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4. FLUID DYNAMICS
CHAPTER: 04. FLUID DYNAMICS 1. Definition: The fluid dynamics is the study of fluid motion with the forces causing the flow. The dynamic behavior of the fluid floe is analyzed by Newton’s Second law of motion which relates acceleration of the fluid flow with the applied force on the fluid. The fluid is assumed to be (1) Incompressible (2) Non-viscous while applying this law . According to this law the net Force “F” acting on a fluid element multiplied by the acceleration in the direction of the applied force . F=mXa The forces acting on fluid flow are I. Gravity Force (Fg) II.
viscous Force (Fv)
III.
Pressure Force (Fp)
IV.
Turbulence Force (Ft)
V.
Compressibility Force (Fc) Total Force (Ft) = Fg + Fv + Fg+Ft+ Fv+ Fc In case compressibility force Fc is neglecting then above equation become Total Force (Ft) = Fg + Fv + Fg+Ft+ Fv equation of motion” In case turbulence Stroke’s equation”
This equation is called “Reynold’s
force Ft is neglecting; then equation is called” Navier
In case Viscous force Fv is neglecting; then equation is called” Euler’s Equation of motion” 2. EULER’S EQUATION OF MOTION In Euler’s equation of motion, forces due to Gravity and pressure are taken into consideration. The following assumption made in the derivation of Euler’s Equation a. Fluid is ideal i.e. Non-viscous b. Flow is compressible c. Flow is non-turbulence d. Flow is steady
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
e. Flow is irrigational
4. FLUID DYNAMICS
f. Flow is one dimensional i.e along stream line g. Velocity is uniform over the cross sectional area Euler’s Equation is derived considering the motion of fluid element along a stream line. Considering a stream line in which flow takes places in S-direction as shown below. Take a cylindrical cross section area dA and length dS on the stream line. The forces acting on the cylindrical elements are a) Pressure Force = PdA
b) Pressure Force = (P +
{ which is acting in the direction of flow }
ds)x dA
{Which is acting opposite to the direction of
flow } c) The weight of the element =
(P + dp .ds ) dA ds
Stream Line
ds
PdA
dS
������
dz
The resultant force on the S-direction = [The mass of fluid x Acceleration in Sdirection (as)]
PdA - (P +
ds) x dA-
cos
=
x as
________________(1)
Where ‘as ‘ is the acceleration in the direction of s as = =
where v is a function of s and t +
=
If the flow is steady ;
+ =0
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
as =
substitute the value of as in equation (1) we get -
ds x dA -
cos
=
x
Dividing by +
cos
+
From fig Cos +
=0
=
+
=0
+ gdz + vdv = 0
____________________(2)
This equation is known as Euler’s Equation of motion
Bernoulli’s Equation from Euler’s Equation (For Ideal Fluid) Bernoulli’s Theorem: it states that “in an ideal, incompressible and steady flow, the total energy at any point of the fluid is remains constant.” total energy consists of (i) pressure energy (ii)KE and (iii) PE Bernoulli’s Equation is obtained by integrating the Euler’s Equation of motion from Equation (2) as +
+
=constant
if the flow is incompressible ‘ ’ is constant then the equation
become + gz + + where :
= constant
+ z = constant
= Pressure energy per unit weight or Kinetic Head = KE per unit weight or kinetic head Z = PE per unit weight or Potential head
Bernoulli’s Equation from Euler’s Equation (For Real Fluid) Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
The real fluids are viscous which offer resistance 4. FLUID DYNAMICSto flow. Due to this, losses in head or energy are always imminent in real fluids. But Bernoulli’s theorem is based on the assumption that the fluid in non-viscous and hence frictionless. In order to apply Bernoulli’s theorem for real fluids, it is necessary to incorporate all such losses under one term known as head loss , hL Applying Bernoulli’s theorem between two points 1 and 2 in a flow field for real fluid , we have +
+ z1
=
+
+ z2 + hL
Force exerted by a flowing fluid on a pipe-Bend V2sin0 0
p2 A2
Fy
2
V2
V2cos0
1 p1A1 Fx
Consider two sections (1) and (2) as shown below Let: v1 = Velocity of flow at section (1) P1 = pressure intensity at section (1) A1 = area of cross-section of pipe at section (1) Similarly v2, p2 and A2 are corresponding velocity, pressure and area at section (2) Let Fx and Fy be the components of the forces exerted by the flowing fluid on the bend in X-and Y- direction respectively. Then the forces exerted by the bend on the fluid in the Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
direction of x and y will be equal Fx and Fy but in opposite directions. Hence component 4. to FLUID DYNAMICS of the forces exerted by the bend on the fluid in the X-direction = - F x and in the direction of Y = - Fy . The other external forces acting on the fluid are p1A1 and p2A2 on the section (1) and (2) respectively. then momentum equation in x-direction is given by Net forces acting on the fluid in the direction of x = rate of change of momentum in x-direction p1A1 – p2A2 cos
-Fx = (mass per sec) (Change of velocity
=
Q (Final velocity in the direction of x – Initial velocity in the direction of x)
=
Q(v2 cos
Fx =
- v1 )
Q (v1 - v2 cos
) + p1A1 – p2A2 cos
Similarly the momentum equation in Y-direction Fy
=
Q
(-v2
sin
)–
p2A2
sin
Now the resultant forces (FR) acting on the bend = And the angle made by the resultant force with horizontal direction is given by tan
=
1. Bernoulli’s Equation From Euler’s Equation (For Ideal Fluid) +
+ z = constant
where:
= Pressure energy per unit weight or Kinetic Head = KE per unit weight or kinetic head Z = PE per unit weight or Potential head
2. Bernoulli’s Equation From Euler’s Equation (For Real Fluid) +
+ z1
=
+
+ z2 + hL
3. Forces exerted by a flowing on a pipe bend Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
a. Forces along x-axis Fx =
Q (v1 - v2 cos
4. FLUID DYNAMICS
) + p1A1 – p2A2 cos
b. Forces along y-axis Fy =
Q (v1 - v2 cos
) + p1A1 – p2A2 cos
Resultant forces F = = Which acts at an angle of tan
with the horizontal
=
4. Continuity equation A1V1 = A2V2 = Q
PROBLEMS 1. Determine the velocity head and rate of discharge at section 1 and 2 for a pipe of 1 2 D1 = 22cm
D2 =12cm
V1 = 5m/s
V1 =? 2
dimensions as shown in fig
1
2. The water is flowing through a pipe having dia 20cm and 10cm at section 1 and 2 respectively. The rate of flow through a pipe is 35liters/s. the section 1 is 6m above the datum and section 2 is 4m above the datum. if the pressure at section 1 is 39.24N/cm 2 . Find the intensity of pressure at section 2 .
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
3. The water flowing through a taper pipe DYNAMICS of length 100m having dia 600mm at the upper 4. FLUID end and 300mm at lower end, at the rate of 50liters/s . The pipe has a slop of 1 in3 0 . Find the pressure at lower end if the pressure at higher level is 19.62N/cm2. 4. A non-uniform part of a pipe line 5m long is laid at a slop of 2 in 5 . Two pressure gauges each fitted at upper and lower ends read 20N/cm 2 and 12,5 N/cm2. If the diameters are at the upper and lower ends are 15cm and 10cm respectively. Determine the quantity of water flowing per second. 5. A pump has a tapering pipe running full of water. The pipe is placed vertically with the diameters at the base and top being 1.2m and 0.6m respectively. The pressure at the upper end is 240mm of Hg vacuum, while the pressure at the lower end is 15KN/m. Assume head loss to be 20% of difference of velocity head. Calculate the discharge, the flow is vertically up ward and difference in elevation is 3.9m.
6. A gasoline of relative density 0.8 is flowing up wards through a vertical pipe line with tapers from 300mm to 150mm. a gasoline mercury differential manometer is connected between 300mm and 150mm pipe section to measure the rate of flow. the distance between manometer taping is 1m and gauge reading is 50cm of Hg a. Find the differential gauge reading in of gasoline head b. Using Bernoulli’s equation and continuity equation, find the rate of flow. Neglect friction and other loss between tapping. 7. A pipe line carrying oil of sp.gr 0.87, changes in dia 200mm dia at a position A to 500mm dia at position B which is 4m at higher level. if the pressure at A and B are 9.81N/cm 2 and 5.886N/cm2 . Respectively and discharge is 200 liters/sec. determine the loss of head and direction of flow 8. a conical tube of length 2 m is fixed vertically with its smaller end up wards. the velocity of flow at the smaller ends is 5m/s. while at the lower end its 2m/s. the pressure head at smaller end 2.5m of liquids the losssof head in the tube is
where v1 is the
velocity at the smaller end and v2 at the lower end respectively. 9. A drainage pump is connected by a section pipe of 0.8m ID and 0.4m OD respectively. It he inlet and out let of the pipe 2m below and 4m above the free surface of water the vacuum gauge connected to the outlet pipe reads 300mm of Hg. if the loss of head in the pipe Is 1/10th of the kinetic head at the outlet of the pipe, calculate the discharge and pressure head at inlet .
10.Find the air pressure in the chamber to discharge 20litersof water through a 6cm dia hose pipe , to a height of 20m from the level of water in the chamber assume the head loss in the pipe line 4m . Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
11.A 450 reducing bend is connected in a pipe line the diameters at inlet and outlet of bend being 60cm and 30cm respectively. find force exerted by the water on the bend if the intensity of pressure at inlet to bend is 9.5N/cm2 and rate of floe of water is 0.620 m3 /s
12.A 300mm dia pipe carries water and head of 20m with a velocity of 3.5m/s. if the axis of the pipe is turns through 450 , find the magnitude and direction of resultant force at the bend
13.250liters per sec of water is flowing in A pipe having a dia 30cm. if the pipe is bent by 135 0 , (i.e change from initial to final direction is135 0) , find the magnitude and direction of resultant force on the bend. The pressure of water flowing is 40N/cm2.
14.A pump of 20KW capacity and 90% efficiency is required to discharge crude oil of relative density 0.85 to above head tank the oil level in over head tank is 20m above the ground level and oil level in storage tank is 5m above the ground level at 40Kpa. if head loss due to friction is 1.5m , determine the rate to flow
15.Water flows through right angle bend of 300mm o f dia at 0.40m 3/s . if the pressure at the inlet and outlet of the bend is 25N/cm2 and 24N/cm2 , determine the magnitude and direction of force exerted by the bend
16.A pipe of 50mm dia is fitted with a nozzle of 25mm dia at one end. find the force exerted by the nozzle on the water flowing at 1.3m3/min in the pipe
17.The diameter of pipe gradually reduced from 1m to 0.7m as shown in fig. the pressure intensity at the center-line 1m section 7.848 kN/2 and rate of flow of water through the pipe is 600 liters/sec. find the intensity of pressure at the center line of 0.7m section. Also determine the force exerted by the flowing water on transition of the pipe.
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
CHAPTER: 04. FLUID DYNAMICS 1. Definition: The fluid dynamics is the study of fluid motion with the forces causing the flow. The dynamic behavior of the fluid floe is analyzed by Newton’s Second law of motion which relates acceleration of the fluid flow with the applied force on the fluid. The fluid is assumed to be (1) Incompressible (2) Non-viscous while applying this law . According to this law the net Force “F” acting on a fluid element multiplied by the acceleration in the direction of the applied force . F=mXa The forces acting on fluid flow are I. Gravity Force (Fg) II.
viscous Force (Fv)
III.
Pressure Force (Fp)
IV.
Turbulence Force (Ft)
V.
Compressibility Force (Fc) Total Force (Ft) = Fg + Fv + Fg+Ft+ Fv+ Fc In case compressibility force Fc is neglecting then above equation become Total Force (Ft) = Fg + Fv + Fg+Ft+ Fv equation of motion” In case turbulence Stroke’s equation”
This equation is called “Reynold’s
force Ft is neglecting; then equation is called” Navier
In case Viscous force Fv is neglecting; then equation is called” Euler’s Equation of motion” 2. EULER’S EQUATION OF MOTION In Euler’s equation of motion, forces due to Gravity and pressure are taken into consideration. The following assumption made in the derivation of Euler’s Equation a. Fluid is ideal i.e. Non-viscous b. Flow is compressible c. Flow is non-turbulence d. Flow is steady
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
e. Flow is irrigational
4. FLUID DYNAMICS
f. Flow is one dimensional i.e along stream line g. Velocity is uniform over the cross sectional area Euler’s Equation is derived considering the motion of fluid element along a stream line. Considering a stream line in which flow takes places in S-direction as shown below. Take a cylindrical cross section area dA and length dS on the stream line. The forces acting on the cylindrical elements are a) Pressure Force = PdA
b) Pressure Force = (P +
{ which is acting in the direction of flow }
ds)x dA
{Which is acting opposite to the direction of
flow } c) The weight of the element =
(P + dp .ds ) dA ds
Stream Line
ds
PdA
dS
������
dz
The resultant force on the S-direction = [The mass of fluid x Acceleration in Sdirection (as)]
PdA - (P +
ds) x dA-
cos
=
x as
________________(1)
Where ‘as ‘ is the acceleration in the direction of s as = =
where v is a function of s and t +
=
If the flow is steady ;
+ =0
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
as =
substitute the value of as in equation (1) we get -
ds x dA -
cos
=
x
Dividing by +
cos
+
From fig Cos +
=0
=
+
=0
+ gdz + vdv = 0
____________________(2)
This equation is known as Euler’s Equation of motion
Bernoulli’s Equation from Euler’s Equation (For Ideal Fluid) Bernoulli’s Theorem: it states that “in an ideal, incompressible and steady flow, the total energy at any point of the fluid is remains constant.” total energy consists of (i) pressure energy (ii)KE and (iii) PE Bernoulli’s Equation is obtained by integrating the Euler’s Equation of motion from Equation (2) as +
+
=constant
if the flow is incompressible ‘ ’ is constant then the equation
become + gz + + where :
= constant
+ z = constant
= Pressure energy per unit weight or Kinetic Head = KE per unit weight or kinetic head Z = PE per unit weight or Potential head
Bernoulli’s Equation from Euler’s Equation (For Real Fluid) Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
The real fluids are viscous which offer resistance 4. FLUID DYNAMICSto flow. Due to this, losses in head or energy are always imminent in real fluids. But Bernoulli’s theorem is based on the assumption that the fluid in non-viscous and hence frictionless. In order to apply Bernoulli’s theorem for real fluids, it is necessary to incorporate all such losses under one term known as head loss , hL Applying Bernoulli’s theorem between two points 1 and 2 in a flow field for real fluid , we have +
+ z1
=
+
+ z2 + hL
Force exerted by a flowing fluid on a pipe-Bend V2sin0 0
p2 A2
Fy
2
V2
V2cos0
1 p1A1 Fx
Consider two sections (1) and (2) as shown below Let: v1 = Velocity of flow at section (1) P1 = pressure intensity at section (1) A1 = area of cross-section of pipe at section (1) Similarly v2, p2 and A2 are corresponding velocity, pressure and area at section (2) Let Fx and Fy be the components of the forces exerted by the flowing fluid on the bend in X-and Y- direction respectively. Then the forces exerted by the bend on the fluid in the Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
direction of x and y will be equal Fx and Fy but in opposite directions. Hence component 4. to FLUID DYNAMICS of the forces exerted by the bend on the fluid in the X-direction = - F x and in the direction of Y = - Fy . The other external forces acting on the fluid are p1A1 and p2A2 on the section (1) and (2) respectively. then momentum equation in x-direction is given by Net forces acting on the fluid in the direction of x = rate of change of momentum in x-direction p1A1 – p2A2 cos
-Fx = (mass per sec) (Change of velocity
=
Q (Final velocity in the direction of x – Initial velocity in the direction of x)
=
Q(v2 cos
Fx =
- v1 )
Q (v1 - v2 cos
) + p1A1 – p2A2 cos
Similarly the momentum equation in Y-direction Fy
=
Q
(-v2
sin
)–
p2A2
sin
Now the resultant forces (FR) acting on the bend = And the angle made by the resultant force with horizontal direction is given by tan
=
1. Bernoulli’s Equation From Euler’s Equation (For Ideal Fluid) +
+ z = constant
where:
= Pressure energy per unit weight or Kinetic Head = KE per unit weight or kinetic head Z = PE per unit weight or Potential head
2. Bernoulli’s Equation From Euler’s Equation (For Real Fluid) +
+ z1
=
+
+ z2 + hL
3. Forces exerted by a flowing on a pipe bend Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
a. Forces along x-axis Fx =
Q (v1 - v2 cos
4. FLUID DYNAMICS
) + p1A1 – p2A2 cos
b. Forces along y-axis Fy =
Q (v1 - v2 cos
) + p1A1 – p2A2 cos
Resultant forces F = = Which acts at an angle of tan
with the horizontal
=
4. Continuity equation A1V1 = A2V2 = Q
PROBLEMS 1. Determine the velocity head and rate of discharge at section 1 and 2 for a pipe of 1 2 D1 = 22cm
D2 =12cm
V1 = 5m/s
V1 =? 2
dimensions as shown in fig
1
2. The water is flowing through a pipe having dia 20cm and 10cm at section 1 and 2 respectively. The rate of flow through a pipe is 35liters/s. the section 1 is 6m above the datum and section 2 is 4m above the datum. if the pressure at section 1 is 39.24N/cm 2 . Find the intensity of pressure at section 2 .
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
3. The water flowing through a taper pipe DYNAMICS of length 100m having dia 600mm at the upper 4. FLUID end and 300mm at lower end, at the rate of 50liters/s . The pipe has a slop of 1 in3 0 . Find the pressure at lower end if the pressure at higher level is 19.62N/cm2. 4. A non-uniform part of a pipe line 5m long is laid at a slop of 2 in 5 . Two pressure gauges each fitted at upper and lower ends read 20N/cm 2 and 12,5 N/cm2. If the diameters are at the upper and lower ends are 15cm and 10cm respectively. Determine the quantity of water flowing per second. 5. A pump has a tapering pipe running full of water. The pipe is placed vertically with the diameters at the base and top being 1.2m and 0.6m respectively. The pressure at the upper end is 240mm of Hg vacuum, while the pressure at the lower end is 15KN/m. Assume head loss to be 20% of difference of velocity head. Calculate the discharge, the flow is vertically up ward and difference in elevation is 3.9m.
6. A gasoline of relative density 0.8 is flowing up wards through a vertical pipe line with tapers from 300mm to 150mm. a gasoline mercury differential manometer is connected between 300mm and 150mm pipe section to measure the rate of flow. the distance between manometer taping is 1m and gauge reading is 50cm of Hg a. Find the differential gauge reading in of gasoline head b. Using Bernoulli’s equation and continuity equation, find the rate of flow. Neglect friction and other loss between tapping. 7. A pipe line carrying oil of sp.gr 0.87, changes in dia 200mm dia at a position A to 500mm dia at position B which is 4m at higher level. if the pressure at A and B are 9.81N/cm 2 and 5.886N/cm2 . Respectively and discharge is 200 liters/sec. determine the loss of head and direction of flow 8. a conical tube of length 2 m is fixed vertically with its smaller end up wards. the velocity of flow at the smaller ends is 5m/s. while at the lower end its 2m/s. the pressure head at smaller end 2.5m of liquids the losssof head in the tube is
where v1 is the
velocity at the smaller end and v2 at the lower end respectively. 9. A drainage pump is connected by a section pipe of 0.8m ID and 0.4m OD respectively. It he inlet and out let of the pipe 2m below and 4m above the free surface of water the vacuum gauge connected to the outlet pipe reads 300mm of Hg. if the loss of head in the pipe Is 1/10th of the kinetic head at the outlet of the pipe, calculate the discharge and pressure head at inlet .
10.Find the air pressure in the chamber to discharge 20litersof water through a 6cm dia hose pipe , to a height of 20m from the level of water in the chamber assume the head loss in the pipe line 4m . Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
11.A 450 reducing bend is connected in a pipe line the diameters at inlet and outlet of bend being 60cm and 30cm respectively. find force exerted by the water on the bend if the intensity of pressure at inlet to bend is 9.5N/cm2 and rate of floe of water is 0.620 m3 /s
12.A 300mm dia pipe carries water and head of 20m with a velocity of 3.5m/s. if the axis of the pipe is turns through 450 , find the magnitude and direction of resultant force at the bend
13.250liters per sec of water is flowing in A pipe having a dia 30cm. if the pipe is bent by 135 0 , (i.e change from initial to final direction is135 0) , find the magnitude and direction of resultant force on the bend. The pressure of water flowing is 40N/cm2.
14.A pump of 20KW capacity and 90% efficiency is required to discharge crude oil of relative density 0.85 to above head tank the oil level in over head tank is 20m above the ground level and oil level in storage tank is 5m above the ground level at 40Kpa. if head loss due to friction is 1.5m , determine the rate to flow
15.Water flows through right angle bend of 300mm o f dia at 0.40m 3/s . if the pressure at the inlet and outlet of the bend is 25N/cm2 and 24N/cm2 , determine the magnitude and direction of force exerted by the bend
16.A pipe of 50mm dia is fitted with a nozzle of 25mm dia at one end. find the force exerted by the nozzle on the water flowing at 1.3m3/min in the pipe
17.The diameter of pipe gradually reduced from 1m to 0.7m as shown in fig. the pressure intensity at the center-line 1m section 7.848 kN/2 and rate of flow of water through the pipe is 600 liters/sec. find the intensity of pressure at the center line of 0.7m section. Also determine the force exerted by the flowing water on transition of the pipe.
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
4. FLUID DYNAMICS
Er Ravichandra Koti, Dept of Mech Engg, SIET, Tumkur.
Page No:
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