Worked example: a glycol cooling loop
Written by Haimi Jordaan, MEng (Mechanical), University of Pretoria. Seven years in a specialist engineering analysis and design group across CFD, FEA and DEM. He wrote the solver behind Fluid Network Studio.Published . Last updated .
A closed cooling loop has to do two jobs at once: move the coolant and carry the heat. This worked example is a 30 per cent ethylene-glycol loop solved with heat transfer switched on, so the hydraulics and the temperatures are solved together rather than one after the other.
The setup
The fluid is 30 per cent ethylene glycol, denser and more viscous than water and with a lower specific heat (density 1038 kg/m3, kinematic viscosity 2.1 x 10^-6 m2/s, specific heat 3650 J/kg.K). A pump circulates it from a cold header supplied at 15 degrees Celsius, with a head-flow curve from 25 m at shut-off down to 14 m at about 7.5 L/s and an efficiency of 65 per cent. The loop runs in DN50 pipe:
- A heat-exchanger drop, modelled as a fitting with a loss coefficient of 6, standing in for the cooler the coolant passes through.
- A process load that injects 8 kW of heat into the stream.
- A 40 m bare return line that sheds heat to a 15 degree ambient through a direct overall coefficient of 8 W per metre of pipe per kelvin.
The physics and the method
This is the Advanced heat-transfer mode. Fluid Network Studio solves the loop hydraulics and the heat transfer together, so the flow that sets the residence time is the flow the network actually produces. The 8 kW load raises the coolant temperature where it is injected, and the bare return line then sheds part of that heat to the ambient along its length, using the direct overall coefficient you set rather than a built-up film. An energy balance is reported as a residual on every solve, so conservation is checked, not assumed.
Glycol changes the numbers against plain water in two ways at once. Its higher viscosity costs pump head, and its lower specific heat means a given heat load moves the temperature more for the same flow. Colouring the network by temperature shows both the rise across the load and the partial recovery down the return.
The solved result
The pump settles at 6.363 L/s and 16.89 m, and the 8 kW load raises the glycol by 0.33 K:
| Quantity | Value |
|---|---|
| Circulating flow | 6.363 L/s |
| Velocity in the DN50 bore | 2.942 m/s |
| Pump head at duty | 16.89 m |
| Efficiency (entered as a fixed value) | 65 % |
| Hydraulic power | 1.094 kW |
| Shaft power | 1.684 kW |
| NPSHa at the pump suction | 39.81 m |
| Energy-balance residual | 9.9 x 10^-12 |
Where the pump head goes, and where the heat goes:
| Element | Head loss (m) | Temperature in (degC) | Temperature out (degC) |
|---|---|---|---|
| 20 m of DN50 to the heat exchanger | 3.799 | 15.00 | 15.00 |
| Heat-exchanger drop, K = 6 | 2.649 | 15.00 | 15.00 |
| 15 m of DN50 to the load | 2.849 | 15.00 | 15.00 |
| The 8 kW process load | - | 15.00 | 15.33 |
| 40 m bare return line to the header | 7.598 | 15.33 | 15.33, the drop is under 0.01 K |
Two results are worth pausing on. First, the cold header and the return sit at the same head, so all 16.89 m of pump head is spent on friction and fittings, and the 40 m return line alone takes 7.598 m of it. At 2.942 m/s the DN50 bore is running briskly, and friction climbs with the square of velocity, so a size up on the return would buy back a large share of the 1.684 kW of shaft power.
Second, the thermal result is the quiet lesson. Eight kilowatts sounds like a lot until it meets 6.363 L/s of glycol, and the stream comes out only 0.33 K warmer. The bare return line then sheds just 105.5 W of that 8 kW, a little over one per cent, because at a third of a degree above the 15 degree ambient there is almost no driving temperature difference to work with. A bare pipe is not a heat rejection device unless the fluid is well above ambient. If this loop has to reject its load rather than carry it away, the heat has to leave through the exchanger, not through the pipework.
Every figure comes from the coupled solve, not from a hand estimate. Open the example, press Solve, and check the temperature rise against the load and the flow.
What you learn
Solving the example gives the pump duty on its curve, the flow around the loop, and the temperature at every node. Colour by temperature to see where the heat goes, then raise the flow or swap the fluid back to water and re-solve to see how the temperature swing changes. The physics of pipe heat loss is set out in full on the pipe heat loss application page.
Heat transfer is part of the Advanced plan, A$39/month or A$390/year. The glossary explains the overall heat-transfer coefficient and the resistances behind it.
Open this example in FNS and colour the network by temperature to trace the heat around the loop.