Tutor profile: Amanda G.
When a viscous fluid is flowing through a pipe of constant diameter, why does the velocity of flow stay constant? Doesn't the fluid lose energy due to viscosity?
Good question! You're absolutely right to remember that a viscous fluid will lose energy as it flows through the pipe. Let's break this question this down to the basics: When fluid flows through the pipe, one critical thing is always conserved: the mass of the fluid. If we then assume that the fluid is relatively non-compressible, usually a good/fair simplification, we can also assume that the volume of the fluid flowing in and out of the pipe is conserved. We also know the continuity equation (CE): ρ1 A1 v1 = ρ2 A2 v2 It tells us that the product of the density of the fluid, cross sectional area of the pipe, and velocity of the fluid is always constant, regardless of viscosity, mass, or height. So by the CE, as this viscous fluid flows through the pipe, it has constant ρ (density) and constant cross-sectional area. Therefore, the velocity must also stay constant! You might be asking yourself, if the fluid loses energy but the velocity is constant, where does the energy go? This is when it's very important to distinguish between kinetic energy and potential energy, KE vs PE. Velocity is a form of KE, but there are many ways for the fluid to store PE, for example in the form of pressure or height. So, let's imagine that the pipe in your question flows horizontally, with no change in height. Then the viscosity of the fluid would absolutely be decreasing the total energy as it flows! This energy is lost in the form of pressure, potential energy.
Why do induced dipole- induced dipole intermolecular forces (IDID IMFs) increase with the size of atoms or molecules?
Let's start with a little background: IDID, also called van Der Waals forces, occur among all molecules of all polarities, but are best illustrated in two nonpolar molecules. First, let's think of a fluorine ion (F-) versus a bromine ion (Br-). F- is very electronegative (EN) atom on the periodic table, and holds its electrons very tightly. Br- is in the same column as F- but is much larger (both in # of protons and in # of shells). Because Br has so many shells, the e- in its outer shells are held much looser than the outer e- in F-. The inner shells "shield" the negative outer e- from the Coulombic attraction of the positive nucleus, so the outer e- of Br- are much easier to pull off and move around. For these reasons, Br- is more "polarizable" than F-, which is a term used to describe how "squishy" the e- cloud of at atom or molecule is. You might note that IDID strength increases with polarizability. She "squishier" and e- cloud is, the easier it is to move e- around w/in an atom, and the easier it is for another atom or molecule to induce a charge. So that's why if you have a very large, polarizable molecule, it generally has stronger IDID interactions than a smaller, non-polarizable molecule. However, be careful to distinguish between size in terms of # of shells and # of protons! Polarizability (increases IDID) increases down a column, but EN (decreases IDID) increases across a period!
What is the curl of a vector field F (conceptually) and how might we apply this?
The curl of F, in a conceptual sense, is a measure of how non-conservative a vector field F is. It measures the failure of F to be conservative. To apply this concept, a problem might ask: Find the work done along a path A , a circle, through the vector field F where curl F = 0. Let's list what we know: curl F = 0. This tells us that the "nonconservative-ness" of F is zero; in other words, F is perfectly conservative. We also know that in perfectly conservative vector fields, line integrals and "work" is path-independent (the quantities do not depend on the path you take). So in a perfectly conservative field, the work depends only on the start and end points. Take path A: a closed circle. Here, the start and end points are the same, so the work done is: (value at the end) - (value at the start) = 0 because the start and end points are the same in a closed path. For a more thorough understanding of curl, let's also go over a more quantitative take: The curl F of some vector field F <P, Q> is calculated by taking the cross product. curl F = Qx - Py = ∂Q/∂x - ∂P/∂y = Partial of Q w/ respect to x minus partial of P with respect to y The curl of a three dimensional vector field F = <P, Q, R> can be found by taking a cross product of del and the vector F, with ∇ = <∂/∂x, ∂/∂y, ∂/∂z> where del is a vector quantity curl F = ∇ x F
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