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vodomira [7]
1 year ago
5

technician a says that dirt bypassing the filter on many common rail injectors can cause an injector to stick open and continuou

sly fuel a cylinder. technician b says that any dirt particles may potentially damage the injector, but would not allow an injector to full fuel or the engine to run away. who is correct?
Engineering
1 answer:
NNADVOKAT [17]1 year ago
3 0

Technician a is correct because he says that Many common rail injectors filters can be bypassed by dirt, which can lead to an injector sticking open and continuously fueling a cylinder.

Coalescence is used to separate the water and fuel. To the fuel injector cleaning kit, fasten your air compressor. Diesel engines run at compression ratios that are greater than those of gasoline engines. greater ratio compared to gasoline engines. increased thermal expansion as a result. more fuel energy that is transformed into usable power. The great benefit of using a dry cylinder sleeve is that by quickly installing new sleeves, the cylinder block can be quickly restored to its original specifications. Vacuum drying can be used to get rid of small amounts of water. A nozzle is used to spray the fuel into the vacuum chamber of engines. Air and unsolved free water are taken out of the oil. The fuel is evenly dispersed, which facilitates efficient drying.

Learn more about injectors here:

brainly.com/question/27969202

#SPJ4

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Select the correct answer.
Genrish500 [490]
A is the correct answer
5 0
2 years ago
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pu
marishachu [46]

The complete question is;

The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.

The image of this system is attached.

Answer:

Velocity = 11.8 ft/s

Explanation:

Since the wheel at A rotates about a fixed axis, then;

v_c = ω•r_c

r_c is 4.5 in. Let's convert it to ft.

So, r_c = 4.5/12 ft = 0.375 ft

Thus;

v_c = 0.375ω

Now the mass moment of inertia about of wheel A about it's mass centre is given as;

I_a = m•(k_a)²

The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug

Also, let's convert ka from inches to ft.

So, ka = 6/12 = 0.5

So,I_a = 1.5528 × 0.5²

I_a = 0.388 slug.ft²

The kinetic energy of the system would be;

T = Ta + Tc

Where; Ta = ½•I_a•ω²

And Tc = ½•m_c•(v_c)²

So, T = ½•I_a•ω² + ½•m_c•(v_c)²

Now, m_c is given as 200 lb.

Converting to slug, we have;

m_c = (200/32.2) slugs

Plugging in the relevant values, we have;

T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)

This now gives;

T = 0.6307 ω²

The system is initially at rest at T1 = 0.

Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.

Whereas at B, M does positive work and at C, W_c does negative work.

When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π

While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b

Where, rb = 3 in = 3/12 ft = 0.25 ft

ra = 7.5 in = 7.5/12 ft = 0.625 ft

So, θ_a = (0.25/0.625) × 10π

θ_a = 4π

Thus, we can say that the crate will have am upward displacement through a distance;

s_c = r_c × θ_a = 0.375 × 4π

s_c = 1.5π ft

So, the work done by M is;

U_m = M × θ_b

U_m = 50lb × 10π

U_m = 500π

Also,the work done by W_c is;

U_Wc = -W_c × s_c = -200lb × 1.5π

U_Wc = -300π

From principle of work and energy;

T1 + (U_m + U_Wc) = T

Since T1 is zero as stated earlier,

Thus ;

0 + 500π - 300π = 0.6307 ω²

0.6307ω² = 200π

ω² = 200π/0.6307

ω² = 996.224

ω = √996.224

ω = 31.56 rad/s

We earlier derived that;v_c = 0.375ω

Thus; v_c = 0.375 × 31.56

v_c = 11.8 ft/s

3 0
3 years ago
Which permission do you need to shoot on the owner’s property?
Elena L [17]

Answer:

filming permit,

( MARK ME BRAINLIEST!!)

4 0
3 years ago
Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of
aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid T=500\ K

Einstein Temperature T_E=300\ K

Heat Capacity in the Einstein model is given by

C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}

e^{\frac{3}{5}}=1.822

Substitute the values

C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})

C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}

C_v=0.97\times (3R)            

6 0
3 years ago
What is the function rule for the line? f(x)=−32x−2f(x)=−23x−2f(x)=32x−2f(x)=−32x+2A coordinate grid with x and y axis ranging f
murzikaleks [220]

Answer:

f(x)=23x−2

Explanation:

still trying to figure that out

7 0
3 years ago
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