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3 years ago

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A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.0

6 ft*lb b. 19.24 ft*lb ? c)-31.17 ftlb d) 119.07 ftlb e)-9.92 ft*lb

1 answer:

harina [27]3 years ago

3 0

Answer:

Load carried by shaft=9.92 ft-lb

Explanation:

Given: Power P=4.4 HP

P=3281.08 W

<u><em>Power: </em></u>Rate of change of work with respect to time is called power.

We know that P= $Torque\times speed$

$\omega=\frac{2\pi N}{60}$ rad/sec

So that P= $\dfrac{2\pi NT}{60}$

So 3281.08= $\dfrac{2\pi \times 2329\times T}{60}$

T=13.45 N-m (1 N-m=0.737 ft-lb)

So T=9.92 ft-lb.

Load carried by shaft=9.92 ft-lb

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What are the available motor sizes for 2023 ariya ac synchronous drive motor systems in kw?.

Anika [276]

The available motor sizes for 2023 Ariya AC synchronous drive motor systems are:

40 kW.

62 kW.

160 kW.

<h3>What is a synchronous motor?</h3>

A synchronous motor refers to an alternating current (AC) electric motor in which the rotational speed of the shaft is directly proportional (equal) to the frequency of the supply current, especially at a steady state.

In Engineering, the available motor sizes for 2023 Nissan Ariya AC synchronous drive motor systems include the following:

40 kW.

62 kW.

160 kW.

Read more on synchronous motor here:

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2 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

goldenfox [79]

Answer:

what?

Explanation:

8 0

3 years ago

Consider a drug-eluting balloon catheter deployed into a blood vessel. The balloon is inflated to perfectly adhere to the vessel

GaryK [48]

Answer:

a) Cr = Co - Fx / D

b) ΔC / Δx = ( CR - Cr ) / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X = D ( Co - Cr ) / X ------ 1

where : F = flux , D = drug diffusion coefficient

X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface = ΔC / Δx = CR - Cr / xR - xRo

6 0

3 years ago

Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you

viktelen [127]

Answer:

$\theta_1=15^o\\\theta_2=75^o$

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

Where $\theta$ is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

$V_y=V_{oy}-gt=V_osin\theta-gt$

The horizontal and vertical distances are, respectively:

$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

$\displaystyle t_f=\frac{2V_osin\theta}{g}$

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

Let's solve for $\theta$

$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

$\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}$

$\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}$

Or equivalently:

$\theta_2=90^o-\theta_1$

Given Vo=37 m/s and R=70 m

$\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}$

$\theta_1=15^o$

And

$\theta_2=90^o-15^o=75^o$

5 0

3 years ago