EME 811
Solar Thermal Energy for Utilities and Industry

5.3. Pumping Power Considerations


Pumping energy is one piece of the puzzle that is important to consider. Pumping power is calculated as the volume of the fluid per unit time (flow capacity) times the density of the fluid times the gravitational constant times the pumping head (vertical distance to be pumped). Pumping energy is simply power multiplied across time. 100kW of power for one hour is 100kWh of energy. Units must be tracked carefully to ensure the correct answer. Friction within the pipe, particularly for pumping over horizontal distances, can be calculated using the Darcy-Weisbach equation (noted in the second video) to relate friction and fluid speed. The head loss due to friction, and as such the required pumping power, is proportional to the square of the fluid speed. As such, this is an important calculation because if your system is designed in a way that requires high pumping speeds, you will have very high pumping energy costs. Additionally, if you pump too slow, you risk damaging your fluid and system components because of high temperatures gained from the solar radiation and not moving that energy through the system, away from the collectors, fast enough. This optimization problem is key to designing a good system.

The video below explains how we can estimate the pumping power required to move the heat-transfer fluid in the system. This kind of calculations becomes handy when you need to determine the cost of using one or another type of fluid for a specific system design. So, watch and see what parameters of the system and fluid need to be taken into account.

Example of calculating pumping power
Click for transcript.

PRESENTER: OK, so this is going to be an example of pumping power. And we're going to use water for now. But essentially, the only difference is the density. So for this equation-- and so if you had a fluid that was denser or less dense, for example, like oil, it would be pretty comparable to this calculation. So power is equal to basically the volumetric flow per unit time times the density of the fluid, times earth's acceleration due to gravity-- acceleration due to gravity on earth, which is 9.81 meters per second, you may recall from physics class, potentially even in high school-- times the head of the pipe that you're in. The distance. And that's essentially it. And you have to just be careful of the units. So if, for example, we have fluid that's traveling 1 meter cubed per hour-- that's the volumetric flow rate-- and the density of our fluid is water, so 1,000 kilograms per meter cubed. And we know that g is 9.81 meters per second squared, acceleration. And let's pick a distance of 10 meters. We want to pump our fluid up 10 meters, essentially. What we end up with, 1 meter cubed per hour times the density is 1,000 kilograms per meter cubed-- and those cancel-- times 9.81 meters per second squared and-- oops. Meters did not cancel with anything yet. Sorry about that. And our distance is 10 meters. So you can see here in the numerator we have kilograms, meters squared, per second squared. There's kilograms, meters squared, per second squared. Kilogram, meters squared, per second squared is equal to a joule unit of energy. We also have time still in the denominator over here. So if we have joules per time, that's power, but we need a conversion factor. We need to say that one hour is 3,600 seconds. So then we can cancel hours, and we're left with seconds, and a joule per second. So a joule per second is equal to a watt. So we get it in watts if we do that. So when we multiply all of those pieces together-- handy-dandy calculator here. 1,000 times 9.81 times 10 meters, divided by 3,600. Divided by 3,600. That would be the error I just typed in my calculator. For all I know, you've actually already done this quicker than I have. You end up with 27.25 watts as your pumping power to achieve that flow rate of 1 meter cubed of fluid per hour. That's the continuous pumping power required to do that. So just real quick here. If we wanted to know the energy, say, over a four-hour period, 27 and 1/4 watts for four hours is 109 watt hours, or 0.11 kilowatt hours of energy. So in this case, we weren't pumping too fast. 1 meter cubed per hour is not that high of rate. We weren't pumping it very far either. Only 10 meters up against gravity. So it's pretty low-cost, low-energy result. But as you scale that up over, say, an entire multiple acres of collectors or whatever, you would definitely see high pumping costs. Thanks.

The example above shows how we can estimate the pumping energy for pumping fluid upwards over a certain vertical distance. The video below shows the case when the fluid moves through horizontal pipes, which is quite common of solar applications. How much energy would be needed in this case? Watch as see how this calculation is done using the Darcy-Weisbach equation.

Example of calculating HeadLoss
Click for transcript.

PRESENTER: So in the previous example, calculated the pumping energy and power requirements for pumping water up 10 meters. In most cases in a solar array, there's horizontal piping, not vertical piping. You're piping the fluid through the collector across a big horizontal surface area. So that still requires energy because of the friction in the pipe. So how do you translate that to an equivalent head loss, essentially? And we do this with the Darcy-Weisbach, which says-- let's see here-- Darcy-Weisbach equation, which says that the head loss due to friction is equal to Darcy friction factor, which can be looked up based on your various fluid parameters, such as the Reynolds number, whether your flows turbulent or laminar, things like that, times the length of your pipe. So if you're pumping it 1,000 meters, that would go there and then the internal diameter of your pipe. Oh, back to the Darcy friction factor, that also has to do with the roughness of the pipe. So if you have a very smooth pipe, then you would get a lower friction factor. Times the average velocity of the fluid squared. Divide by 2 times the acceleration on Earth, 9.81 meters per second squared. So you can see already that the head over here is-- h sub f, the thing we're calculating, is affected by the square of velocity, which shows that as speed increases in your fluid, you're going to have a much higher head lost, which from the previous calculation shows you have much higher energy. So if we have a Darcy friction factor of say, 0.2 and that's unitless and we want to say, do a pipe that's 1,000 meters long with an internal diameter of about an inch, 0.03 meters, we're going to do it for two different philosophies here. So let's say that the average velocity for the first round is 3 meters per second. And we know that g is 9.81 meters per second squared. And we plug all these values in-- 0.02 times 1,000 meters, diameter of 0.03 meters. Velocity is 3 meters per second squared. 2 times 9.81 meters per second squared. So you can see some of these units cancel out, meters, meters, seconds squared, seconds squared. We have meters squared, meters in the denominator, so we're going to end up with meters, because there's two meters up top, one below. So in the end, this is equivalent to units of meters. Once we crunch the numbers, times 1,000 divided by 0.03 times 3 squared divided by 2 divided by 9.81, we end up with 306 meters of head loss. So what this says is that one pipe that's 1,000 meters long and one inch in diameter at 3 meters per second has the same essentially energetic requirements as pumping fluid with no friction losses but straight up 306 meters vertically on Earth. For on the moon, it would take less energy, just as a side note, because the moon has 1/6 the gravitational constant. So it does matter these types of things, though I doubt you'll be installing a solar collector on the moon for now. But just a little side note there, all these little details do matter. And so let's see how this changes if instead of 3 meters per second average velocity, we have say, 1/2 a meter per second average velocity. So up here, this 3 would change to be 0.5 and we would run that calculation again. And we would get-- let's see here-- 0.2 times 1,000 times 0.5 squared instead of 3 squared divided by 2 and 9.81. Hold on. I forgot to divide by 0.3-- 0.03, I mean. There we go. That's a better number. There we get about 8.5 meters for the second version of this calculation. And so you can see that by reducing the flow speed to 0.5, we have much less head loss or much less pumping energy required. And so the implications of this are really that you can save a lot of energy by pumping slower. At the same time, in a solar thermal energy collection application, that means your fluid would get hotter a lot faster. And so it could be a good thing under certain circumstances and in others, it could be bad, because your fluid could be overheating if it's going too slow through your collector. So this becomes an optimization problem where you have to balance all these different things happening at once, the fluid flow, as well as the rate of energy being absorbed, as well as what those maximum temperature thresholds are for good operation of your system without damaging the fluid or any other components, as well. That's sort of a fine line that has to be walked to ensure that your system is working correctly. So hopefully, that gives you a little bit of an insight into that on a more technical level. And thanks for listening.