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saveliy_v [14]
3 years ago
6

In the fluid-flow analogy for electrical circuits, what is analogous to a conductor?

Engineering
1 answer:
Vedmedyk [2.9K]3 years ago
8 0
I am going to say a pump
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Two balls are chosen randomly from an urn containing 8 white 4 black, and orange balls. Suppose that we win $ 2 for each black b
Scorpion4ik [409]
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8 0
3 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
jok3333 [9.3K]

Answer:

Q=4.98\times 10^{-3}\ m^3/s.

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

We know that

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where Q is volume flow rate

L is length of tube

d_i is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s.

6 0
3 years ago
A well-insulated rigid vessel contains 3 kg of saturated liquid water at 40oC. The vessel also contains an electrical resistor t
user100 [1]

Answer:

The final temperature is 111.66°C

Explanation:

The given conditions :-

i) Well insulated means no heat loss.

ii) Rigid vessels means volume remains same.

iii) Initial temperature ( T₁ ) = 40°C. = 273 + 40 = 313 K

iv ) Mass of water in vessel = 3 kg.

v) current drawn by resistor ( i ) = 10 ampere.

vi) Voltage applied ( V ) = 50 volts.

vii) The time for which resistor operating ( t ) = 30 minute = 30 * 60 = 1800 seconds.

Now we have to calculate heat developed by resistor in vessel.

Q = V * i * t  = 50 * 10 * 1800 = 900,000 J = 900 KJ.

Since it is a rigid container so the work done is zero.

Q = du    ( du - change in internal energy)

Q = m * C * dT      ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

T₂ - 313 = 71.6674

T₂ = 384.6674 K

T = 384.6674 - 273 = 111.66°C

So the final temperature is 111.66°C.

3 0
3 years ago
The compressed-air tank has an inner radius r and uniform wall thickness t. The gage pressure inside the tank is p and the centr
Sedaia [141]

Answer:

Explanation:

Given that:

The Inside pressure (p) = 1402 kPa

= 1.402 × 10³ Pa

Force (F) = 13 kN

= 13 × 10³ N

Thickness (t) = 18 mm

= 18 × 10⁻³ m

Radius (r) = 306 mm

= 306 × 10⁻³ m

Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)

Then;

the state of the plane stress can be expressed as follows:

(\sigma_ x)  = \dfrac{Pd}{4t}+ \dfrac{F}{2 \pi rt}

Since d = 2r

Then:

(\sigma_ x)  = \dfrac{Pr}{2t}+ \dfrac{F}{2 \pi rt}

(\sigma_ x)  = \dfrac{1402 \times 306 \times 10^3}{2(18)}+ \dfrac{13 \times 10^3}{2 \pi \times 306\times 18 \times 10^{-3} \times 10^{-3}}

(\sigma_ x)  = \dfrac{429012000}{36}+ \dfrac{13000}{34607.78467}

(\sigma_ x)  = 11917000.38

(\sigma_ x)  = 11.917 \times 10^6 \ Pa

(\sigma_ x)  = 11.917 \ MPa

\sigma_y = \dfrac{pd}{2t} \\ \\ \sigma_y = \dfrac{pr}{t} \\ \\  \sigma _y = \dfrac{1402\times 10^3 \times 306}{18} \ N/m^2 \\ \\ \sigma _y = 23.834 \times 10^6 \ Pa \\ \\ \sigma_y = 23.834 \ MPa

When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.

Thus;

\tau _{xy} =0

3 0
3 years ago
A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of 10 m ´ 8 m, an effective filtration rate of 7.70 m/h.
galben [10]

Answer:

Explanation:

given data

loading rate = 8.00 m/h

filtration rate = 7.70 m/h

dimensions = 10 m × 8 m

filter cycle duration = 52 h

time = 20 min

to find out

flow rate  and  volume of water is used for back washing plus rinsing the filter  

solution  

we consider here production efficiency is 96%

so here flow rate will be  

flow rate = area × rate of filtration  

flow rate = 10 × 8 × 7.7  

flow rate = 616 m³/h

and  

we know back washing generally 3 to 5 % of total volume of water per cycle so  

volume of water is = 616 × 52

volume of water is  32032 m³

and  

volume of water of back washing is = 4% of 32032  

volume of water of back washing is 1281.2 m³

8 0
3 years ago
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