A pump can move a small amount of liquid at very high pressure.
Another pump can move a huge amount of liquid at relatively low pressure.
Which one requires more power?
You cannot answer that from pressure alone.
You cannot answer it from flow rate alone either.
Pump power depends on both.
The basic relationship is simple:
Hydraulic power = pressure difference × flow rate
In common US field units, one of the most useful equations is:
Hydraulic horsepower = PSI × GPM ÷ 1714
This calculation appears in pumping operations throughout the oil and gas industry.
It can be used when thinking about:
Water injection pumps
Chemical pumps
Transfer pumps
Pipeline pumps
Hydraulic fracturing pumps
Cementing equipment
Produced water pumps
High pressure test pumps
Hydraulic power units
The equation is simple enough to calculate by hand, but interpreting the answer correctly requires understanding the difference between hydraulic horsepower, shaft horsepower, motor horsepower, pressure, head, and efficiency.
What Is Hydraulic Horsepower?
Hydraulic horsepower is the rate at which useful hydraulic energy is being transferred to a liquid.
It depends primarily on:
Pressure increase
Flow rate
If you increase pressure while keeping flow constant, hydraulic horsepower increases.
If you increase flow while keeping pressure constant, hydraulic horsepower also increases.
Double either one and the theoretical hydraulic power doubles.
What Is the Basic Hydraulic Horsepower Formula?
When pressure is expressed in PSI and flow is expressed in US gallons per minute:
HHP = PSI × GPM ÷ 1714
Where:
HHP = hydraulic horsepower
PSI = pressure difference across the pump
GPM = gallons per minute
The 1714 constant handles the required unit conversions.
Why Is Pressure Difference Used Instead of Discharge Pressure?
A pump does not create useful pressure based only on the discharge gauge reading.
It raises pressure from its suction condition to its discharge condition.
Therefore:
Pump differential pressure = discharge pressure minus suction pressure
That differential is what should normally be used in the hydraulic power calculation.
Suppose:
Suction pressure = 100 psi
Discharge pressure = 1,100 psi
Pump differential pressure:
1,100 minus 100 = 1,000 psi
The pump is adding approximately 1,000 psi.
Using 1,100 psi would overstate the pressure increase produced by the pump.
Example: Calculate Hydraulic Horsepower
Suppose a pump delivers:
500 GPM
at a differential pressure of:
1,000 psi
Use:
HHP = 1,000 × 500 ÷ 1714
Multiply:
1,000 × 500 = 500,000
Divide:
500,000 ÷ 1714 = approximately 292
The pump is delivering approximately:
292 hydraulic horsepower
That is the useful hydraulic power transferred to the liquid.
It does not mean a 292 horsepower motor is sufficient.
Efficiency still needs to be considered.
What Is Pump Efficiency?
A real pump cannot convert every unit of shaft power into useful liquid power.
Some energy is lost through:
Mechanical friction
Internal leakage
Fluid recirculation
Disk friction
Bearing losses
Seal losses
Other hydraulic losses
Pump efficiency describes how effectively shaft power becomes hydraulic power.
The basic relationship is:
Efficiency = hydraulic horsepower ÷ shaft horsepower
Therefore:
Shaft horsepower = hydraulic horsepower ÷ efficiency
Efficiency must be entered as a decimal.
For example:
80 percent = 0.80
Example: Calculate Required Shaft Horsepower
From the previous example:
Hydraulic horsepower = 292 HP
Suppose pump efficiency is:
80 percent
Convert efficiency:
80 percent = 0.80
Now:
Shaft horsepower = 292 ÷ 0.80
Shaft horsepower = 365 HP
The pump therefore requires approximately 365 horsepower at the shaft to produce 292 hydraulic horsepower under these assumptions.
Why Is Shaft Horsepower Higher Than Hydraulic Horsepower?
Because the pump loses some energy.
If 365 horsepower enters the pump shaft and only 292 horsepower becomes useful hydraulic power:
Efficiency = 292 ÷ 365
Efficiency = approximately 0.80
That is:
80 percent
The remaining power is lost through the various inefficiencies of the machine.
Is Motor Horsepower the Same as Shaft Horsepower?
Not necessarily.
An electric motor also has efficiency losses.
The motor consumes electrical power and converts most of it into mechanical shaft power.
If the motor is 95 percent efficient, more electrical power must enter the motor than leaves its shaft.
The full energy path is:
Electrical power
Motor shaft power
Pump shaft power
Hydraulic power
Every conversion can involve losses.
How Do You Include Motor Efficiency?
Suppose:
Hydraulic horsepower = 292 HP
Pump efficiency = 80 percent
Motor efficiency = 95 percent
First calculate required pump shaft power:
292 ÷ 0.80 = 365 HP
Then estimate electrical input equivalent:
365 ÷ 0.95 = approximately 384 HP
So approximately 384 horsepower worth of electrical input power is required under these simplified conditions.
How Do You Convert Horsepower to Kilowatts?
Use:
1 horsepower = approximately 0.746 kilowatt
Therefore:
Kilowatts = horsepower × 0.746
For example:
384 HP × 0.746 = approximately 286 kW
The estimated electrical input in the previous example is therefore roughly:
286 kilowatts
How Do You Convert Kilowatts to Horsepower?
Use:
Horsepower = kilowatts × 1.341
For example:
200 kW × 1.341 = approximately 268 HP
These conversions are useful when pump information is provided in horsepower but motors or electrical systems are rated in kilowatts.
Why Does Flow Rate Matter So Much?
Imagine two pumps both operating at:
1,000 psi differential pressure
Pump A moves:
100 GPM
Pump B moves:
1,000 GPM
Pump A hydraulic horsepower:
1,000 × 100 ÷ 1714 = approximately 58 HP
Pump B hydraulic horsepower:
1,000 × 1,000 ÷ 1714 = approximately 583 HP
The pressure is identical.
The second pump requires ten times the hydraulic power because it moves ten times as much liquid.
This is why a pressure gauge alone tells you very little about pump power.
Why Does Pressure Matter So Much?
Now keep flow constant.
Suppose both pumps move:
500 GPM
Pump A differential pressure:
500 psi
Pump B differential pressure:
2,000 psi
Pump A:
500 × 500 ÷ 1714 = approximately 146 HHP
Pump B:
2,000 × 500 ÷ 1714 = approximately 583 HHP
The second pump requires four times the hydraulic power because it develops four times the differential pressure.
What Happens If Both Pressure and Flow Double?
Suppose the original condition is:
500 psi
500 GPM
Hydraulic horsepower:
500 × 500 ÷ 1714 = approximately 146 HP
Now double both:
1,000 psi
1,000 GPM
Hydraulic horsepower:
1,000 × 1,000 ÷ 1714 = approximately 583 HP
Both pressure and flow doubled, but power increased by a factor of four.
This becomes important when operators try to increase both rate and pressure simultaneously.
Equipment can reach its power limit quickly.
How Do You Calculate Hydraulic Horsepower From Barrels Per Minute?
Oilfield pumping rates are often reported in barrels per minute rather than gallons per minute.
One barrel contains:
42 US gallons
Therefore:
GPM = barrels per minute × 42
Substitute that into the normal equation:
HHP = PSI × BPM × 42 ÷ 1714
This simplifies to approximately:
HHP = PSI × BPM ÷ 40.81
This is a convenient field equation.
Example: Hydraulic Horsepower at 10 Barrels Per Minute
Suppose:
Pump rate = 10 barrels per minute
Differential pressure = 5,000 psi
Use:
HHP = 5,000 × 10 ÷ 40.81
HHP = approximately 1,225
The theoretical hydraulic horsepower is approximately:
1,225 HP
That is already a substantial amount of power.
Example: Hydraulic Horsepower During a High Rate Pumping Operation
Suppose pumps are delivering:
60 barrels per minute
at:
8,000 psi differential pressure
Use:
HHP = 8,000 × 60 ÷ 40.81
Multiply:
8,000 × 60 = 480,000
Divide:
480,000 ÷ 40.81 = approximately 11,762
The operation is delivering approximately:
11,800 hydraulic horsepower
That number helps explain why high pressure, high rate pumping operations require large amounts of installed power.
Does That Mean 11,800 Engine Horsepower Is Enough?
No.
Hydraulic horsepower represents useful fluid power.
The engines, transmissions, pumps, and drive systems have losses.
If total conversion efficiency were:
85 percent
Required input horsepower would be:
11,762 ÷ 0.85
This is approximately:
13,838 HP
Actual equipment selection requires appropriate operating margins and manufacturer limits as well.
How Do You Calculate Horsepower From Barrels Per Day?
For continuous liquid systems, flow may be reported in barrels per day.
First convert BPD to GPM.
Use:
GPM = BPD × 42 ÷ 1,440
This simplifies to:
GPM = BPD × 0.02917
Then use:
HHP = PSI × GPM ÷ 1714
Example: Pipeline Pump Horsepower From BPD
Suppose an oil pipeline moves:
20,000 barrels per day
The pump adds:
500 psi
First convert flow:
20,000 × 0.02917 = approximately 583.4 GPM
Now calculate hydraulic horsepower:
500 × 583.4 ÷ 1714
HHP = approximately 170 HP
If pump efficiency is:
75 percent
Required shaft horsepower:
170 ÷ 0.75 = approximately 227 HP
The motor must then be selected based on the complete equipment and design requirements.
Can You Rearrange the Equation to Find Flow Rate?
Yes.
Start with:
HHP = PSI × GPM ÷ 1714
Rearrange:
GPM = HHP × 1714 ÷ PSI
Suppose you have:
500 hydraulic horsepower
and need:
2,000 psi
The theoretical flow is:
500 × 1714 ÷ 2,000
GPM = approximately 428.5
Convert to barrels per minute:
428.5 ÷ 42 = approximately 10.2 BPM
Can You Rearrange the Equation to Find Pressure?
Yes.
Use:
PSI = HHP × 1714 ÷ GPM
Suppose:
Available hydraulic power = 300 HP
Flow = 400 GPM
Pressure:
300 × 1714 ÷ 400
Pressure = approximately 1,286 psi
This is the theoretical pressure increase corresponding to that flow and hydraulic power.
Why Can a Pump Not Produce Maximum Pressure and Maximum Flow at the Same Time?
This question depends on pump type and driver power, but power provides part of the explanation.
Imagine a driver with limited horsepower.
At low pressure, the pump may be able to move a relatively large flow.
As required pressure increases, the power needed at the same flow also increases.
Eventually the driver reaches its power limit.
If pressure must increase further, flow may need to decrease.
This is especially intuitive for positive displacement pumping systems.
What Is a Positive Displacement Pump?
A positive displacement pump moves a relatively fixed volume of fluid for each cycle or revolution.
Examples include:
Reciprocating pumps
Plunger pumps
Gear pumps
Progressive cavity pumps
Diaphragm pumps
The actual flow can vary because of speed, leakage, slip, and other factors, but the operating principle is based on displacement.
Pressure develops in response to resistance downstream.
Does a Positive Displacement Pump Create Pressure?
A positive displacement pump primarily creates flow.
Pressure develops when the system resists that flow.
Imagine pumping into an open tank.
Resistance is low, so discharge pressure may remain modest.
Now close a valve downstream.
The pump continues trying to displace liquid.
Pressure can rise rapidly.
That is why positive displacement systems require appropriate pressure protection.
What Is a Centrifugal Pump?
A centrifugal pump transfers energy to liquid using a rotating impeller.
Its flow and developed head depend on the interaction between the pump and the system.
Unlike a simple positive displacement pump, a centrifugal pump does not move one fixed volume with each revolution.
Its operating point is determined by the pump curve and system curve.
Does the Same Horsepower Formula Work for Centrifugal Pumps?
The pressure and flow relationship still describes hydraulic power.
However, centrifugal pump performance is commonly expressed using head rather than PSI.
A useful equation is:
Hydraulic horsepower = GPM × head × specific gravity ÷ 3960
Where:
Head is in feet
Specific gravity describes liquid density relative to water
This is another common pump power equation.
What Is Pump Head?
Head expresses the energy added to a liquid as an equivalent height of fluid.
It is commonly measured in:
Feet of liquid
or:
Meters of liquid
Head is useful because a centrifugal pump produces approximately the same head for liquids of different densities at the same operating point, while the corresponding pressure changes with fluid density.
Why Are Head and Pressure Not the Same Thing?
Pressure depends on both:
Fluid head
Fluid density
A column of dense saltwater creates more pressure than the same height of a lighter hydrocarbon liquid.
The pressure gradient can be expressed approximately as:
Pressure gradient = 0.433 × specific gravity
in PSI per foot.
Therefore:
Pressure = 0.433 × specific gravity × head
Example: Convert 100 Feet of Water Head to PSI
For water:
Specific gravity = approximately 1
Pressure:
0.433 × 1 × 100
Pressure = approximately 43.3 psi
So 100 feet of water head corresponds to roughly:
43 psi
What If the Fluid Has a Specific Gravity of 0.8?
Use:
Pressure = 0.433 × 0.8 × 100
Pressure = approximately 34.6 psi
The pump can produce the same 100 feet of head while generating less pressure because the fluid is lighter.
Why Does Pumping a Denser Fluid Require More Horsepower?
Consider the head based equation:
HHP = GPM × head × specific gravity ÷ 3960
If flow and head remain constant, increasing specific gravity increases hydraulic horsepower.
A denser liquid has more weight per unit volume.
More power is required to raise or accelerate the same volumetric flow through the same head.
Example: Compare Water and Light Hydrocarbon Liquid
Suppose a pump moves:
500 GPM
against:
200 feet of head
For water:
SG = 1.0
HHP:
500 × 200 × 1 ÷ 3960
HHP = approximately 25.3 HP
Now use a lighter hydrocarbon:
SG = 0.75
HHP:
500 × 200 × 0.75 ÷ 3960
HHP = approximately 18.9 HP
Same flow.
Same head.
Lower hydraulic horsepower because the fluid is lighter.
Does Specific Gravity Appear in the PSI Equation?
Not directly.
The equation:
HHP = PSI × GPM ÷ 1714
already uses pressure.
Fluid density has already influenced the relationship between head and pressure.
If you know actual differential pressure, you do not multiply by specific gravity again.
Doing so would effectively apply the density correction twice.
When Should You Use the Head Equation?
Use the head equation when you know:
Flow in GPM
Head in feet
Specific gravity
The equation is:
HHP = GPM × head × SG ÷ 3960
Use the pressure equation when you know:
Flow in GPM
Differential pressure in PSI
The equation is:
HHP = PSI × GPM ÷ 1714
Both describe the same basic hydraulic power when the inputs are consistent.
Example: Prove the Two Equations Give Similar Results
Suppose:
Flow = 500 GPM
Head = 200 feet
SG = 1
Head method:
HHP = 500 × 200 × 1 ÷ 3960
HHP = approximately 25.25 HP
Convert head to pressure:
Pressure = 0.433 × 1 × 200
Pressure = 86.6 psi
Pressure method:
HHP = 86.6 × 500 ÷ 1714
HHP = approximately 25.26 HP
The tiny difference comes from rounding.
The two methods describe the same physical power.
What Is Brake Horsepower?
Brake horsepower commonly refers to the mechanical power required at the pump shaft.
For a pump:
BHP = HHP ÷ pump efficiency
Suppose:
HHP = 100 HP
Pump efficiency = 70 percent
BHP:
100 ÷ 0.70 = approximately 143 HP
The driver must provide roughly 143 shaft horsepower under those conditions, before considering additional design margins or drive losses.
Why Does Pump Efficiency Change?
Pump efficiency is not necessarily constant.
For centrifugal pumps, efficiency changes with operating point.
A pump usually has a region where it operates most efficiently.
Move too far away from that region and efficiency can decrease.
For positive displacement pumps, leakage and mechanical losses can also change with:
Pressure
Speed
Viscosity
Wear
Temperature
Pump condition
Using one efficiency value for every operating condition can therefore create inaccurate estimates.
Why Is Best Efficiency Point Important?
A centrifugal pump has a best efficiency point where hydraulic efficiency is near its maximum for a particular configuration and speed.
Operating reasonably near this region generally supports better performance.
Operating far away can increase problems involving:
Recirculation
Vibration
Hydraulic forces
Seal wear
Bearing loads
Energy consumption
The lowest power calculation is not the only concern.
Reliability matters too.
How Does Pump Wear Affect Power and Performance?
A worn pump may lose internal efficiency.
For example, clearances can increase.
More liquid can recirculate internally.
The pump may require more shaft power relative to the useful hydraulic output.
An operator might notice that the same equipment no longer develops the expected pressure or flow under conditions where it previously performed well.
Power, pressure, flow, and equipment condition should be considered together.
Can You Estimate Pump Efficiency From Field Measurements?
Yes, if enough information is available.
Suppose you know:
Differential pressure
Flow rate
Shaft power
First calculate hydraulic horsepower:
HHP = PSI × GPM ÷ 1714
Then:
Pump efficiency = HHP ÷ shaft horsepower
Multiply by 100 to express it as a percentage.
Example: Calculate Pump Efficiency
Suppose:
Flow = 700 GPM
Differential pressure = 600 psi
Measured shaft power = 300 HP
First calculate HHP:
600 × 700 ÷ 1714
HHP = approximately 245 HP
Efficiency:
245 ÷ 300 = 0.817
Multiply by 100:
Efficiency = approximately 81.7 percent
So the estimated pump efficiency is roughly:
82 percent
Why Can Motor Amps Increase When Pump Load Increases?
An electric motor must supply the mechanical power demanded by the pump.
If the pump requires more shaft power, motor electrical load generally increases.
Operators may see this as increased:
Current
Kilowatts
Motor load percentage
A sudden change in motor load can therefore provide useful information about pump or process conditions.
Does Higher Discharge Pressure Always Mean Higher Motor Load?
Not necessarily.
Flow also matters.
Remember:
Power depends on pressure difference × flow.
Suppose pressure rises while flow falls dramatically.
The total power requirement may increase, decrease, or remain similar depending on the combination.
This is why pressure alone should not be used to predict motor load.
Example: Higher Pressure but Lower Power
Condition A:
Pressure = 500 psi
Flow = 1,000 GPM
HHP:
500 × 1,000 ÷ 1714 = approximately 292 HP
Condition B:
Pressure = 800 psi
Flow = 500 GPM
HHP:
800 × 500 ÷ 1714 = approximately 233 HP
Pressure increased from 500 to 800 psi.
But hydraulic horsepower decreased because flow was cut in half.
Why Can Closing a Discharge Valve Affect Different Pumps Differently?
For a centrifugal pump, throttling the discharge valve changes the system resistance and moves the operating point along the pump curve.
Flow normally decreases.
The resulting shaft power behavior depends on the pump design.
For a positive displacement pump, closing the discharge path can cause pressure to rise rapidly because the pump continues trying to move its displacement.
These machines therefore require different operating considerations.
Why Is a Relief Valve Important on Positive Displacement Pumps?
A positive displacement pump can continue moving fluid against increasing resistance.
If the discharge becomes blocked, pressure can rise until:
A relief device opens
The driver stalls
A component fails
Another protective system acts
The relief system provides a controlled path to limit excessive pressure.
It should never be treated as an optional accessory.
Can You Calculate Hydraulic Horsepower Across a Control Valve?
The same pressure and flow relationship can estimate hydraulic power being dissipated across a liquid pressure drop.
Suppose a valve drops:
500 psi
at:
300 GPM
Hydraulic power associated with that pressure drop is:
500 × 300 ÷ 1714
This is approximately:
87.5 HP
That does not mean the valve is producing 87.5 useful horsepower.
It means substantial hydraulic energy is being dissipated through the restriction, mainly through turbulence and related losses.
This helps explain why severe pressure reduction can create vibration, noise, erosion, and flashing concerns.
What Happens to Hydraulic Energy Across a Restriction?
Energy is not simply destroyed.
Useful pressure energy is converted into other forms.
Much of it ultimately becomes heat and turbulent motion.
Depending on the fluid and conditions, a large pressure drop can also cause:
Flashing
Cavitation
Gas release
Noise
Vibration
Erosion
A large pressure drop at high flow represents a significant amount of power being dissipated in a small region.
How Much Hydraulic Power Is Lost Through 100 PSI at 1,000 GPM?
Use:
HHP = 100 × 1,000 ÷ 1714
HHP = approximately 58.3 HP
So a 100 psi pressure loss at 1,000 GPM represents about:
58 hydraulic horsepower
At continuous operation, unnecessary pressure losses can translate into meaningful energy cost.
Why Does Pipeline Pressure Loss Increase Pumping Cost?
A pump must replace the energy lost through the system.
Pressure can be lost through:
Pipe friction
Valves
Filters
Strainers
Meters
Heat exchangers
Restrictions
Elevation changes
Other equipment
If a strainer becomes plugged and adds another 50 psi of pressure drop, the pump must supply that additional differential pressure to maintain the same flow.
That requires more power.
Example: Cost of an Extra 50 PSI Pressure Drop
Suppose:
Flow = 2,000 GPM
Additional pressure drop = 50 psi
Extra hydraulic horsepower:
50 × 2,000 ÷ 1714
HHP = approximately 58.3 HP
If pump efficiency is:
80 percent
Extra shaft horsepower:
58.3 ÷ 0.80 = approximately 72.9 HP
That extra load can operate continuously until the restriction is corrected.
This is why pressure differential across equipment can have an energy consequence as well as an operating consequence.
How Do You Calculate Power for a Water Injection Pump?
Suppose an injection pump takes water at:
100 psi suction pressure
and discharges at:
2,500 psi
Flow is:
1,500 GPM
First calculate differential pressure:
2,500 minus 100 = 2,400 psi
Hydraulic horsepower:
2,400 × 1,500 ÷ 1714
HHP = approximately 2,100 HP
If pump efficiency is:
85 percent
Shaft horsepower:
2,100 ÷ 0.85 = approximately 2,471 HP
This shows why high pressure injection systems can require very large drivers.
Why Can Injection Pressure Rise Over Time?
Possible reasons include:
Formation injectivity changes
Scale
Plugging
Filter problems
Tubing restrictions
Changing fluid properties
Increasing required reservoir pressure
Valve restrictions
If flow remains constant while required differential pressure rises, pump horsepower requirement also rises.
The driver may eventually become the limiting factor.
How Do You Know Whether a Pump Is Power Limited?
Suppose the pump needs to increase pressure but the driver is already near maximum load.
The system may not be able to maintain the same flow at the higher differential pressure.
Power is approximately constrained by:
Pressure × flow
If available horsepower is fixed, demanding more pressure reduces the flow that can theoretically be supported at that power.
Actual behavior depends on pump type and performance characteristics.
Example: Estimate Flow at a Power Limit
Suppose a system can provide:
1,000 hydraulic horsepower
Required pressure increases to:
4,000 psi
Calculate flow:
GPM = 1,000 × 1714 ÷ 4,000
GPM = approximately 428.5
Convert to barrels per minute:
428.5 ÷ 42 = approximately 10.2 BPM
If pressure rises to:
5,000 psi
at the same 1,000 HHP limit:
GPM = 1,000 × 1714 ÷ 5,000
GPM = approximately 342.8
That is:
342.8 ÷ 42 = approximately 8.16 BPM
Higher pressure means less available flow at the same hydraulic power.
How Do You Calculate Energy From Horsepower?
Horsepower is power, not total energy.
Energy depends on how long that power is used.
For electrical systems:
Energy in kilowatt hours = kilowatts × hours
Suppose a pump consumes:
250 kW
for:
10 hours
Energy:
250 × 10 = 2,500 kilowatt hours
This becomes useful when estimating pumping energy cost.
How Do You Estimate Pump Electricity Cost?
Suppose:
Electrical demand = 250 kW
Operating time = 24 hours
Electricity price = $0.10 per kilowatt hour
Daily energy:
250 × 24 = 6,000 kilowatt hours
Daily energy cost:
6,000 × $0.10 = $600
If an unnecessary restriction increases power demand, that additional cost continues for every hour the system operates.
Actual utility billing can include other charges, but the calculation provides a useful first estimate.
Why Is Hydraulic Horsepower Useful for Troubleshooting?
It connects three measurements that operators already watch:
Pressure
Flow
Power
Suppose motor load rises.
Check whether:
Flow increased
Differential pressure increased
Both increased
Pump efficiency may have deteriorated
Mechanical problems developed
The hydraulic horsepower calculation provides a quick way to determine whether the increased load is expected from process conditions.
Example: Motor Load Increased After a Flow Increase
Original condition:
Pressure differential = 400 psi
Flow = 600 GPM
HHP:
400 × 600 ÷ 1714 = approximately 140 HP
New condition:
Pressure differential = 450 psi
Flow = 800 GPM
HHP:
450 × 800 ÷ 1714 = approximately 210 HP
Hydraulic demand increased from approximately:
140 HP
to:
210 HP
A substantial motor load increase would therefore be expected.
What If Motor Load Rises but Hydraulic Horsepower Does Not?
That can justify investigation.
Possible causes include:
Mechanical friction
Bearing problems
Pump damage
Alignment issues
Changing efficiency
Instrumentation error
Motor problems
The calculation does not diagnose the failure by itself.
It tells you whether the useful hydraulic workload explains the increased power demand.
What Are the Most Common Hydraulic Horsepower Mistakes?
The first is using discharge pressure instead of differential pressure.
The second is forgetting pump efficiency when estimating shaft power.
The third is applying efficiency in the wrong direction.
To find shaft horsepower:
Divide hydraulic horsepower by efficiency.
Do not multiply.
The fourth is mixing BPM and GPM.
Remember:
1 barrel = 42 gallons
The fifth is using a pressure based equation and then multiplying by specific gravity again.
If actual pressure differential is already known, density is already reflected in that pressure.
Why Does Dividing by Efficiency Increase Required Power?
Suppose:
HHP = 100
Efficiency = 80 percent
If you multiplied:
100 × 0.80 = 80 HP
That would imply the inefficient pump needs less input power than the useful power it produces.
That cannot be correct.
Instead:
100 ÷ 0.80 = 125 HP
The pump requires 125 shaft horsepower to deliver 100 hydraulic horsepower.
A Complete Pump Power Example
Suppose a produced water pump operates with:
Suction pressure = 50 psi
Discharge pressure = 650 psi
Flow = 1,200 GPM
Pump efficiency = 78 percent
Motor efficiency = 94 percent
First calculate differential pressure:
650 minus 50 = 600 psi
Hydraulic horsepower:
600 × 1,200 ÷ 1714
HHP = approximately 420 HP
Now calculate shaft horsepower:
420 ÷ 0.78 = approximately 538 HP
Now estimate electrical input equivalent:
538 ÷ 0.94 = approximately 572 HP
Convert to kilowatts:
572 × 0.746 = approximately 427 kW
So the simplified power chain is:
Hydraulic output = approximately 420 HP
Pump shaft requirement = approximately 538 HP
Electrical input equivalent = approximately 572 HP
Electrical power = approximately 427 kW
This example shows why the motor rating must exceed the theoretical hydraulic horsepower.
What Is the Simplest Way to Remember Pump Power?
Remember two words:
Pressure and flow.
For US field units:
HHP = PSI × GPM ÷ 1714
If you use barrels per minute:
HHP = PSI × BPM ÷ 40.81
Then, if pump efficiency is known:
Shaft HP = HHP ÷ efficiency
Those equations answer a surprising number of practical pumping questions.
Frequently Asked Questions
What is hydraulic horsepower?
Hydraulic horsepower is the useful power transferred by a pump to a liquid.
What is the formula for hydraulic horsepower?
When pressure is in PSI and flow is in GPM:
HHP = PSI × GPM ÷ 1714
Should I use discharge pressure or differential pressure?
Normally use the pressure increase across the pump:
Differential pressure = discharge pressure minus suction pressure
How do you calculate hydraulic horsepower from barrels per minute?
Use:
HHP = PSI × BPM ÷ 40.81
How many gallons are in one barrel?
One oilfield barrel contains 42 US gallons.
Is hydraulic horsepower the same as motor horsepower?
No. Hydraulic horsepower is useful fluid power. Pump and motor inefficiencies mean the required input power is higher.
How do you account for pump efficiency?
Use:
Shaft horsepower = hydraulic horsepower ÷ pump efficiency
Enter efficiency as a decimal.
What does 80 percent efficiency mean?
It means approximately 80 percent of the shaft power becomes useful hydraulic power under the stated condition.
How do you calculate pump efficiency?
Use:
Efficiency = hydraulic horsepower ÷ shaft horsepower
Multiply by 100 for percent.
How do you convert horsepower to kilowatts?
Use:
kW = HP × 0.746
How do you convert kilowatts to horsepower?
Use:
HP = kW × 1.341
What is brake horsepower?
In pump applications, brake horsepower generally refers to the mechanical power required at the pump shaft.
What is pump head?
Head expresses fluid energy as an equivalent height of liquid.
How do you calculate hydraulic horsepower from head?
A common equation is:
HHP = GPM × head in feet × specific gravity ÷ 3960
Why does specific gravity appear in the head equation?
The same height of a denser liquid represents more pressure and requires more hydraulic power at the same volumetric flow.
Do you multiply by specific gravity when using PSI?
Not if you are already using the actual differential pressure in PSI. The density effect is already represented in that pressure.
Does higher pressure always require more horsepower?
Only if the other variables are considered. Power depends on both pressure and flow.
Can pressure increase while horsepower decreases?
Yes. If flow decreases enough, hydraulic horsepower can fall even while pressure rises.
Why does a plugged filter increase pumping power?
A restriction creates additional pressure drop. Maintaining the same flow through that added resistance requires more pump differential pressure and therefore more power.
Why are high pressure pumping operations so power intensive?
They combine very high pressure with high flow. Since hydraulic power is proportional to both, required horsepower becomes large quickly.
What is the most useful pump power equation to remember?
For common US field units:
HHP = PSI × GPM ÷ 1714
Then remember that the driver must provide more power than the hydraulic output because real pumps are not 100 percent efficient.