pressure drop is:

Control valve DP = Pump discharge pressure (13.1 bar(g) ) - Static Head (50 ft * 0.43352 * spec grav 1.0 = 21.6 psi = 1.49 bar) - Piping Line Losses (15 psi = 1.03 bar)

Destination 1 Control valve DP = 10.6 bar

Minimum flow bypass Orifice Plate Pressure drop is:

Orifice DP = Pump discharge Pressure (13.1 bar(g) )- Vessel Operating Pressure (1.5 bar(g) ) - Line Losses (15 psi = 1.03 bar(g) )

Orifice DP = 10.6 bar(g)

3. Determine the pump differential pressurePump DP = Pump discharge Pressure (13.1 bar(g) - Pump Suction Pressure (1.5 bar(g) )

PumpDP = 11.6 bar4. Determine the pump hydraulic horsepower

HHP = Pump DP(11.6 bar) * Rated Flow rate (98.8 m3/hr)

HHP = 31.6 kW

5. Determine the Shaft Input Power

Shaft Power = Hydraulic Horsepower (31.6 kW) / hydraulic and motor efficiency (50%)

Shaft Power = 63.2 kW6. Check to ensure a centrifugal pump is appropriate

Hydraulic Head = pump differential pressure / rho g = 11.6 bar / (1.0 gm/cm3 * 9.8 m/sec2) = 118 m

According to the heuristics this head is in reason for a "single stage" centrifugal pump

Flow = 98.8 m3/hr = 1.5 m3/min

According to the heuristics this flow is reasonable for a "single stage" centrifugal pump.

Thus we can assume a centrifugal pump is an appropriate choice for the application.

7. Input for Capcost

Pump Type: Centrifugal

Shaft Power: 63.2 kW

Pressure At Suction: 1.5 bar(g)

Number of Spares: 1 (chosen due to the service being a reflux pump)

Summary:

That's it, pump sizing in a nutshell! Clearly we haven't covered every possible situation you might encounter in sizing a pump, but the principals are here. We've avoided the need to discuss NPSH because CAPCOST doesn't require this as input. Expectations:

I expect to see your calculations documented in some manner. Copy the relevant information from the spreadsheet into the equipment list.

For the Octene pump you should hand in detailed calculations (by hand or using Mathcad). The calculations should include a sketch like the one above.

Grading:

I will be looking for accuracy in your calculations (or reasonableness in your assumptions) and enough documentation that I can follow what you have done. In the end though, the pump calculation spreadsheet should be complete, and the equipment list should contain the necessary information.

Compressor and Blower (compressible flow) Sizing

Background:

The CapCost program requires that you first determine the type of compressor, blower or fan. Then, depending on whether the unit is a "fan" or a "compressor" you will need

gas flow rate (m3/s) and pressure rise (kPa), or

Shaft Power (kW), and

the materials of construction.

Purpose:

To determine enough information that CAPCOST can be used to estimate the cost of the compressor or fan/blower.

Expectations:

Use your process simulator and the efficiencies in the rules of thumb to develop the kW requirement for the compressor.

EMBED Excel.Sheet.8

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