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Cell phones require powerful batteries in orde to work effectively. Which activity is best described as an engineering endeavor

natta225 [31]

Where are the options?

8 0

3 years ago

Line.

Veronika [31]

Air supplied to a pneumatic system is supplied through the C. Actuator

Explanation

Pneumatic systems are like hydraulic systems, it is just that these systems uses compressed air rather than hydraulic fluid. Pneumatic systems are used widely across the industries. these pneumatic systems needs a constant supply of compressed air to operate. This is provided by an air compressor. The compressor sucks in air at a very high rate from the environment and stores it in a pressurized tank. the Air is supplied thereafter with the help of a actuator valve that is a more sophisticated form of a valve.

From the above statement it is clear that Air supplied to a pneumatic system is supplied through the Actuator

7 0

4 years ago

A soil has the following Green-Ampt parameters Effective porosity 0.400 Initial volumetric moisture content-15% Hydraulic Conduc

Rudik [331]

Answer:

The graphs are attached below

Explanation:

Ans) We know,

F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]

where K = hydraulic conductivity =0.1 cm/hr

z = capillary suction =20cm

Δ∅ = ∅ (1 - Se)

∅ = effective porosity , Se = initial moisture content

Δ∅ = 0.40(1 - 0.15)

Δ∅ = 0.34

Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]

F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]

Also, infiltration rate (f),

f = K [(zΔ∅ + F)/F]

f = 0.40 [6.8 + F]/F

Condition of ponding,

Ponding time tp = K zΔ∅ i(i-k)

where, i = rainfall rate (1cm/hr)

tp = 0.40(6.8) / 0.60

tp= 4.53 hours

Now, cumulative infiltraton at ponding Fp = i tp

Fp = 1 x 4.53 or 4.53 cm

For infiltration at time less then ponding time , infiltration rate = rainfall rate

For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm

For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm

For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm

For t = 1 hr , f = 1cm/hr ; F = 1cm/hr

For t > tp ,

Equivalent time origin(to),

to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)

to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)

to = 4.53 - 2.5 ( 4.53 - 3.44)

to = 1.82 hr

Hence,

F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]

Solving above equation for t by assuming F, and further solving equation for infiltration rate

Ans (a) Following is required curve :

O.S 1-5 2 time Chr)ホー Lt 2 time Chr)

Ans (c) Following is required table :

Ans (d) For time step t = 0.25 hrs

At t < tp ; i = f = 1 cm/hr

F = i x t

F = 1 x 0.25

F = 0.25 cm

5 0

4 years ago

A tungsten matrix with 20% porosity is infiltrated with silver. Assuming that the pores are interconnected, what is the density

daser333 [38]

Answer:

15.4 g/cm³, 17.4 g/cm³

Explanation:

The densities can be calculated using the formula below

ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)

fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8

a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³

b) density after infiltration with silver

fraction occupied by silver = 20 / 100 = 0.2

density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³

5 0

3 years ago

A balanced bank of delta-connected capacitors is connected in parallel with the load which complex power associated with each ph

Sergio [31]

Answer:

77.2805 μF

Explanation:

Given data :

V = 2460 V

Q = 191 Kva

<u>Calculate the size of Each capacitor </u>

first step : calculate for the value of Xc

Q = V^2/ Xc

Xc ( capacitive reactance ) = V^2 / Q = 2460^2 / ( 191 * 10^3 ) = 31.683 Ω

Given that 1 / 2πFc = 31.683

∴ C ( size of each capacitor ) = $\frac{1}{2\pi *65 *31.683}$ = 77.2805 μF

8 0

3 years ago