Overview

In this topic, it is explained in detail how to evaluate and tackle the impact of the heating on the pump systems components.

Controller, motor and fluid heat up will be addressed separately, and cooling strategies will be described for each. Example calculations and indications on where to find the missing information will complete the topic.

Controller (non-integrated pump drives)

In the case of non-integrated controllers, it is reasonable to imagine that on customer site they will be installed in electrical cabinets, together with the rest of the electrical equipment.

It can happen that the cabinet ventilation is undersized with respect to the heat load of the electrical components. If this happens, the air temperature could increase to the point that the controller cannot efficiently dissipate the heat anymore, so its components will warm up. If the controller temperature exceeds the safety limit, it will automatically stop the pump to self preserve.

Controller Cooling Strategies

The cabinet ventilation is key for an effective controller cooling, but there are still installation pitfalls which can affect the efficiency of the ventilation. In particular, the controller should be rail-mounted in such way that the venting slits are facing upwards, letting the warm air rise and exit the controller case by buoyancy. Furthermore, around the controller (and in particular in proximity of the slits in the casing) there should be sufficient space (the cabinet shouldn’t be completely crammed).

The controller heating depends on the power transferred through it to the motor, therefore the working point of the pump has an impact on the controller temperature. Knowing the air temperature in the cabinet, the working point of the pump and the thermal characteristic of the controller can give a good estimate of the controller temperature.

The thermal characteristic and the formula to extrapolate the temperature at different environmental conditions are present in the pump system manual (as shown below, in Figures 1 and 2).

Example

Let’s assume there is a BPS-2000 pump operating at 8000 rpm, delivering 80 lpm. The controller is placed in a cabinet, where the air reaches the temperature of 40 degC. If the air was at 25 degC, it would be possible to read out the controller temperature directly from Figure 1 (and the Tc would be equal to 35 degC).

Since the characteristic was determined at 25 degC, Tc = 35 degC must be fed into the equation in Figure 2 together with Ta = 40 degC. The resulting controller temperature will be Tc = 35 + (40 – 25) = 50 degC.

The controller temperature limit is 70 degC (see the red dashed line in Figure 1), therefore there is no need for additional ventilation.

This calculation was performed assuming that the cabinet was already designed and appropriately sized, but there could be occasions when the customer requires the heat emitted by the controller as a design parameter for the cabinet. In these cases, the controller losses must be looked up in the pump system characterization file.

In this excel file there are listed the controller losses for each working point of the pump. Since the cabinet must be able to dissipate the heat in all possible working conditions, the maximum value of that column can be provided to the customer as a “worst case scenario” thermal load.

Motor

Iron and copper losses dissipate part of the electrical power as heat, which can increase the motor temperature. The better the motor cooling, the smaller the temperature increase. On the contrary, inefficient working points for the motor, combined with poor cooling, will cause a rise in the temperature. The drive unit contains the motor itself, plus some delicate sensors and electronics. If the motor temperature increases above the operational limit, it might affect the lifetime of the components and of the drive unit.

In this section there are described the options available to avoid motor overheating.

Motor Cooling Strategies

The larger the motor size, the smaller is the ratio between surface and volume. Since the heat generation depends on the volume, while the heat emission towards the environment depends on the surface area, there is a general tendency of having thermal challenges for large motor sizes. To increase the heat dissipating surface, Levitronix designs the casings for large motors with fins. As a result, if both the ambient air and process fluid are not too warm, the motor is able to passively cool by natural convection.

The heat generation in the motor depends on the working point, and its dissipation depends on the fluid and air temperature. By knowing these parameters, one can estimate the motor temperature using the thermal characteristic and extrapolation formula (which can be found in the pump system manual, Figures 4 and 5).

In case the natural convection is not sufficient, Levitronix offers cooling modules to enhance the heat dissipation capacity. Most motor sizes can mount either an Air Cooling Module (which distributes a compressed air supply around the motor), a Fan Cooling Module (electrically powered, blows air around the motor), or a Water Cooling Module (it requires a water supply). Certain motors feature an integrated water cooling loop running inside the motor to extract heat from the core. Each solution has benefits and drawbacks compared to the others:

• The Air Cooling Module requires compressed air, which is expensive to produce and not available in all every production facility. The module is simple to operate and robust, but less efficient compared to the alternatives, and the exhaust air flow might displace suspended particles in the environment.
• The Fan Cooling Module runs on electrical power, which is normally available in every production facility. It is also more efficient in the cooling, but the particle displacement drawback persists.
• The Water Cooling Module requires a continuous supply of water, which might not be always available. However, it has many benefits, among which it is very quiet and effective, and it doesn’t displace ambient particles.

Their performances can be compared by using the respective motor thermal characteristics in the manual.

Example

Let’s assume there is a BPS-2000 pump operating at 8000 rpm, delivering 40 lpm. The ambient air surrounding the motor reaches the temperature of 30 degC, while the process fluid is at 50 degC. If the air and fluid were at 25 degC, it would be possible to read out the motor temperature directly from Figure 4 (and the Tm would be equal to ~72 degC).

Since the characteristic was determined at 25 degC, Tm = 72 degC must be fed into the first equation in Figure 5 together with Ta = 30 degC and Tl = 50 degC. The resulting motor temperature will be Tm = 72 + (50 – 25)*0.28 + (30 – 25) = 84 degC.

The motor recommended temperature limit is 80 degC (see the red dashed line in Figure 4), therefore the cooling is insufficient.

If the calculations are repeated but considering a water cooled drive unit (Figure 6, assuming coolant temperature 25 degC) and the third formula in Figure 5, the new motor temperature becomes 42 degC, which is acceptable.

Fluid

The hydraulic losses in the pump head cause a temperature increase between inlet and outlet, which depends on the flow rate and the fluid properties.

The formula correlating all these elements is:

Where P_hydr.loss is the hydraulic loss in W, rho is the fluid density, Q is the volumetric flow rate in m3/s, C is the specific heat in J/kg*K, and DeltaT is the temperature difference between outlet and inlet.

By knowing the fluid properties (in particular density and specific heat) and the flow rate, the hydraulic power loss is the only unknown element of the equation.

The hydraulic power loss can be looked up in the characterization file by knowing the working point (either speed and flow rate or pressure and flow rate). Typically, the hydraulic loss can be calculated as the difference between mechanical power (developed by the motor) and the hydraulic power (developed by the impeller), or by knowing the hydraulic power and the hydraulic efficiency.

At working points characterized by low flow rate Q and high pressure, the hydraulic efficiency is low. As a consequence, most of the mechanical power is dissipated as heat (P_hydr.loss is high). The combination of low flow rate and high hydraulic losses causes the temperature increase (DeltaT) to be high. On the contrary, at large flow rates the efficiency is typically higher, so the hydraulic losses are lower, and they are impressed on a large flow rate, resulting in low temperature increase in the fluid.

The impact of temperature on the fluid can be important for the process application case, in particular when handling unstable chemistries (e.g. out-gassing fluids, reactive mixtures) or biological media (living cell cultures which can die at non-optimal temperatures, proteins which can de-naturate and unfold if heated up). While controller and motor temperature impact the lifetime and uptime of the pump system (a matter of production efficiency), the fluid temperature can have a strong impact on the quality of the product (a matter of effectiveness of our products in tackling the customers’ challenge), and therefore it has the utmost importance.

Fluid Thermal Management

Fluid heating comes from the hydraulic losses in the pump head, which might be high in setups with strong restrictions. Therefore, decreasing the system load is the most obvious step to reduce the fluid heat up.

This can be achieved in multiple ways:

• By removing the system restrictions (increasing the tube diameter, removing unnecessary obstacles in the fluid path)
• By operating at a correct working point. For example in TFF processes, even though the filter can sustain high pressures, it is desirable to limit the pressure and operate at the Optimum Operating Point (see Figure 7 below).

If high pressures are required for the process, the heat can still be extracted from the fluid in various ways:

• The hydraulic loop can be exposed to a cold environment (for example by placing part of the tube in a bucket of water and ice, of by running the process in a refrigerated room)
• By using a cooled tank (e.g. double jacketed)
• By installing a heat exchanger in series with the pump
• During diafiltration processes, the bio-reactor can be replenished by adding cold buffer solution, with the consequence that the average fluid bulk temperature will remain low.
• In certain bio-processes, low temperature might have less priority than fast processing, so the focus of the process optimization should shift towards high cross-membrane pressures instead of fluid temperature.

Example

Let’s assume a customer application uses a BPS-2000 hybrid to pump 40 lpm at 1.5 bar. The fluid is water, and at the inlet of the pump it has a temperature of 30 degC. From the characterization file we can read that the hydraulic efficiency will be 37%, and the mechanical power will be 240 W. Therefore the hydraulic loss will be (100%-37%) * 240 W = 151 W.

The fluid density is 1000 kg/m3, while the specific heat is 4180 J/kg*K. Combining all these elements the temperature increase across the pump will be DeltaT = 151 / (1000 * 4180 * 40/60000) = 0.05 degC.

To show the different impact of operating at inefficient working points, let’s assume the same pump is operating at 20 lpm and 3 bar (same hydraulic power as before, but more due to the pressure than to the flow rate). At this working point the hydraulic efficiency is 20%, and the mechanical power 490 W, hence the hydraulic loss will be (100% – 20%) * 490 W = 392 W.

Therefore the temperature increase in the fluid will be 392 / (1000 * 4180 * 20/60000) = 0.28 degC (six-fold increase compared to before, even though it’s still a negligible effect for a single pass process).

Hydraulic Power Dissipation in the Hydraulic Loop

We need to distinguish between fluid heat up within the pump head, and viscous dissipation in the circuit. The fluid heat up in the pump head is characteristic of the pump used and it’s efficiency, and as explained above it is calculated starting from the hydraulic loss in the pump head (mechanical power – hydraulic power).

The fluid heat up due to viscous dissipation in the circuit depends on the circuit and flow rate, independently of the pump technology, and it is calculated assuming as a “worst case scenario” that the whole hydraulic power is dissipated (this assumption is not always true, as explained below).

The hydraulic power delivered by the pump is calculated as the product of Flow and Pressure. The pressure decreases the tubes, fittings and components due to viscous dissipation. From the energy balance point of view, this means that the hydraulic power impressed by the pump to the fluid is slowly dissipated into heat as the fluid flows (the flow rate remains the same, but the pressure decreases).

In the System Load Curve lesson, it is explained how the curve is composed of two terms: a static term, and a dynamic term. The power related to the dynamic term is dissipated into heat due to viscosity friction. The power to overcome the static term of the system load is stored as potential energy (as it happens in hydro power plants).

Let’s compare as an example two hydraulic system: the first (Figure 5) is an unrestricted open loop with a height difference between the down- and up-stream tanks, the second (Figure 6) is a closed loop with high restriction.

The first system will have a gradual system load curve, offset by the static jump (Figure 7), while the second system will have a steep system load curve (Figure 8).

Let’s suppose both pump systems operate at the same working points (same speed, flow and pressure). The open and unrestricted system will dissipate less hydraulic power than the closed and restricted one.

For this reason, when calculating the fluid heat up over the pump head and circuit together, the mechanical power should be considered as entirely dissipated as heat. Therefore the fluid heat-up formula becomes

Fundamentals of Convection and Heat Dissipation to the Environment

As established in the previous section, the fluid will dissipate mechanical and hydraulic power as heat, which might cause fluid heat up. In reality though, the fluid temperature might not increase, which means that the heat is subsequently emitted towards the surrounding environment.

The amount of thermal power that the fluid can transmit by forced convection (for example to a pipe) depends on the thermal gradient driving the heat transfer (the temperature difference between fluid and pipe surface), the heat exchange surface (e.g. calculated using the pipe length and diameter, the tank surface, etc.), convection coefficient (the most complex parameter to calculate, since it depends on the flow regime and fluid properties). The conduction of the heat from inside of the tube to the outside depends on the geometry of the tube (mainly the thickness), its material and properties, and the gradient temperature. Very often, the pipe is exposed on the outside to natural convection only. As a result, the convection inside and outside of the pipe are modeled in different ways.

Overall, we can extract a few practical rules from the heat conduction models:

• If the tube is insulated or double contained, the heat transferred will be negatively affected by the material thickness and low conductivity.
• Low fluid velocities will also reduce the convection, as the heat transfer coefficient will lower.
• The larger the temperature difference between fluid and ambient air, the larger will be the heat transfer.
• The larger the surface (i.e. tube length), the more effective the heat transfer
• The more conductive the tube materials, the more effective the heat transfer
• To dissipate effectively the fluid heat-up, the heat transfer should be maximized.