Answer:
R = 31.9 x 10^(6) At/Wb
So option A is correct
Explanation:
Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A
Thus; R = L/μA,
Now from the question,
L = 4m
r_1 = 1.75cm = 0.0175m
r_2 = 2.2cm = 0.022m
So Area will be A_2 - A_1
Thus = π(r_2)² - π(r_1)²
A = π(0.0225)² - π(0.0175)²
A = π[0.0002]
A = 6.28 x 10^(-4) m²
We are given that;
L = 4m
μ_steel = 2 x 10^(-4) Wb/At - m
Thus, reluctance is calculated as;
R = 4/(2 x 10^(-4) x 6.28x 10^(-4))
R = 0.319 x 10^(8) At/Wb
R = 31.9 x 10^(6) At/Wb
Answer:
are you wht
didn't understand the question
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)](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p%5Cnu%29)
![E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p_%7Batm%7D%20%5Cfrac%7B1%7D%7B%5Crho%7D%29)
![57.5 \times 10^3 = \dot m ( \frac{20^2}{2} + 25 \times 9.81 + 101325 \frac{1}{780})](https://tex.z-dn.net/?f=57.5%20%5Ctimes%2010%5E3%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7B20%5E2%7D%7B2%7D%20%2B%2025%20%5Ctimes%209.81%20%2B%20101325%20%5Cfrac%7B1%7D%7B780%7D%29)
solving for flow rate
![\dot m = 99.977we know that [tex]\dot m = \rho AV](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%2099.977%3C%2Fp%3E%3Cp%3Ewe%20know%20that%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdot%20m%20%20%3D%20%5Crho%20AV)
![\dot m = 780 \frac{\pi}{4} D^2\times 16](https://tex.z-dn.net/?f=%5Cdot%20m%20%20%3D%20780%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20D%5E2%5Ctimes%2016)
solving for d
![99.97 = 780 \times \frac{\pi}{4} D^2\times 16](https://tex.z-dn.net/?f=99.97%20%3D%20780%20%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20D%5E2%5Ctimes%2016)
d = 0.090 m
so radius = 0.045 m
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](https://tex.z-dn.net/?f=Torque%5Ctimes%20speed)
rad/sec
So that P=![\dfrac{2\pi NT}{60}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Cpi%20NT%7D%7B60%7D)
So 3281.08=![\dfrac{2\pi \times 2329\times T}{60}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%5Cpi%20%5Ctimes%202329%5Ctimes%20T%7D%7B60%7D)
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