# How much of pump power goes to heating the fluid?

• RobertD
I think what you mean is that the motor uses as much power as it needs to to move the required volume of fluid through the heat exchanger, and that any extra power would just be converted to heat and stay in the fluid. This is a good point. It would be nice to know how much power the motor actually uses to move the required volume of fluid to dissipate 1300 Watts in the heat exchanger. You would need to know the design point flow rate and the actual flow rate.
RobertD
Let's assume you have a perfectly insulated closed liquid loop with a pump who's motor is cooled by the liquid its pumping. Assume steady state operation, so the pump is operating at some pressure difference and flow rate based on intersection of pump curve and back pressure curve. Does this mean that essentially all of the electrical power input/draw goes into heating the fluid? I first thought that only the inefficiencies of the motor and pump went to heating the fluid and the rest went to doing work on the fluid (maintaining flow against back pressure), but it seems that this work on the fluid goes right back into the fluid as heat due to friction losses in the boundary layer. It seems even if you had a perfectly efficient motor and pump, to where all the power goes to maintaining flow rate against back pressure, you would still be continuously converting the kinetic energy of the fluid to heat because of the frictional losses in boundary layer. Is this correct? Should I assume all pump power turns to heat and ignore efficiency?

Hello Robert,

Not clear to me what you mean with a perfectly insulated closed loop: the liquid also does no work on the outside world so all the energy losses from friction go into heating it ? And you also cool the pump motor with that liquid ? I'd say that yes, apart from a little kinetic energy and the heat content of the pump itself all energy supplied is turned into heat and the only other place it can sit is in the ever increasing liquid. No steady state solution.

 sorry: in the ever increasing heat content of the liuid

Last edited:
RobertD
You can get a handle on the effect of the amount of frictional heating on the temperature rise in the following way: Suppose you have a fluid (like water) flowing through a pipe, and the volumetric flow rate and frictional pressure drop are Q and ##\Delta P## respectively. If you apply the open system version of the first law of thermodynamics to this situation, the change in enthalpy of the fluid through the pipe is zero. Therefore, you get :
$$Q\rho C_p\Delta T=Q\Delta P$$ or
$$C_p\Delta T=\frac{\Delta P}{\rho}$$
Try this equation out for water to see how tiny the temperature rise of the water is for typical pressure drops.

RobertD
BvU said:
Hello Robert,

Not clear to me what you mean with a perfectly insulated closed loop: the liquid also does no work on the outside world so all the energy losses from friction go into heating it ? And you also cool the pump motor with that liquid ? I'd say that yes, apart from a little kinetic energy and the heat content of the pump itself all energy supplied is turned into heat and the only other place it can sit is in the ever increasing liquid. No steady state solution.

Insulated, meaning no heat transfer to or from the loop or pump from the outside. Closed loop, meaning all the fluid that leaves the pump returns to the pump. By steady state operation I meant not during pump start up, rather during nominal operation. I guess its more of a quasi steady state. Agreed, that the liquid temperature would just keep increasing since everything is insulated.

Apply conservation of energy to the system. There is known electrical energy going in. If perfectly insulated no energy leaves the system. So the system must get hotter and hotter.

RobertD
Chestermiller said:
You can get a handle on the effect of the amount of frictional heating on the temperature rise in the following way: Suppose you have a fluid (like water) flowing through a pipe, and the volumetric flow rate and frictional pressure drop are Q and ##\Delta P## respectively. If you apply the open system version of the first law of thermodynamics to this situation, the change in enthalpy of the fluid through the pipe is zero. Therefore, you get :
$$Q\rho C_p\Delta T=Q\Delta P$$ or
$$C_p\Delta T=\frac{\Delta P}{\rho}$$
Try this equation out for water to see how tiny the temperature rise of the water is for typical pressure drops.

I used Q=deltaP*Vol.FR, then plug into Q=mdot*cp*deltaT and solve for deltaT. Looks like this can be simplified to the equation you have. I'm not too concerned with the temperature 'rise', but rather the max temperature the fluid gets too. I'm dealing with a pump that draws about 1300 Watts, but the liquid-to-air heat-exchanger is sized to remove about 3500 Watts from electronics heat going into the fluid and then dump into air. Since the heat entering due to pump is a large percentage of the total system heat load, it has a substantial impact on the temperature that the fluid settles too. I'm wondering if I should assume all 1300 Watts goes into the liquid, or say half that if the pump has 50% efficiency. I think its all 1300 Watts, because the losses due to motor, bearing friction, recirculation, etc go into the liquid as heat, but so does the rate of work to continuously overcome frictional backpressure of flow.

It seems like we are all in agreement that all of the electrical power drawn by the pump goes into the fluid. Thank you all for commenting and helping out, much appreciated!

The rule of thumb we used to use was 2.5 C temperature rise for every 1000 psi frictional pressure drop. So, its typically a very small effect in practice.

The power dissipated by the heat exchanger probably depends on the temperature of the fluid (and ambient air). So to work out the max fluid temperature you need to find out the temperature at which the heat exchanger dissipates 1300W.

The fully closed fully insulated system originally described has now acquired a heat exchanger .

Motor may be rated 1300 W but actual running power might be much less than that . If the system is just sloshing water around with minimal pressure differences then the motor power could well be delivering 100W or even less and heating effect on water could be trivial

In any case any frictional heating of the water would just add a little to the total heat to be dissipated by the relatively massive heat exchanger .

Chestermiller

## 1. How does the amount of pump power affect the heating of the fluid?

The amount of pump power directly affects the heating of the fluid. The more power that is used to operate the pump, the more heat is generated in the fluid. This is because the energy used by the pump is converted into heat as it moves the fluid through the system.

## 2. What is the relationship between pump power and fluid heating efficiency?

The relationship between pump power and fluid heating efficiency is inverse. This means that as pump power increases, the efficiency of heating the fluid decreases. This is because a higher pump power requires more energy input and can lead to more energy loss in the form of heat.

## 3. How can I calculate the amount of pump power that goes to heating the fluid?

The amount of pump power that goes to heating the fluid can be calculated by using the formula P = Q x ΔP, where P is the pump power, Q is the flow rate, and ΔP is the pressure difference across the pump. This calculation will give you the total amount of power used by the pump, but keep in mind that not all of this power will go towards heating the fluid.

## 4. Are there any ways to reduce the amount of pump power that goes to heating the fluid?

Yes, there are several ways to reduce the amount of pump power that goes to heating the fluid. One way is to choose a pump with a higher efficiency, which will require less power to operate. Another way is to reduce the flow rate, as this will decrease the amount of energy needed to pump the fluid. Additionally, using insulation or heat exchangers can help to minimize heat loss in the system.

## 5. How can I optimize the amount of pump power used for fluid heating in my system?

To optimize the amount of pump power used for fluid heating, it is important to consider the specific needs of your system. This includes factors such as the required flow rate, the desired temperature of the fluid, and the type of pump being used. It is also important to regularly maintain and monitor the pump system to ensure it is operating at its most efficient level. Consulting with a professional engineer can also help in determining the best strategies for optimizing pump power and fluid heating in your system.

• Mechanical Engineering
Replies
38
Views
995
• Mechanical Engineering
Replies
8
Views
787
• Mechanical Engineering
Replies
4
Views
2K
• Mechanical Engineering
Replies
19
Views
1K
• Mechanical Engineering
Replies
3
Views
2K
• Mechanical Engineering
Replies
1
Views
2K
• Mechanical Engineering
Replies
15
Views
1K
• Mechanical Engineering
Replies
6
Views
1K
• Mechanical Engineering
Replies
2
Views
819
• Mechanical Engineering
Replies
8
Views
3K