Class Notes (1,100,000)
US (490,000)
UC-Davis (10,000)
ECH (20)
Lecture 5

ECH 140 Lecture Notes - Lecture 5: Chemical Engineering

Engineering Chemical
Course Code
ECH 140
Peter Streove

This preview shows half of the first page. to view the full 1 pages of the document.
Fall 2012
University of California at Davis
Department of Chemical Engineering and Materials Science
ECH 140
HW #4
Problems (10 points unless otherwise indicated)
#1) Consider steady-state heat transfer in a wall. The wall divides two environments: one at
temperature T1 and the other at temperature T2. In the x-direction the wall is 1 foot thick, in the y
and z directions the wall is 100 ft long. Starting with Laplace’s equation for steady-state heat
transfer in 3 directions, perform order of magnitude analysis to determine if you can neglect any
terms in Laplace’s equation.
#2) Solve the ODE:
(d2y/dx2) + x(dy/dx) = ax
#3) Species mass transfer in a fluid flowing between a parallel plates conduit can be given in terms
of the governing differential equation known as the convective-diffusion equation:
vz(C/z)=D[ (2C/y2) + (2C/z2) ]
,where vz is the velocity in the z direction, C is species concentration and D is the species
diffusivity. The term on the left-hand side (LHS) accounts for convective mass transfer in the flow
direction z (the axial direction). The terms on the right hand side (RHS) account for diffusion in the
y direction and diffusion in the z direction, respectively. Determine when diffusion in the axial
direction can be neglected relative to the convective term.
#4) 30 points. Consider unsteady state diffusion of a species into a slab. The thickness of the slab
is b. At time before zero, the slab is at a uniform concentration Co. At time t greater than zero the
surfaces of the slab are maintained at concentration C1. The PDE of concern is;
(C/t)=D(2C/x2) where the LHS term accounts for the time dependency and
the RHS term for diffusion in the slab. Since the problem is symmetric we can solve for half the
slab, using the following BCs, after placing the coordinate system at the middle of the slab:
#1) t<0, C=Co, all x
#2) t>0, C=C1, x=b/2
#3) any t, (C/x)=0, x=0
Using separation of variables obtain the solution for C.
You're Reading a Preview

Unlock to view full version