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3 years ago

Does the Diesel engine have engine knock or detonation problem? Why?

Engineering

1 answer:

Luda [366]3 years ago

6 0

Explanation:

Yes Diesel engine have problem of knocking.

We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.

Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.

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What was the reason alloys were used instead of metals like copper, tin, or iron?

lidiya [134]

Alloys were stronger and more durable

8 0

2 years ago

100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t

LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.

Using the given mass m, molar mass M, we can get the following equation:

pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

U=-1.848*10^6 J

Note, that the energy was taken away from the system.

5 0

4 years ago

In a website browser address bar, what does “www” stand for?

Ludmilka [50]

Answer:

www stands for world wide web

Explanation:

It will really help you thank you.

3 0

3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$

$w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um$

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

3 years ago

During an expansion process, the pressure of a gas changes from 15 to 140 psia according to the relation P = aV + b, where a = 5

Dominik [7]

The work done during the process is 359 btu

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<u>Explanation:</u>

Given-

P1 = 15psia

P2 = 140 psia

V1 = 7ft³

a = 5 psia/ft³

b = C

P = aV +b

Work done, W = ?

P1 = aV1 + b

15 = 5 (7) + b

b = -20 psia

P2 = aV2 + b

140 = 5 ( V2) - 20

V2 = 32 ft³

The work done by the process is the area under the curve which is trapezoidal.

Therefore,

Work done, W = area of trapezoid

= (P2 + P1 / 2) (V2 - V1)

= ( 140 + 15 / 2 ) ( 32 - 7)

= 1937.5 psia ft³

= 1937.5/ 5.4039 = 359 btu

Therefore, the work done during the process is 359 btu

5 0

3 years ago