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Anna35 [415]
3 years ago
3

BHE:FLI:JPM next sequence

Engineering
1 answer:
Natasha_Volkova [10]3 years ago
0 0

Answer:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

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A fan draws air from the atmosphere through a 0.30-mdiameter round duct that has a smoothly rounded entrance. A differential man
marusya05 [52]

Answer:

V = 50 ft³/s

H.P = 0.466

Explanation:

Given that

Diameter of the duct, D = 0.3 m

Vacuum Pressure of the duct, Z = 0.025 m

P(w) = pressure of water, 1000 kg/m³

P(a) = pressure of air, 1.22 kg/m³

To find the pressure change we use the formula

ΔP = P(w) * g * Z

ΔP = 1000 * 9.8 * 0.025

ΔP = 245 Pa.

We need the area, do we find that too

A = πd²/4

A = π * 0.3² * 1/4

A = 0.071 m²

Recall the energy equation to be

1/2v² = ΔP/p(a) , so that if we rearrange, we have

v² = 2ΔP/p(a)

v = √(2ΔP/p(a)), on substituting the values, we have

v = √(2 * 245)/1.22

v = √490/1.22

v = √401.64

v = 20.04

The volume flow rate is then equal to

Velocity * Area.

V = 20.04 * 0.071

V = 1.42 m³/s, converting to ft³/s, we have 50 ft³/s

Horsepower output is gotten using

P = ΔP * V

P = 245 * 1.42

P = 347.9 w, converting this to HP, we have 0.466 HP

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3 years ago
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Answer:

Fully Automated

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Answer:

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Explanation: Check the attachment

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3 years ago
Explain the LWD process why is it important in drilling operations?
barxatty [35]

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.

Explanation:

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5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

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This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

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