Typical DO Sag Curve David Reckhow CEE 577 #16 2 1 CEE 577 - - PDF document

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Typical DO Sag Curve David Reckhow CEE 577 #16 2 1 CEE 577 - - PDF document

CEE 577 Lecture #16 10/23/2017 Updated: 23 October 2017 Print version Lecture #16 Streeter Phelps: Reaeration & Dams (Chapra, L22 & L23) David Reckhow CEE 577 #16 1 Typical DO Sag Curve David Reckhow CEE 577 #16 2 1 CEE


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SLIDE 1

CEE 577 Lecture #16 10/23/2017 1

Lecture #16 Streeter‐Phelps: Reaeration & Dams

(Chapra, L22 & L23)

David Reckhow CEE 577 #16 1

Updated: 23 October 2017

Print version

Typical DO Sag Curve

David Reckhow CEE 577 #16 2

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SLIDE 2

CEE 577 Lecture #16 10/23/2017 2

Streeter Phelps Equation

David Reckhow CEE 577 #16 3

This equation can be solved by separation of variables and integration, or by use of an integrating factor. The boundary condition is t = 0 @ D = Do. This yields the DO sag where D = stream deficit at time t, [mg/L] Do = initial oxygen deficit (@ t = 0), [mg/L]

V dD dt k VL k VD

d a

  L L e

  • k t

d

 

D D e k L k k e e

  • k t

d

  • a

r k t k t

a r a

   

  

David Reckhow CEE 577 #16 4

 

D D e k L k k e e

  • k xU

d

  • a

r k xU k xU

a r a

   

  

And recognizing that: t=x/U

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SLIDE 3

CEE 577 Lecture #16 10/23/2017 3

Critical Time

 The most stress is placed on the aquatic life in a stream when

the DO is at a minimum, or the deficit, D, is a maximum. This

  • ccurs when dD/dt = 0. We can obtain the time at which the

deficit is a maximum by taking the derivative of the DO sag equation with respect to t and setting it equal to zero, then solving for t. This yields,

David Reckhow CEE 577 #16 5

tcrit = time at which maximum deficit (minimum DO)

  • ccurs, [days]

crit a r a r

  • a

r d

  • t

= 1 k - k k k D k k k L ln ( 1             

Special Case for D and tc

 When ka and kr are

equal:

         

  • r

c

L D k t 1 1

 

t k

  • r
  • r

e D t k L D

 

David Reckhow CEE 577 #16 6

From Davis & Masten, page 290-291

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SLIDE 4

CEE 577 Lecture #16 10/23/2017 4

Critical concentration

 Once the critical time is known, you can

calculate the cmin

 this is an abbreviated form of the full equation,

which is only valid for tcrit

David Reckhow CEE 577 #16 7

crit rt

k a d

  • s

e k k L c c

 

min

This differs from Chapra’s equation 21.14 on page 397, Why??????

Estimating Reaeration Rates (d‐1)

 O’Connor‐Dobbins formula

 based on theory  verified with some deep waters

 Churchill formula

 Tennessee Valley  deep, fast moving streams

 Owens formula

 British  shallow streams

David Reckhow CEE 577 #16 8

5 . 1 5 .

9 . 12 H U ka 

(U in ft/s; H in ft)

67 . 1

6 . 11 H U ka 

85 . 1 67 .

6 . 21 H U ka 

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SLIDE 5

CEE 577 Lecture #16 10/23/2017 5

The method of Covar (1976)

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Values for ka are in units of d-1

Using the Covar approach

 Determine proper Domain

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H (ft) U (ft/s) Formula <2 Any Owens >2 <1.2H0.34 O’Connor- Dobbins >2 >1.2H0.34 Churchill

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SLIDE 6

CEE 577 Lecture #16 10/23/2017 6

More on ka estimation

 Tsivoglou & Wallace (1972) method

 ka = 0.88US, for Q = 10‐300 cfs  ka = 1.8US, for Q = 1‐10 cfs

 Temperature correction

 =1.024

David Reckhow CEE 577 #16 11

k k

T C T C

 20 20

Dam Reaeration

 Butts and Evans (1983):

David Reckhow CEE 577 #16 12

r abH H T     1 0 38 1 011 1 0 046 . ( . )( . )

Ratio of deficit above and below the dam Difference in water elevation Temperature (oC) Empirical coefficients which relate to water quality and dam type (Table 20.2)

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SLIDE 7

CEE 577 Lecture #16 10/23/2017 7

Wind dependent reaeration formulas

David Reckhow CEE 577 #16 13

 To next lecture

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