Ask question

Ask question

All categories

3 years ago

6

A completely mixed activated-sludge process is being designed for a wastewater flow of 10,000 m3/d (2.64 mgd) using the kinetics

equations. The influent BOD5 of 180 mg/l is essentially all soluble, and the design effluent BOD5 is 20 mg/L. The effluent biosolids concentration is 15 mg/l, of which 80% is volatile and 65% is biodegradable. For sizing the aeration tank, the SRT is selected to be 8 d and the MLVSS 3000 mg/L. The kinetic constants are as follows:
Y=0.6 mg VSS/mg BOD, ka = 0.06 d-1, Ks = 60 mg/L of BOD, and k = 5.0 day-1.
Determine the efficiency of soluble BOD removal, using the formula
E=S0-Se/S0 X 100
Here, S0 is influent soluble BOD concentration and Se is influent soluble BOD concentration.

1 answer:

Elan Coil [88]3 years ago

5 0

Answer:

The efficiency of soluble BOD removal is 88.8889%

Explanation:

For this question not all the given parameters will be used, only the following:

BODi = influent soluble concentration = 180 mg/L

BODe = effluent soluble concentration = 20 mg/L

Question: Determine the efficiency of soluble BOD removal, η = ?

The efficiency of soluble BOD removal can be calculated using the following formula:

$\eta =\frac{BOD_{i}-BOD_{e}}{BOD_{i} } *100=\frac{180-20}{180} *100=88.8889$ %

You might be interested in

Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an

Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

LC = 472.728

R = 1225 ft

Explanation:

See the attached file for the calculation.

8 0

2 years ago

How much does 1 gallon of water weigh in pound given that the density of water is 1gram/ cm3

MAXImum [283]

Explanation:

There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.

6 0

2 years ago

Technician A says a solenoid is not used on modern automobiles. Technician B says fuel injectors and starter motor solenoids are

julia-pushkina [17]

Answer:

ON THE BOTTOM

Explanation:

red cord plus black cord

5 0

2 years ago

Give two disadvantages of a moving coil Meter

Temka [501]

1.Only suitable for dc
2.more expensive than moving iron type
3. Easily damaged

7 0

2 years ago

An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

Allushta [10]

Answer:

(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0

3 years ago