Hydraulic Learning

Calculating Hydraulic System Heat Dissipation

Hydraulic System Heat Load Calculation

Calculating Hydraulic System Heat Dissipation

Context: Power HydraulicsThe field of engineering dealing with the generation, control, and transmission of power using pressurized fluids..

This exercise guides you through calculating the total heat generated by a typical hydraulic system. Understanding this heat load is critical for selecting the correct cooler (heat exchanger) to maintain a stable operating temperature and prevent fluid degradationThe breakdown of oil properties (like viscosity) due to excessive heat, leading to poor lubrication and system failure., component wear, and system failure. We will analyze a system's power input and its useful work output to determine the waste heat that must be dissipated.

Pedagogical Note: This exercise teaches you to apply the fundamental principles of energy conservation (First Law of Thermodynamics) to a real-world power hydraulic circuit. You will learn that all power not used for mechanical work is converted into heat.

Learning Objectives

  • Understand the sources of heat generation in a hydraulic system.
  • Calculate the total power input to the system (in HPHorsepower, a unit of power.).
  • Calculate the useful power output (work) performed by the system.
  • Determine the waste heat (in HP) and convert it to the required cooling capacity (in BTU/hrBritish Thermal Units per hour, a standard unit for heat exchanger capacity.).

Study Data

We will analyze a hydraulic power unit (HPU) running a single hydraulic motor. The system data is provided below. Assume standard atmospheric pressure.

System Specifications
Characteristic Value
System Type Open-Loop Motor Circuit
Hydraulic Fluid ISO VG 46 Mineral Oil
Target Operating Temp 140°F (60°C)
System Schematic (HPU and Motor)
HPU (Pump) Valves Motor Reservoir (Tank) Input Power p_max: 3000 PSI Q_p: 20 GPM Useful Work p_motor: 2500 PSI Q_motor: 18 GPM Work
Parameter Symbol Value Unit
Pump Flow Rate \(Q_p\) 20 GPM
Max System Pressure (Relief Setting) \(p_{\text{max}}\) 3000 PSI
Motor Operating Pressure \(p_{\text{motor}}\) 2500 PSI
Motor Operating Flow \(Q_{\text{motor}}\) 18 GPM

Questions to Address

  1. Calculate the total input power (in Horsepower, HP) supplied to the fluid.
  2. Calculate the useful power (in HP) being consumed by the hydraulic motor to perform work.
  3. Determine the total waste heat (in HP) generated by the system.
  4. Convert the waste heat from HP to the standard cooling unit: BTU/hr.
  5. If the system has a natural (passive) heat dissipation of 5,000 BTU/hr, what is the additional cooling capacity required from a heat exchanger?

Fundamentals of Hydraulic Power & Heat

In hydraulics, power is the product of flow (GPM) and pressure (PSI). Heat is a byproduct of inefficiency. Any pressure drop not associated with performing useful work (i.e., any inefficiency) generates heat.

1. Calculating Hydraulic Horsepower (HP)
The standard formula for hydraulic horsepower in the US customary system is: \[ \text{HP} = \frac{p \times Q}{1714} \] Where 'p' is pressure in PSIPounds per Square Inch., 'Q' is flow in GPMGallons Per Minute., and 1714 is the conversion constant.

2. Power, Work, and Heat
The First Law of Thermodynamics states energy is conserved. In our system: \(P_{\text{in}} = P_{\text{work}} + P_{\text{heat}}\). Therefore, the heat to be dissipated is the total power put into the system minus the power used for actual work. \[ \text{HP}_{\text{heat}} = (\text{HP}_{\text{in}}) - (\text{HP}_{\text{work}}) \]

3. Converting HP to BTU/hr
To select a cooler, we must convert horsepower to British Thermal Units (BTU) per hour. \[ 1 \text{ HP} = 2545 \text{ BTU/hr} \]

Solution: Calculating Hydraulic System Heat Dissipation

Question 1: Calculate the total input power (in Horsepower, HP) supplied to the fluid.

Principle

The total input power is the maximum power the system can draw, which is defined by the pump's full flow rate at the maximum pressure (set by the relief valve). This represents the total energy being put *into* the fluid.

Mini-Lesson

This step uses the core hydraulic power formula. Power is a rate of doing work. In US customary units, this rate is defined as Pressure (force/area) times Flow (volume/time). The constant 1714 is a 'fudge factor' that combines all the necessary unit conversions (e.g., in³ to gallons, minutes to seconds, ft-lbs to HP).

Pedagogical Note

We use the maximum pressure (3000 PSI) and full pump flow (20 GPM) to determine the total input power. This represents the 'worst-case' energy input the system is designed to handle.

Norms

Per ANSI/NFPA (National Fluid Power Association) standards, system power calculations are fundamental for component sizing and safety analysis.

Formula(s)

We use the main Hydraulic Horsepower formula:

\[ \text{HP}_{\text{in}} = \frac{p_{\text{max}} \times Q_p}{1714} \]
Hypotheses

We make the following assumptions for this calculation:

  • The pump is 100% volumetrically efficient (i.e., it truly delivers 20 GPM at 3000 PSI).
  • The '1714' conversion constant is accurate.
  • The relief valve is the component that sets the maximum pressure.
Data

From the problem statement:

ParameterSymbolValueUnit
Pump Flow Rate\(Q_p\)20GPM
Max System Pressure\(p_{\text{max}}\)3000PSI
Tips

A common mistake is using the motor's operating pressure (2500 PSI) to find the total input power. The input power is always based on the pump's total output capability (max flow at max pressure).

Schematic (Before Calculations)

Visualizing the input power source. The power comes from the pump at the relief valve setting.

Input Power Source
P Pump M RV Set: 3000 PSI To System Q_p: 20 GPM
Calculation(s)

We apply the hydraulic horsepower formula using max pressure and full flow.

Step 1: State the formula

\[ \text{HP}_{\text{in}} = \frac{p_{\text{max}} \times Q_p}{1714} \]

This is the standard formula for hydraulic horsepower.

Step 2: Substitute values and solve

We substitute \(p_{\text{max}} = 3000 \text{ PSI}\) and \(Q_p = 20 \text{ GPM}\) from the data table.

\[ \begin{aligned} \text{HP}_{\text{in}} &= \frac{3000 \text{ PSI} \times 20 \text{ GPM}}{1714} \\ \text{HP}_{\text{in}} &= \frac{60000}{1714} \\ \text{HP}_{\text{in}} &= 35.006 \text{ HP} \end{aligned} \]

This result is rounded to 35.0 HP for simplicity in the next steps.

Schematic (After Calculations)

The result of our calculation:

Calculated Input Power
Total Input Power (HP_in) 35.0 HP
Analysis

The total power being put into the hydraulic fluid by the pump is 35.0 HP (rounded). This is the total energy budget we have. This energy will be consumed as either useful work or waste heat.

Cautionary Points

This is hydraulic horsepower (fluid power), not the electric horsepower at the motor. The electric motor's HP rating will be higher to account for pump inefficiency (e.g., if the pump is 90% efficient, the electric motor needs to provide 35.0 / 0.90 ≈ 39 HP).

Key Takeaways
  • Input Power is based on the system's *maximum* capabilities: \(p_{\text{max}}\) and \(Q_p\).
  • The formula is \(\text{HP} = (P \times Q) / 1714\).
Did You Know?

The constant 1714 is specific to PSI and GPM. If you use SI units (Pascals and m³/s), the formula is simply \(P = p \times Q\) (Power in Watts). 1714 is a "fudge factor" that combines all unit conversions (gallons to in³, minutes to seconds, ft-lbs to HP).

FAQ

[Common follow-up questions]

Why not use the motor's 2500 PSI?

2500 PSI is the pressure needed for the work. The system has to be able to supply 3000 PSI (the relief setting), so the total energy input is based on this 3000 PSI maximum.

What if the pump flow was variable?

We would still use the *maximum* sustained flow and pressure for the worst-case heat calculation. This ensures the cooler is sized for the most demanding part of the work cycle.

Final Result
The total input power is **35.0 HP**.
Your Turn

What would the input power be if the relief valve was set to 2000 PSI (and pump flow was still 20 GPM)?

Memo Card

Question 1 Summary:

  • Concept: Total System Input Power
  • Formula: \(\text{HP} = (p_{\text{max}} \times Q_p) / 1714\)
  • Data: 3000 PSI, 20 GPM

Question 2: Calculate the useful power (in HP) being consumed by the hydraulic motor to perform work.

Principle

Useful power (work) is the power consumed by the actuator (in this case, the hydraulic motor) to move the load. This is calculated using the actual pressure and flow consumed by the motor during operation.

Mini-Lesson

This is the same formula as Q1, but applied to a different part of the circuit. We are isolating the 'work' component. Any power not used for work (e.g., flow over a relief valve, pressure drops in hoses, friction) is 'lost' and becomes heat.

Pedagogical Note

Notice the data we are using: it's *only* the pressure and flow that the *motor* is actually consuming (2500 PSI, 18 GPM). This isolates the "useful" part of the power transmission.

Norms

While no specific norm *defines* 'work' in this exact way, ISO 4409 (Hydraulic fluid power — Positive-displacement pumps, motors and integral transmissions — Methods of testing and presenting basic steady-state performance) outlines the procedures for measuring this useful power output.

Formula(s)

We use the same HP formula, but with the motor's operating parameters:

\[ \text{HP}_{\text{work}} = \frac{p_{\text{motor}} \times Q_{\text{motor}}}{1714} \]
Hypotheses

We assume:

  • The 18 GPM and 2500 PSI are measured correctly at the motor's work ports.
  • This represents the motor's average sustained load.
Data

From the problem statement:

ParameterSymbolValueUnit
Motor Operating Pressure\(p_{\text{motor}}\)2500PSI
Motor Operating Flow\(Q_{\text{motor}}\)18GPM
Tips

Notice we are using the motor's actual operating parameters, not the system maximums. This is the power doing real work. The 2 GPM difference (20 GPM in - 18 GPM used) is a key source of heat!

Schematic (Before Calculations)

Visualizing the useful work. This is the power consumed *only* by the motor.

Useful Work Component
M Motor 18 GPM @ 2500 PSI Useful Work Load
Calculation(s)

Apply the formula with the motor's data:

Step 1: State the formula

\[ \text{HP}_{\text{work}} = \frac{p_{\text{motor}} \times Q_{\text{motor}}}{1714} \]

This is the same formula as Q1, but applied to the 'work' part of the circuit.

Step 2: Substitute values and solve

We substitute the motor's specific parameters: \(p_{\text{motor}} = 2500 \text{ PSI}\) and \(Q_{\text{motor}} = 18 \text{ GPM}\).

\[ \begin{aligned} \text{HP}_{\text{work}} &= \frac{2500 \text{ PSI} \times 18 \text{ GPM}}{1714} \\ \text{HP}_{\text{work}} &= \frac{45000}{1714} \\ \text{HP}_{\text{work}} &= 26.254 \text{ HP} \end{aligned} \]

This result is rounded to 26.25 HP for the next step.

Schematic (After Calculations)

The result of our calculation:

Calculated Useful Power
Useful Work Power (HP_work) 26.25 HP
Analysis

This 26.25 HP is the power being converted from hydraulic (fluid) power into mechanical power (torque and speed) at the motor's shaft. (This assumes a 100% efficient motor; in reality, the mechanical HP would be slightly less).

Cautionary Points

This is the hydraulic work. The *mechanical* work at the shaft will be slightly less due to the motor's own mechanical and volumetric inefficiencies (e.g., if the motor is 95% efficient, the final mechanical power is 26.25 * 0.95 = 24.9 HP).

Key Takeaways
  • Work Power (Useful Power) is based on the load's actual operating parameters: \(p_{\text{load}}\) and \(Q_{\text{load}}\).
Did You Know?

The 2 GPM "lost" (20 GPM in - 18 GPM used) is likely flowing over the pressure-compensated pump's compensator or a relief valve. This flow, dropping from 3000 PSI to 0 PSI (tank), generates a large amount of heat by itself! (2 GPM * 3000 PSI / 1714 = 3.5 HP of heat).

FAQ

[Common follow-up questions]

Where did the other 2 GPM go?

It is bypassed back to the tank, likely through the pump's compensator or a system relief valve. This is a major, and intentional, source of heat generation in this circuit.

Why did the pressure drop from 3000 to 2500?

This 500 PSI "loss" is the pressure drop across the directional valve, hoses, and fittings. This pressure drop is also converted 100% into heat. (18 GPM * 500 PSI / 1714 = 5.25 HP of heat).

Final Result
The useful power output (work) is **26.25 HP**.
Your Turn

If the motor stalls (load is too high), its flow (GPM) becomes 0. What is its useful power (HP)?

Memo Card

Question 2 Summary:

  • Concept: Useful Work Power
  • Formula: \(\text{HP} = (p_{\text{load}} \times Q_{\text{load}}) / 1714\)
  • Data: 2500 PSI, 18 GPM

Question 3: Determine the total waste heat (in HP) generated by the system.

Principle

Waste heat is the total inefficiency of the system. Based on the First Law of Thermodynamics (Energy Conservation), it's the total input power (what we put in) minus the useful work power (what we got out).

Mini-Lesson

The First Law of Thermodynamics states energy cannot be created or destroyed, only transformed. The 35.0 HP of hydraulic power *must* go somewhere. 26.25 HP was transformed into mechanical work. The "missing" 8.75 HP was transformed into thermal energy (heat).

Pedagogical Note

This is the most crucial step for thermal management. You've identified the *problem*: 8.75 HP of power is constantly heating your oil, like a giant 6.5 kW toaster coil submerged in your tank.

Norms

This calculation is a direct application of the First Law of Thermodynamics. Sizing of components based on this heat load is governed by ISO/TR 16562 (Hydraulic fluid power — Thermal load capacity of coolers).

Formula(s)

The relationship is: \(P_{\text{total}} = P_{\text{work}} + P_{\text{heat}}\). Rearranging for heat:

\[ \text{HP}_{\text{heat}} = \text{HP}_{\text{in}} - \text{HP}_{\text{work}} \]
Hypotheses

We assume:

  • The only two significant outputs are work and heat (e.g., no significant sound energy, etc.).
  • Our calculations for \(\text{HP}_{\text{in}}\) and \(\text{HP}_{\text{work}}\) are correct.
Data

From our previous answers:

ParameterSymbolValueUnit
Input Power\(\text{HP}_{\text{in}}\)35.0HP
Useful Work\(\text{HP}_{\text{work}}\)26.25HP
Tips

A 20-30% heat load ( (8.75 / 35.0) = 25% ) is very common for standard open-loop hydraulic systems. High-efficiency systems might be 10-15%, while a system in full relief (0 work) is 100% heat!

Schematic (Before Calculations)

Visualizing the energy balance. We know the input and the work output.

Energy Balance (Unknown Heat)
System Input Power 35.0 HP Useful Work 26.25 HP Waste Heat ? ? ?
Calculation(s)

We apply the energy balance formula using our results from Q1 and Q2.

Step 1: State the formula

\[ \text{HP}_{\text{heat}} = \text{HP}_{\text{in}} - \text{HP}_{\text{work}} \]

This formula represents the core energy balance, stating that total power is the sum of work power and heat power.

Step 2: Substitute known values

We substitute \(\text{HP}_{\text{in}} = 35.0 \text{ HP}\) (from Q1) and \(\text{HP}_{\text{work}} = 26.25 \text{ HP}\) (from Q2).

\[ \begin{aligned} \text{HP}_{\text{heat}} &= 35.0 \text{ HP} - 26.25 \text{ HP} \\ \text{HP}_{\text{heat}} &= 8.75 \text{ HP} \end{aligned} \]

This shows that 8.75 HP of the input power is not used for work and is therefore converted to heat.

Schematic (After Calculations)

The final energy balance is solved.

Energy Balance (Solved)
System Input Power 35.0 HP Useful Work 26.25 HP Waste Heat 8.75 HP
Analysis

This 8.75 HP of power isn't 'lost'; it is converted directly into heat. This heat comes from all system inefficiencies: the 2 GPM of flow not used by the motor (dumping over the relief valve) and the pressure drops across all valves, hoses, and fittings.

Cautionary Points

This calculation is an *average*. If the motor stalls (0 HP work), the *entire* 35.0 HP becomes heat. You must size your cooler for the *worst-case sustained* heat load, not just the average.

Key Takeaways
  • Heat (HP) = Input (HP) - Work (HP).
  • All inefficiency becomes heat. This is the First Law of Thermodynamics in action.
Did You Know?

On a mobile machine like an excavator, this waste heat is why the hydraulic oil cooler (radiator) is often just as large as the engine's water cooler (radiator)!

FAQ

[Common follow-up questions]

Why is there so much heat? Where does it come from?

We can break it down: 1. Heat from 'lost' flow: 2 GPM (20-18) flowing over the 3000 PSI relief: (2 * 3000) / 1714 = 3.5 HP 2. Heat from pressure drop: 18 GPM of work flow had to drop from 3000 PSI (pump) to 2500 PSI (motor). This 500 PSI drop is heat: (18 * 500) / 1714 = 5.25 HP 3. Total Heat: 3.5 HP + 5.25 HP = 8.75 HP. The math matches!

Final Result
The system is generating **8.75 HP** of waste heat.
Your Turn

If the input power was 40 HP and useful work was 30 HP, what is the waste heat in HP?

Memo Card

Question 3 Summary:

  • Concept: Waste Heat (HP)
  • Formula: \(\text{HP}_{\text{heat}} = \text{HP}_{\text{in}} - \text{HP}_{\text{work}}\)
  • Data: 35.0 HP, 26.25 HP

Question 4: Convert the waste heat from HP to the standard cooling unit: BTU/hr.

Principle

To select a heat exchanger (cooler), we must use the standard thermal unit, which in the US is BTU/hr (British Thermal Units per hour). This is a simple unit conversion.

Mini-Lesson

HP is a unit of mechanical power. BTU/hr is a unit of thermal power. They describe the same thing (energy per unit time) but in different contexts. The conversion \(1 \text{ HP} = 2545 \text{ BTU/hr}\) is a fundamental constant of US customary units.

Pedagogical Note

This is a simple unit conversion, but it's the most important one in hydraulic cooling. All cooler manufacturers (e.g., Parker, Eaton, Thermal Transfer) rate their products in BTU/hr (or sometimes HP, assuming this conversion).

Norms

This conversion is based on the standard definition of the British Thermal Unit (BTU) and Horsepower, derived from the mechanical equivalent of heat.

Formula(s)

The standard conversion is:

\[ \text{Cooling (BTU/hr)} = \text{HP}_{\text{heat}} \times 2545 \]
Hypotheses

This assumes the conversion factor 1 HP = 2545 BTU/hr is sufficiently precise. This is the standard accepted conversion for this type of engineering calculation.

Data

From our previous answer:

ParameterSymbolValueUnit
Waste Heat\(\text{HP}_{\text{heat}}\)8.75HP
Conversion Factor-2545BTU/hr per HP
Tips

A "rule of thumb" often used is just `HP * 2500` because it's easier to do mental math, but 2545 is the correct number to use for an exam. In the field, 2500 is often close enough.

Schematic (Before Calculations)

Visualizing the unit conversion.

Unit Conversion
8.75 HP × 2545 ? ? ? BTU/hr
Calculation(s)

We apply the conversion formula to our result from Q3.

Step 1: State the formula

\[ \text{Cooling (BTU/hr)} = \text{HP}_{\text{heat}} \times 2545 \]

This is the standard conversion factor from mechanical horsepower to thermal (heat) power in BTU/hr.

Step 2: Substitute known values

We substitute \(\text{HP}_{\text{heat}} = 8.75 \text{ HP}\) (from Q3).

\[ \begin{aligned} \text{Cooling} &= 8.75 \text{ HP} \times 2545 \\ \text{Cooling} &= 22,268.75 \text{ BTU/hr} \end{aligned} \]

This is the total thermal load the system must dissipate per hour to maintain its temperature.

Schematic (After Calculations)

The result of the conversion.

Conversion Result
8.75 HP = 22,269 BTU/hr
Analysis

The system is dumping 22,269 BTUs into the oil every hour. To maintain a stable temperature (like the 140°F target), we must remove 22,269 BTUs every hour.

Cautionary Points

Do not confuse this with `kW`. 1 HP is approx 0.746 kW. 1 kW is approx 3412 BTU/hr. Using the wrong conversion factor is a very common error in thermal calculations.

Key Takeaways
  • The key conversion factor is: **1 HP = 2545 BTU/hr**.
Did You Know?

A typical home furnace for a 2000 sq ft house might be 80,000 BTU/hr. This hydraulic system is generating as much heat as a small home heater running constantly!

FAQ

[Common follow-up questions]

Where does the 2545 number come from?

It's derived from the mechanical equivalent of heat (Joule's experiment) and the definitions of HP (33,000 ft-lbf/min) and BTU (the heat to raise 1 lb of water by 1°F).

Final Result
The system is generating **22,269 BTU/hr** (rounded) of heat.
Your Turn

How many BTU/hr is 10 HP of waste heat?

Memo Card

Question 4 Summary:

  • Concept: Power to Heat Conversion
  • Formula: \(\text{BTU/hr} = \text{HP} \times 2545\)
  • Data: 8.75 HP

Question 5: If the system has a natural (passive) heat dissipation of 5,000 BTU/hr, what is the additional cooling capacity required from a heat exchanger?

Principle

The total heat generated (22,269 BTU/hr) must be removed. The system components (tank, hoses, blocks) will naturally dissipate some of this heat to the surrounding air. We only need to buy a cooler big enough to handle the difference.

Mini-Lesson

This is a simple heat balance equation. For the temperature to be stable, `Heat In` must equal `Heat Out`.
`Heat_In` = \(\text{HP}_{\text{heat}} \times 2545\) = 22,269 BTU/hr.
`Heat_Out` = `Passive_Dissipation` + `Active_Cooling` (the cooler)
Therefore: `Active_Cooling` = `Heat_In` - `Passive_Dissipation`.

Pedagogical Note

We can use the environment to help us! A large reservoir in a cool, well-ventilated area can dissipate a lot of heat naturally, reducing the size (and cost) of the required cooler. A system in a hot, enclosed room has almost 0 passive dissipation.

Norms

Cooler manufacturers (e.g., Parker, Eaton) provide selection charts based on this "Net Cooling Required" number, often called "Heat Load to be Dissipated".

Formula(s)

The required cooler size is the total heat minus what the system removes on its own.

\[ \text{Cooler Size} = \text{Total Heat} - \text{Passive Dissipation} \]
Hypotheses

We assume:

  • The 5,000 BTU/hr passive dissipation is a reliable, consistent number (e.g., the ambient air isn't 120°F).
  • This passive dissipation occurs at our target operating temperature of 140°F.
Data

From our previous answer and the question:

ParameterValueUnit
Total Heat Generated22,269BTU/hr
Passive Dissipation5,000BTU/hr
Tips

Always be conservative. If you calculate 17,269 BTU/hr, you would select the *next size up* (e.g., a 20,000 BTU/hr cooler), not the next size down. This adds a safety factor.

Schematic (Before Calculations)

Visualizing the final heat balance. We need to solve for the active cooler.

Net Cooling Calculation
Hot System Total Heat In 22,269 BTU/hr Passive Out 5,000 BTU/hr Cooler Out ? ? ?
Calculation(s)

We apply the net cooling formula using our total heat from Q4 and the given passive dissipation.

Step 1: State the formula

\[ \text{Cooler Size} = \text{Total Heat} - \text{Passive Dissipation} \]

This formula balances the total heat generated against the heat removed, both passively (by the system) and actively (by the cooler).

Step 2: Substitute known values

We substitute \(\text{Total Heat} = 22,269 \text{ BTU/hr}\) (from Q4) and \(\text{Passive Dissipation} = 5,000 \text{ BTU/hr}\) (from the problem statement).

\[ \begin{aligned} \text{Cooler Size} &= 22,269 \text{ BTU/hr} - 5,000 \text{ BTU/hr} \\ \text{Cooler Size} &= 17,269 \text{ BTU/hr} \end{aligned} \]

This final value is the minimum size for the heat exchanger, before adding a safety factor.

Schematic (After Calculations)

The final cooling requirement is known.

Net Cooling (Solved)
Required Cooler Size (Net) 17,269 BTU/hr
Analysis

The system generates 22,269 BTU/hr, but 5,000 BTU/hr "leaks" out naturally through the reservoir and hoses. We must therefore install a heat exchanger (cooler) capable of removing at least* the remaining 17,269 BTU/hr to prevent the system from overheating.

Cautionary Points

The 5,000 BTU/hr passive dissipation is an *estimate*. It depends heavily on ambient temperature, reservoir size, and airflow. Do not overestimate this value! If you are in a hot environment, you might assume this is 0 for safety.

Key Takeaways
  • Net Cooling = Total Heat - Passive Dissipation.
  • Always add a safety factor (e.g., +15% to +25%) to your final number before selecting a cooler.
Did You Know?

Many coolers are rated based on a "TD" (Temperature Difference) between the oil and the ambient air (e.g., 140°F oil - 80°F air = 60°F TD). A cooler rated for 20,000 BTU/hr at 100°F TD might only provide 12,000 BTU/hr at 60°F TD.

FAQ

[Common follow-up questions]

What if I don't add a cooler?

The system would still try to balance. The oil temperature would keep rising, which increases passive dissipation. The temperature would stabilize when Passive Dissipation = 22,269 BTU/hr. This might happen at 250°F, by which point your seals are fried and your oil is "cooked".

What's a good cooler to pick?

You would look for a cooler rated for *at least* 17,269 BTU/hr. A good safe choice would be a 20,000 or 25,000 BTU/hr model to add a safety margin.

Final Result
An additional cooling capacity of **17,269 BTU/hr** is required.
Your Turn

If total heat is 30,000 BTU/hr and passive dissipation is 10,000 BTU/hr, what cooler size is needed (in BTU/hr)?

Memo Card

Question 5 Summary:

  • Concept: Net Cooling Required
  • Formula: \(\text{Cooler} = \text{Heat}_{\text{total}} - \text{Heat}_{\text{passive}}\)
  • Data: 22,269 BTU/hr, 5,000 BTU/hr

Interactive Tool: Heat Load Calculator

Use this tool to see how changing system pressure and motor flow affects the required cooling. (Note: This simulation makes assumptions about flow losses for simplicity).

Input Parameters
3000 PSI
18 GPM
Key Requirements
Total Input Power (HP) -
Required Cooling (BTU/hr) -

Final Quiz: Test Your Knowledge

1. In US hydraulics, what is the constant used to convert (PSI × GPM) to HP?

2. If a system's input power is 50 HP and its useful work output is 40 HP, how much *heat* (in HP) is generated?

3. What is the conversion factor from 1 HP to BTU/hr?

Glossary

HP (Horsepower)
A unit of power in the US customary system. 1 HP = 33,000 ft-lbf/min.
BTU/hr (British Thermal Unit per Hour)
A unit of heat (thermal power). The amount of heat required to raise one pound of water by one degree Fahrenheit in one hour.
PSI (Pounds per Square Inch)
A unit of pressure, common in the US system.
GPM (Gallons Per Minute)
A unit of volumetric flow rate, common in the US system.
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