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Alex_Xolod [135]
3 years ago
9

For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the

length is Δp1. If the length of the pipe is then doubled, what is the relation of the new pressure drop Δp2 to the original pressure drop Δp1 at the original mass flow rate?
delta P2=?
Engineering
1 answer:
Goshia [24]3 years ago
8 0

Answer:

\Delta p_{2} = 2\cdot \Delta p_{1}

Explanation:

The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.

\Delta p_{2} = 2\cdot \Delta p_{1}

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What the different methods to turn on thyrister and how can a thyrister turned off​
myrzilka [38]

Answer:

forward voltage triggering

temperature triggering

dv/dt triggering

light triggering

gate triggering

Then turning off;

Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals

Explanation:

hope it helps

8 0
3 years ago
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length
EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2l

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL  = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

so

D₃ = D( ΔP₁₋₂ / ΔP₂₋₃ )^{\frac{1}{4}

we substitute

D₃ = D( 1.657 )^{\frac{1}{4}

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

8 0
3 years ago
A stream of liquid n-pentane flows at a rate of 50.4 L/min into a heating chamber, where it evaporates into a stream of air 15%
allochka39001 [22]

Answer:

(a) the fractional conversion of pentane achieved in the furnace is  90% conversion

(b) the volumetric flow rates (Umin) of the feed air  is 256 x 10³ 1/m

(c) the volumetric flow rates (Lmin) of the gas leaving the condenser is 404.9  x 103 l/min

Explanation:

a)   Molecular weight of pentane = 72.15 g/mol

density of liquid pentane = 626 kg/m3

Flow rate of feed liquid nitrogen = 50.4 l/min

                                                  = 626*50.4*10-3

                                                   = 31.55 kg/min

                                                   = 31.55/72.15 kmol/min

                                                    = 0.4372 kmol/min

Pentane existng the burner = 3.175 kg/min

Fractional conversion = (31.55 - 3.175)/ 31.55

= 0.9 = 90% conversion

b)

C₂H₅ + 8O₂ ----------->   5CO₂ + 6H₂O

From the Stoichiometric reaction,

8 mol of O2 are used for combustion of 1 mol of pentane

for 0.4372 kmol/min of pentane = 8 * 0.4372 kmol/min of Oxygen will be required

                                                  = 3.49 kmol/min of O2

amount of air will be = 3.49/0.21 = 16.62 kmol/min

15% excess air = 16.62*1.15 = 19.12 kmol/min

assuming air to be ideal gas

V = nRT / P.........(1)

V = 19.12 X 8.314 X 336 / 208.6

  = 256 m³ = 256 x 10³ 1/m

c)  Oxygen:

Amount of pentane consumed = (31.55 - 3.175) = 28.375 kg/min = 28.375/72.15 = 0.3932 kmol/min......(2)

Amount of O2 consumed = 8*0.3932 = 3.146 kmol/min

Amount of O2 fed by air = 0.21*19.12 = 4.0215 kmol/mim

unused O2 left = 4.0215 - 3.146

= 0.8755 kmol/min = 19.75*103 l/min............ (using (1))

Carbon Dioxide:

1 mol of pentane = 5 mol of CO2

0.3932 kmol/min of pentane = 5*0.3932 kmol/min of CO2..................(from (2)

                                                    = 1.966 kmol/min

= 44.362*103 l/min..........................(using (1))  

Nitrogen:

v = 0.79 x 19.12 x 8.314 x 275 / 100.325

   = 340.8 m³ / min = 340.8 x 10³ 1/min

Total volumetric flow rate of gases leaving the condenser = (340.8 + 44.36 + 19.75) x 103 l/min

= 404.9  x 103 l/min

6 0
3 years ago
Match the types of AI to the correct automation tasks.
Maru [420]

Answer:

Theory of Mind : A robotic head has a face that recognizes and simulates emotions.

Self aware : A robot tries to protect itself from harm

Purely Reactive : A door automatically opens when a person steps in front of it.

Limited Memory : A personal assistant software tracks a persons travel routes and suggests shorter routes.

Explanation:

Artificial intelligence is simply a technology which enables automation. It enables system perform task without being explicitly controlled. Purely reactive systems do not store Data in memory, it simply observes what going on at the moment which is what it was programmed to do and takes a step. One the machine detects someone approaching up to a certain distance, it opens.

Limited Memory systems store information about the past and this enhances its Decison making, prediction engines, self driving cars use this kind of artificial intelligence.

Theory of Mind : Here, systems are trained to detect, understand and replicate what is understood. Once the robot identifies an emotion, it replicates it.

Self - Aware : An advanced level of AI, where systems will not only be able to replicate what they see, but also make conscious decisions as to which action to take in different circumstances.

5 0
3 years ago
The net potential energy EN between two adjacent ions, is sometimes represented by the expression
Anastaziya [24]

Answer:

as answered in the attached file.

Explanation:

The detailed steps, derivation and appropriate differentiation is as shown in the attachment

3 0
3 years ago
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