Answer:
1.693 feet
Explanation:
We have given Q = 90 gpm
We know that 1 cubic feet per second = 443.833 gpm
So 90 gpm will be equal to ![\frac{90}{443.833}=0.202\frac{ft^3}{sec}](https://tex.z-dn.net/?f=%5Cfrac%7B90%7D%7B443.833%7D%3D0.202%5Cfrac%7Bft%5E3%7D%7Bsec%7D)
Let d is the diameter of the pipe then ![V_{avg}=\frac{Q}{A}=\frac{0.2}{\frac{\pi }{4}d^2}=\frac{0.255}{d^2}](https://tex.z-dn.net/?f=V_%7Bavg%7D%3D%5Cfrac%7BQ%7D%7BA%7D%3D%5Cfrac%7B0.2%7D%7B%5Cfrac%7B%5Cpi%20%7D%7B4%7Dd%5E2%7D%3D%5Cfrac%7B0.255%7D%7Bd%5E2%7D)
We know that for pipe flow critical Reynolds number =2300
So ![\frac{V_{avg}d}{\nu }=2300](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bavg%7Dd%7D%7B%5Cnu%20%7D%3D2300)
Value of
is
given in question
So ![\frac{0.255\times d}{d^2\times 6.5\times 10^{-5}}=2300](https://tex.z-dn.net/?f=%5Cfrac%7B0.255%5Ctimes%20d%7D%7Bd%5E2%5Ctimes%206.5%5Ctimes%2010%5E%7B-5%7D%7D%3D2300)
d=1.693 feet
Answer:
Explanation:
Given that:
Torque T = 2300 lb - ft
Bending moment M = 1500 lb - ft
axial thrust P = 2500 lb
yield points for tension σY= 100 ksi
yield points for shear τY = 50 ksi
Using maximum-shear-stress theory
![\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}](https://tex.z-dn.net/?f=%5Csigma_A%20%3D%20%5Cdfrac%7BP%7D%7BA%7D%2B%5Cdfrac%7BMc%7D%7BI%7D)
where;
![A = \pi c^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20c%5E2)
![I = \dfrac{\pi}{4}c^4](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B%5Cpi%7D%7B4%7Dc%5E4)
![\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}](https://tex.z-dn.net/?f=%5Csigma_A%20%3D%20%5Cdfrac%7BP%7D%7B%5Cpi%20c%5E2%7D%2B%5Cdfrac%7BMc%7D%7B%20%5Cdfrac%7B%5Cpi%7D%7B4%7Dc%5E4%7D)
![\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}](https://tex.z-dn.net/?f=%5Csigma_A%20%3D%20%5Cdfrac%7B2500%7D%7B%5Cpi%20c%5E2%7D%2B%5Cdfrac%7B1500%2A12c%7D%7B%20%5Cdfrac%7B%5Cpi%7D%7B4%7Dc%5E4%7D)
![\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}](https://tex.z-dn.net/?f=%5Csigma_A%20%3D%20%5Cdfrac%7B2500%7D%7B%5Cpi%20c%5E2%7D%2B%5Cdfrac%7B72000c%7D%7B%5Cpi%20c%5E3%7D%7D)
![\tau_A = \dfrac{T_c}{\tau}](https://tex.z-dn.net/?f=%5Ctau_A%20%3D%20%5Cdfrac%7BT_c%7D%7B%5Ctau%7D)
where;
![\tau = \dfrac{\pi c^4}{2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20%5Cdfrac%7B%5Cpi%20c%5E4%7D%7B2%7D)
![\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}](https://tex.z-dn.net/?f=%5Ctau_A%20%3D%20%5Cdfrac%7BT_c%7D%7B%5Cdfrac%7B%5Cpi%20c%5E4%7D%7B2%7D%7D)
![\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}](https://tex.z-dn.net/?f=%5Ctau_A%20%3D%20%5Cdfrac%7B2300%2A12%20c%7D%7B%5Cdfrac%7B%5Cpi%20c%5E4%7D%7B2%7D%7D)
![\tau_A = \dfrac{55200 }{\pi c^3}}](https://tex.z-dn.net/?f=%5Ctau_A%20%3D%20%5Cdfrac%7B55200%20%7D%7B%5Cpi%20c%5E3%7D%7D)
![\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B%5Csigma_x%2B%5Csigma_y%7D%7B2%7D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B%28%5Csigma_x%20-%20%5Csigma_y%29%5E2%7D%7B2%7D%2B%20%5Ctau_y%5E2%7D)
![\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B2500%2B72000%7D%7B2%20%5Cpi%20c%20%5E3%7D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B%282500%20%2B72000%29%5E2%7D%7B2%20%5Cpi%20c%5E3%7D%2B%20%5Cdfrac%7B55200%7D%7B%5Cpi%20c%5E3%7D%7D%20%5C%20%5C%20%5C%20%5C%20%5C%20------%281%29)
Let say :
![|\sigma_1 - \sigma_2| = \sigma_y](https://tex.z-dn.net/?f=%7C%5Csigma_1%20-%20%5Csigma_2%7C%20%20%3D%20%5Csigma_y)
Then :
![2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)](https://tex.z-dn.net/?f=2%5Csqrt%7B%28%20%20%20%5Cdfrac%7B2500c%20%2B%2072000%7D%7B2%20%5Cpi%20c%5E3%7D%29%5E2%2B%20%28%20%5Cdfrac%7B55200%7D%7B%5Cpi%20c%5E3%7D%29%5E2%20%7D%20%3D%20100%2810%5E3%29)
![(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6](https://tex.z-dn.net/?f=%282500%20c%20%2B%2072000%29%5E2%20%2B%28110400%29%5E2%20%3D%2010000%2A10%5E6%20%5Cpi%5E2%20c%5E6)
![6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0](https://tex.z-dn.net/?f=6.25c%5E2%20%2B%20360c%2B%2017372.16-10%2C000%5C%20%5Cpi%5E2%20c%5E6%20%3D0)
According to trial and error;
c = 0.75057 in
Replacing c into equation (1)
![\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B2500%2B72000%7D%7B2%20%5Cpi%20%280.75057%29%20%5E3%7D%20%5Cpm%20%5Csqrt%7B%5Cdfrac%7B%282500%20%2B72000%29%5E2%7D%7B2%20%5Cpi%20%280.75057%29%5E3%7D%2B%20%5Cdfrac%7B55200%7D%7B%5Cpi%20%280.75057%29%5E3%7D%7D)
![\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}](https://tex.z-dn.net/?f=%5Csigma_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B2500%2B72000%7D%7B2%20%5Cpi%20%280.75057%29%20%5E3%7D%20%2B%20%20%5Csqrt%7B%5Cdfrac%7B%282500%20%2B72000%29%5E2%7D%7B2%20%5Cpi%20%280.75057%29%5E3%7D%2B%20%5Cdfrac%7B55200%7D%7B%5Cpi%20%280.75057%29%5E3%7D%7D%20%20%5C%20%5C%20%5C%20%20OR%20%5C%5C%20%5C%5C%20%5C%5C%20%20%20%5Csigma_%7B1%2C2%7D%20%3D%20%5Cdfrac%7B2500%2B72000%7D%7B2%20%5Cpi%20%280.75057%29%20%5E3%7D%20-%20%20%5Csqrt%7B%5Cdfrac%7B%282500%20%2B72000%29%5E2%7D%7B2%20%5Cpi%20%280.75057%29%5E3%7D%2B%20%5Cdfrac%7B55200%7D%7B%5Cpi%20%280.75057%29%5E3%7D%7D)
![\sigma _1 = 22193 \ Psi](https://tex.z-dn.net/?f=%5Csigma%20_1%20%3D%2022193%20%5C%20Psi)
![\sigma_2 = -77807 \ Psi](https://tex.z-dn.net/?f=%5Csigma_2%20%3D%20-77807%20%5C%20Psi)
The required diameter d = 2c
d = 1.50 in or 0.125 ft
Answer:
Explanation:
Usar motores eléctricos en aviones ofrece numerosas ventajas reales. A diferencia de los motores de combustión interna los motores eléctricos no necesitan aire para funcionar, lo que significa que pueden mantener toda su capacidad y potencia incluso a altitudes elevadas donde el aire es más tenue.
Answer:Vb=-6i-(-0.1ωab+8)j m/s
Explanation:
Va=V0+Va0
Va=V0+(ra0 x ωao)
ω=Angular velocity of link A0
Using r0a=0.1m;
Va=V0+(0.1i x ω0a K)
Va=0
ixk=j
Va=0+0.1ω0aj
Calculating te velocity of using te equation below
Vb=Va+Vba
Vb=Va+ωab x rba
ωab=40rad/s
rab=-0.21i+0.15j
Va=0.1ω0aj
Vb=Va+ωabxrba
Vb=0.1ω0aj+40k x -(0.21i+0.15j)
Vb=0.1ω0aj-8j-6i
Vb=-6i-(-0.1ωab+8)j m/s