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aleksklad [387]
3 years ago
15

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

Engineering
1 answer:
natta225 [31]3 years ago
7 0

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

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A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and deca
dexar [7]

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

5 0
3 years ago
1. In an assembly two flanges are held together by a 1/2"" bolt that is threaded into one of the flanges. (a) If the clearance h
creativ13 [48]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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3 years ago
Natural ventilation uses primarily
Zolol [24]

Natural ventilation unlike fan forced ventilation uses the natural forces of wind and buoyancy to deliver fresh air into buildings

8 0
2 years ago
Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 20 C.?
SIZIF [17.4K]

Answer:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

Explanation:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

The saturation temperature of steam is 45.81⁰ C at the pressure of 10 KPa which is higher than 20⁰C of river water. So river water at 20⁰C can be used to maintain the condenser pressure to 10 KPa.

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3 years ago
Distinguish and describe the stage of the product development life cycle reflected in the following scenario.
Lelu [443]

Answer:Very productive

Explanation:

7 0
3 years ago
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