How much of pump power goes to heating the fluid?

[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?

 

[BvU]

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.

[edit] sorry: in the ever increasing heat content of the liuid

Any comments from @Chestermiller ?

 

[Chestermiller]

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]

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.

Any comments from @Chestermiller ?

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.

 

[CWatters]

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]

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.

 

[RobertD]

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!

 

[Chestermiller]

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.

 

[CWatters]

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.

 

[Nidum]

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 .