This is my first draft at this, any corrections or comments would be appreciated.
Helcul
Incompressible Friction Flow
The flow in a computer water cooling system is incompressible flow with friction losses. The flow is laminar (low velocity, smooth flowlines) except in the blocks and heat exchanger where turbulence is encouraged. I will assume 4 lpm flow for the example calculations.
Re = Reynolds Number = D v / nu
D = diameter = 12.7 mm
v = velocity
v@ 4 liters per minute ~ 4e6 mm^3 / min / Pi / (6.35 mm)^2 * min/60s = 8.8 mm/s
nu = kinematic viscosity, for water:
at 20C ~ 1 mm^2 / s
at 60C ~ 5 mm^2 / s
Re ~ 12.7mm * 8.8 mm/s / 2 mm^2/s = 56 [unitless]
Head Loss due to tube friction
Head loss (per mm of tube) = friction factor * Length * velocity^2 / 2 / D / g
Darcy Equation for laminar flow (Re < 2100), friction factor = 64 / Re = 64 / 56 = 1.14
Head loss = 1.14 * 1 mm * (8.8 mm / s )^2 / 2 / 12.7 mm / (9810 mm /s^2) = 3.6e-4 mm per mm length
So if you have 10 ft of tubing (3048mm length), head loss = 3.6e-4 * 3048 mm = 1.08 mm [very small loss]
Head Loss due to restrictions
Head loss due to restrictions = K * velocity^2 / 2 / g
Examples:
Sharp Pipe entrance, K = 0.50
Head loss will be ~ 0.5 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 2.1e-3 mm
Rounded Pipe entrance, K = 0.04 to 0.28
Head loss will be ~ 0.12 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 5.1e-4 mm
90deg elbow loss ~ 30 Le/D
Head loss ~ 30 * 12.7 mm * 3.6e-4 mm per mm length = 0.14 mm [small loss but equivalent to about 15 inches of tubing]
Line kink reducing flow area to 1/3 ~ K = 0.85
Head loss will be ~ 0.85 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 3.6e-3 mm [small loss but equivalent to about 10 inches of tubing]
Head Loss due to waterblocks and radiator
Here are the major flow restrictions in the system. The flow through the blocks needs to be turbulent in order to get the heat transfer coefficients higher. Turbulent flow results in significant pressure drops. The pressure drop is a complicated function of the flowrate best solved by experiment (unless you have the time and money to do a CFD analysis).
Here is some experimental data of Swiftech blocks:
MCW6000 CPU waterblock, data from: "Swiftech MCW6000 Waterblock"
Flowrate lpm headloss mmH20
4 600
6 1100
8 2750
MCW20-R chipset waterblock, data from: "Swiftech MCW20-R Chipset Waterblock "
Flowrate lpm headloss mmH20
4 163
Here is some experimental data, "Heat exchanger pressure drops vs flow rates"
For the heat exchanger called Big Momma (similar to one I have):
Flowrate lpm headloss mmH20
1.9 35
3.8 211
5.7 668
7.6 1125
Motive Power
A pump will produce a flowrate necessary to overcome the head losses of a system (or it won’t flow).
Here is some data from a Swiftech pump:
Swiftech MCP650 Pump
Flowrate lpm head mmH20
4 2300
6 2000
8 1800
10 1500
12 1200
You can see by comparing the pumping power of this pump vs the pump losses of the MCW6000 waterblock that the likely flowrate of a system including these 2 parts must be less than 8 lpm. In other words, at 8 lpm the pump generates 1800 mmH20 which isn’t enough head for the MCW6000 which has a loss of about 2750 mmH20 at that flowrate.
System Analysis
Now to put it all together. This is an iterative process but it is not hard as there aren’t too many variables and most of the relationships are straightforward.
Problem Statement: If we have a system with 4 feet of tubing, a Big Momma radiator, a MCW20-R waterblock, a MCW6000 waterblock, driven by a MCP650 pump, what will the resulting flowrate be?
The first thing to do is to assume a flow rate and then iterate until the pressure provided by the pump equals the losses in the system. Starting at 4 lpm, the losses in the system will be:
Tubing + fittings Small losses
Radiator 259 mm (interpolated)
CPU block 600 mm
Chipset block 163 mm
Total Losses 1022 mm
Motive Power 2300 mm
Net Pressure 1278 mm
Lots of excess pressure available, the system will flow faster, try 6 lpm:
Tubing + fittings Small losses
Radiator 740 mm (interpolated)
CPU block 1100 mm
Chipset block 326 mm (estimate)
Total Losses 2166 mm
Motive Power 2000 mm
Net Pressure -166 mm
The net pressure is close to zero, so the system will flow at about 5.9 lpm.
At a 5.9 lpm flowrate, the radiator can remove about 295 W from the flow. This is way more power than the CPU will be generating so the radiator will be able to cool the flow to ambient with lots of margin.
A northbridge chipset which is putting out 15W (estimate) will result in a 3.3C degree rise in water temperature (the chipset waterblock thermal resistance is about 0.22 C / W at this flowrate).
A CPU which is putting out 100W will result in a 19C degree rise in water temperature (the CPU waterblock thermal resistance is about 0.193 C / W at this flowrate).
So expect the water to rise about 22.3C above ambient and a resulting CPU temperature somewhat under that. The actually temperature reading depends on the sensor location and other variables, but at worst case, your sensor will read ambient + 22.3C which will be lower than the ambient + 35C that you would expect from a CPU putting out 100W with air cooling.
Helcul
Incompressible Friction Flow
The flow in a computer water cooling system is incompressible flow with friction losses. The flow is laminar (low velocity, smooth flowlines) except in the blocks and heat exchanger where turbulence is encouraged. I will assume 4 lpm flow for the example calculations.
Re = Reynolds Number = D v / nu
D = diameter = 12.7 mm
v = velocity
v@ 4 liters per minute ~ 4e6 mm^3 / min / Pi / (6.35 mm)^2 * min/60s = 8.8 mm/s
nu = kinematic viscosity, for water:
at 20C ~ 1 mm^2 / s
at 60C ~ 5 mm^2 / s
Re ~ 12.7mm * 8.8 mm/s / 2 mm^2/s = 56 [unitless]
Head Loss due to tube friction
Head loss (per mm of tube) = friction factor * Length * velocity^2 / 2 / D / g
Darcy Equation for laminar flow (Re < 2100), friction factor = 64 / Re = 64 / 56 = 1.14
Head loss = 1.14 * 1 mm * (8.8 mm / s )^2 / 2 / 12.7 mm / (9810 mm /s^2) = 3.6e-4 mm per mm length
So if you have 10 ft of tubing (3048mm length), head loss = 3.6e-4 * 3048 mm = 1.08 mm [very small loss]
Head Loss due to restrictions
Head loss due to restrictions = K * velocity^2 / 2 / g
Examples:
Sharp Pipe entrance, K = 0.50
Head loss will be ~ 0.5 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 2.1e-3 mm
Rounded Pipe entrance, K = 0.04 to 0.28
Head loss will be ~ 0.12 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 5.1e-4 mm
90deg elbow loss ~ 30 Le/D
Head loss ~ 30 * 12.7 mm * 3.6e-4 mm per mm length = 0.14 mm [small loss but equivalent to about 15 inches of tubing]
Line kink reducing flow area to 1/3 ~ K = 0.85
Head loss will be ~ 0.85 * (8.8 mm / s )^2 / 2 / (9180 mm/s^2) = 3.6e-3 mm [small loss but equivalent to about 10 inches of tubing]
Head Loss due to waterblocks and radiator
Here are the major flow restrictions in the system. The flow through the blocks needs to be turbulent in order to get the heat transfer coefficients higher. Turbulent flow results in significant pressure drops. The pressure drop is a complicated function of the flowrate best solved by experiment (unless you have the time and money to do a CFD analysis).
Here is some experimental data of Swiftech blocks:
MCW6000 CPU waterblock, data from: "Swiftech MCW6000 Waterblock"
Flowrate lpm headloss mmH20
4 600
6 1100
8 2750
MCW20-R chipset waterblock, data from: "Swiftech MCW20-R Chipset Waterblock "
Flowrate lpm headloss mmH20
4 163
Here is some experimental data, "Heat exchanger pressure drops vs flow rates"
For the heat exchanger called Big Momma (similar to one I have):
Flowrate lpm headloss mmH20
1.9 35
3.8 211
5.7 668
7.6 1125
Motive Power
A pump will produce a flowrate necessary to overcome the head losses of a system (or it won’t flow).
Here is some data from a Swiftech pump:
Swiftech MCP650 Pump
Flowrate lpm head mmH20
4 2300
6 2000
8 1800
10 1500
12 1200
You can see by comparing the pumping power of this pump vs the pump losses of the MCW6000 waterblock that the likely flowrate of a system including these 2 parts must be less than 8 lpm. In other words, at 8 lpm the pump generates 1800 mmH20 which isn’t enough head for the MCW6000 which has a loss of about 2750 mmH20 at that flowrate.
System Analysis
Now to put it all together. This is an iterative process but it is not hard as there aren’t too many variables and most of the relationships are straightforward.
Problem Statement: If we have a system with 4 feet of tubing, a Big Momma radiator, a MCW20-R waterblock, a MCW6000 waterblock, driven by a MCP650 pump, what will the resulting flowrate be?
The first thing to do is to assume a flow rate and then iterate until the pressure provided by the pump equals the losses in the system. Starting at 4 lpm, the losses in the system will be:
Tubing + fittings Small losses
Radiator 259 mm (interpolated)
CPU block 600 mm
Chipset block 163 mm
Total Losses 1022 mm
Motive Power 2300 mm
Net Pressure 1278 mm
Lots of excess pressure available, the system will flow faster, try 6 lpm:
Tubing + fittings Small losses
Radiator 740 mm (interpolated)
CPU block 1100 mm
Chipset block 326 mm (estimate)
Total Losses 2166 mm
Motive Power 2000 mm
Net Pressure -166 mm
The net pressure is close to zero, so the system will flow at about 5.9 lpm.
At a 5.9 lpm flowrate, the radiator can remove about 295 W from the flow. This is way more power than the CPU will be generating so the radiator will be able to cool the flow to ambient with lots of margin.
A northbridge chipset which is putting out 15W (estimate) will result in a 3.3C degree rise in water temperature (the chipset waterblock thermal resistance is about 0.22 C / W at this flowrate).
A CPU which is putting out 100W will result in a 19C degree rise in water temperature (the CPU waterblock thermal resistance is about 0.193 C / W at this flowrate).
So expect the water to rise about 22.3C above ambient and a resulting CPU temperature somewhat under that. The actually temperature reading depends on the sensor location and other variables, but at worst case, your sensor will read ambient + 22.3C which will be lower than the ambient + 35C that you would expect from a CPU putting out 100W with air cooling.
