Answer:
Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible
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Explanation:
We need to rearrange the following formula for the values given in parenthesis.
(1) x+xy = y, (x)
taking x common in LHS,
x(1+y)=y
![x=\dfrac{y}{1+y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7By%7D%7B1%2By%7D)
(2) x+y = xy, (x)
Subrtacting both sides by xy.
x+y-xy = xy-xy
x+y-xy = 0
x-xy=-y
x(1-y)=-y
![x=\dfrac{-y}{1-y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-y%7D%7B1-y%7D)
(3) x = y+xy, (x)
Subrating both sides by xy
x-xy = y+xy-xy
x(1-y)=y
![x=\dfrac{y}{1-y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7By%7D%7B1-y%7D)
(4) E = (1/2)mv^2-(1/2)mu^2, (u)
Subtracting both sides by (1/2)mv^2
E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2
E-(1/2)mv^2 =-(1/2)mu^2
So,
![2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}](https://tex.z-dn.net/?f=2%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%3D-mu%5E2%5C%5C%5C%5Cu%5E2%3D%5Cdfrac%7B-2%7D%7Bm%7D%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%5C%5C%5C%5Cu%3D%5Csqrt%7B%5Cdfrac%7B-2%7D%7Bm%7D%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%7D%5C%5C%5C%5Cu%3D%5Csqrt%7B%5Cdfrac%7B2%7D%7Bm%7D%28%5Cdfrac%7B1%7D%7B2%7Dmv%5E2-E%29%7D)
(5) (x^2/a^2)-(y^2/b^2) = 1, (y)
![\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%3D%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%5C%5C%5C%5Cy%5E2%3Db%5E2%28%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%29%5C%5C%5C%5Cy%3Db%5Csqrt%7B%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%7D)
(6) ay^2 = x^3, (y)
![y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}](https://tex.z-dn.net/?f=y%5E2%3D%5Cdfrac%7Bx%5E3%7D%7Ba%7D%5C%5C%5C%5Cy%3D%5Csqrt%7B%5Cdfrac%7Bx%5E3%7D%7Ba%7D%7D)
Hence, this is the required solution.
Answer:
The minimum diameter is 1.344 in
Explanation:
The angular speed of the driveshaft is equal to:
![w=\frac{2\pi N}{60}](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B2%5Cpi%20N%7D%7B60%7D)
Where
N = rotational speed of the driveshaft = 2900 rpm
![w=\frac{2\pi *2900}{60} =303.69rad/s](https://tex.z-dn.net/?f=w%3D%5Cfrac%7B2%5Cpi%20%2A2900%7D%7B60%7D%20%3D303.69rad%2Fs)
The torque in the driveshaft is equal to:
![\tau=\frac{P}{w}](https://tex.z-dn.net/?f=%5Ctau%3D%5Cfrac%7BP%7D%7Bw%7D)
Where
P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s
![\tau=\frac{73700}{303.69} =242.68lb*ft](https://tex.z-dn.net/?f=%5Ctau%3D%5Cfrac%7B73700%7D%7B303.69%7D%20%20%3D242.68lb%2Aft)
The minimum diameter is equal to:
![d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}](https://tex.z-dn.net/?f=d_%7Bmin%7D%20%3D%28%5Cfrac%7B16T%7D%7B%5Cpi%20%2A%5Ctau%20%7D%20%29%5E%7B1%2F3%7D)
Where
T = shear stress = 6100 psi
τ = 242.68 lb*ft = 2912.16 lb*in
![d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in](https://tex.z-dn.net/?f=d_%7Bmin%7D%20%3D%28%5Cfrac%7B16%2A2912.16%7D%7B%5Cpi%20%2A6100%7D%20%29%5E%7B1%2F3%7D%20%3D1.344in)
Answer:
Check the explanation
Explanation:
Electrical current can be measured according to the rate of electric charge flow in any electrical circuit:
![i(t) = dQ(t) / dt](https://tex.z-dn.net/?f=i%28t%29%20%3D%20dQ%28t%29%20%2F%20dt)
the derivative of the electric charge by time determines the momentary current.
i(t) will be the momentary current I at time t in amps (A).
Q(t) will also be the momentary electric charge in coulombs (C).
t is the time in seconds (s).
so if the current is constant:
I = ΔQ / Δt
I will be the current in amps (A).
ΔQ will be the electric charge in coulombs (C), which is expected to flows at time duration of Δt.
Δt is the time duration in seconds (s).
Kindly check the attached image below to get the step by step explanation to the question above.