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3 years ago

Please help ASAP!!

Engineering

2 answers:

MakcuM [25]3 years ago

5 0

Answer:

Option A

Explanation: Voltage regulator, any electrical or electronic device that maintains the voltage of a power source within acceptable limits.

liberstina [14]3 years ago

4 0

Answer:

The answer you are looking for is D

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Air is compressed in the compressor of a turbojet engine. Air enters the compressor at 270 K and 58 kPa and exits the compressor

tia_tia [17]

Answer:

Recall the Entropy Balance (Second Law of Thermodynamics) equations for the system below

And for ideal gas, we know that

Change in Entropy of the air kJ/Kg.K, ΔS = S₁ - S ₂ = Cp Ln (T₂/T₁) - R Ln(P₂/P₁)

where R is gas constant =0.287kJ/kg.K and given

Cp=1.015 kJ/ kg.K

T₂ = 465k T₁=270k, P1=58kPa, P2=350kPa

substituting these values into the eqn above, we have

S = (1.015x 0.544) - (0.287x1.798)

S = 0.0363kJ/Kg.K

Hence, the mass specific entropy change associated with the compression process = 0.0363kJ/Kg.K

8 0

4 years ago

. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight soli

Lostsunrise [7]

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:

8 0

3 years ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0

4 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

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7 0

3 years ago

By the end of the day, a 45 ft x 25 ft x 8 ft office has a carbon dioxide air level of 1,100 ppm. Determine the time in hours ne

Julli [10]

The total hours that are necessary to reduce carbon dioxide level to at least 325 ppm are 4 hours.

<h3>How does carbon dioxide concentration change over time?</h3>

First hour:

1,100 - 20 %
1,100 - 220 = 880

880 - 20%
880 - 176 = 704

Second hour:

704 - 20%
704 - 140.8 =663.2

663.2 - 20%
663.2 - 132.64 = 530.56

Third hour:

530.56 - 20%
530.56 - 106.112 = 424.44

424.44 - 20%
424.44 - 84.88 = 339.55

Fourth hour:

339.55 - 20%
339.55 - 67.91 = 271.64

Learn more about carbon dioxide in: brainly.com/question/3049557

#SPJ1

6 0

2 years ago