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telo118 [61]
3 years ago
7

Friction factor for fluid flow in pipe does not depend upon the A. pipe length. B. pipe roughness. C. fluid density & viscos

ity. D. mass flow rate of fluid.
Chemistry
1 answer:
Feliz [49]3 years ago
8 0

Answer:

C. fluid density & viscosity

Explanation:

In 1850, Darcy-Weisbach experimentally deduced an equation to calculate shear losses ("friction"), in a tube with permanent flow and constant diameter:

hf = (f x L x V^2) / (D x 2g)

where:

hf: shear losses

f:  shear loss factor (pipe roughness)

g: gravity acceleration

D: tube diameter

L: tube length  

V: fluid average speed in the tube

To calculate the loss factor “f” in the Poiseuille laminar region, he proposed in 1846 the following equation:

f = 64 / Re

Where:

Re: Reynolds number

The influence of the parameters on f is quantitatively different according to the characteristics of the current.

In any straight pipeline that transports a liquid at a certain temperature, there is a critical speed below which the regimen is laminar. This critical value that marks the transition between the two regimes, laminar and turbulent, corresponds to a Re = 2300, although in practice, between 2000 and 4000 the situation is quite inaccurate. Thus:

Re <2000: laminar regimen

2000 <Re <4000: critical or transition zone

Re> 4000: turbulent regime

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4 0
3 years ago
Using the picture below, compare and contrast the plant cell and the animal cell (What do they have different? What do they have
STALIN [3.7K]

Answer:

Animal cell lacks chloroplasts and a cell wall

Explanation:

Unlike animal cells, plant cells have chloroplasts and a cell wall. In a plant cell, the chloroplasts work to use photosynthesis in order to turn light into energy for the plant cell. Animal cells do not utilize light energy, and therefore do not carry out photosynthesis and do not have chloroplasts.

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6 0
3 years ago
A chemist wants to make 6.5 L of a 0.340 M CaCl2 solution. Part A What mass of CaCl2 (in g) should the chemist use? Express your
katovenus [111]

Answer:

The answer to your question is 245 grams

Explanation:

Data

Volume 6.5 L

Molarity = 0.34

mass of CaCl₂ = ?

Process

1.- Calculate the molar mass of CaCl₂

molar mass = (1 x 40) + (2 x 35.5)

                   = 40 + 71

                   = 111 g

2.- Convert the grams to moles

                       111 g of CaCl₂ -------------- 1 mol

                         x                   ---------------0.34 mol

                         x = (0.34 x 111) / 1

                         x = 37.74 g

3.- Calculate the total mass

                        37.74 g ------------------ 1 L

                            x        ------------------ 6.5 L

                            x = (6.5 x 37.74) / 1

                            x = 245.31

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Ans

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