CHEN 3005 Lecture Notes - Lecture 9: Momentum, Saurer, Newtonian Fluid

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9 Sep 2014

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Why is this linear? z x pressure: acts uniformly viscous stress (newtonion fluidsfluids w/o microstructure) Unknowns (4 unknown things): vx, vy, vz, p (pressure) Equations: conservation of mass, conservation of linear momentum { vx, vy, vz}all = momentum density. Problem: fix bottom plate at 0, top plate moves and finding velocity profile. Iz: infinity extended in y-direction for left/right for top/bottom. V = x y z mass: m = v = x y z dm/dt = x y z(d /dt) Volumetric flow rate: v. da (this corresponds to flow rate outsince it is in positive x direction) mass flow rate: v. da put it all together. X y z(d /dt) = vx(x,z) y z - vx(x + x, z) y z on right side, 1st term is in, 2nd term is out d /dt = (vx(x,z) - vx(x + x, z))/ x. Change in momentum = convection (convective terms) + surface terms + body terms.

Related Questions

1. Equation shows the relation between T and V when an ideal gas is adiabatically and reversibly expanded or compressed from T₁V₁ to T₂V₂,

a) Prepare a table in an Excel spreadsheet showing the values of V, T, and p (i.e., three columns) when 200 g. CH₄(g) is reversibly and adiabatically expanded starting at T₁ = 350K, V₁ = 0.010 m³ to a final volume of 200 L. Assume ideal gas behaviour. Needed heat capacity data can be found in the Lecture Notes Appendix.

In the first column, use values of V starting from V₁ = 0.010 m³ to 200 L (0.200 m³) in steps of 0.010 m³ . In column 2 calculate each new T (eq. 2.42) and in column 3 the corresponding new pressure. Remember eq. 2.29.

b) Next prepare two separate graphs, one of T (y-axis) vs. V (L or m³, your choice) and one of p (y-axis) against V using the values from the columns in a). Present your table and graphs on the same page. Make sure your graphs include a title for each, axis legends (x-values V (m³), y-values T (K) or p (Pa).

Components of U

The internal energy, U, is the total energy of a system. For any isolated system, the internal energy is constant. U is a state function, meaning that any path used in calculating ?U will result in the same answer.

For any pure substance or fixed mixture of substances, the internal energy, U, can be determined from any two of the variables P, V, and T. It is often most convenient to choose V and T as the variables. It is helpful, then, to examine how U depends on V and T independently. Because U is a state function, ?U only depends on Vi,Ti and Vf,Tf.

The cyclic rule can be used to rewrite the differential in terms of the information available. Recall that the cyclic rule uses each variable in the numerator once, in the denominator once, and as the fixed value once. So for variables x, y, and z, it would be written as

(?x?y)z(?z?x)y(?y?z)x=?1

Part A - The dependence of U on V

Given the relationship

(?U?V)T=T(?P?T)V?P,

use the cyclic rule to write (?U/?V)T in terms of the measurable quantities P, ?, T, and ?. Recall that ? and ? are the isobaric volumetric thermal expansion coefficient and the isothermal compressibility, respectively, defined by

?=1V(?V?T)Pand?=?1V(?V?P)T

Express your answer in terms of P, ?, T, and ?.

CHEN 3005 Lecture Notes - Lecture 9: Momentum, Saurer, Newtonian Fluid

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