Chemistry, 25.09.2019 03:00, ianukog2481

Awater main made out of galvanized iron has an inner pipe diameter of 0.1 meters, an inlet pressure of 1000 kpa, outlet pressure of 500 kpa, water flow rate of 5 m3/min, and 10 90 degree elbows. what length galvanized iron line should be used to reach the desired outlet pressure?

Answers: 1

Show answers

Answers

answered: trevorhenyan51

Explanation:

The given data is as follows.

Diameter = 0.1 m, $P_{1}$ = 1000 kPa

$P_{2}$ = 500 kPa

Change in pressure $\Delta P$ = 1000 kPa - 500 kPa = 500 kPa

Since, 1000 Pa = 1 kPa. So, 500 kPa will be equal to $500 \times 10^{3}$ .

Q = 5 $m^{3}/min$ = $\frac{5}{60} m^{3}/sec$ = 0.0833 $m^{3}/sec$

It is known that Q = $A \times V$

where, A = cross sectional area

V = speed of the fluid in that section

Hence, calculate V as follows.

V = $\frac{Q}{A}$

= $\frac{Q \times 4}{\pi \times d^{2}}$

= $\frac{0.0833 \times 4}{3.14 \times (0.1)^{2}}$

= 10.61 m/sec

Also it is known that Reynold's number is as follows.

Re = $\frac{\rho \times V \times d}{\mu}$

= $\frac{1000 \times 10.61 \times 0.1}{10^{-3}}$

= 1061032.954

As, it is given that the flow is turbulent so we cannot use the Hagen-Poiseuille equation as follows. Therefore, by using Blasius equation for turbulent flow as follows.

$\Delta P = \frac{0.241 \times \rho^{0.75} \times \mu^{0.25} \times L}{D^{4.75}} \times Q^{1.75}$

$500 \times 10^{3} = \frac{0.241 \times (1000)^{0.75} \times (10^{-3})^{0.25} \times L}{(0.1)^{4.75}} \times (0.0833)^{1.75}$

$500 \times 10^{3} = 5535.36 \times L$

L = $\frac{500 \times 10^{3}}{5535.36}$

= 90.328 m

Thus, we can conclude that 90.328 m length galvanized iron line should be used to reach the desired outlet pressure.

answered: Guest

explanation:

the largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. the number of potential organic compounds has been estimated at 1060—an astronomically high number. the existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. the simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. in addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. we use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. the familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. we can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. this leads to differences in geometries and in the hybridization of the carbon orbitals.

answered: Guest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.