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2 years ago

13

What is the main purpose of the alternator?

1 answer:

Serhud [2]2 years ago

7 0

To power the cars electrical system

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A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

The end of a large tubular workpart is to be faced on a NC vertical boring mill. The part has an outside diameter of 38.0 in and

nata0808 [166]

Answer:

(a) the cutting time to complete the facing operation = 11.667mins

b) the cutting speeds and metal removal rates at the beginning= 12.89in³/min and end of the cut. = 8.143in³/min

Explanation:

check attached files below for answer.

5 0

2 years ago

If he wants to keep the height the same, what could the other dimensions be for him to get the volume he wants?

Fiesta28 [93]

tbm queria saber essa pergunta

8 0

2 years ago

Oil of density 780 kg/m3 is flowing at a velocity of 20 m/s at the atmospheric pressure in a horizontal cylindrical tube elevate

Soloha48 [4]

Answer:

radius = 0.045 m

Explanation:

Given data:

density of oil = 780 kg/m^3

velocity = 20 m/s

height = 25 m

Total energy is = 57.5 kW

we have now

E = kinetic energy+ potential energy + flow work

$E = \dot m ( \frac{v^2}{2] + zg + p\nu)$

$E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})$

$57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})$

solving for flow rate

$\dot m = 99.977we know that [tex]\dot m = \rho AV$

$\dot m = 780 \frac{\pi}{4} D^2\times 16$

solving for d

$99.97 = 780 \times \frac{\pi}{4} D^2\times 16$

d = 0.090 m

so radius = 0.045 m

3 0

3 years ago

A) Total Resistance<br> b) Total Current<br> c) Current and Voltage through each resistor

Varvara68 [4.7K]

C my friend 20 characters suck

3 0

2 years ago