Homework _6.key - PGE 323 Homework#6 KEY May 2 2006 Pope 1...

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

View Full Document Right Arrow Icon
PGE 323 Homework#6 KEY May 2, 2006 Pope 1. Accumulation mass = mass in - mass out ( 29 ( 29 ( 29 w w w w w w w w t t t r r r 2 r rh S S 2 rh u 2 rh u t +∆ +∆ π ∆ φρ - φρ = - π ρ - π ρ (a) Dividing by r r t ∆ ∆ and taking the limits gives the mass balance: ( 29 ( 29 w w w w S r u 2 h 2 h t r r φρ ρ π π = - Α ssuming constant φ and ρ w gives: ( 29 w w ru S 2 h 2 h t r r G G π π φ = - Use definition of fractional flow w w u f u = to get: ( 29 w w ruf S 2 h 2 h t r r G G π π φ = - q 2 hru cons tan t = π = ( 29 w w qf S 1 2 h t r r G G - π φ = w w S f q t 2 h r r G - = π φ (b) Mass balance in terms of water saturation: w w w w S df S q t 2 h r dS r G - = π φ w S Sw dr t Sw dt r G = - w w w w S f q t S 2 h r S r G G G - = π φ (c) The velocity of a front of constant water saturation is:
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 2

Homework _6.key - PGE 323 Homework#6 KEY May 2 2006 Pope 1...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online