8. FLUID FLOW: mechanical energy equation CH EN 170

Fluid Flow What’s a fluid? Anything that flows—practically, a liquid or a gas. So far, you have probably studied physics in relation to solid bodies (blocks sliding down slopes, balls banging into each other) But physics can also be applied to study the way fluids flow Understanding the way fluids flow is one of the most important skills for a chemical engineer

Why is this important? In industrial chemical engineering, understanding fluid flow helps us understand the behavior of pipes, pumps, and valves. Can you think why fluid flow is important for biomedical engineering?

Why is this important? Modeling fluid flow in biomedical devices: Artificial Heart Valve Oxygen Mask

Fluid Flow Schedule Over the next three classes, you are going to learn the basics of analyzing fluid flow Today: Discussing the theory of fluid flow Understanding the mechanical energy equation Tuesday Short exam review Examples of working problems with the mechanical energy equation Next Thursday Pumps Efficiency, pump curves, power

+ = = Conservation of mass REMEMBER THIS? Rate of mass flowing in. Rate of mass flowing out. Rate of accumulation of mass. = + REMEMBER THIS? A total mass balance like this also helps us solve fluid flow problems. For this class everything we do will be steady state: Rate of mass flowing in. Rate of mass flowing out. = 𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡 What useful things does this tell us?

Imagine a fluid flowing through a nozzle (like on a garden hose): Outlet Inlet 1 2 𝐷1 𝐷2 𝐷1 𝐷2 1 2 Outlet 𝑚 𝑖𝑛 = 𝑚 𝑜𝑢𝑡 Inlet 𝑚 1 = 𝑚 2 𝜌 1 𝑉 1 = 𝜌 2 𝑉 2 What happens to this equation if the fluid is incompressible (it’s density doesn’t change)? 𝑉 1 = 𝑉 2 We can almost always treat a fluid as incompressible if it is: A liquid A gas flowing at low speeds

𝑙𝑒𝑛𝑔𝑡ℎ 3 𝑡𝑖𝑚𝑒 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑡𝑖𝑚𝑒 × 𝑙𝑒𝑛𝑔𝑡ℎ 2 𝑉 = 𝑣 𝑎𝑣𝑔 𝐴 Inlet Outlet 𝐷1 𝐷2 1 2 𝐷1 𝐷2 If we take a slice of our nozzle at any point, the volumetric flow rate can be expressed as: The average velocity of the fluid flowing through that slice multiplied by the area 𝑙𝑒𝑛𝑔𝑡ℎ 3 𝑡𝑖𝑚𝑒 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑡𝑖𝑚𝑒 × 𝑙𝑒𝑛𝑔𝑡ℎ 2 𝑉 = 𝑣 𝑎𝑣𝑔 𝐴 So if we take the equation we just derived for a single stream of incompressible fluid: 𝑉 1 = 𝑉 2 𝑣 1 𝐴 1 = 𝑣 2 𝐴 2

𝑣 1 𝐴 1 = 𝑣 2 𝐴 2

Example: Nozzle A liquid enters a horizontal nozzle with an initial diameter of 2 cm and a final diameter of 1 cm. If the fluid enters the pipe at 1 m/s, what is the velocity of the fluid leaving the pipe? Life hack: solve symbolically first. Life hack 2: What’s the area of a circle in terms of D (instead of R for radius)? =𝜋 𝐷 2 2 = 𝜋 4 𝐷 2 𝑣 1 𝐴 1 = 𝑣 2 𝐴 2 𝐴=𝜋 𝑅 2 Inlet Outlet 𝐷1 𝐷2 1 2 = 𝑣 1 𝜋 4 𝐷 1 2 𝜋 4 𝐷 2 2 = 𝑣 2 𝐷 1 𝐷 2 2 = 1 𝑚 𝑠 2 𝑐𝑚 1 𝑐𝑚 2 𝑣 2 = 𝑣 1 𝐴 1 𝐴 2 =4 𝑚 𝑠

Conservation of energy Just like conservation of mass allows us to do a mass balance, conservation of energy allows us to do an energy balance. To understand fluid flow, we use a mechanical energy balance For steady-state incompressible flow, it looks like this: 𝑃 2 𝜌 + 1 2 𝛼 2 𝑣 2 2 +𝑔 𝑧 2 − 𝑃 1 𝜌 + 1 2 𝛼 1 𝑣 1 2 +𝑔 𝑧 1 = 𝑤 𝑠 − 𝑤 𝑓

Understanding the mechanical energy Equation 𝑃 2 𝜌 + 1 2 𝛼 2 𝑣 2 2 +𝑔 𝑧 2 − 𝑃 1 𝜌 + 1 2 𝛼 1 𝑣 1 2 +𝑔 𝑧 1 = 𝑤 𝑠 − 𝑤 𝑓 Energy per mass being added or taken away from the fluid stream by work. Energy per mass of the fluid stream at point 2 (leaving the system) Energy per mass of the fluid stream at point 1 (entering the system.)

𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 A different form 𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 𝑃 1 , 𝑣 1 , 𝑧 1 𝑃 2 , 𝑣 2 , 𝑧 2 pressure, velocity, height 𝑤 𝑠 - energy added by pump 𝑤 𝑓 - energy lost to friction I want you to understand what this equation means.

1 2 𝛼 𝑣 2 2 − 𝑣 1 2 Kinetic energy term v is an average velocity 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 v is an average velocity For math reasons, this means we have to add the kinetic energy correction factor: 𝜶 For “laminar” (smooth) flow: 𝛼≅2 Laminar flow is typically thick and slow (think honey spilling out of a jar) For “turbulent” (rough) flow: 𝜶≅𝟏 We can usually assume that flow is turbulent!

𝑔( 𝑧 2 − 𝑧 1 ) Potential energy term z = 0 ft z = 10 ft z = 20 ft z = -10 ft z = 0 ft z = 10 ft z is height or elevation compared to some reference height It’s important to use the same reference for both z values Up is always positive (going up adds PE) g is positive in this equation 𝒈=𝟗.𝟖𝟏 𝒎 𝒔 𝟐 =𝟑𝟐.𝟐 𝒇𝒕 𝒔 𝟐

Pressure term 𝑃 2 − 𝑃 1 𝜌 We can think of pressure as a way of storing mechanical energy Think of a dam: The weight of the water behind the dam (potential energy!) increases the pressure on the dam If the dam breaks, the pressure will be gone, and the kinetic energy of the water will increase Read section 7.2 in your book (about pressure) before Tue. Make sure you understand the meaning of: absolute pressure, gauge pressure, atmospheric pressure

𝑤 𝑠 Shaft work This stands for shaft work. It represents pumps or turbines (they have a shaft that spins) Remember from physics—work adds energy to a system or takes it away. A pump moves a fluid. It adds energy to a fluid stream ( 𝒘 𝒔 is positive) A turbine generates uses a fluid stream to generate electricity. It takes energy from a fluid stream ( 𝒘 𝒔 is negative) Pump Turbine

Friction − 𝑤 𝑓 Friction In physics class, you’ve probably learned that two surfaces rubbing against each other produce friction When a fluid flows against a surface (like the inside of a pipe) this also produces friction The fluid loses energy from this (often in the form of heat) 𝒘 𝒇 will always be positive, but the minus in the equation makes the whole term negative. Sometimes (for example a smooth pipe) we can assume that friction is negligible. Friction

𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 1 2 Pump Turbine If one of the terms on the left is negative that indicates a decrease in that type of energy. In which of the systems pictured above does potential energy decrease? Remember this is a balance. If one type of energy decreases, that energy has to go somewhere. Where might that potential energy go? Increasing velocity, increasing pressure, out of the fluid (friction, as heat), out of the fluid (turbine)

𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 1 2 Pump Turbine Remember, 𝑤 𝑠 for a turbine is negative because the turbine takes energy out of the fluid. Friction also takes energy away—so the entire right side of the equation is negative. What does that mean mathematically for the left side? What does that mean physically for the energy of the fluid?

𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 1 2 Pump Turbine A pump is the opposite. 𝑤 𝑠 is positive, and the pump puts energy into the fluid. Physically, where might that energy go? Replacing energy lost by friction, increasing pressure, increasing velocity, raising elevation Do you see how this makes sense mathematically? As you study engineering, you should gain an intuitive feel for how energy shifts around and the kinds of physical effects this can have on systems.

Example: open water tank The examples above showed fluid flowing up or down hill and hooked to a pump or turbine. But we can adapt the mechanical energy equation to describe lots of different systems. Let’s look at a simple system: a tank of water open to the atmosphere. We can easily measure: ℎ, the height of the water 𝑃 𝑎𝑡𝑚 , the pressure of the atmosphere pressing down at the top of the water And we can look up the density of water, 𝜌 Let’s find the pressure at the bottom of the tank in terms of ℎ, 𝑃 𝑎𝑡𝑚 , and 𝜌

Example: open water tank First, we need to choose where to put Point 1 and Point 2. A good idea is: one point where we have information, one point where we need information. Second: lay out the information we have about each point Third: simplify the mechanical energy equation and solve. 𝑃 1 = 𝑃 𝑎𝑡𝑚 𝑣 1 =0 𝑃 2 − 𝑃 1 𝜌 + 1 2 𝛼 𝑣 2 2 − 𝑣 1 2 +𝑔( 𝑧 2 − 𝑧 1 )= 𝑤 𝑠 − 𝑤 𝑓 𝑧 1 =ℎ 1 𝑃 2 − 𝑃 1 𝜌 =𝑔 𝑧 1 Potential energy at top goes to pressure at bottom 𝑃 2 = 𝑃 1 +𝜌𝑔 𝑧 1 = 𝑃 𝑎𝑡𝑚 +𝜌𝑔ℎ 2 𝑧 2 =0 𝑣 2 =0 𝑃 2 =?