1

CEE 330, Spring 2011

Solution for HW#3 Solution 1

Solution 2

a)

b) first order decay

3 33

0

10 4000 25 4000 ( ) (20, 000 ) (0.25 ) ( )

11

in out rxn

in out out

out out

out

dm m m mdt

QC QC VkC

m mg m C m Cday L day day

mgCL

= − ±

= − −

= × − × − × ×

=

( )

3

2

10.20 0.6

distance 50, 000 152, 083 0.6velocity 86, 4009.6

10

4.7

4.2

ktt o

daysday

t

t

C C em dayt s days

m ss

m

C e

mgCL

−

⎡ ⎤⎛ ⎞−⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

=

= = = × =

=

=

/

3 3 3

(0) ( )

0.9 50 8.7 0.9

4.7

u s d d total down

down

down

Q Q C Q C

m mg m m Cs L s s

mgCL

+ =

⎛ ⎞× = + ×⎜ ⎟

⎝ ⎠

=

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  2

Solution 3

a) The downstream chloride concentration equals:

Yes, this is above the 20 mg/L standard.

b)

Solution 4

The 600MW value the output for the power plant:

Apply energy balance for this system. The heat input is converted into electricity, wasted heat to stack and to cooling water. In this case, cooling water is NOT returned to the river but evaporate. The wasted heat absorbed by cooling water is enough to cause liquid water to change to vapor.

3 3

3 3

(12 10 ) ( 40 5 )21

10 5

mg m mg mmgL s L s

m m Ls s

× + ×=

+

3 3

3 3

3

3 3 6

(12 10 ) ( 5 )20 36

10 5

36 5 86, 400 1, 000 1 1 161 10 10

mg m mg mX mg mgL s L s Xm m L Ls s

mg m s L g ton tonsMax daily massL s day m mg g day

× + ×= → =

+

= × × × × × =

15% Stack Waste heat

85% cooling water

36% eff

600 MWe

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  3

3

3

600nergy added from the power plant= 16670.36

Heat flux to cooling water =0.85 (1667 600) 907 907 10

heat flux to cooling water=

(2257 4.18 / 85 ) 2612

907 10

vap

o o

MWE MW

MW MW kW

m C m C T

m kJ kJ C C m kJ

m kW

• •

• •

•

=

× − = = ×

+ Δ

= + × = ×

= × ×3

33

1 / 0.347 /2612 10

kJ s kg m m skW kJ kg

× × =

 

Solution 5.

a) V = 1.0 × 106 gal, Cin = 100 mg/L, Cout = 25 mg/L, Q = 500 gal/min. The mass balance on BOD in the CMFR is given as follows, assuming steady state conditions.

b)

c) θ = V/Q, so if θ doubles and Q remains constant, V must double. 2 x (1.0 x 106 gal) = 2 × 106 gal Solution 6.

For this PFR, θ = 30 min, Cin=100 pathogens/L, Cout=1 pathogen/L, and Q = 1,000gal/min.

6

0

( ) (500 / min)(100 / 25 / ) min60(1.0 10 )(25 / )

0.090 /

in out rxn

in out

out

in out in out

out out

dm m m mdt

QC QC VkCC C

QC QC Q C C gal mg L mg LkVC VC gal mg L h

k h

= − −

= − −

=− − −

= = = ××

=

3

63

(2 10 min)

4

1.0 10 2 10 min500 / min

25 / 100 /

6.93 10 / min 60 min/ 0.042 /

ktt in

kout in

k

C C edC k Cdt

C C e

V galQ gal

mg L mg L e

k h h

θ

θ

−

−

− ×

−

= ×

= −

= ×

×= = = ×

= ×

= × × =

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  4

a) For first-order decay in a plug flow reactor

b) The hydraulic residence time equals 30 min; therefore,

c) For a CMFR

d) The CMFR should be better handling if your concern is handling variable loadings. e) The chlorine residual=0.20mg/L and the chlorine demand=0.15mg/L. Therefore, the total chlorine added equals,

Solution 7.

a)

b)

(30min)1 / 0.15 / min100 /

ktt

o

k

C eC

pathogen L e kpathogens L

−

−

=

= ⇒ =

V V30 min V 30, 000 galQ 1, 000 gal / min

θ = = = ⇒ =

1

100 /1 / 660, 0000.15 / min11, 000 / min

inout

CC kVQ

pathogens Lpathogen L V galVgal

=+

= ⇒ =×

+

33

1, 000 3.78 60 min 24(0.20 0.15 ) 1.9 10min 10

mg mg gal L g h gL L gal mg h day day

+ × × × × × = ×

3

3650 0.90722

V m hmQh

θ = = =

33

3 3

0

0 00 0

(250 )(3600 )1.25 10 1.25

1, 000722

in out rxn

in

rxn in

dm m m mdt

QC E QCbecause m and QC

pCi sE pCi m pCis hC

mQ m L Lh

= = − ±

= + − ±= =

= = = × × =

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