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3 years ago

Ứng dụng của lưu chất trong máy nén thủy lực

Engineering

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

Hầu hết các công việc quan trọng sử dụng nguyên lý thủy lực. Trợ lực lái, giảm xóc, kính chắn gió và phanh là những ứng dụng phổ biến của hệ thống thủy lực trên xe. Thang máy hai trụ và thang nâng bốn trụ được sử dụng trong ngành công nghiệp ô tô để nâng các phương tiện phục vụ công tác bảo dưỡng và kiểm tra.

Explanation:

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What is the purpose of O-ring and valve seals in a cylinder head?

Andrews [41]

Answer:

its to show the shape is flat and only flat at the botom and top and you can set it up ther way and it wlll still look the same.

Explanation:

8 0

3 years ago

You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from

Aleksandr [31]

Answer:

a) $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

b) attached below

c) type zero system

d) k > $\frac{g}{200}$

e) The gain K increases above % error as the steady state speed increases

Explanation:

Given data:

Motor voltage = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; $\frac{K1}{1+St}$ be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= $\frac{Ws}{Es} = \frac{200}{1+1.2s}$

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > $\frac{g}{200}$

7 0

3 years ago

What should be your strongest tool be for gulding your ethical decisions making process

valkas [14]

Answer:

Recognize that there is a moral dilemma.

Determine the actor. ...

Gather the relevant facts. ...

Test for right versus wrong issues. ...

Test for right versus right paradigms. ...

Apply the resolution principles. ...

Investigate the trilemma options. ...

Make the decision.

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3 years ago

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

sveta [45]

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

$\epsilon_{car}=1-\frac{T_c}{T_H}$

Where $T_c$ is the atmospheric temperature and $T_H$ is the maximum burn temperature.

For the case (B), the efficiency we will use is:

$\epsilon_{b}=0.4\epsilon_{car}$

The work done by the engine can be calculated as:

$W=\epsilon Q=\epsilon H_v\cdot m_{fuel}$ where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

$m=\frac{P\cdot t}{\epsilon H_v}$

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

$TC=m\cdot c$

8 0

3 years ago

How would you describe what would happen to methane if the primary bonds were to break?

erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0

3 years ago