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pickupchik [31]
3 years ago
10

In water and wastewater treatment processes a filtration device may be used to remove water from the sludge formed by a precipit

ation reaction. The initial concentration of sludge from a softening reaction (Chapter 4) is 3 2 percent (20,000 mg/L) and the volume of sludge is 100 m . After filtra- tion the sludge solids concentration is 35 percent. Assume that the sludge does not change density during filtration, and that liquid removed from the sludge contains no sludge. Using the mass balance method, determine the volume of sludge after filtration.
Engineering
1 answer:
PtichkaEL [24]3 years ago
6 0

Answer:

The volume of sludge after filtration is 0.914 m

Explanation:

Solution

Given that:

We have to find the Volume of sludge after filtration

Now,

The  Sludge concentration is = 32%,

The sludge volume = 100 m,

The sludge concentration after filtration = 35%

Then,

The mass balance equation is stated below

Cin∀in = Cout∀out

Now,

We Solve for ∀out

∀out =Cin∀in/Cout

By substituting the values

∀out = (0.32)(100 m)/(0.35) = 0.914 m

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I need more details to your question
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2 years ago
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How did Atlantis benefit from lessons learned in construction of earlier orbiters?
Alenkasestr [34]

Answer:

Atlantis benefited from lessons learned in the construction and testing of Enterprise, Columbia and Challenger. ... The Experience gained during the Orbiter assembly process also enabled Atlantis to be completed with a 49.5 percent reduction in man hours (compared to Columbia).

Explanation:

8 0
3 years ago
Write torsion equation and explain the importance of each components.<br>​
Elanso [62]
The equations are based on the following assumptions

1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.

Nomenclature

T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
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G Modulus of rigidity (N/m^2)
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4 0
3 years ago
1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu
lianna [129]

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

5 0
3 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
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