Ask question

Ask question

All categories

3 years ago

8

Consider a single bacterial cell as a discrete particle with a diameter of 1x10-6m and a specific gravity of 1.01. Assuming lami

nar flow conditions, calculate how long would it take for the cell the settle 1 foot in water at 20 oC?For a flow rate of 4MGD, what surface area would be required to settle individual cells? Is sedimentation a practical method for removing individual bacterial cells?

1 answer:

Alinara [238K]3 years ago

6 0

Answer:

Sedimentation is not a good method.

Explanation:

We need to apply Stoke laws and assume that is valid here.

So,

$V_s= 418(G-1)d^2(\frac{3T+70}{100})$

Replacing the values,

$V_s= (418)(1.01-1)(10^-3)^3*(\frac{3*20+70}{100})\\V_s=5.434*10^-6mm/s$

Here then we calculate the time,

$t_{req}=\frac{x}{v}$

Where x= Distance, v= velocity

$t_{req}=\frac{1foot}{5.434*10^{-6}} = \frac{304.8}{5.434*10^{-6}}\\t_{req}=649.2days$

To calculate the surface required we need first to calculate the volume through the volume,

So,

$Q=4MGD=4*10^6*0.135 (ft^3/day)$

Then,

$V_{req} = Q*t\\V_{req}=4*10^6*0.134*649.2\\V_{req}=34.79*10^7ft^3$

Here we can calculate the surface

$S_{req}= \frac{Volume}{Distance}=\frac{34.79*10^7}{1}\\S_{req}=7987.8 Acres$

<em>So, the requeriment of Area of tank and settlement time is huge, it's not a practical method.</em>

You might be interested in

Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$

8 0

3 years ago

Engineers create a new metal that is stronger than steel but much lighter. This material is also significantly cheaper than what

zysi [14]

The best step for the engineers to make next is option D. Begin to design an airplane using this metal.

<h3>What is the metallic is plane parts?</h3>

Aluminum and its alloys are nevertheless very famous uncooked substances for the production of business planes, because of their excessive electricity at exceedingly low density. Currently, excessive-electricity alloy 7075, which includes copper, magnesium and zinc, is the only used predominantly withinside the plane industry.

The solution is D, due to the fact even as it's far crucial to marketplace the fabric and ensure humans are inquisitive about buying, they first want to attempt to layout aircraft the usage of this fabric. There isn't anyt any use promoting an aircraft constituted of this material_ if a aircraft can not be built.

Read more about the aircraft:

brainly.com/question/5055463

#SPJ1

7 0

2 years ago

g A pedometer treats walking 2,000 steps as walking 1 mile. Write a program whose input is the number of steps, and whose output

Nataly [62]

Answer:

# Program is written in Python Programming Language

# Comments are used for explanatory purpose

# Program starts here

# Accept input

Steps = input (Number of Steps: ")

# Calculate distance

distance = float(2000) * float(steps)

#Print Formatted Result

print('%0.2f' % distance)

# End of Program

.--------

The above program converts number of steps to miles.

At line 5, the number of steps is inputted and stored in variable named Steps.

At line 6, the number of miles is calculated by multiplying 2000 by the content of variable Steps

The result is printed at line 8

8 0

3 years ago

Read 2 more answers

How many parts (screws and bolts included) does the average car have?

Leokris [45]

A single car has about 30,000 parts, counting every part down to the smallest screws

5 0

3 years ago

Read 2 more answers

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago