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nignag [31]
4 years ago
9

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

and a reaction rate constant, k, of 0.20/d. The values of ultimate BOD and the reaction rate constant of the original wastewater are most nearly _____.
Engineering
1 answer:
zzz [600]4 years ago
3 0

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
3 years ago
The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin
IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel =  4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

\dot{m}=\rho AV

Putting all the value to get the velocity of the flow

\frac{\dot{m}}{\rho A} = V

V = \frac{4000}{1000*4*2}

V = 0.5 m/s

4 0
3 years ago
-1 1/6 divided by 2 1/3
pantera1 [17]

Answer:

-\frac{1}{2}

Explanation:

-1\frac{1}{6}:2\frac{1}{3}=\\  \\=-\frac{7}{6}:\frac{7}{3}\\  \\=-\frac{7}{6}*\frac{3}{7}\\  \\=-\frac{1}{2}

4 0
4 years ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
D
TEA [102]

Answer:

true

Explanation:

4 0
3 years ago
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