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Sergeeva-Olga [200]
3 years ago
15

(2 points) A perfectly mixed aeration pond with no recycle serves as the biological reactor for a small community. The pond rece

ives 30 m3 /d of influent with a BOD of 350 mg/L that must be reduced to 20 mg/L before discharge. It has been found that the kinetic constants for this system are: Ks = 100 mg BOD/L, kd = 0.10 day-1 , μm = 1.6 day-1 , Y = 0.60 mg SS/mg BOD. Assume Xo = 0. a. What must the hydraulic retention time (θ = V/Q) be in the aeration pond? b. What mass of microbes will be produced in the pond each day?
Engineering
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

a)  t = 165 days

b) 9.9 kg/day

Explanation:

Given data:

final lelvel of BOD is 20 mg/l

Ks = 100 mg BOD/L,

kd = 0.10 day-1 ,

μm = 1.6 day-1 ,

Y = 0.60 mg SS/mg BOD

a) we know that

c_{out} = \frac{c_{in}}{1 + kt}

20 = \frac{350}{1 + 0.1 t}

solving for t

t  =  165 days

b) mass of microbes = Q(mlD) × C(mg/l)

= 30\times 10^{-3} (350-20) = 9.9 kg/day

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Does the Diesel engine have engine knock or detonation problem? Why?
Luda [366]

Explanation:

Yes Diesel engine have problem of knocking.

We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.

Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.

6 0
3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
3 years ago
Can some help me with this !!! Is 26 points!!
Aleonysh [2.5K]
Third one
15,000,000 ohms because M=10^6
8 0
3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
n the Spring of 2015, three utility companies in the Ukraine received email purporting to come from Ukraine's parliament, the Ra
Tresset [83]

Answer:

Trojan horse

Explanation:

A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.

5 0
3 years ago
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