KEVOS® MAKEOS · Hydraulics
Liquid Throttling Temperature Rise (Valve and Orifice Heating, Joule–Thomson for Liquids)
How much does hydraulic oil, water or another liquid warm as it passes through a relief valve, flow control, orifice or partly closed valve that drops its pressure without doing work?
Temperature rise of a liquid throttled at constant enthalpy across a valve, orifice or other restriction, from its density, expansion coefficient and specific heat and the pressure dropped (the Joule–Thomson relation applied to a liquid): the temperature rise, the pressure energy turned to heat per kilogram, the simpler estimate that ignores thermal expansion and how far the expansion term lowers it, the temperature after the restriction, and a check of the rise against your limit.
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How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Liquid properties | ||
| Density | kg/m³ | Density of the liquid at the working temperature, from the supplier’s data sheet or property tables. |
| Volumetric expansion coefficient | ×10⁻⁶ /K | Fractional volume increase per kelvin at constant pressure, from fluid data. |
| Specific heat at constant pressure | kJ/(kg·K) | Specific heat at constant pressure, from fluid data. |
| Pressure drop | ||
| Liquid temperature before the restriction | °C | Temperature of the liquid arriving at the restriction. |
| Pressure dropped across the restriction | bar | Pressure upstream minus pressure downstream of the restriction. |
| Largest acceptable temperature rise | K | Your limit on temperature rise for this duty; it only moves the pass or fail line. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
Liquid properties
What this means: Density
- What it is
- The mass of one cubic metre of the liquid at the temperature and pressure where it reaches the restriction.
- Units and format
- Enter a number in kg/m³: kilograms per cubic metre. The form accepts 1 kg/m³ to 25000 kg/m³. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 870 kg/m³ as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A denser liquid has a smaller specific volume, so it warms less for the same pressure drop.
- Get it right
- Mineral hydraulic oils are about 850–890 kg/m³; take the value at the working temperature from the data sheet.
What this means: Volumetric expansion coefficient
- What it is
- How much the liquid’s volume grows for each kelvin it warms at constant pressure, as a fraction of its volume; it sets how much of the pressure energy is not turned into a temperature rise.
- Units and format
- Enter a number in ×10⁻⁶ /K: ×10⁻⁶ /K. The form accepts 0 ×10⁻⁶ /K to 10000 ×10⁻⁶ /K. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 700 ×10⁻⁶ /K as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A larger value slightly lowers the temperature rise.
- Get it right
- Use the volumetric, not the linear, coefficient; mineral oils are several times more expansive than water.
What this means: Specific heat at constant pressure
- What it is
- The heat needed to warm one kilogram of the liquid by one kelvin at constant pressure.
- Units and format
- Enter a number in kJ/(kg·K): kilojoules per kilogram per kelvin. The form accepts 0.01 kJ/(kg·K) to 50 kJ/(kg·K). Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1.9 kJ/(kg·K) as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The temperature rise falls in inverse proportion to it.
- Get it right
- Mineral oils are near 1.9–2.1 kJ/(kg·K) and water near 4.18; a low value gives a larger temperature rise.
Pressure drop
What this means: Liquid temperature before the restriction
- What it is
- The temperature of the liquid as it arrives at the valve, orifice or other restriction.
- Units and format
- Enter a number in °C: degrees Celsius. The form accepts -200 °C to 1000 °C. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 40 °C as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A hotter liquid has a slightly smaller rise, because the expansion term T β grows; the outlet temperature rises with it.
- Get it right
- The absolute temperature enters the expansion term, so use the real liquid temperature, not a design ambient.
What this means: Pressure dropped across the restriction
- What it is
- The pressure lost across the restriction without doing useful work: upstream pressure minus downstream pressure.
- Units and format
- Enter a number in bar: bar (1 bar = 100 kPa). The form accepts 0.000001 bar to 20000 bar. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 300 bar as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The temperature rise is in proportion to it.
- Get it right
- For a relief valve passing flow back to tank, this is the full set pressure above the tank pressure.
What this means: Largest acceptable temperature rise
- What it is
- The largest temperature rise you can accept across this restriction, set by the fluid, seals, process or downstream equipment.
- Units and format
- Enter a number in K: kelvin (a difference of 1 K equals 1 °C). The form accepts 0 K to 1000 K. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 10 K as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Moves only the pass or fail line; the calculated rise does not change.
- Get it right
- This is your own limit for the pass or fail line, not a fluid rating.
Method and formulas
- A throttle does no work and, if it is short and insulated, exchanges little heat, so the liquid’s enthalpy is the same before and after it (Callen §6.3, equation 6.37).
- At constant enthalpy dT = (v ÷ cp)(T β − 1) dP (Callen equation 6.41), with v = 1 ÷ ρ and T in kelvin. For a pressure drop Δp the rise is ΔT = v (1 − T β) Δp ÷ cp; for liquids T β is well below 1, so they warm.
- Ignoring expansion gives the simpler v Δp ÷ cp, the pressure energy v Δp turned to heat; the expansion term lowers it by the share T β.
- Example: oil at 40 °C (870 kg/m³, β 700 × 10⁻⁶/K, cp 1.9 kJ/(kg·K)) dropping 300 bar warms 14.17 K; the simple estimate of 18.15 K is 21.9 % high.
Limitations
- Properties are taken at the inlet and held constant; for large drops use values at the mean temperature.
- The liquid alone with no heat loss; in a circuit the heat spreads into pipes, tank and cooler, so the measured rise is usually smaller.
- The liquid must stay liquid: if the outlet pressure falls below its vapour pressure it flashes or cavitates and this result does not apply.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Estimate oil heating across a relief valve, flow control or pressure-reducing valve
- Find the temperature after an orifice or partly closed valve in a water or process line
- Check whether a throttled circuit needs more cooling
Who it is for
- Hydraulic engineers
- Process and plant engineers
- Maintenance engineers chasing hot oil
What you get
- Temperature rise across the restriction
- Pressure energy turned to heat per kilogram
- Simple estimate ignoring expansion, and the share the expansion term removes
- Temperature after the restriction
Interpreting the result
- With the example oil, every 10 bar dropped across a valve without doing work warms it by about 0.47 K; that heat must leave through the cooler.
- For liquid compressed quickly rather than throttled, use the liquid adiabatic compression tool: most compression heating returns on decompression, while throttling heat is a true loss.
Before relying on the result
- Check that the outlet pressure stays above the liquid’s vapour pressure at the outlet temperature.
References
- Herbert B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd edition (John Wiley & Sons, 1985), sections 3.9 and 6.3, printed pp. 84–86 and 160–165. Method reference; no text, worked examples, problems or tables are reproduced.
- NIST Chemistry WebBook, SRD 69, Thermophysical Properties of Fluid Systems (https://webbook.nist.gov/chemistry/fluid/), a public source of fluid property values to enter.
Learn the Concepts Behind This Tool
These KEVOS® Library readings explain the method this tool uses, with worked examples.
Method explained here · KEVOS® original book
Viscous heating and the oil's heat balance
What it covers: Turn pressure drops and leaks into heat, calculate the temperature rise across a relief valve, build a hydraulic unit's heat balance with the feedback through viscosity, and size a cooler.
Useful alongside · KEVOS® original book
Throttling, mixing and friction
What it covers: Three losses with no heat and no visible waste: pressure regulators, tempering valves and friction in pipes and filters, their entropy generation, their cost in kilowatts and how to reduce them.
Useful alongside · KEVOS® original book
Throttling: the Joule–Thomson effect at the regulator
What it covers: Why gas cools through a regulator, LN-1's drop from 15 °C to −20 °C on a full pack, the inversion temperature, why hydrogen warms, CO₂ regulators that freeze, and a cold-gas alarm.
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