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

When adding two 8 bit binary numbers, which of the following statements is true?

diamong [38]

Answer:

The result might require 9 bits to store

4 0

2 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ $\frac{P_{3} }{P_{2} }$ = $\frac{T_{3} }{T_{2} }$

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

You should always engage the parking brake whenever using any type of lift.

Shtirlitz [24]

Answer:

I would yes every time so it would not role off the lift

Explanation:

8 0

3 years ago

Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat

PtichkaEL [24]

Answer:

Detailed solution is attached below in three simple steps the problem is solved.

7 0

3 years ago

A rectangular steel alloy A-36 (structural steel) plate is hanging vertically and supporting a hanging weight of 90 kN. The plat

KengaRu [80]

Answer:

a) Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b)Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm\

c) % Reduction in area = (450-449.7391/450 ) = 0.58 %.

Explanation:

a) Change in length = Pl /AE = 90*1000*0.5*1000/75*207*6*10^3

= 0.4838mm Expansion.

Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b )Change in width = - μpt/AE = -(0.3*90*1000*6/ 75*207*6*10^3)

= -1.739 * 10^-3 mm

Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm

c )

New C/s area = 74.97827 *5.998260 = 449.7391 mm^2

% Reduction in area = (450-449.7391/450 ) = 0.58 %.

3 0

4 years ago