Engineering Materials

https://drive.google.com/open?id=0B0pAhHvwE1leN251cWQ1VGRsaE0

Read More

Ceramics Industry

https://drive.google.com/open?id=0B0pAhHvwE1leWkdGNXF0bTJtQ0E

Read More

Soap Industry

https://drive.google.com/open?id=0B1hJ4E_kmH3_MDJ5ZVJsTUM4Wmc

Read More

Cement Industry

Cement Industry is very important in economy of a country. It play an important role in the building materials of country.

Here are slides on cement industry available on following link

https://drive.google.com/open?id=0B2sdJ0meBCuCUVFDVjIzTzd6SHM  

Read More

Water Treatment

Water is a universal solvent

Water is a universal solvent. It has ability to dissolve some organic and mostly inorganic materials.
water is very important for the existence of life on earth.Major part(about 75%) of our earth is contained by water.

Uses

Water is used by every live thing on earth. For example, plants consume water to grow and prepare their food.

Animals get water through rivers, and open water resources.

Human beings use water for drinking, bathing, in industries.

As stated earlier water can dissolve at most every material. It can dissolve impurities and hazardous chemicals in it.

To prepare pure water, it water is treated through different procedures.

https://drive.google.com/open?id=0B2sdJ0meBCuCdU43ZXhDZTFkY2s

Following link has PDF file on water treatment:

Read More

Hydrostatic Equilibrium In Centrifugal Field

Centrifugal field is also an application of Hydrostatic Equilibrium it works on the fact that the more the fluid dense, more it farther from the axis of rotation .The lighter (with comparatively low density) liquid will form a layer on heavy liquid.

In a rotatory centrifuge the liquid is thrown outward from the axis of rotation. Surface above the liquid take a shape of a curve or parabolic shape when the centrifuge is rotated about its axis.But at very high speed(In industrial centrifuge) it is rotated at very high speed & surface above liquid take a shape of cylindrical layer.
                                     
                                      r1 is the radial distance from the axis of rotation to the free liquid surface.

                                     r2 is the radius of the centrifugal bowl.

 The whole mass of the liquid rotates like a rigid body.

 F = ma.........(1)

                         Taking Differential

 dF = adm......(2)

  The acceleration of the fluid is given as; a = ω 2 r

Substituting  a = ω 2 r in eq (1) becomes

dF = ω 2 rdm......(3)

If ρ is the density of the given liquid, and b is the breadth of the ring, thus, the mass of the element can be written as; m = πr2 b 

 differentiating equation (3)  leads to  dm = d(πr2 b · ρ) = 2πrbρ · dr 

 Final equation  of  Centrifugal field become

Read More

Applications of Hydrostatic Equilibrium

When we study some thing we look for its application in our daily day life. The hydrostatic equilibrium has wide applications in engineering & other fields of science.

Some Applications of  Hydrostatic Equilibrium are here

1. Barometric Equation
2. Manometers
3. Centrifuge & Decanters

Barometric Equation

Barometric Equation states that Pressure(P) decreases with increase in Height(h)

It is  given as follow:

P=Cexp(−MgRTh).

Derivation Considering an ideal gas with density  ρ is compressedPPP through pressure  P then according to ideal gas law: PV=mMRT,⇒P=mVMRT=ρMRT. Here T is the absolute temperature, R is the universal gas constant equal to 8.314JK⋅mol, M is the molar mass, which is for air equal to 0.029kgmol. It follows from here that the density is given by the formula ρ=MPRT. Putting this into the differential relation for dP gives: dP=−ρgdh=−MPRTgdh,⇒dPP=−MgRTdh. We obtain a differential equation describing the gas pressure P as a function of the altitude h. Integrating gives the equation: ∫dPP=−∫MgRTdh,⇒lnP=−MgRTh+lnC. Getting rid of the logarithms, we obtain the so-called barometric formula P=Cexp(−MgRTh). The constant of integration C can be determined from the initial condition P(h=0)=P0, where P0 is the average sea level atmospheric pressure. Thus, dependency of the barometric pressure on the altitude is given by the formula P=P0exp(−MgRTh). Substituting the known constant values (see Figure 2 above), we find that the dependence P(h) (in kilopascals) is expressed by the formula: P(h)=101.325exp(−0.02896⋅9.8078.3143⋅288.15h)=101.325exp(−0.00012h)[kPa], where the height h above sea level is expressed in meters. If the pressure is given in millimeters of mercury (mmHg), the barometric formula is written in the form: P(h)=760exp(−0.00012h)[mmHg]. In case when the height h is given in feet, and pressure in inches of mercury (inHg), this formula is written as P(h)=29.92exp(−0.00039h)[inHg]. The barometric formula is often used for estimating the air pressure under different conditions, although it gives slightly higher values compared with the real ones.

Example 1

Determine at what altitude the air pressure is twice less than on the sea level? Solution. To estimate the altitude, we use the barometric formula P(h)=P0exp(−0.00012h). When h=0, the pressure P(h) is equal to the average atmospheric sea level pressure P0. At a certain altitude H, the pressure is twice less: P(H)=P02=P0exp(−0.00012H). It follows from here that exp(−0.00012H)=12. Taking logarithms of both sides, we find the altitude H: ln12=−0.00012H,⇒ln2=0.00012H,⇒H=ln20.00012≈5780m.

Read More