{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW2 2009 Solution

# HW2 2009 Solution - SPRING 2009 V PF = Qt d = Q k 1 S 1 S e...

This preview shows pages 1–3. Sign up to view the full content.

CEE 4310 WATER QUALITY ENGINEERING SPRING 2009 Problem 3 Solution

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

View Full Document
CEE 4310 WATER QUALITY ENGINEERING SPRING 2009 Problem 1 Solution For a PF system, V Q k ln S S PF e 0 = - For a CM system, V Q S S 1 k CM 0 e = - Taking the ratio of the above two equations: V V Q S S 1 k Q k ln S S S S 1 ln S S CM PF 0 e e 0 0 e e 0 = - - = - - The table below contains the results of the calculations for the removals specified above. Removal, % 50 75 90 95 99 S e /S 0 0.50 0.25 0.10 0.05 0.01 V CM /V PF 1.44 2.16 3.91 6.34 21.5 Problem 2 Solution Making a mass balance for an elemental volume of a PF reactor: QS - Q(S + S) - kS 2 A x = 0 Q A S x kS 2 = - Taking the limit and applying the chain rule, v dS dx dS dx dx dt dS dt kS 2 = = = - Performing the integration, dS S k dt 2 0 t S S d 0 e = - 1 S 1 S kt e 0 d - = t d = V PF /Q
CEE 4310 WATER QUALITY ENGINEERING
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SPRING 2009 V PF = Qt d = Q k 1 S 1 S e- For a CM reactor, QS-QS e-kS V e 2 CM = 0 ( 29 V Q S S kS CM e e 2 =-Taking the ratio, ( 29 ( 29 ( 29 e e e e 2 e e e 2 2 e e e e 2 e e PF CM S S S S S S S S S S S S S S S S S S S S 1 S 1 k Q kS S S Q V V =--=--= --= --= The table below contains the results of the calculations for the removals specified above. Removal, % 50 75 90 95 99 S e /S 0.50 0.25 0.10 0.05 0.01 V CM /V PF 2.00 4.00 10.0 20.0 100 Problem 4 Solution...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

HW2 2009 Solution - SPRING 2009 V PF = Qt d = Q k 1 S 1 S e...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online