p0226

# p0226 - 158 CHAPTER 2. DIMENSIONAL ANALYSIS 2.26 Chapter 2,...

This preview shows pages 1–2. Sign up to view the full content.

158 CHAPTER 2. DIMENSIONAL ANALYSIS 2.26 Chapter 2, Problem 26 Problem: Once transients have settled out, the radial pressure gradient, dp/dr ,ina rotating tank of liquid is a function of the density, ρ , angular velocity, ,andrad ia l distance from the center of the tank, r . Deduce a formula for dp/dr as a function of ρ , and r using dimensional analysis. Solution: The dimensional quantities and their dimensions are [ dp/dr ]= F L 2 · 1 L = MLT 2 L 3 = M L 2 T 2 [ ρ M L 3 , [ 1 T , [ r L There are 4 dimensional quantities and 3 independent dimensions ( M,L,T ) , so that the number of dimensionless groupings is 1. The appropriate dimensional equation is [ dp/dr ]=[ ρ ] a 1 [ ] a 2 [ r ] a 3 Substituting the dimensions for each quantity yields ML 2 T 2 = M a 1 L 3 a 1 T a 2 L a 3 = M a 1 L 3 a 1 + a 3 T a 2 Thus, equating exponents, we arrive at the following three equations. 1= a 1 2= 3 a 1 + a 3 a 2 We can solve the first and third equations immediately for a 1 and a 2 ,v iz . , a 1 =1 and a 2 =2 Substituting into the second equation, we find

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

### Page1 / 2

p0226 - 158 CHAPTER 2. DIMENSIONAL ANALYSIS 2.26 Chapter 2,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online