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2 years ago

5

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

ed that the amplitude of motion should not exceed 0.1 mm per 1 N of unbalance force. The mass of the motor is 120 kg and the operating speed is 720 rpm Use the concept of transfer function to determine the required stiffhess coefficient of each mount.

1 answer:

Artyom0805 [142]2 years ago

8 0

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

$\frac{A}{F}$ = $\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}$

Formula Used:

$A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}$

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

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

so,

$\omega = \frac{2 \pi\times 720}{60}$ = 75.39 rad/s

Using the given formula:

Damping is negligible, so, $\zeta = 0$

$\frac{A}{F}$ will give the tranfer function

Therefore,

$\frac{A}{F}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

$0.1\times 10^{-3}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

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g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

2 years ago

Independent auto lots usually have ____ finance rates than dealerships

wel

Independent auto lots usually have <u>higher</u> finance rates than dealerships

<u>Explanation:</u>

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

But this negotiation and lowering of the finance rates is not possible with the independent auto lots and thus they charge higher rates compared to the dealerships.

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2 years ago

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

padilas [110]

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

or

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

6 0

2 years ago

Read 2 more answers

What factors are likely to promote the formation of stress corrosion cracks?

Naddika [18.5K]

Answer:

Stress corrosion cracking

Explanation:

This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.

Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.

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3 years ago