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bekas [8.4K]
3 years ago
9

A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl

ow to the pond has a flow rate of Q = 4,000 m3/day and a BOD1 concentration of Cin = 25 mg/L. The volume of the pond is 20,000 m3. The purpose of the pond is to allow time for the decay of BOD to occur before discharge into the environment. BOD decays in the pond with a first-order rate constant equal to 0.25/day. What is the BOD concentration at the outflow of the pond, in units of mg/L?
Engineering
1 answer:
german3 years ago
8 0

Answer:

BOD concentration at the outflow = 17.83 mg/L

Explanation:

given data

flow rate of Q = 4,000 m³/day

BOD1 concentration of Cin = 25 mg/L

volume of the pond = 20,000 m³

first-order rate constant equal = 0.25/day

to find out

What is the BOD concentration at the outflow of the pond

solution

first we find the detention time that is

detention time t = \frac{volume}{flow rate}

detention time t = \frac{20000}{4000}

detention time = 5 days

so

BOD concentration at the outflow of pond is express as

BOD concentration at the outflow = Cin ( 1 - e^{-kt} )

here k is first-order rate constant and t is detention time and Cin is BOD1 concentration

so

BOD concentration at the outflow = 25 ( 1 - e^{-0.25(5)} )

BOD concentration at the outflow = 17.83 mg/L

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vazorg [7]

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

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Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

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3 years ago
The total floor area of a building, including below-grade space but excluding unenclosed areas, measured from the exterior of th
alex41 [277]

Answer:

Gross building area

Explanation:

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6 0
3 years ago
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Dominik [7]

Answer:

0

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)=\frac{s}{(s+3)^2}

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3 0
3 years ago
6. What types of injuries can occur in an electronics lab and how can they be prevented?
marysya [2.9K]

Answer:

The most common injuries in a chemistry lab is making a fire, heat burns, chemical burns, cuts and scrapes, contamination, inhalation, and spills and breaks.

1.) You can prevent making a fire by making sure you close and seal flammable materials.

2.) You can prevent heat burns by teaching the students how to properly use tongs,water baths, and other cooling equipment. 

3.) You can prevent chemical burns by treating the chemicals with caution, measure carefully, and use the approved containers.

4.) You can prevent cuts and scrapes by telling the students how to use the blades safely, and also when they are disposing broken or sharp items they should know how to wrap them up so no one else will get hurt. 

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8 0
3 years ago
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
Anni [7]

Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

7 0
3 years ago
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