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3 years ago

2. What is the main job of a cylinder head?

Engineering

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder

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5. Bar stock of initial diameter = 90 mm is drawn with a draft = 10 mm. The draw die has an entrance angle = 18, and the coeffi

yulyashka [42]

Answer:

a) 0.2099

b) 46.5 MPa

c) 233765 N

d) 3896 W

Explanation:

r = (A'' - A') / A'', where

A'' = 1/4 * π * D''²

A'' = 1/4 * 3.142 * 90²

A'' = 6362.55 mm²

D' = D'' - d = 90 - 10 = 80 mm

A' = 1/4 * π * D'²

A' = 1/4 * 3.142 * 80²

A' = 5027.2 mm²

r = (A'' - A') / A'

r = (6362.55 - 5027.2) / 6362.55

r = 1335 / 6362.55

r = 0.2099

Draw stress = σd

Y' = k = 105 MPa

Φ = 0.88 + 0.12(D/Lc), where

D = 0.5 (90 + 80) = 85 mm

Lc = 0.5 [(90 - 80)/sin 18] = 16.18 mm

Φ = 0.88 + 0.12(85/16.18) = 1.51

σd = Y' * (1 + μ/tan α) * Φ * In(A''/A')

σd = 105 * (1 + 0.08/tan18) * 1.51 * In(6362.55/5027.2)

σd = 105 * 1.246 * 1.51 * 0.2355

σd = 46.5 MPa

F = A' * σd

F = 5027.2 * 46.5

F = 233764.8 N

P = 233764.8 (1 m/min)

P = 233764.8 Nm/min

P = 3896.08 Nm/s

P = 3896.08 W

6 0

4 years ago

A radio antenna 326 m in diameter receives a radio signal from a very distant object at perpendicular incidence. The radio signa

Fed [463]

Assume that the antenna absorbs all the radiation that falls on the disk and calculate the power in Watts received by the antenna are the (a) 3.0e-8.

<h3>What is perpendicular with examples?</h3>

Two awesome traces intersecting every different at 90° or a proper perspective are referred to as perpendicular traces. Here, AB is perpendicular to XY due to the fact AB and XY intersect every different at 90°. The traces are parallel and do now no longer intersect every different. They can in no way be perpendicular to every different.

Read more about the perpendicular :

brainly.com/question/1202004

#SPJ1

6 0

2 years ago

A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulu

Naya [18.7K]

Answer:

The correct answer to the following question will be "1.23 mm".

Explanation:

The given values are:

Average normal stress,

$\sigma=200 \ MPa$

Elastic module,

$E = 77 \ GPa$

Length,

$L = 570 \ mm$

To find the deformation, firstly we have to find the equation:

⇒ $\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}$

⇒ $=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}$

On taking " $\frac{PL}{Ebt}$ " as common, we get

⇒ $=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}$

⇒ $=\frac{5PL}{HEbt}$

Now,

The stress at the middle will be:

⇒ $\sigma=\frac{P}{A}$

⇒ $=\frac{P}{(\frac{2}{3})bt}$

⇒ $=\frac{3P}{2bt}$

⇒ $\frac{P}{bt} =\frac{2 \sigma}{3}$

Hence,

⇒ $\delta=\frac{5 \sigma \ L}{6E}$

On putting the estimated values, we get

⇒ $=\frac{5\times 200\times 570}{6\times 77\times 10^3}$

⇒ $=\frac{570000}{462000}$

⇒ $=1.23 \ mm$

8 0

3 years ago

A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para

mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c) 54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

$V_p=\frac{3810.5}{\sqrt{3} }=2200 V$

The complex power S is given as:

$S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA$

$where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA$

The line current I is given as:

$I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A$

The phase voltage at the sending end is:

$V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV$

The magnitude of the line voltage at the source end of the line ( $V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V$

b) The Total real and reactive power loss in the line is:

$S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000$

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

$S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000$

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0

4 years ago

Unwanted resistance is being discussed.

krok68 [10]

I don’t know if I’m right but I’m guessing B

6 0

4 years ago