This post will discuss how to size a Galin Pump for a given application - what constraints exist, and how to navigate them.

A Galin Pump has the following components:

A pump is usually selected for either: moving fluid from one place to another, or controlling fluid flow/volume. Depending on which of these two functions you want, you choose either a centrifugal pump or a positive displacement pump. What if you could have both? Let us discuss this possibility below with Galin Pump.

We have the whole world of possibilities presented to us in terms of choice of sizing of the pump chamber, and electrical machines etc. This blank slate makes it difficult to present the flexibility of the Galin Pump in a tangible way. Let us start with what we have on hand - two D5065 motors from ODrive. We will use the performance constraints of these motors, and explore the performance options of Galin Pump using them.

The important characteristics of these motors for Galin Pump are:

  • maximum torque: 2 Nm,

  • rated speed: 1700 RPM,

  • maximum current: 65 A,

  • maximum voltage: 32 V,

  • maximum power output: 1.8 kW.

The maximum torque that the motor is capable of developing constrains the maximum pressure (P) difference (between discharge and suction chambers) of liquid that Galin Pump is capable of handling. However, we have a wiggle room in sizing the chamber to accomodate a range of pressures because: torque = pressure x area x lever arm - refer to figure 1. How exactly does this affect the pump chamber sizing? Let us say, we have 5000 PSI of pressure that we need to handle, can we do this with 2 Nm of torque?

area: \(A = (R - r) \cdot d \),

lever arm: \(l = (R + r) / 2 \), therefore:

torque: \(\tau = \frac{1}{2} \cdot P \cdot d \cdot (R^2 - r^2) \).

For a given maximum torque, we can play with the size of the chamber height (d), outer radius (R) and inner radius (r) to support the required maximum pressure. We will designate the product \(d \cdot (R^2 - r^2)\) as ‘\(S\)’ - a sort of scaling factor, that will frequently appear in the discussion to follow. Obviously, there are practical limits to how small or large we can make the chamber, but hopefully, you can appreciate the flexibility we have available to us.

Before we begin to select some numbers, let us also remember the flow/dosing volume (V) is an additional parameter that is important in pump selection. Galin Pump will have 8 suction and discharge cycles per shaft revolution, and each cycle will deliver a volume of fluid equal to: \(V = \pi \cdot (360^\circ - 4 \cdot \theta)/360^\circ \cdot S\) (where \(\theta\) is the vane angle). Let us constrain the vane angle to \(40^\circ\) (for historical reasons, and to constrain our optionality a bit in order to keep things a bit sane). Therefore, \(V = \pi \cdot (200^\circ/360^\circ) \cdot S\). We notice a similarity between the equation for maximum torque, and volume delivered:

torque: \(\tau = \frac{1}{2} \cdot P \cdot S\), and

volume: \(V = 8 \cdot \pi \cdot \frac{200^\circ}{360^\circ} \cdot S\).

To minimise torque (\(\tau\)) for a given pressure (P) we want to minimise \(S\), but to maximise volume (V) we want to maximise \(S\). We’ve previously selected our pressure as 5000 PSI (or 34 MPa in SI units) therefore, the largest value \(S\) can be is: \(1.14 \times 10^{-7} m^3\), which means the largest volume per shaft revolution we can achieve is: \(1.6 \times 10^{-6} m^3\) (1.6 mL). This means that the whole chamber’s internal volume is: 0.36 mL (a metric teaspoon contains 5 mL).

Can we machine parts this small? Yes. Can we couple the chamber to motors the size of D5065 to it? Maybe. Is this a solvable problem - yes. We are not breaking laws of physics here.

Now we are getting somewhere. The motor can spin at a maximum speed of 1700 RPM and sustain a torque of 2 Nm, therefore, we calculate the maximum flow/dosing output of the pump to be: 45 mL/second (or 2.7 L/min, 163 L /hr). This is crazy high for a dosing pump. The flow capacity of Galin Pump is comparable to centrifugal pumps, e.g. here, here, here. These are all selected for the same power consumption of 1.8 kW - keep in mind that these centrifugal pumps are pumping quite low pressure (8, 9, 8 bar respectively) compared to what we just sized our pump for. Remember also, that we are using the maximum speed at the maximum torque (or maximum pressure). If we operate at a lower pressure, we could increase the speed, and hence increase flow.

The dosing/metering function of Galin Pump is performed by changing the speed of rotation of the vanes. What kind of flow rate accuracy can we achieve with Galin Pump? This all depends on the accuracy of moving the vanes to the required positions at the right time, as well as any slippage. Let us ignore slippage for now, and just focus on our ability to control positions of the vanes of the pump. This, in fact, is quite an involved process, because within the motor drivers we are using FOC to control the torque delivery of the motors. However, we can get a back-of-the-envelope lower limit to the flow rate accuracy by considering the accuracy of the encoder. Let us use a standard popular encoder AMT-102 whose position accuracy is reported as \(0.25^\circ\). This means that we can control the volume of the chamber to within \(2 \cdot 0.25^\circ = 0.5^\circ\) (we have two shafts/motors/encoders). How can we propagate this to flow control accuracy? Let us assume we set the flow rate to be \(V_{set}/t\), our error in volume will be:

\(V_{error} = \pm 0.5^\circ / t\), therefore:

flow rate accuracy (in percent): \(\frac{V_{error}}{V_{set}} \times 100 = \frac{\pm 0.5^\circ}{200^\circ} \times 100 = \pm 0.25\%\).

From the above equation, we can see that the flow rate accuracy gets better as the chamber volume increases, so decreasing the vane width (\(\theta\)) serves to increase the possible flow rate accuracy. But decreasing the vane width will lead us to have to decrease the size of the inlet & outlet ports, which will affect the ease of flow of fluid in and out of the chambers.

Consequently, we have an accurate dosing pump that is capable of the same fluid delivery volumes as a centrifugal pump.

How big will such a Galin Pump be? Let us assume we make the whole chamber out of a lump of steel, it will be in fact insignificant in weight compared to the motors, their drivers and battery. The two motors are about 400 g each, adding to this 100 g for both motor drivers, but it is the battery that blows us out of the water at about 5 kg. What do we need the battery for? To absorb the regenerative power generated by the motors during the pump operation - can we avoid using a battery? Yes, we could use supercapacitors, or if we don’t care about the efficiency boost our pump gets we can just dump the energy as heat - but yes, heating or rather, cooling becomes an issue to solve. (It is of course solvable.)

The above calculations demonstrate the flexibility and performance of the Galin Pump architecture. The parameter space is very large, even when we place constraints on the pressure and flow requirements of the pump. This is advantages to the end user in many ways. For example, say, you have a Galin Pump with a certain chamber size, by increasing or decreasing the power of the attached motors you can increase and decrease the maximum pressure&flow at which the pump can operate. Because we have motor drivers controlling the motors which in turn control the pump shafts, we have digital/real-time control of the flow delivered by Galin pump.

Direct comparison to an existing dosing pump:

Let us compare the performance of Galin Pump to an existing piston dosing pump (TAP15-25). (This pump was chosen because the manufacturer provided adequate specifications of their pump, enabling appropriate and fair calculations to be made during the comparison process.)

Let us first describe the TAP15-25 in some detail:

  • a piston dosing pump, which has a maximum stroke length of 15 mm, and piston head diameter (bore) of 25 mm. Manually adjusting the stroke length from 0 mm to 15 mm allows us to adjust the flow rate from 0 to 100%,

  • powered by either a 0.18 kW 400VAC 3-phase motor, or 0.25 kW 230VAC single phase motor,

  • the motor has a fixed speed gearbox that pushes the piston at either: 58 / 78 or 116 strokes per minute,

  • depending on the materials the pump is made from affects the maximum operating pressure, which can either be 10 bar or 20 bar.

We’ll choose the highest specification options, i.e. a TAP15-25 that has:

  • operating pressure: 20 bar (2000 kPa),

  • operating speed: 116 strokes per minute (i.e. 116 revolutions of the crank shaft per minute),

  • flow rate: 0 to 50 L/hr.

Given the above data, we make the following calculations, assumptions:

  • stroke length (l) = 15 mm,

  • crank radius (a) = 7.5 mm,

  • bore (b) = 25 mm,

  • therefore, maximum cylinder volume: \( V_{max} = \pi \cdot (\frac{b}{2})^2 \cdot l = \) 7.4 mL. (Sanity check: if the pump crank shaft spins at 116 RPM, then we can calculate the flow per hour as: 7. 4 mL x 116 x 60 = 51.5 L/hr - checks out!),

  • in order to get 1% dosing accuracy, how accurately would we need to be able to adjust the stroke length?

\(V_{actual} / V_{set} \times 100 = (\pi\cdot\frac{b}{2}^2\cdot (l\pm x)) / (\pi\cdot\frac{b}{2}^2\cdot l) \times 100 = \frac{l\pm x}{l}\times 100 = 1\%,\),

simplifying, we get: \(\pm x = (1 - 0.01)\cdot l = 0.99 \cdot l = 0.15 \) mm. That’s a very fine level of manual adjustment (obviously this changes as the stroke length changes).

(Sanity check: adjusting the stroke length down by 0.15 mm from 15 mm leads to a change in flow rate from 51.5 L/hr to 50.7 L/hr (a 1% change).)

Existing piston, or rather, positive displacement (power) pumps are generally stated to have constant torque requirements. This is not correct. Yes, the pressure (and thereby force) seen by the head of the piston is constant (plus/minus some minor extra forces due to inlet/outlet valves) during the suction, and then also constant (but possibly different if the suction/discharge pressures are different) during the discharge strokes - but torque? NO. The torque that needs to be applied changes with the lever arm of the crank-shaft mechanism. It is certainly not constant. The torque varies considerably during a single shaft revolution.

Let us look into this. We have a simple simulation for the operation of a piston pump which we setup with the above parameters in order to calculate the torque load on the electrical machine:

Figure 1 - Galin Pump chamber with relevant dimensions labelled: R - outer chamber radius, r - shaft radius, d - chamber height.

Figure 2 - estimated torque load seen by the electrical machine tasked with rotating the crankshaft at constant speed for our example pump TAP15-25.

The maximum (peak) torque that the electrical machine must be capable of developing at 116 RPM is 8.3 Nm, the average torque is 4.7 Nm.

Using the data presented in figure 2, we can estimate the total work done by the electrical machine pushing fluid in and out at 20 bar pressure: work = torque x distance = 29.5 J. We can verify this calculation by considering that work also equals to: work = force x distance (as seen by the piston head) = 29.5 J. This is purely the work done to move the fluid at pressure the stroke length of the cylinder. These calculations use the torque, work, etc output from the simulation which calculates these values during 0 to 180 degree revolution of the crankshaft. Please contact Galin Engine to obtain a copy of the simulation code.

We can also calculate the power output of the electrical machine over a stroke: power = torque x speed, although torque is not constant. Figure 3 below plots the instantaneous power as a function of time, and crank angle. The peak and average power output are 100.5 W, and 56.9 W respectively. However, these are theoretical values. Hence, we see that it is practical to use a 180 W motor to drive the pump.

Figure 3 - instantaneous power requested from electrical machine as a function of time, and crank angle. The maximum requested power is 100.5 W, and average power requested is 56.9 W.

What would a Galin Pump look like constrained by these same performance parameters:

  • maximum fluid pressure: 20 bar (2000 kPa),

  • flow: 0 - 50 L/hr.

Let us size the chamber so that both the Galin Pump and piston pump have the same output volume at discharge. We’ll keep the vane angle at 40\(^\circ\), therefore the chamber volume is given by: \(V_c = \pi \cdot \frac{200^\circ}{360^\circ} \cdot d \cdot(R^2 - r^2) = 7.4\) mL. This means that the product: \(d\cdot(R^2-r^2) = 4.2\) mL. Before going ahead trying to figure out how to proportion d, R and r, let us consider the torque acting on the vanes.

As previously mentioned, existing piston/power pumps are NOT constant torque machines. This is because most commonly these pumps are coupled with a rotary machine, and a crankshaft mechanism that translates rotary motion to linear motion. The lever arm of a crankshaft mechanism is not constant.

Galin Pump is a constant torque machine, because the lever arm is constant. The load torque seen by the electrical machine is equal to: \(\tau = 2 \cdot P \cdot A \cdot l\), where P is the pressure of the fluid (2000 kPa), A is the internal area of the vane (\(d \cdot (R-r)\)), l is the lever arm (\((R+r)/2\)). The factor of 2 accounts for torque acting on both vanes of the shaft. Summarising, and simplifying we have: \(\tau = P \cdot d \cdot (R^2-r^2)\). From our earlier chamber sizing calculations we already know that \(d\cdot(R^2-r^2) = 4.2\) mL, therefore: \(\tau = 8.4\) Nm.

Next, we calculate the power requirement for our shaft driving motors. There are 8 chamber volume discharge events per shaft revolution. Therefore, the motor rotational speed can be lowered by a factor of 8, resulting in 116 RPM (speed of comparison piston pump motor) being reduced to 14.5 RPM. This results in driving motor power rating of: \(P_{motor} = 8.4 \cdot (14.5/60\cdot 2 \cdot\pi) = 14.5\) W. We have two motors driving each shaft of the Galin Pump, and thereby the total power requirement is 29 W.

A huge advantage of the Galin Pump is that it can be directly driven, i.e. no gearing is required to modify the driving motor speed and output torque. Because the discharge rate is high in Galin Pump, we can utilise the plethora of high torque, low RPM motors.

We still haven’t sized the chamber, so let us do that now. Our existing constraint is that: \(d\cdot(R^2-r^2) = 4.2\) mL. How to decide on the actual values of d, R and r? Well, we could start by keeping the area of the vane a square, therefore, d = (R-r), and say we want the lever arm to equal the chamber height to keep everything simple, i.e. l = d. Under these constraints we get: R = 3r, d = 2r, and r = 6.4 mm.

Galin pump specifications that match the output performance of the TAP15-25:

  • maximum fluid pressure: 20 bar (2000 kPa),

  • r = 6.4 mm, R = 19.2 mm, d = 12.8 mm,

  • power of electrical machines (motors) x 2: 15 W x 2

  • average speed of electrical machines: 14.5 RPM

  • flow: 0 - 50 L/ hr,

  • how do we regulate the output flow? by adjusting the average speed of rotation electronically, so in order to get a 1% change in flow, we would decrease it by 1%.

Added benefits:

  • digital real-time control of the position of both shafts, and real-time adjustment of flow,

  • reversible flow at the push of a button,

  • increasing the power capacity of the electrical machines allows for higher speeds, and therefore a metering pump can turn into a fluid transfer pump,

  • pressure and flow software observers (no need for external sensors).

What does Galin Pump look like?

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