Vena contracta is the point in a fluid stream where the diameter of the stream is the least, and fluid velocity is at its maximum.

How is vena contracta formed in fluid flow?

The reason for the phenomenon Vena contracta is that the fluid streamline cannot abruptly changes direction  and they are unable to closely follow the sharp angles in orifices, nozzles and pipes. When fluid passes through the orifices, the gradual contraction of fluid takes place. The fluid stream begins to converge and passes through orifice.  The streamline continues to converging in smooth path even after it passes through the orifices. Fluid converges to the minimum section in downstream of flow and then it diverges again. This minimum cross section is known as vena contracta.

Types of orifices:

Orifice is defined as the small opening on side or bottom of a tank through which any kind of fluid is flowing.

Orifices are classified into small orifice and large orifice depending upon the size of orifice and head of fluid in that orifice.

 Small orifice is the one in which has the head of fluid from the centre of orifice is more than five times the depth of orifice. 

Also, the large orifice is the one which has the head is less than five times the depth of orifice.

Based on shape of orifice they are classified as following:

• Rectangular orifice

• Circular orifice

• Triangular orifice

• Square orifice

Hydraulic Co-efficients

There are mainly 3 hydraulic coefficients, they are as follows:

• Co efficient of contraction, Cc

• Coefficient of discharge, Cd

• Coefficient of velocity, Cv

Coefficient of Contraction (Cc)

It is the ratio of area of water jet at vena-contracta to the area of the orifice.

`C_c=a_c/a`

Where,

`a_c` = Area of jet at vena-contracta

a = Area of the orifice

Coefficient of Velocity (Cv)

It is the ratio of velocity of water jet at vena-contracta to the theoretical velocity.

Coefficient of discharge (Cd)

It is the ratio of actual discharge to the theoretical discharge.

The relation between the coefficient of contraction, coefficient of velocity and coefficient of discharge can be expressed as,

`C_d=C_c*C_v`

Vena contracta is the point in a fluid stream where the diameter of the stream is the least, and fluid velocity is at its maximum.

How is vena contracta formed in fluid flow?

The reason for the phenomenon Vena contracta is that the fluid streamline cannot abruptly changes direction  and they are unable to closely follow the sharp angles in orifices, nozzles and pipes. When fluid passes through the orifices, the gradual contraction of fluid takes place. The fluid stream begins to converge and passes through orifice.  The streamline continues to converging in smooth path even after it passes through the orifices. Fluid converges to the minimum section in downstream of flow and then it diverges again. This minimum cross section is known as vena contracta.

Types of orifices:

Orifice is defined as the small opening on side or bottom of a tank through which any kind of fluid is flowing.

Orifices are classified into small orifice and large orifice depending upon the size of orifice and head of fluid in that orifice.

 Small orifice is the one in which has the head of fluid from the centre of orifice is more than five times the depth of orifice. 

Also, the large orifice is the one which has the head is less than five times the depth of orifice.

Based on shape of orifice they are classified as following:

• Rectangular orifice

• Circular orifice

• Triangular orifice

• Square orifice

Hydraulic Co-efficients

There are mainly 3 hydraulic coefficients, they are as follows:

• Co efficient of contraction, Cc

• Coefficient of discharge, Cd

• Coefficient of velocity, Cv

Coefficient of Contraction (Cc)

It is the ratio of area of water jet at vena-contracta to the area of the orifice.

`C_c=a_c/a`

Where,

`a_c` = Area of jet at vena-contracta

a = Area of the orifice

Coefficient of Velocity (Cv)

It is the ratio of velocity of water jet at vena-contracta to the theoretical velocity.

Coefficient of discharge (Cd)

It is the ratio of actual discharge to the theoretical discharge.

The relation between the coefficient of contraction, coefficient of velocity and coefficient of discharge can be expressed as,

`C_d=C_c*C_v`

SOLUTION:

Here, the area of the cylindrical tank is,

`A=pi/4**0.6^2=0.2827 m^2`

And, the area of the orifice is,`a=0.0012 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of upper orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+1.5))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+1.5))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=1 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))int_{1}^{0}(dh)/(root ()(h)+root ()(h+1.5))`

`=-A/(C_d*a*root ()(2g))int_{1}^{0} (root ()(h)-root ()(h+1.5))/(-1.5)dh`

Solving this using calculator,we get,

`T_1=43.96 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to drop level of water from CD to EF (i.e, empty tank) can be obtained by integrating this equation from limit h=1.5 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g)) int_{1.5}^{0}1/root ()(h)dh`

Solving this using calculator, we get,

`T_2=217.12 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1`

`T=261.08 sec`

SOLUTION:

Here, the area of the cylindrical tank is,

`A=pi/4**0.6^2=0.2827 m^2`

And, the area of the orifice is,`a=0.0012 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of upper orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+1.5))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+1.5))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=1 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))int_{1}^{0}(dh)/(root ()(h)+root ()(h+1.5))`

`=-A/(C_d*a*root ()(2g))int_{1}^{0} (root ()(h)-root ()(h+1.5))/(-1.5)dh`

Solving this using calculator,we get,

`T_1=43.96 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to drop level of water from CD to EF (i.e, empty tank) can be obtained by integrating this equation from limit h=1.5 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g)) int_{1.5}^{0}1/root ()(h)dh`

Solving this using calculator, we get,

`T_2=217.12 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1`

`T=261.08 sec`

SOLUTION:

Radius of the tank, R = 1.25 m

Area of cylinder,`A=pi**1.25^2=4.908 m^2`

Area of orifice,`a=pi**0.15^2/4=0.01767 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to empty the cylindrical  tank can be obtained by integrating this equation from limit h=1.25+3=4.25 to h=1.25 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_1=-A/(C_d*a*root ()(2g))\int_{4.25}^{1.25}root ()(h)dh`

Solving this using calculator, we get,

`T_1=190.85 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT. Let x be the radius of the liquid surface.

Area of liquid surface is,`A=pix^2`

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`dT=-(pix^2 *dh)/(C_d*a*root ()(2gh))`

From figure,

`US=x=root ()(R^2-(R-h)^2)`

or,`x^2=2Rh-h^2`

Now,

The total time required to empty the hemispherical tank can be obtained by integrating this equation from limit h=1.25 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_2=-pi/(C_d*a*root ()(2g))*\int_{1.5}^{0} (2Rh-h^2)/root ()(h)dh`

Solving this using calculator, we get,

`T_2=105.6 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1=190.85+105.6`

`T=261.08 sec`

SOLUTION:

Radius of the tank, R = 1.25 m

Area of cylinder,`A=pi**1.25^2=4.908 m^2`

Area of orifice,`a=pi**0.15^2/4=0.01767 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to empty the cylindrical  tank can be obtained by integrating this equation from limit h=1.25+3=4.25 to h=1.25 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_1=-A/(C_d*a*root ()(2g))\int_{4.25}^{1.25}root ()(h)dh`

Solving this using calculator, we get,

`T_1=190.85 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT. Let x be the radius of the liquid surface.

Area of liquid surface is,`A=pix^2`

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`dT=-(pix^2 *dh)/(C_d*a*root ()(2gh))`

From figure,

`US=x=root ()(R^2-(R-h)^2)`

or,`x^2=2Rh-h^2`

Now,

The total time required to empty the hemispherical tank can be obtained by integrating this equation from limit h=1.25 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_2=-pi/(C_d*a*root ()(2g))*\int_{1.5}^{0} (2Rh-h^2)/root ()(h)dh`

Solving this using calculator, we get,

`T_2=105.6 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1=190.85+105.6`

`T=261.08 sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.0025pi \ m^2`

 Let us consider  any instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`. Let r be the radius of the liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`... (i)

Now from figure,

`0.4/x=1/(2+x)`

or,`0.8+0.4x=x`

`x=4/3`

Also,

`R/r=(2+x)/(x+h)`

`1/r=(10/3)/((4+3h)/3)`

`r=(4+3h)/10`

Now from equation  (i),

`-pi(4+3h)^2/100 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(100C_d*a*root ()(2g))*((4+3h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(100C_d*a*root ()(2g))\int_{2}^{0}((4+3h)^2)/(root ()(h))dh `

Solving this using calculator, we get, 

`T=160.211 Sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.0025pi \ m^2`

 Let us consider  any instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`. Let r be the radius of the liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`... (i)

Now from figure,

`0.4/x=1/(2+x)`

or,`0.8+0.4x=x`

`x=4/3`

Also,

`R/r=(2+x)/(x+h)`

`1/r=(10/3)/((4+3h)/3)`

`r=(4+3h)/10`

Now from equation  (i),

`-pi(4+3h)^2/100 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(100C_d*a*root ()(2g))*((4+3h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(100C_d*a*root ()(2g))\int_{2}^{0}((4+3h)^2)/(root ()(h))dh `

Solving this using calculator, we get, 

`T=160.211 Sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.00785\ m^2`

Area of cylindrical portion,`A=pi**2^2=12.57 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to empty the cylindrical  tank can be obtained by integrating this equation from limit h=14+2=4.25 to h=4 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_1=-A/(C_d*a*root ()(2g))\int_{6}^{4}root ()(h)dh`

Solving this using calculator, we get,

`T_1=541.65 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT. Let r be the radius of the conical liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`...(i)

Now from figure,

`1/x=2/(4+x)`

or,`x+4=2x`

`x=4`

Also,

`1/r=4/(x+h)`

`4r=h+4`

`r=(h+4)/4`

Now from equation  (i),

`-pi(h+4)^2/16 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(16C_d*a*root ()(2g))*((4+h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=4 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}((4+h)^2)/(root ()(h))dh `

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}(h^(3/2)+8h^(1/2)+16h^(-1/2))`

`T=1124.42 Sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1=541.65+1124.42`

`T=1666.07 sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.1^2)/4=0.00785\ m^2`

Area of cylindrical portion,`A=pi**2^2=12.57 m^2`

Let,

`T_1` be the time required to empty the cylindrical portion .

`T_2` be the time to empty the hemispherical portion 

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to empty the cylindrical  tank can be obtained by integrating this equation from limit h=14+2=4.25 to h=4 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T_1=-A/(C_d*a*root ()(2g))\int_{6}^{4}root ()(h)dh`

Solving this using calculator, we get,

`T_1=541.65 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t and let the level of water falls by an amount dh in small time interval dT. Let r be the radius of the conical liquid surface.

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

`-pir^2dh=C_d*a*root ()(2gh)*dT`...(i)

Now from figure,

`1/x=2/(4+x)`

or,`x+4=2x`

`x=4`

Also,

`1/r=4/(x+h)`

`4r=h+4`

`r=(h+4)/4`

Now from equation  (i),

`-pi(h+4)^2/16 dh=C_d*a*root ()(2gh)*dT`

`dT=-pi/(16C_d*a*root ()(2g))*((4+h)^2)/(root ()(h))dh`

The total time required to empty the given tank can be obtained by integrating this equation from limit h=4 to h=0 since h is assumed to be a  depth of water surface above the orifice i.e,bottom of tank.

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}((4+h)^2)/(root ()(h))dh `

`T=-pi/(16C_d*a*root ()(2g))\int_{4}^{0}(h^(3/2)+8h^(1/2)+16h^(-1/2))`

`T=1124.42 Sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1=541.65+1124.42`

`T=1666.07 sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.4^2)/4=0.1256\ m^2`

Area of pool,`A=31**10=300 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of upper orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+3))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+3))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))\int_{2}^{0}(dh)/(root ()(h)+root ()(h+3))`

`=-A/(C_d*a*root ()(2g))\int_{2}^{0} (root ()(h)-root ()(h+3))/(-3)dh`

Solving this using calculator,we get,

`T_1=630.3 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t within the triangular element and let the level of water falls by an amount dh in small time interval dT.

Width of water surface is,

`=h/3**10`

And the area of water surface is,

`A=(h/3**10)**30=100h`

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to drop level of water from CD to E (i.e, empty tank) can be obtained by integrating this equation from limit h=3 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g))\int_{3}^{0}1/(root ()(h))dh`

`=-100/(C_d*a*root ()(2g))\int_{3}^{0}root ()(h)dh`

Solving this using calculator, we get,

`T_2=1037.7 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1`

`T=630.3+1037.7`

`T=1668 sec`

SOLUTION:

Here,

Area of orifice,`a=(pi**0.4^2)/4=0.1256\ m^2`

Area of pool,`A=31**10=300 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of upper orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+3))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+3))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))\int_{2}^{0}(dh)/(root ()(h)+root ()(h+3))`

`=-A/(C_d*a*root ()(2g))\int_{2}^{0} (root ()(h)-root ()(h+3))/(-3)dh`

Solving this using calculator,we get,

`T_1=630.3 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t within the triangular element and let the level of water falls by an amount dh in small time interval dT.

Width of water surface is,

`=h/3**10`

And the area of water surface is,

`A=(h/3**10)**30=100h`

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to drop level of water from CD to E (i.e, empty tank) can be obtained by integrating this equation from limit h=3 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g))\int_{3}^{0}1/(root ()(h))dh`

`=-100/(C_d*a*root ()(2g))\int_{3}^{0}root ()(h)dh`

Solving this using calculator, we get,

`T_2=1037.7 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1`

`T=630.3+1037.7`

`T=1668 sec`

SOLUTION:

Here,

Area of orifice,`a=150 **10^(-4) m^2`

Area,`A_1=10 m^2`

Area,`A_2=20 m^2`

Let, 

h be the difference between water level on either side of the profile.

y be the depth of water in A at any instant.

dy be the change in depth of A during the small time interval dT.

Now, decrease in volume of water in A is,

`=A_1*dy=-10dy`

Similarly increase in volume of water in B is,

`=A_2*dy_2=20dy_2`

Since the decrease in volume of A corresponds to the increase in volume of B,

`dy_2=10/20*dy=1/2*dy`

Now the total change in level of water in small time dt is,

`dh=dy-(-dy_2)=3/2dy`

`dy=2/3dh`

Discharge through orifice in time dT is,

`=C_d*a*root ()(2gh) dT`.... (i)

Thus the decrease in volume of water in time dT is,

`=A_1*dy`

`=20**2/3dh`... (ii)

Thus from (i) and (ii),

`-10**2/3**dh=C_d*a*root ()(2gh) dT`

or,`\int_{0}^{T}dT=-20/(3**C_d**a**root ()(2g))*\int_{h}^{0} h^(-1/2) dh`

or,`T=20/(3**0.62**150**10^(-4)**root ()(2g))*2root ()(h)` ,where h=2

`T=457.7 Sec`

SOLUTION:

Here,

Area of orifice,`a=150 **10^(-4) m^2`

Area,`A_1=10 m^2`

Area,`A_2=20 m^2`

Let, 

h be the difference between water level on either side of the profile.

y be the depth of water in A at any instant.

dy be the change in depth of A during the small time interval dT.

Now, decrease in volume of water in A is,

`=A_1*dy=-10dy`

Similarly increase in volume of water in B is,

`=A_2*dy_2=20dy_2`

Since the decrease in volume of A corresponds to the increase in volume of B,

`dy_2=10/20*dy=1/2*dy`

Now the total change in level of water in small time dt is,

`dh=dy-(-dy_2)=3/2dy`

`dy=2/3dh`

Discharge through orifice in time dT is,

`=C_d*a*root ()(2gh) dT`.... (i)

Thus the decrease in volume of water in time dT is,

`=A_1*dy`

`=20**2/3dh`... (ii)

Thus from (i) and (ii),

`-10**2/3**dh=C_d*a*root ()(2gh) dT`

or,`\int_{0}^{T}dT=-20/(3**C_d**a**root ()(2g))*\int_{h}^{0} h^(-1/2) dh`

or,`T=20/(3**0.62**150**10^(-4)**root ()(2g))*2root ()(h)` ,where h=2

`T=457.7 Sec`

SOLUTION:

Let,

`F_1` be the force due to nozzle 1

`F_2` be the force due to nozzle 2

For `D_1`,

`F_1=d/(dt)(mV)`

`=rho*Q*V`

`=rho*A_1*(V_1)^2`

`=(pi rho)/4*(D_1)^2*(V_1)^2`... (i)

Similarly the force due to nozzle 2 is,

`F_2=(pi rho)/4*(D_2)^2*(V_2)^2`... (ii)

Also,

`V_1=root ()(2gH_1)`

And `V_2=root ()(2gH_2)`

Since `H_2=H_1/2`,

`V_2=root ()(gH_1)`

According to questions,

`F_1=F_2`

`(D_1)^2*(V_1)^2=(D_2)^2*(V_2)^2`

`(D_1)^2*(2gH_1)=(D_2)^2*(gH_1)`

`D_2=root ()(2)*D_1`

`D_2=1.414D_1`

SOLUTION:

Let,

`F_1` be the force due to nozzle 1

`F_2` be the force due to nozzle 2

For `D_1`,

`F_1=d/(dt)(mV)`

`=rho*Q*V`

`=rho*A_1*(V_1)^2`

`=(pi rho)/4*(D_1)^2*(V_1)^2`... (i)

Similarly the force due to nozzle 2 is,

`F_2=(pi rho)/4*(D_2)^2*(V_2)^2`... (ii)

Also,

`V_1=root ()(2gH_1)`

And `V_2=root ()(2gH_2)`

Since `H_2=H_1/2`,

`V_2=root ()(gH_1)`

According to questions,

`F_1=F_2`

`(D_1)^2*(V_1)^2=(D_2)^2*(V_2)^2`

`(D_1)^2*(2gH_1)=(D_2)^2*(gH_1)`

`D_2=root ()(2)*D_1`

`D_2=1.414D_1`

SOLUTION:

Here,

Discharge through submerged portion is,

`Q_1=C_d**b (H_2-H)root ()(2gH)`.. (i)

And the discharge through the free portion is,

`Q_2=2/3C_d*b*root ()(2g)(H^(3/2) -(H_1)^(3/2)`.. (ii)

Thus the total discharge is,

`Q=Q_1+Q_2`

`=C_d**b (H_2-H)root ()(2gH)+2/3C_d*b*root ()(2g)(H^(3/2) -(H_1)^(3/2)`

`=5.1+2.44`

`=7.54m^3/s`

SOLUTION:

Here,

Discharge through submerged portion is,

`Q_1=C_d**b (H_2-H)root ()(2gH)`.. (i)

And the discharge through the free portion is,

`Q_2=2/3C_d*b*root ()(2g)(H^(3/2) -(H_1)^(3/2)`.. (ii)

Thus the total discharge is,

`Q=Q_1+Q_2`

`=C_d**b (H_2-H)root ()(2gH)+2/3C_d*b*root ()(2g)(H^(3/2) -(H_1)^(3/2)`

`=5.1+2.44`

`=7.54m^3/s`

SOLUTION:

Given,

`r_A=0.4m`

`r_B=0.6 m`

Velocity relative to the nozzle,`V_A=V_B=8`m/s

Let `omega` be the angular velocity of the sprinkle.

Now,

Absolute velocity,`V_1=V_A+omega*r_A`

`V_1=8+0.4omega` (positive because the tangential velocity is acting in the same direction as flow velocity )

Similarly, absolute velocity at point B is,

`V_2=V_B-omegar_B`

`V_2=8-0.6omega` (negative because the tangential velocity is acting in opposite direction to the flow velocity )

Now the torque exerted by water coming out at sprinkler A is,

`=rho*Q*V_1*r_1`

This acts in clockwise direction.

Similarly, the torque exerted by water coming out at sprinkler B is,

`=rho*Q*V_2*r_2`

This acts in anticlockwise direction.

Since the moment of the momentum of flow entering is zero and no external torque is applied on the sprinkle, the resultant torque must be zero.i.e,

`T_r=0`

or,`rho*Q (V_2r_2-V_1r_1)=0`

or,`V_2*r_2=V_1*r_1`

or,`(8+0.4omega)**0.4=(8-0.6omega)**0.6`

or,`omega=3.077`rad/s

or,`omega=3.077**60/(2pi)`

or`omega=29.4 r p m`

SOLUTION:

Given,

`r_A=0.4m`

`r_B=0.6 m`

Velocity relative to the nozzle,`V_A=V_B=8`m/s

Let `omega` be the angular velocity of the sprinkle.

Now,

Absolute velocity,`V_1=V_A+omega*r_A`

`V_1=8+0.4omega` (positive because the tangential velocity is acting in the same direction as flow velocity )

Similarly, absolute velocity at point B is,

`V_2=V_B-omegar_B`

`V_2=8-0.6omega` (negative because the tangential velocity is acting in opposite direction to the flow velocity )

Now the torque exerted by water coming out at sprinkler A is,

`=rho*Q*V_1*r_1`

This acts in clockwise direction.

Similarly, the torque exerted by water coming out at sprinkler B is,

`=rho*Q*V_2*r_2`

This acts in anticlockwise direction.

Since the moment of the momentum of flow entering is zero and no external torque is applied on the sprinkle, the resultant torque must be zero.i.e,

`T_r=0`

or,`rho*Q (V_2r_2-V_1r_1)=0`

or,`V_2*r_2=V_1*r_1`

or,`(8+0.4omega)**0.4=(8-0.6omega)**0.6`

or,`omega=3.077`rad/s

or,`omega=3.077**60/(2pi)`

or`omega=29.4 r p m`