In the fluid-flow analogy for electrical circuits, what is analogous to a conductor?

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

Artyom0805 [142]

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

8 0

3 years ago

Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago

A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).

Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress $\sigma_m = 90MPa$

stress amplitude $\sigma_a = 190MPa$

a) $\sigma_m =\frac{\sigma_max+\sigma_min}{2}$

$90 =\frac{\sigma_{max}+\sigma_{min}}{2}$ --------------1

$\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}$

$190 = \frac{\sigma_{max}-\sigma_{min}}{2}$ -----------2

solving 1 and 2 equation we get

$\sigma_{max} = 280MPa$

b) $\sigma_{min} = - 100MPa$

c)

stress ratio $=\frac{\sigma_{min}}{\sigma_{max}}$

$=\frac{-100}{280} = -0.35$

d)magnitude of stress range

$=(\sigma_{max} -\sigma_{min})$

= 280 -(-100) = 380 MPa

3 0

3 years ago

What is electricity defined as

antoniya [11.8K]

Answer:

Exact answer: a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current.

4 0

3 years ago

A gas cylinder is connected to a manometer that contains water. The other end of the manometer is open to the atmosphere, which

horrorfan [7]

Answer: the absolute static pressure in the gas cylinder is 82.23596 kPa

Explanation:

Given that;

patm = 79 kPa, h = 13 in of H₂O,

A sketch of the problem is uploaded along this answer.

Now

pA = patm + 13 in of H₂O ( h × density × g )

pA= 79 + (13 × 0.0254 × 9.8 × 1000/1000)

pA = 82.23596 kPa

the absolute static pressure in the gas cylinder is 82.23596 kPa

4 0

3 years ago