Fluid Dynamics for Brewing Lecture 2 Fluid Dynamics Lecture 2- P 1

Problem set #2 will cover these topic areas Learning Objective – Goals for Today Head Loss System differential pressure concept System flow characteristics Head Loss due to elevation Head Los due to pipe friction Head Loss due to system components Overall system head loss and pressure drop Pump sizing Problem set #2 will cover these topic areas Fluid Dynamics Lecture 1- P 2

Head Loss in Piping Systems System head losses can be calculated in a way that returns a numerical pressure value that has units of pressure expressed as a height-equivalent of water. For example, if a total head loss value is stated as “3 feet”, this means that the pressure that must be overcome by the pump in order to move liquid through the system is a pressure that is equivalent to the pressure exerted by a column of water that is 3 feet high. Pressures expressed in units of “feet of water” (ftH2O) or “inches of water” (inH2O) can be converted to other pressure units using conversion factors. 1 psi = 0.4335 ftH2O = 0.03613 inH2O

Head Loss in Piping Systems The pressure that is exerted by a column of water increases with increasing column height Think of a time when you were in a swimming pool and dove down. When you dove deeper, you had a larger “column” of water above you than when you were only slightly below the surface, so you experienced a larger pressure when you were deeper. Water P1 P2 P3 P3 > P2 > P1

Head Loss in Piping Systems – hLelev Let’s now discuss how we calculate the various components of head loss within a piping system. Elevation head loss, hLelev, is easily calculated by determining the total height differential between the pump discharge and the liquid discharge point within the system to which the liquid is to be pumped. The total height is then equal to the total head in units of ft (ft of H2O). If we use the diagram from before and assume a height of 20 ft we see: Tank #1 Tank #2 Height = 20 ft hLelev = 20 ft Fluid Dynamics Lecture 2- P 5

Head Loss in Piping Systems – hLpipe Calculating head loss due to pipe friction, hLpipe, is a bit more complicated The amount of friction and viscosity-induced resistance to flow within a pipe is dependent upon many variables including: Velocity of fluid in the pipe Viscosity of the fluid Density of the fluid Geometry of the pipe (length & diameter) Relative smoothness of the pipe etc…. Fortunately, dedicated engineers and scientists have done the work to develop equations and correlations that allow us to account for the many factors that contribute to head loss due to pipe friction. Fluid Dynamics Lecture 2- P 6

Head Loss in Piping Systems – hLpipe To use the Moody Diagram, first determine relative roughness, select the appropriate curve, determine the Reynolds number, then read off the value of f from the vertical axis. f 3 1 2 Fluid Dynamics Lecture 2- P 7

Head Loss in Piping Systems – hLpipe As an alternative to using the Moody diagram, there are many correlations that have been developed over the years in order to mitigate the need to use a reference chart, and to have an equation, or set of equations, that can be readily used to calculate f using computer programs or spreadsheets. One such set of equations are referred to as the Churchill Equations: The Churchill Equations are explicit in the variables e/D and NRe and are valid across all ranges of those variables of interest for real-world applications. Fluid Dynamics Lecture 2- P 8

Head Loss in Piping Systems The need to lift fluid against the force of gravity is the overwhelmingly dominant contributor to the overall system head loss in our example. Admittedly, our example did not include very much pipe (only 80’), so you would probably intuitively expect head loss from pipe friction to be relatively small. But, in piping systems, especially relatively small piping systems, in which a relatively low-viscosity, Newtonian fluid (e.g. water, beer…) is being transported under turbulent flow conditions, it is very common that the need to lift the liquid against gravity is the dominant contributor to head loss. In a piping system that is transporting a high-viscosity fluid (e.g. thick lubricating grease, honey, ketchup….) or a non-Newtonian, shear thickening substance, head loss due to pipe and system-component friction becomes a much more dominant contributor to overall total head loss.

Pump Sizing Now that we have evaluated the total head loss in our example piping system, we are in a position to address one of the most important components within a fluid transport system: the Pump There are many important things to consider when specifying and selecting a pump for a particular application (and we will discuss some of those things later) but, as a pump user, the most fundamental thing that must be considered is an answer to the very general question “how big does the pump need to be to do the job?” In order to understand how big the pump needs to be, we need to first do a better job of properly specifying what we mean by “do the job”. Let’s talk in a very general way about how pumps are characterized Fluid Dynamics Lecture 2- P 10

Pump Sizing Pumps are characterized by the liquid flow rate that can be achieved when pumping against a certain amount of pressure For example, a specification by a pump manufacturer might state that a pump is “capable of 3.5 gpm flow at 14 feet of head”. This means that this particular pump is able to pump a particular liquid at a rate of 3.5 gallons/minute against a back-pressure that is equivalent to the pressure exerted by a column of water that is 14 feet high. Fluid Dynamics Lecture 2- P 11

Pump Sizing In order for a pump to move liquid, it must do work on the liquid, and work requires energy. Doing more work requires more energy. The rate at which work is done (work per unit time) is the definition of power. Moving more liquid per unit time requires more power. Moving liquid faster requires more power. Moving liquid against a greater resistance (pressure) requires more power. Fluid Dynamics Lecture 2- P 12

Pump Sizing The power to drive the pumps used in breweries (and most everywhere else) is supplied by an electric motor. The amount of power required to move a liquid is dependent upon the liquid flow rate, the back pressure within the system and the efficiency of the motor. An equation that shows this relationship is: 𝑃𝑜𝑤𝑒𝑟= 𝑄 ∆ 𝑃 𝑡𝑜𝑡𝑎𝑙 1714 𝜂 Where: Power = Motor Drive Power (h.p.) Q = Flow rate (gallons/minute or gpm) DPtotal = total system back pressure or head (pounds/in2 or psi) h = overall pump efficiency

Pump Sizing Overall pump efficiency, h, is expressed as the decimal form of a % (e.g. 85% = 0.85) Pump drive motor efficiencies vary depending upon the size of the motor. Larger drive motors usually have higher efficiency ratings. Here is a chart showing data for a particular type of pump drive supplied by this particular vendor: Fluid Dynamics Lecture 2- P 14

Pump Sizing Here’s the same data with the chart adjusted to focus on the region between 0 – 15 h.p. on the x-axis: Pump drive efficiencies between 60-75% are typical for 0.5-2 h.p. pump drives. Pump drive efficiencies between 75-85% are typical for 2-15 h.p. pump drives,

Pump Sizing Let’s look at our piping system example that we have been using and try to estimate the size of the pump that we will need to obtain our desired flow within the system. Tank #1 Tank #2 Pump Valve Fluid Dynamics Lecture 2- P 16

Pump Sizing Here’s a reminder of what we have, and what we know so far that is relevant to sizing the pump: Liquid flow rate is 30 gpm hLtot = 31 ft Efficiency can be estimated using vendor data: Power requirements are given by: 𝑃𝑜𝑤𝑒𝑟= 𝑄 ∆ 𝑃 𝑡𝑜𝑡𝑎𝑙 1714 𝜂 Tank #1 Tank #2 Pump Valve Fluid Dynamics Lecture 2- P 17

Pump Sizing There are two things we need to do: Assume a value for pump drive efficiency Convert the value for total system head loss, hLtot, into units of psi Use vendor data, if available, to help you make assumptions about a reasonable value for pump efficiency. Based on the data we have, and my “engineering judgment” I will assume a value of h = 65% for pump drive efficiency (based on experience, I think it will be a relatively small drive and therefore relatively inefficient….). Next, convert the units of hLtot = 31 ft into units of psi using the conversion factor: 1 ftH2O = 0.4335 psi 31 𝑓𝑡𝐻2𝑂 1 𝑥 0.4335 𝑝𝑠𝑖 1 𝑓𝑡𝐻2𝑂 =13.4 𝑝𝑠𝑖 hLtot =13.4 psi Fluid Dynamics Lecture 2- P 18

Pump Sizing We now have everything that we need in the correct units: Liquid flowrate = Q = 30 gpm DPtotal = hLtot = 13.4 psi Efficiency = 65% = 0.65 Plugging these values into the equation: Gives: 𝑃𝑜𝑤𝑒𝑟= 𝑄 ∆ 𝑃 𝑡𝑜𝑡𝑎𝑙 1714 𝜂 𝑃𝑜𝑤𝑒𝑟= (30 𝑔𝑝𝑚) (13.4 𝑝𝑠𝑖) 1714 (0.65) =0.36 ℎ.𝑝. To pump 30 gpm in our example system, we need at least a 0.36 h.p. pump.

Pump Sizing Another example for practice: You are the head brewer at a large, regional brewery. The brewery is expanding and installing 4 new lauter tuns. A piping system with a single pump provides sparge water to the spray nozzle headers in all 4 lauter tuns. The overall head loss associated with the system, including all 4 lauter tuns, is 95 psi . You need a pump that can provide at least 30 gpm to the spray nozzle header in each lauter tun. It is possible that all four lauter tuns might need to be sparging at the same time. What size pump is required? Fluid Dynamics Lecture 2- P 20

To pump 120 gpm in this system, we need at least an 8.3 h.p. pump. Pump Sizing Here’s what we know from the problem statement, and can assume: We need to be able to deliver 4 x 30 gpm = 120 gpm sparge water Total system pressure is 95 psi This is likely a bigger system than in our previous example, so assume efficiency of 80% (0.80) Use: Plug in the values: 𝑃𝑜𝑤𝑒𝑟= 𝑄 ∆ 𝑃 𝑡𝑜𝑡𝑎𝑙 1714 𝜂 𝑃𝑜𝑤𝑒𝑟= (120𝑔𝑝𝑚) (95𝑝𝑠𝑖) 1714 (0.80) =8.3 ℎ.𝑝. To pump 120 gpm in this system, we need at least an 8.3 h.p. pump.

Head Loss & Pump Sizing – Questions? Do you have any questions specifically about system head loss determination or pump sizing? Fluid Dynamics Lecture 1- P 22