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3 years ago

Please help will give brainliest please answer all 3

Engineering

1 answer:

larisa [96]3 years ago

8 0

Answer: For #1 I'm going to go with A because that has to do with biology

For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).

For #3 I'll say marine/maritime engineer (you can just say marine)

Hope it helps!

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

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