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Lyrx [107]
3 years ago
14

In a reversible process both the system and surrondings can be returned to their initial states. a)-True b)-False

Engineering
1 answer:
Nuetrik [128]3 years ago
6 0

Answer:

a)-True

Explanation:

When the direction of a process is reversed through infinitesimal changes to the system. The infinitesimal changes are done by the surroundings. After the process has been reversed the system and surroundings go back to their original state.

The changes take an infinite amount of time so, it is not possible practically.

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Applications of attenuators.
agasfer [191]

Answer:

Attenuators are electrical components designed to reduce the amplitude of a signal passing through the component, without significantly degrading the integrity of that signal. They are used in RF and optical applications

Explanation:

6 0
3 years ago
1. Select the punch that is used to make a mark in metal
charle [14.2K]

Answer:

D center punch

Explanation:

it has a fine point so it is good with accuracy

3 0
4 years ago
A flow of 2.6 MGD leaves a primary clarifier with a BOD of 131 mg/L. Determine the aeration period (hydraulic detention time) of
SashulF [63]

Answer:

8.5 days

Explanation:

Given data :

Flow ( Q ) = 2.6 MGD = 11819.834 m^3/day

BOD = 131 mg/L

BOD loading rate = 35 Ibs/1000 ft^3   per day = 0.5606 kg/m^3/day

<u>Calculate the sludge age of the facility </u>

Given  the BOD  applied to the aeration tank  = 11819.834 m^3/day * 131mg/l

                                                                            = 1548.398 kg/day

first calculate the volume of the aeration tank

V = BOD applied / BOD loading rate

V = 1548.398 / 0.5606 = 2762.03 m^3

Hydraulic Detention time = V / Q

                                          = 2762.03 / 11819.834  = 0.2336 day = 5.6 hour

next : determine the mass rate of the waste

= 44% * 0.5606 kg/m^3/day

= 0.2466 kg/m^2/day

finally determine the sludge age

= V * Xt / ∅w * R

= ( 2762.03 m^3 * 2100 * 10^-3 ) / ( 0.2466 * 2762.03 kg/day )

= 8.5 days

8 0
3 years ago
Please help me. I have no idea what I'm doing.​
Helen [10]
A.an internationally standard used to measure a physical quality
5 0
3 years ago
Read 2 more answers
A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w
Alexxandr [17]

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P = \frac{230}{6}

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor = \frac{300}{V}

V = \frac{300}{38.33}

V = 7.826 ml

6 0
3 years ago
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