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Brrunno [24]
3 years ago
7

Soap is a very interesting chemical. We even discussed it on the discussion board. How does it work, exactly?

Engineering
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:

Saponification is a process in which soap is formed from mixtures of  sodium or potassium salts of fatty acids. These fatty acids are reacted high temperature of At 80°C-100°C with alkali to extract salt. These alkali can be sodium hydroxide or potassium hydroxide.

Soap has both polar (ionic) and non polar molecules due to which it has characteristics of both hydrophilic substance (having tendency to mix with water) and hydrophobic substance (have tendency to mix with oils) and due to this nature it can act as an emulsifier.

An emulsifier has tendency to diffuse one liquid into another   liquid which is incapable of mixing with homogeneous liquid like water.

Cleansing action takes place due to presence of ionic and non-polar properties at same time, in combination with solubility principles. The ionic end of soap molecule is the salt end. It is hydrophilic (water soluble) in nature. The non-polar end cotains long hydrocarbon chains and is hydrophobic (water repellent).

When immiscible liquids like grease or oil mixed with soap water, non polar end (hydrophobic end) absorbs the dirt which means the soap will form the micelles and trap the dirt in it. As micelles is soluble in water it will remove the dirt with it.

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Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

5 0
3 years ago
For this problem, calculate the following by hand and show the procedure for how you obtained the results. Subsequently, solve p
Gemiola [76]

Answer:wat

Explanation:

4 0
3 years ago
Give me the description of - Feedforward control loops with an example.
ValentinkaMS [17]

Answer:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop

Explanation:

Feedforward basically configured and used mainly to avoid errors in a control system entering or disrupting a control loop. Although Feedforward control seems to be a very attractive idea, it imposes a high responsibility on both the system developer and the operator to examine and consider mathematically the effect of disruptions on the process concerned.

example of feedforward is  

Shower

which consist of following control points

Hear toilet flush (measurement)  

Customize water to compensate  

feedback refers to that point when water turns hot before the configuration changes

5 0
4 years ago
A design that is found to have flaws and must be redesigned and retested is an example of what
Aleksandr [31]

Answer:

A. prototype

Explanation:

7 0
4 years ago
A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t
Minchanka [31]

Answer:

Q=7.3\times 10^{-3} m^3/s

Explanation:

Given that

At topd_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm

\rho =900\dfrac{Kg}{m^3}

We know that

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2

A_1V_1=A_2V_2

\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2

\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2

V_2=8.02V_1

Z_2=12 sin60^{\circ}

\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}

So V_1=1.30m/s

We know that flow rate Q=AV

Q=A_1V_1

By putting the values

A_1=\dfrac{\pi}{4}d^2

Q=7.3\times 10^{-3} m^3/s

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

4 0
3 years ago
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