Subject: Fluid Mechanics B.E. (Civil) Year/Part: II/I
Solution of Board exam question of TU, IOE (New Course) - Dr. K. N. Dulal
2068, Baishakh (Regular exam) 1. Define stream line with its drawing equation. A steady, three dimensional velocity field is given by V = (0.657+1.73x+0.984y+az)i+(2.61+cx+1.91y+bz)j+(-2.73x-3.66y-3.64z)k. Calculate constants a, b, c such that the flow field is irrotational. Solution: Stream line is an imaginary curve drawn through the flow field in such a way that the tangent to it at any point indicates the direction of velocity at that point. As stream lines join points of equal velocity, these are velocitycontours.It is useful to visualize the flow pattern. Equation of streamline is
u = (0.657+1.73x+0.984y+az), v= (2.61+cx+1.91y+bz) and w= (-2.73x-3.66y-3.64z) Rotational components: (Curl of V)/2
(
),
(
),
(
)
For irrotational flow , , ( ( (
)
(
)
(
)
(
) (
)) )
a = 2.73, b = 3.66, c = 0.984 2. Define the term pressure and skin friction drag with neat sketch. Describe briefly, with neat sketches, the changes in flow pattern and the drag coefficient with the variation of Reynold’s number when a thin circular cylinder of infinite length is placed in a fluid stream. Solution: Consider an arbitrary shaped solid body placed in a real fluid, which is flowing with a uniform velocity V in a horizontal direction. Consider a small elemental area dA on the surface of the body. The forces
acting on surface dA are: pressure force acting perpendicular surface, shear force acting tangential to the surface. Let θ be the angle made by the pressure force with horizontal direction.
FL PdA τ0dA
FR
PdAcosθ
PdA FD
V
θ
τ0dA
θ PdAsinθ
τ0dAsinθ θ τ0dAcosθ
stationary body
Drag force Drag force on elemental area = Total force in the direction of motion Total drag is ∫
∫
The first term is known as pressure drag and the second term is known as friction drag. Variation of flow around a cylinder with different Re
Re = 2 to 30
Re = 40 to 70
Re = 90
For Re<0.2, the inertia force is negligible and the flow pattern is symmetrical. With the increase of R e,the flow pattern becomes unsymmetrical with respect to the axis perpendicular to the direction of flow. At Re from 2 to 30, very weak vortices are formed on the downstream of the cylinder. It is the initial stage for the development of the wake. At Re from 40 to 70, the wake as well as a pair of vortices become quite distinct. With further increase in the value of Re, the vortices become more and more elongated in
the direction of flow. At Re = 90, these vortices become cylindrical, they leave the cylinder and slowly move in the downstream direction. Coefficient of drag (CD) of cylinder for different Reynold no. (Re) a. For Re<1, b. For Re between 1-2000: CD decreases and reaches a minimum value of 0.95 at Re = 2000. c. For Re = 2000-30000: CD increases and attains a maximum value of 1.2 at Re = 30000. d. For Re = 30000-300000: CD decreases. At Re = 300000, CD = 0.3 e. For Re>300000: CD increases and it becomes equal to 0.7 in the end.
3. Derive the expression for center of pressure on a plane inclined surface immersed in liquid. Also show pressure diagram for such surface.
O θ y yp
z
̅ F
̅
CG
O
CP zp
Consider a plane surface of area A totally immersed in a liquid of specific weight and inclined an angle θ to the free surface. Consider an element of area dA at a vertical distance y from the free surface. P = pressure on each element, ̅ = vertical distance of CG from free surface, yp = vertical distance of CP from the free surface, F = resultant force, z = distance of element from O-O, ̅= distance of CG from O-O and zp = distance of CP from O-O. Finding resultant force Force on element (dF) = Summing the forces, ∑ Assuming to be constant
∑ Here, ∑ Hence,
= first moment of area about an axis through O-O = ̅
̅
̅
̅
Finding the position center of pressure Taking moment of force on element (dM) about O dM= dF x z = ∑ ∑ Total moment of all forces (M) = ∑ ∑ = second moment of area about an axis through O-O = moment of inertia = I0 (a) Moment of resultant force F about O = Equating a and b
̅
(b) ̅
̅ From parallel axis theorem ̅ where IG = M.I. about an axis through CG. Substituting for I0 ( ̅ ) ̅ [ (̅ ) ] ̅ ̅
Pressure diagram for inclined surface
̅
OR The swing check valve in figure covers a 22.86cm diameter circular opening in a slanted wall. The hinge is 15cm from the center line. The valve will open when the hinge moment is 50Nm. Find the value of h for the water to cause this condition.
Air 22.86cmdia
Water h Hinge O
A
B ycp 600
15cm F 30cmdia
Solution: Hinge moment = 50 Nm Area (A) = ( ) = 0.041m2 Location of CG ( ̅) = h Resultant force on the valve is ̅ 9810x.041h = 402.21h Vertical distance of Center of pressure from the free surface ̅
̅
ycp Position of CP from CG (BO) = Location of F from A (AB) = Taking moment about A, FxAB = 50 (
)
h = 0.82m 4. Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in water (SG = 1) with its axis vertical. Show that the body is stable is
[
(
)]
.
Solution: Length of cylinder = L Radius of cylinder = R Specific gravity of cylinder = SG h = depth of immersion Weight of cylinder = Weight of water displaced
M G B
h = SGxL
0
Position of center of buoyancy (OB) = SGxL/2 Position of CG (OG) = L/2 BG = OG-OB = (
)
(
GM = MB-BG =
)
For stable equilibrium, GM
0
[
(
)
(
)]
OR A U-tube shown in figure is filled with a liquid of specific gravity 1.25 to a height of 15cm in both the limbs. It is rotated about a vertical axis 15cm from one limb and 30cm from the other. If the speed of rotation is 60rpm, find the difference in the liquid levels in the two limbs. Also find the pressure at points M and N at the base of U-tube. 30cm
15cm
Z1 Z2 Z0 N
M S 60rpm
Solution: Distance from the axis of rotation to left side (r1) = 0.3m Distance from the axis of rotation to right side (r2) = 0.15m Speed of rotation (N) = 60 rpm Angular velocity ( )
= 6.28 rad/s = =
0.181m 0.045m
Difference in level = 0.181-0.045 = 0.136m Z1+ Z2 = 0.3 (the total height should be same until no spilling) Z1 - Z2 = 0.136 Solving Z1 = 0.218m, Z1 = 0.082m Pressure at N = Pressure at M =
= 1.25x9810x0.218 = 2673 Pa = 1.25x9810x0.082 = 1005.5 Pa
5. Write precisely how the boundary layer theory helped in simplifying the complex motion of fluid. Why does the boundary layer increase with distance from the upstream edge? A simplistic laminar boundary layer model is: √ (
for √ )
(√
)for
Does this expression satisfy boundary conditions applicable to the laminar boundary layer velocity profile? Solution: Boundary layer is a thin layer of the flow in the vicinity of the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary. In the boundary layer velocity gradient (du/dy) is large and the shear stress acts. Outside the boundary, velocity is constant and velocity gradient is zero and hence shear stress is zero. Hence, there are two regions of flows: one is the boundary layer zone close to the boundary where the effect of viscosity is mostly confined and another region outside theboundary layer zone where the flow is inviscid. Thus, the boundary layer theoryhelped in simplifying the complex motion of fluid by considering the region where the velocity gradient exists. At the leading edge of the fluid, the thickness of the boundary layer is zero. On the downstream, for the fluid in contact with the boundary, the velocity of flow is zero and at some distance from the boundary the velocity is u.Hence a velocity gradient is set up which retards the motion of fluid due to the shear
resistance. Near the leading edge, the fluid is retarded in thin layer. At subsequent point downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded. Boundary conditions for laminar flow: (a) At y = 0, u = 0, (b) At
, (c) At
and in
between y = 0 to , velocity gradient exists. The general equation of velocity profile for laminar boundary layer is u = ay+by2+cy3+dy4. First equation √
for √
For this equation, At y = 0, u = 0 At
,
. Velocity gradient is constant i.e. the velocity profile is linear.
√
Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile. Second equation (
√ )
(√
)for (
At At
,
√
√ )
.Velocity gradient is constant i.e. the velocity profile is linear.
and
Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile. 6. Two water tanks are connected to each other through a mercury manometer with inclined tubes, as shown in the figure below. The pressure difference between two tanks is 20Kpa. Calculate “a” and “θ”. Tank A
A
a 2a X 26.8cm
θ
Solution:
B
a X Mercury SG = 13.6
Tank B
PB-PA = 20 Kpa (As Hg is deflected towards A, PB>PA) Equating the pressure at XX
-20x1000+13.6x9810x2a = 0 a = 0.075 m 2a = (26.8/100)Sinθ 2x0.075 = 0.268Sinθ θ = 340
7. A tank of area A is provided with an orifice 40mm in diameter at its bottom. Water flows into the tank at a uniform rate and is discharged through the orifice. It is found that when the head of water over the orifice is 0.68m, the water surface rose at a rate of 0.0014m/Sec. But, when the head of water is 1.24m, the water surface rose at 0.00062m/sec. Find the rate of flow and the cross-sectional area of the tank. Take Cd = 0.62. Solution: Diameter of orifice (d) = 40mm = 0.04m = 0.001257 m2
Area of orifice (a) =
Coeff.of discharge (Cd) = 0.62 =0.00345 √ √ Inflow rate (Qi) =? Cross-sectional area of tank (A) =? First case: Head (h) = 0.68m, dh/dt = 0.0014m/s Second case: Head (h) = 1.24m, dh/dt = 0.00062m/s General equation for tank with inflow (Qi) and outflow (Qo) ( ) √
Substituting values for both cases √
Qi=0.0014A+0.002845
(a)
√
Qi=0.00062A+0.003842 Solving a and b A =1.278 m2 Qi = 0.00463 m3/s
(b)
OR A tank is in the form of frustum of a cone having top diameter of 2m, a bottom diameter of 0.8m and height 2m and is full of water. Find the time of emptying the tank through an orifice 100mm in diameter provided at the bottom. Take Cd = 0.625. 1m
2m
0.4m
H0
Solution: Cd = 0.625 Area of orifice (a) =
= 0.00785 m2
H1 = 2m, H2 = 0m, R1 = 1m, R0 = 0.4m From similar triangles,
H0 = 1.33m √
(
)
(
√
)
= 13.03
Time of emptying the tank is [ ( [ (
) )
( (
) )
( (
)] )] = 160 Sec
8. For the two orifices shown in the figure below, determine Y2 such that
2m
.
orifice1
10m orifice2
Y2 X2
X1
Solution: H1 = 2m, Y1 = 10-2 = 8m, H2 = 10-Y2 Y2= ? Coefficient of velocity for orifice 1 (
)
Coefficient of velocity for orifice 2 (
)
√ √
Since the two orifices are identical Cv1 = Cv2 √
√
(
) (
)
Solving for Y2 Y2 = 1, 9 As Y1 = 8m, Y2 = 9 (>Y1) is not feasible. Hence Y2 = 1m
9. The diameter of a pipe bend is 30cm at inlet and 15cm at outlet and the flow is turned through 1200 in a vertical plane. The axis at inlet is horizontal and the center of the outlet section is 1.5m below the center of the inlet section. Total volume of water in the bend is 0.9m 3. Neglecting friction, calculate the magnitude and direction of the force exerted on the bend by water flowing through it at 250lps and when the inlet pressure is 0.15N/mm2.
Y
W
Z1
P1 V1
X 120 R
1
0
Ry
Rx
F Z2 P2 V2
2
Solution: Diameter at section 1 (d1) = 30cm = 0.3m = 0.07068m2
Area at section 1 (A1) =
Diameter at section 2(d2) = 15cm = 0.15m = 0.01767m2
Area at section 2 (A2) =
Discharge (Q) = 250 lps = 0.25 m3/s Volume of water within control volume (Vol) = 0.9m3 Weight of water within control volume (W) = = 9810x0.9 = 8829N Velocity at section 1 (V1) =Q/A1 = 0.25/0.07068 = 3.54 m/s Velocity at section 2 (V2) =Q/A2 = 0.25/0.01767=14.15 m/s Z2 = 0, Z1 = 1.5m θ= 180-120 = 600 Pressure at 1 (P1) = 0.15N/mm2 = 0.15x106N/m2 Resultant force exerted by the water on the bend = ? Applying Bernoulli’s equation between 1 and 2
P2 = 70870 N/m2 ∑ ( ( ( ( = 13882 N
) ) )
( ( (
) ) ) )
(
)
∑ ( (
) )
= 4681N Resultant force ( )
√
=14650N
Resultant force exerted by the water on the bend = 14650N (to the right and downward) Direction of resultant force =
=18.60
=
10. The pipe flow in the figure is driven by the pump. What gauge pressure is needed to be supplied by the pump to provide water flow rate of Q = 60m3/h? Neglect head loss from A to B. Head loss from C to D=
; Head loss from D to E =
; dAB (diameter of pipe AB) = dCD = 5cm; dDE = 2cm. where VCD =
velocity in pipe CD and VDE = velocity in pipe DE. D
1
80m
10m B A
C
Pump
Solution: Discharge (Q) = 60m3/h = 60/3600 m3/s = 0.0167 m3/s Diameter of pipe AB and CD = 5cm = 0.05m C/s Area of pipe AB and CD (AAB, ACD) =
= 0.001963m2
Diameter of pipe DE =2cm = 0.02m C/s Area of pipe DE (ADE) =
= 0.000314m2
E
Q
Velocity of flow through AB and CD (VAB = VCD) =Q/AAB = 0.0167/0.001963 = 8.5m/s Velocity of flow through DE (VDE) = Q/ADE = 0.0167/0.000314 = 53.2m/s Head loss between A and B (hLAB) = 0 Head loss from C to D (hLCD) =
= 110.5m
Head loss from D to E (hLDE) =
= 73.65m
Total head loss (hL) = 0+110.5+73.65 = 184.15m Hp = Head supplied by the pump At point 1, V1 = 0, P1 = 0 (atm) At point E, PE = 0 (atm) Applying Bernoulli’s equation between 1 and E (Taking datum through A)
hp =398.4m Applying Bernoulli’s equation between 1 and C
Pc = 3970107 Pa = 3970 Kpa
11. Express the kinematic viscosity in stokes for a liquid with specific gravity 0.95 and dynamic viscosity 0.011 poise. A U-tube is made up of two capillarities of diameters 1.0mm and 1.5mm respectively. The U-tube is kept vertically and partially filled with water of surface tension 0.0075kg/m and zero contact angle. Calculate the difference in the level of the miniscii caused by capillarity.
h
d1 d2
Solution: Specific gravity = 0.95 Density of fluid (ρ) = 0.95x1000 = 950 kg/m3 Dynamic viscosity ( ) = 0.011 poise = 0.1x0.011 = 0.0011 NS/m2
Kinematic viscosity ( ) =? = 1.158x10-6 m2/s = 1.158x10-6x104 Stokes = 0.01158 Stokes Diameter of tube 1 (d1) = 1mm = 1x10-3 m Diameter of tube 2 (d2) = 1.5mm = 1.5x10-3 m Surface tension (σ) = 0.0075kg/m = 0.073575 N/m Contact angle (θ) = 0 Capillary rise in tube 1 (h1)
= 0.0316m
Capillary rise in tube 2 (h2)
= 0.021m
Difference in level (h) = 0.0316-0.021 = 0.0106m 12. The pressure drop in an air duct depends on the length and diameter of the duct, the mass density, viscosity of the fluid and the velocity of the flow. Obtain an expression for the pressure drop in dimensionless form using Buckingham –π theorem. Estimate the pressure drop in a 20m long air duct if a model of the duct operating with water produces a pressure drop of 10 KN/m2 over 10m length. The scale ratio is 1:50. Given, = 1000 kg/m3, = 1.2kg/m3 = 0.001NS/m2, = 0.0002NS/m2 Solution: ( Total number of variables = 6 No. of fundamental dimensions = 3 No. of π terms = 6-3= 3 ( ) (I) Choose D, V and ρ as repeating variables. First π term (II) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c1+1 =0 a1+b1-3c1-1 = 0 -b1- 2 = 0 c1 = -1, b1 = -2 a1-2+3-1=0
)
a1 = 0 Substituting the values of a1, b1 and c1 in II
Second π term (III) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c2 = 0 a2+b2-3c2+1 = 0 b2 = 0 a2 = -1 Substituting the values of a2, b2 and c2 in III
Third π term (IV) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c3+1 =0 a3+b3-3c3-1 =0 -b3-1 = 0 c3 = -1 b3 = -1 a3-1+3-1 = 0 a3 =-1 Substituting the values of a3, b3 and c3 in (IV)
Substituting the values of π1, π2 and π3 in (I) (
) (
)
Multiplying first πterm by 1/π2 and 1/π3 and expressing the product as a function of 1/π3
( (
)
)
Numerical Scale ratio (Lr) = 1/50 Pressure drop in model (Pm) = 10 KN/m2= 10x103 N/m2 Density of air (ρp) = 1.24 kg/m3 Density of water (ρm) = 1000 kg/m3 Viscosity of water (μm) = 0.001NS/m2 Viscosity of air (μp) = 0.0002NS/m2 Pressure drop in prototype (Pp) = ? As the problem involves both viscous and pressure force, we have to use both Reynolds and Euler model law. From Reynolds’ model law Re)model = Re) Prototype
Lm/Lp = 1/50
Vm = 0.31Vp From Euler model law Eu) model = Eu) prototype √
√
√ √
Pp = 129N/m2 Pressure drop in prototype corresponding to 10 KPa over 10m length in model = 129N/m2 Pressure drop for 20m length = 2x129 = 258N/m2 OR How are the repeating variables selected for dimensional analysis? Prove that the scale ratio for discharge for a distorted model is given as
( )
( )
where Qp = discharge through
prototype, Qm= discharge through model, (Lr)H = horizontal scale ratio, (Lr)V = Vertical scale ratio. Solution:
Repeating variables appear in most of the π groups.They have a large influence on the problem.There is great freedom in choosing these. Some rules which should be followed for selecting repeating variables are as follows: There are n ( = 3) repeating variables. In combination they must contain all of dimensions (M, L, T) The repeating variables must not form a dimensionless group. Dependent variable should not be selected as repeating variable. No two repeating variables should have the same dimensions. They should be measurable in an experiment. They should be of major interest to the designer. Qp = discharge through prototype, Qm= discharge through model, (Lr)H = horizontal scale ratio, (Lr)V = Vertical scale ratio A = C/s Area of flow, V = Velocity, B = width, h = depth of flow (a) ( ) ( ) ( ) and
where
( )
From Froude’s model law √
√
( )
(c)
From a, b and c ( )
( )
(b)
2068, Shrawan (Back exam) 1. State and prove Newton’s law of viscosity. Prove that P is pressure and
, where K is Bulk Modulus of elasticity,
is density.
Solution: Newton’s law of viscosity states that the shear stress is proportional to the rate of deformation or velocity gradient.
Where = shear stress, du/dy = velocity gradient andthe constant of proportionality ( ) = coefficient of viscosity.
A
dx A’
D
D’
τ dφ
dy
τ B
C
Let us consider a fluid confined between two plates, where the bottom plate is stationary and the upper plate is moving. Let ABCD is the fluid at any time t. Due to the application of shear force τ, the fluid deforms to A’BCD’ at time t+dt. Let dy = distance between two layers, AA’ = dx and shear strain = dφ. For small angle, dx = dφ. dy Also dx = du. dt Equating (a) Shear stress is proportional to rate of shear strain. (b) From a and b
Bulk modulus is (c) Considering unit mass of substance, Differentiating w.r.t. ρ (
)
(d) From c and d,
2. An 8cm diameter piston compresses manometer oil into an inclined 7mm diameter tube, as shown in figure below. When a weight W is added to the top of the piston, the oil rises an additional distance of 10cm up the tube. How large is the weight, in N?
W 10cm
D = 8cm Piston
h
Initial level
X
X
dh Y
Y
Final level
Oil, sp gr = 0.827
D = 7mm
15
0
Solution: Diameter of piston (D) = 8cm = 0.08m Diameter of tube (d) = 7mm = 0.007m h = 0.10Sin15 = 0.0258m Due to compression, the fluid in the container moves down by dh and the fluid in the tube moves up by 10cm. Volume of fluid fallen = Volume of fluid risen dh = 0.000766m
Equating the pressure at new level (YY) (
)
(b)
(
)
W=1N
3. A cylinder, 2m in diameter and 3m long weighing 3KN rests on the floor of the tank. It has water to a depth of 0.6m on one side and liquid of sp gr 0.7 to a depth of 1.25m on the other side. Determine the magnitude and direction of the horizontal and vertical components of the force required to hold the cylinder in position.
C 0.6m
A
E O D
Water
Oil
1.25m
B
Solution: OA= OB = OC = 1m, BD = 0.6m OD = 1-0.6 = 0.4m ) = 0.9165m CD = ( (
)= 66.40
OE = 1.25-1 = 0.25m (
)= 75.50
)
̅̅̅̅ ̅̅̅̅ ( ) ( ) = 0.7x9810x1.25x3x1.25/2-9810x0.6x3x0.6/2 = 10797N (left)
Net vertical force (FV)= Weight of volume of water vertically above AB + Weight of volume of water vertically above BC = FvAB (up) + FvBC (up) = ( ) ( ) =
(
= 32917N (up)
)
[
(
)]
The components to hold the cylinder in place are 10797 N to the right and 32917-3000 = 29917N down . 4. An open cylinder tank 0.5m in diameter and 1m height is completely filled with water and rotated about its axis at 240 rpm. Determine the radius up to which the bottom will be exposed and the volume of water spilled out of the tank. Solution: r =0.25m A
B
1m z
C
D
z1
P r1
Radius (r) = 0.25m N = 240rpm Angular velocity ( )
= 25.13 rad/s = 2.01m
Z1 = 2.01-1= 1.01m r1 = ? Volume of water spilled = ?
r1 = 0.18m Volume of water spilled = Volume of paraboloid (APB-CPD) = 0.146 m3
OR Develop an expression for the position of the Metacenter relative to the center of buoyancy. Solution: M A
θ O
E
D
G B
C B’
FB =W
Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is tilted through a small angle θ, the CB changes from B to B’, the position of water line changes to ED and two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and FB is the buoyant force. Consider an element of area dA at a distance x from the CG of plan of vessel on either side. Volume of element (dV) = dA.xθ (xθ = height for small angle) x x Buoyant force on the element of left side (dFB) = Weight of element = where is the sp wt of water. Similarly, Buoyant force on the element of right side (dFB) = Buoyant force on the two elements produces a couple. Moment of couple = ( )= ( ) Total moment due to altered displacement (M) = ∫ ( ) = ∫( ) ) = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia) ∫( (a) Moment due to the movement of CB from B to B’ (M’) = FB BB’ = WBB’ (b) Where V is the volume of water displaced. For equilibrium, M = M’
GM = BM-BG 5. Derive the continuity equation for cylindrical coordinates.
Y
O
VZ T
R
Vθ
S dZ
Q
Vr
dθ P
N M
θ
dr X
r
Consider any point P(r, θ, Z) in space. Let dr, dθ and dZ be small increments in r, θ and Z direction respectively. Let Vr, Vθ and Vz be the velocity in r, θ and Z direction respectively, and ρ be the density of fluid. Mass of fluid entering face PQNM per unit time = (
Mass of fluid leaving face RSTO per unit time = (
[
Gain of mass in z-direction =
) )
Similarly, (
Gain of mass in r-direction
(
Gain of mass in θ-direction (
Net gain of mass [
(
)
(
)
)
) (
)
)
(
(c) )
(
]
Rate of increase of mass per unit time Equating d and e
(b)
(d) (
)
(e)
)
]
(
)
(a)
[
(
)
[
(
(
)
)
(
(
)
)
(
(
] )
)
]
This is the continuity equation in 3D cylindrical polar co-ordinate. For steady flow, (
)
(
)
. The continuity equation is (
)
For incompressible flow, (
. The continuity equation is
)
6. Derive Bernoulli’s equation from Euler’s equation of motion. A venturimeter with a throat diameter of 100mm is fitted in a vertical pipeline of 200mm diameter with oil of sp.gr. 0.88 flowing upwards at a rate of 0.06m/s. The venturimeter coefficient is 0.96. Two pressure gauges calibrated in KN/m2 are fitted at tapping points, one at the throat and the other at the inlet pipe 320mm below the throat. The difference between two pressure gauge readings is 28 KN/m2. Working from Bernoulli’s equation, determine (I) the volume rate of oil through the pipe, (II) the difference in level in the two limbs of mercury manometer if it is connected to the tapping points. Solution: Euler’s equation in 1D is Integrating ∫
∫
∫
Dividing by g
or, This is the Bernoulli’s equation. Diameter at inlet (d1) = 20cm = 0.2m C/s area of inlet (A1) =
= 0.0314m2
Diameter at throat (d2) = 10cm = 0.1m C/s area of throat (A2) =
= 0.00785m2
Sp.gr. of oil (S0) =0.88 Density of oil (ρ) =0.88x1000 = 880 kg/m8 Difference in elevation (Z2-Z1) = 320cm =0.32m Difference in pressure (P1-P2)= 28KN/m2 = 28000N/m2 Sp.gr. of mercury (S) =13.6 Cd =0.96
Throat
2
1 Inlet
Discharge of oil (Q) =? Manometer reading (x) =? (I) Applying Bernoulli’s equation between inlet (1) and throat (2)
(
) (
) (a)
According to continuity equation A1 V1=A2V2 =
=4V1
(b)
Solving a and b (
)
V1 = 1.95m/s Discharge (Q) =A1V1 = 0.0314x1.95 = 0.612 m3/s Actual discharge = Cd Q = 0.96x0.612 = 0.5875 m3/s (
II. (
)=
(
) =2.92
) (
)
x = 0.2m 7. Derive an expression for estimating discharge for partially submerged orifice. Solution: In partially sub-merged orifice, the outlet side of the orifice is partially submerged under liquid. This type of orifice has two portions: free discharging orifice at the upper part and submerged orifice at the lower part.
H1
H2
H
Let H1 = height of liquid above the upper edge of orifice, H2 =height of liquid above the lower edge of orifice, H= Difference in liquid level between two tanks, b = width of orifice, and Cd = coefficient of discharge Discharge through orifice (Q) = Discharge through free portion (Q1) + Discharge through submerged portion (Q2) Free portion is treated as large rectangular orifice. Hence (
√
)
For submerged portion, ( Q = Q1 + Q2 √
)√ (
)
(
)√
8. A 450 reducing bend is connected in a pipe line carrying water. The diameter at inlet andoutlet of the bend is 400mm and 200mm respectively. Find the force exerted by water on the bend if the intensity of pressure at inlet of the bend is 215.8KN/m2. The rate of flow of water is 0.5m3/s. The loss of head in the bend is 1.25m of oil of sp.gr. 0.85. Solution: P2
+Y
2 1 450
Fx
P1 Fy
Solution: Diameter at section 1 (d1) = 400mm = 0.6m Area at section 1 (A1) =
= 0.1256m2
Diameter at section 2(d2) = 200mm = 0.2m Area at section 2 (A2) = Discharge (Q) = 0.5 m3/s
= 0.0314m2
Reactions
+X
Velocity at section 1 (V1) =Q/A1 = 3.98 m/s Velocity at section 2 (V2) =Q/A2 = 15.92 m/s Pressure at section 1 (P1) = 215.8KN/m2 = 215800N/m2 Loss of head = 1.25m of oil of sp gr 0.85, P= 0.85x9810x1.25 N/m2 Loss of head in terms of water (hL) =
=1.0625m
Angle of bend ( )= 450 Resultant force (FR) =? Direction of resultant force =? Applying Bernoulli’s equation between 1 and 2 (Z1 = Z2)
P2 = 86574N/m2 ∑ ( (
) )
( ( = 21544N
)
( ( (
) ) ) )
(
)
∑ ( (
) )
= 7551N Resultant force (
)
√
=22829N
Resultant force exerted by the water on the bend = 22829N (to the right and downward) Direction of resultant force =
=
=19.30
9. Define boundary layer with appropriate sketches. How laminar sub-layer is developed in boundary layer. Differentiate between hydraulically smooth and rough boundary. Solution: When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is
U
stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the velocity gradually increases. Boundary layer stream velocity (U)
Solid body
The term boundary layer is defined as the thin layer of the flow on the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary.
Boundary layer along a thin flat plate Turbulent Boundary layer
U Laminar Boundary layer
Laminar sublayer
Laminar zone
Turbulent zone Transition zone
Consider the flow of fluid, having free stream velocity U, over a smooth thin plate which is flat and parallel to the direction of free stream of fluid. At the leading edge of the fluid, the thickness of the boundary layer is zero. As the fluid moves along the plate, the thickness goes on increasing.The zones formed during boundary layer development are: Laminar boundary layer, Transition zone, turbulent boundary layer. If the plate is very smooth, even in the region of turbulent boundary layer, there is a very thin layer just adjacent to the boundary, in which the flow is laminar. This layer is known as laminar sub-layer. Let k is the average height of irregularities projecting from the surface of a boundary and = thickness of laminar sub layer. If k is much less than , the boundary is called hydrodynamically smooth boundary. In this case eddies cannot reach the surface irregularities. If is much less than k, the boundary is called hydrodynamically rough boundary. In this case eddies come in contact with the surface irregularities and lot of energy will be lost. From Nikuradse’s experiment:
<0.25: Smooth boundary >6: rough boundary 0.25< <6: transition 10. Define aerofoil with accepted terminology with neat sketch. A kite, which may be assumed to be a flat plate and mass 1kg, soars at an angle to the horizontal. The tension in the string holding the kite is 60N when the wind velocity is 50 km/h horizontally and the angle of string to the horizontal direction is 350. The density of air is 1.2 kg/m3. Calculate the drag coefficient for the kite in the given position if the lift coefficient in the same position is 0.45. Both coefficients have been based on the full area of the kite.
Solution:
chord line α
L
An airfoil is a streamlined body which may be either symmetrical or unsymmetrical. Some of the terminologies used to characterize airfoil are as follows. Chord line: It is the line joining the front and rear edge. Angle of attack: It is the angle between the direction of flowing fluid and chord line. Camber: It is the curvature of an airfoil. Span: The overall length of airfoil is called its span (L). Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio. Stall: An airfoil is said to be in stall condition when the angle of attack of an airfoil is greater than the angle of attack at maximum lift.
Kite 0
35
Drag
String T
Solution:
Lift
W
Mass of kite = 1kg Weight of kite (W) = 1x9.81 N = 9.81N Area of kite = A Velocity (V) = 50 km/hr =
=13.88/s
Density of air ( ) = 1.2kg/m3 Tension (T) = 60N Coefficient of lift (CL) = 0.45 Coefficient of drag (CD) = ? Forces in X-dir (Drag) FD = Tcos35 = 60cos35 =49.14N Force in Y-dir (Lift) FL = Tsin35+9.81 = 60sin35+9.81=44.22N
A = 0.85 m2
CD = 0.5 11. Determine the relationship between model and prototype kinematic viscosity if both Reynold’s number and Froude number are to be satisfied. Solution: From Reynolds model law. (Re) model = (Re) prototype
(a) From Froude model law (Fr) model = (Fr) prototype √
√
(
)
From a and b (
)
(b)
12. Show by dimensional analysis that the power P required to operate a test tunnel is given by (
)
where is density of fluid, is viscosity, V is fluid mean velocity, P is the power required and L is the characteristics tunnel length. Solution: ( ) Total number of variables = 5 No. of fundamental dimensions = 3 No. of π terms = 5-3= 2 ( ) (I) Choose ρ, L and V as repeating variables. First π term (II) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T a1+1 =0 -3a1+b1+c1+2 = 0 -c1-3 = 0 Solving a1 = -1, c1 = -3 -3x-1+b1-3+2 = 0 b1 = -2 Substituting the values of a1, b1 and c1 in II
Second π term (III) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T a2+1 = 0 -3a2+b2+c2-1=0 -c2-1 = 0 Solving a2 = -1, c2 = -1 -3x-1+b2-1-1 = 0 b2 = -1 Substituting the values of a2, b2 and c2 in III
Substituting the values of π1 and π2 in I (
) (
) (
)
2068, Chaitra (Regular exam) 1. Define kinematic and dynamic viscosity. A small thin plane surface is pulled through the liquid filled space between two large horizontal planes in the parallel direction. Show that the force required will be minimum if the plate is located midway between the planes. Solution: According to Newton’s law of viscosity, shear stress is proportional to the rate of deformation or velocity gradient.
Where = shear stress, du/dy = velocity gradient andthe constant of proportionality ( ) = coefficient of viscosity. The constant is also called dynamic viscosity or absolute viscosity. Kinematic viscosity is defined as the ratio of dynamic viscosity to density.
Where = Kinematic viscosity and ρ = density
h V y
= shear stress in the upper part = shear stress in the upper part F = pull required to drag the plate
[
]
For F to be minimum, dF/dy = 0 [( (
)
]
)
Thus, the plate is located midway between the planes for the force to be minimum.
2. Define atmospheric, gauge and pressure. A U-tube shown in figure is 10mm in diameter and contains mercury. If 12ml of water is poured into the right hand leg, what are the ultimate heights in the two legs? Solution: The pressure exerted by the atmospheric air on the surface with which it is in contact is known as atmospheric pressure. If the pressure is quoted by taking atmospheric pressure as a datum, it is called gauge pressure. The pressure measured above a perfect vacuum is called absolute pressure.
Water h
y y E
F
120mm
Mercury
Solution: With the addition of 12ml of water, Drop in mercury level in right leg = y Rise in mercury level in left leg = y Depth of water column = h h= Vol of water added/Cross sectional area =
(
) = 0.153m = 153mm
PE = PF 13.6x9810x2y = 9810x0.153 y = 0.0056m = 5.6mm Height of free mercury level in the left leg = 120+5.6 = 125.6mm Height of free water level in the right leg = 120-5.6+153 = 267.4mm
3. Prove that in case of force vortex, rise of liquid level at the end is equal to the fall of liquid level at the axis of rotation. The hemispherical dome shown in figure is filled with water and is attached to the floor by two diametrically opposite bolts. What force in either bolt is required to hold the dome down, if the dome weighs 25KN? Solution:
R
A
B
hr N
M
hf
O
C
D
h P
Q
R = Radius of cylinder MN = Water level at absolute equilibrium (original water level) After rotation, AOB is the profile of the liquid surface. hr = Rise of liquid at end hf = Fall of liquid at the end Volume of liquid before rotation = ( ) Volume of liquid rotation = Volume of cylinder ABQP- Volume of paraboloid AOB (
)
(
)
Volume of liquid before rotation = Volume of liquid after rotation (
)
(
)
(
Hence, rise of liquid level at the end = fall of liquid level at the axis of rotation 4cm
5m
Water
1.5m
Bolt
)
Vertical force on the dome due to water (Fv) = Weight of volume of water (imaginary) vertically above dome = [ ] [
]
= 381325N = 381.325KN (up) Weight of dome = 25 KN (down) Net Vertical force = 318.325-25 = 356.325 (up) Force on either bolt = 356.325/2 = 178.163KN (up)
4. Show that if ‘B’ is the center of buoyancy and ‘M’ is the metacenter of a partially immersed floating body, then BM=I/V, where I is the second moment of the area of the surface of floatation about the longitudinal axis and V is the immersed volume. Solution: dFB
M
A
θ O
E
D
G B
C B’
dFB
FB =W
Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is tilted through a small angle θ, the CB changes from B to B’, the position of water line changes to ED and two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and FB is the buoyant force. Consider an element of area dA at a distance x from the CG of plan of vessel on either side. Volume of element (dV) = dA.xθ (xθ = height for small angle) x x Buoyant force on the element of left side (dFB) = Weight of element = where is the sp wt of water. Similarly, Buoyant force on the element of right side (dFB) = Buoyant force on the two elements produces a couple.
Moment of couple = ( )= ( ) Total moment due to altered displacement (M) = ∫ ( ) = ∫( ) ) = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia) ∫( (a) Moment due to the movement of CB from B to B’ (M’) = FB BB’ = WBB’ (b) Where V is the volume of water displaced. For equilibrium, M = M’
5. Explain Lagrangian and Eulerian methods of describing fluid flow.
[2]
Solution: Lagrangian method deals with the motion of single fluid particle. This approach is more complex and the equations of motion are difficult to solve. Eulerian method deals with the motion of fluid particles at a point in space. This method is simple and widely used as the resulting equations can be solved easily. 6. A flow is described by the stream function . Locate the point at which the velocity vector √ 0 has a magnitude of 6 unit and makes an angle 145 with the X-axis. Solution: √ √ √ Resultant velocity ( )
√
√
=6
(a) √ √
y = 0.7x (b) Solving a and b x = 1.16, y = 0.81
7. Prove that the Cippoletti weir is a trapezoidal weir having side slopes 14.20 with vertical. Water flows over a rectangular sharp crested weir 2m long (divided into two bays of 1m each by pier). The head over the sill is 0.75m. The weir is fitted at the end of the long rectangular channel 3m wide and flow depth
1.5m. Starting from the first principles, determine the rate of discharge over the weir. Consider the velocity of approach and the effect of end contraction. Coefficient of discharge for the weir is 0.7.
Solution:
θ/2
θ/2
H
L
Discharge through Cippoletti weir is √ For a rectangular weir with two end contractions, the discharge is given by (
)√
√
√
From the above expression, it is clear that discharge of a rectangular weir is reduced by due to end contractions, which is compensated by discharge through triangular notch. Discharge through triangular notch = loss of dischargedue to end contraction √
√
Length (L) = 2m 1 bay = 2 end contractions n = 2x2 = 4 Head over sill (H) = 0.75m C/s of channel (A) = 3mx1m = 3m2 Coefficient of discharge (CD) = 0.7 Discharge over weir (Q) = ? Discharge over a rectangular weir without velocity of approach and with end contraction is (
√ √
(
) )
= 2.28 m3/s
Velocity of approach (Va) = Q/A = 2.28/3 = 0.76 m/s
√
Velocity head (
)
= 0.03m
Discharge by considering velocity of approach √
(
(
√ (
)) [(
)
]
)[(
(
)
]=2.38 m3/s
Next trial Velocity of approach (Va) = Q/A = 2.38/3 = 0.79 m/s Velocity head (
)
= 0.032m
Discharge by considering velocity of approach √
(
(
√ (
)) [(
)
]
)[(
(
)
] = 2.39 m3/s
As the difference in Q is negligible, the iteration is stopped. Q=2.39 m3/s 8. A pipe bend placed in a horizontal plane tapers from 50cm diameter at inlet to 25cm diameter at outlet. An oil of density 850kg/m3 enters the reducing bend horizontal and gets turned by 450 clockwise direction. The discharge is 0.45 m3/s and pressure at inlet of 40 KN/m2 drops to 23 KN/m2 at the outlet due to frictional effect. Find out the magnitude and direction of resultant force on the bend. Solution:
+Y 1 P1
Fx
450
Reactions
Fy 2
P2 Diameter at section 1 (d1) = 0.5m Area at section 1 (A1) =
= 0.19635m2
Diameter at section 2(d2) = 0.25m Area at section 2 (A2) =
= 0.04909m2
Discharge (Q) = 0.45 m3/s Velocity at section 1 (V1) =Q/A1 = 2.3 m/s
+X
Velocity at section 2 (V2) =Q/A2 = 9.16m/s Pressure at section 1 (P1) = 40KN/m2 = 40000N/m2 Pressure at section 2 (P2) =23KN/m2 =23000N/m2 Angle of bend ( )= 450 Resultant force (FR) =? Direction of resultant force =? ∑ ( (
) ) (
( ( (
)
) ) ) )
( = 5458N
(
)
∑ ( (
) )
= -3275N Resultant force (
)
√
=6365N
Resultant force exerted by the water on the bend = 6365N (to the right and upward) Direction of resultant force =
=
(
)
=310
9. What is hydrodynamic boundary layer? Workout the shape factor ( distribution given by
( )
) parameter for the velocity
.
Solution: When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the velocity gradually increases. U Boundary layer stream velocity (U)
Solid body
The term boundary layer is defined as the thin layer of the flow on the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary. In the boundary layer velocity gradient (du/dy) is large and the shear stress exerted by the fluid is given by Newton’s law of viscosity as
. Outside the boundary, velocity is constant and
velocity gradient is zero and hence shear stress is zero.
( ) Displacement thickness ( ) ∫ (
)
|
∫ [
( )
]
|
Momentum thickness ( ) ∫
(
)
∫ [( )
( )
|
|
∫ ( )
[
( )
]
]
10. A 2mm diameter spherical ball of unit weight 117.5 KN/m3 is dropped in a mass of fluid of viscosity 15 poise and sp.gr. 0.95. Find out drag force, pressure drag, skin drag and terminal velocity of ball.
Solution: Diameter of ball (D) = 2mm = 0.002m Sp.wt. of fluid ( ) = 0.95x9810 N/m3 Viscosity of fluid ( ) = 15 poise = 15/10 = 1.5 PaS Sp.wt. of ball ( ) = 117.5 kN/m3 Drag force (FD) = ?
Pressure drag and friction drag = ? Terminal velocity (V) = ? Weight of ball (W) = Drag force (FD) + Buoyant force on ball (FB) FD = W – FB = 0.000453N Pressure drag = Friction drag =
= 0.000151 N = 0.000302N
From Stoke’s law,
V = 0.016 m/s Checking the Reynold’s no. = 0.02 As Re<0.2, above expression for FD is valid.
11. Discuss types of model studies. If gravity, viscosity and surface tension are equally important in a model, show that for dynamic similarity the relationship between viscosity ratio and model scale ratio Lr is given by
, surface tension ratio
.
Solution: There are two types of models. a. Undistorted model: An undistorted model is that model which is geometrically similar to its prototype. Result from such model can be directly applied to the prototype. b. Distorted model: A distorted model is that model in which one or more terms of the model are not identical with their counter-parts in the prototype. A distorted model may have either geometrical distortion or material distortion or distortion of hydraulic quantities or a combination of these. Results obtained from the distorted models cannot be directly applied. (Re) model = (Re) prototype
(a) (Fr) model = (Fr) prototype
√
√
√
√
(b)
(Wb) model = (Wb) prototype √
√
√
√
(c)
Equating b and c, and squaring (d) Substituting Vr from b and
from d into a
√
12. Derive an expression representing Euler’s equation of motion. Solution:
ds
B θ
A
(
)
dz
PdA dW
Consider a streamline AB in which flow is taking place in S direction. Consider a cylindrical element of length ds and area dA. Let P be the pressure on left face and ( ) on the right face. Forces acting: pressure force and gravity force Resultant force in S- direction = m as where as = acceleration in s direction (
)
(
)
For steady flow Dividing by ρ As all are function of S,
Or,
2069 Ashadh (Back exam) 1. Give two examples of non-Newtonian fluid. If the velocity distribution of fluid over a plate is given by where u is velocity in m/s at a distance y from the plate, determine the shear stress at y=0.2m. Take dynamic viscosity as 8.45x10-4 NS/m2. Solution: Examples of non-Newtonian fluid: Blood, slurry Dynamic viscosity ( ) = 8.45x10-4 NS/m2 Shear stress ( ) at y=0.2m = ?
(
)
At y = 0.2m, shear stress is (
) = 0 N/m2
2. Find the pressure difference between the container A and B as shown in the figure.
130mm
A
230mm Liquid of sp.gr = 0.8 0
45
230mm
B
550mm Water Mercury E
Solution: Pressure at E = pressure at F
300mm F
= -90901 N/m2 = -90.9 Kpa 3. When a 22N weight is placed at the end of the uniform floating wooden beam as shown in figure below, the beam tilts at an angle θ with its upper right corner at the surface, as shown. Determine (a) the angle θ and (b) the specific gravity of wood. Assume that the point of application of 22N weight cuts at point B. 2.75m
A 22N 10cmx10cm B
θ D
C
Water
Solution: A W= 22N
h E
B
θ
D C FB Wwood
h = 2.75tanθ Weight of wood (
)
Buoyant force (FB)= Weight of water displaced = ( ) = =
10cmx10cm
Wwood + W = FB (a) Taking moment about B ∑ =0 (b) Solving a and b N/m3 Sp. Gr. Of wood= 6656/9810 = 0.7 4. Find the resultant force on ABCDEF of 10m width and its location due to the liquid as shown in figure. -117.72 Kpa
Liquid of sp. gr. = 1.2
5m 2m
2m A 2m
B
2m C
4m
D 2m E r= 1.6m
F 3m G
Solution: Equivalent head of liquid due to -117.72 Kpa pressure =
= -10m
This negative pressure will reduce the fluid surface by 10m. Make calculation by taking imaginary fluid surface (IFL) as free surface.
-117.72 Kpa
Liquid of sp. gr. = 1.2
5m 2m
2m
10m A 2m B
2m
IFL
C C’
4m
yp D’
D
FV
2m E
FH x r= 1.6m F 3m G
Net Horizontal force on curved surface (FH) = ̅
= 5.47m from C’
̅
Vertical force on D’D ( Vertical force on EF (
̅ = 1.2x9810x(8.2x10)x4.1 = 3957746N = 3957.7KN
= 706320N (down)
) )
(
)
Net vertical force (FV) = 473140-706320 = -233180N = -233.18KN Location of FV Taking moment about C’ x = 2.65m
(
)
= 473140N (up)
Resultant force (FR) = √ ( ) = 3964.6KN Resultant force acts at a distance of 5.47m down IFL and 2.65m left from C’D. 5. Consider the following steady, incompressible two dimensional velocity field: ⃗
(
)
(
) m/s. Generate an analytical expression for the flow
streamline and draw several streamlines in the upper right quadrant from x = 0 to 3 and y = 0 to 3. Determine if there are any stagnation points in this flow field and if so, where are they. Solution: u = 4.35+0.656x, v = -1.22-0.656y Equation of streamline
Integrating ∫
∫ (
)
( ) [ ( (let -K1x0.656 = K2) ( ) ( K2 (Replacing e by K3) [( )( -5.307-2.85y+0.8x+0.43xy=K3
(
) ]
) ) )]
0.8(x+0.54xy-3.6y)=K3+5.307 Replacing (K3+5.307)/8 by c X+0.54xy-3.6y=c. This is the equation of streamline. Plotting of streamlines
C=0 x 0 1 2 3
y 0 0.3 0.8 1.5
C=1 x 1 2 3
y
C=2 x
y
0.0 0.4 1.0
2 3
0.0 0.5
Streamlines At stagnation point u =0 and v =0 4.35+0.656x = 0 x = -6.6 -1.22-0.656y = 0 y = -1.9 Stagnation occurs at (-6.6, -1.9). OR A flow has a potential function function
given by:
(
). Derive the corresponding stream
and show that some of the stream lines are straight lines passing through the origin of
coordinates. Find the inclination of these lines. Solution: (
) (a) (
)
(b)
Integrating eq. (a) w.r.t. x (c) Where c= f(y) Differentiating eq. (c) w.r.t. y (d) Equating eq. (b) and (d)
(
)
Integrating w.r.t. y Substituting the value of c in eq.(c) ( ) For √ For , this equation (in the form of y=ax) shows that some of the stream lines are straight lines passing through the origin of coordinates. Inclination of the straight lines √
6. Describe a cippoletti weir. How does it differ from a rectangular sharp crested weir. A cippoletti weir is a trapezoidal weir having side slopes of 4 vertical to 1 horizontal.
θ/2
θ/2
H
L
Discharge through Cippoletti weir is √ For a rectangular weir with two end contractions, the discharge is given by ( √
)√ √
From the above expression, it is clear that discharge of a rectangular weir is reduced by due to end contractions, which is compensated by discharge through triangular notch. Discharge through triangular notch = loss of dischargedue to end contraction √
√
√
7. Find the velocity V1 of the water flowing in a vertical pipe as shown in figure below. The loss of head is two times the velocity head of small pipe. The pipe diameter at section 1 is 15cm and that at section 2 is 6cm.
1 V1 5m
2 0.5m
Hg
Solution: Applying Bernoulli’s equation between 1 and 2 (Taking datum through 2)
(a) From continuity equation, A1V1= A2V2 (b) Writing manometric equation
(
= dynamic head due to stagnation, total head = static +dynamic) (c)
From a, b and c (
)
(
)
0.02m/s 8. Water flows through a 900 reducer bend. The pressure at the inlet is 206 KN/m2 (gauge) where the cross-sectional area is 0.01m2. At the exit section, the area is 0.0025m2 and the velocity is 15m/s. Assume that the bend does not discharge into the atmosphere but 0.0025m2 are conduit continues. Further assume that the head loss in the reducer bend is
, where u is the average velocity at the inlet
and exit of the reducer. Determine the force exerted by the flow on the reducer bend. Solution: P2
+Y 2
1
+X
P1
Fx Fy
Reactions
Area at section 1 (A1) = 0.01m2 Area at section 2 (A2) = 0.0025m2 Velocity at section 2 (V2) =15 m/s Discharge (Q) =A2V2= 0.0025x15 = 0.0375 m3/s Velocity at section 1 (V1) =Q/A1 = 3.75 m/s Pressure at section 1 (P1) = 206KN/m2 = 206000N/m2 u = (3.75+15)/2 = 9.375m/s Head loss (hL) =
= 4.5m
Applying Bernoulli’s equation between 1 and 2 (Z1 = Z2)
= 56386 N/m2
∑ ( (
) )
= 2200 N ∑ ( (
) )
= 703 N Resultant force (
)
√
=2310N
Resultant force exerted by the water on the bend = 2310N (to the right and downward) Direction of resultant force =
= 17.70
=
9. Air at 200c flows parallel to a smooth thin flat plate at 4.75m/s. The plate is 3.23m long. Determine whether the boundary layer on the plate is most likely laminar, turbulent or somewhere in between (transitional). Compute the boundary layer thickness at the end of the plate for two cases: (a) The boundary layer is laminar everywhere, and (b) The boundary layer is turbulent everywhere. Solution: Velocity of air (V) = 4.75 m/s Length of plate (L) = 3.23m Assume kinematic viscosity of air ( ) = 1.5x10-5 m2/s Reynold no. (Re) =
= 10.2x105
As Re>5x105, the boundary layer is turbulent. a. For laminar boundary layer, the upper value of Reynold no. is Re = 3x105 Thickness of laminar boundary layer ( )
√
= 0.029m = 29mm
√
b. Turbulent boundary layer Re = 10.2x105 Thickness of turbulent boundary layer ( )
(
)
= 0.076m = 76mm
10. Draw a neat sketch of aerofoil and define the terms chord line, camber, angle of attack and aspect ratio associated with aerofoil. A wing of a small aircraft is rectangular in plan having a span of 10m and a chord of 1,2m. In straight and level flight of 240km/hr, the total aerodynamic force acting on the wing is 20KN. If the lift/drag ratio is 10, calculate the coefficient of lift and the total weight the aircraft can carry. Assume air density be 1.2 kg/m3. Solution:
chord line α
L
Chord line: It is the line joining the front and rear edge. Angle of attack: It is the angle between the direction of flowing fluid and chord line. Camber: It is the curvature of an airfoil. Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio. Speed of aircraft (V) = 240km/hr = (240x1000)/3600 = 66.67m/s Plan area (A) = 10x1.2 = 12m2 Total aerodynamic force (FR) = 20KN
( ) FD = 1.99KN FL = 10x1.99 = 19.9KN Weight the aircraft can carry = FL = 19.9KN
= 0.62
11. Using Buckingham’s π theorem, show that the velocity through a circular orifice is given by √
[
] where H = head causing flow, D= diameter of orifice,
coefficient of viscosity, g = acceleration due to gravity. Solution: ( ) Total no. of variables (n) = 6 No. of fundamental dimensions (m) = 3 No. of π terms = 6-3 =3 ( ) (a) Take , g, H as repeating variables. First π term ( ) ( ) ( ) Equating the powers of M, L and T a1 = 0 -3a1+b1+c1+1 = 0 -2b1-1 = 0 b1 = -1/2 -3x0+(-1/2)+c1+1 = 0 c1 = -1/2 Substituting the value of a1, b1 and c1 √
Second π term ( ) ( ) ( ) Equating the powers of M, L and T a2+1 = 0 a2 = -1 -3a2+b2+c2-1 = 0 -2b2-1 = 0 b2 = -1/2 -3x-1+(-1/2)+c2-1 = 0 c2 = 3/2 Substituting the value of a2, b2 and c2
Third π term
= specific mass,
=
( ) ( ) ( ) Equating the powers of M, L and T a3 = 0 -3a3+b3+c3+1 = 0 -2b3 = 0 b3 = 0 -3x-0+0+c3+1 = 0 c3 = -1 Substituting the value of a3, b3 and c3
Substituting the values of π1, π2, π3 in eq. a (
)
√
(
√
)
(
√
)
(
√
(
√
√
)
√
) [
]
12. Write down Euler’s equation, Navier-Stoke’s equation and Bernoulli’s equation with appropriate expression of each component. Solution: Euler’s equation
= density of fluid = change in pressure with space = change in velocity with space = change in elevation with space Navier-Stoke’s equation in 1D
First term= local acceleration, second term = convective acceleration, third term = pressure force per unit mass, fourth term= gravity force per unit mass, fifth term = shear force per unit mass Bernoulli’s equation
= pressure head = velocity head Z = elevation head
2069 Chaitra (regular exam) 1. A solid cone of maximum radius R and vertex angle is to rotate at an angular velocity . An oil of viscosity and thickness t fills the gap between the cone and the housing. Derive an expression for the torque required and the rate of heat dissipation in the bearing.
R r Oil
thickness = t
H
ds
2θ θ
Solution: Maximum radius of cone = R Thickness of oil = t Angular velocity = ( ) = Viscosity of oil = Consider an elementary area dA at radius r of the cone. Torque on the element (dT) = Shear force x r (
)
(
)
(
)
Total torque (T) = ∫ Rate of heat dissipation = power utilized in overcoming resistance = Alternative expression in terms of H (
)
dr
2. The pipe and connection B as shown in figure below are full of oil of specific gravity 0.9 under pressure. If the U-tube contains mercury, find the elevation of point A in meters. 5
3x10 N/m A
2
Oil, SG = 0.9
y B 1m
4.12m
1.1m El. 100m Zero of gauge scale Hg
Solution: sp gr of oil = 0.9 Let y = elevation difference between A and B Starting from right side and ending to A ( ) ( ) y=10.65m Elevation of A = 100+2.1+10.65 = 112.75m Alternative method Starting from right side and finding PB ( ) ( ) = 394087 N/m2 Applying Bernoulli’s equation at section A and B (V = 0: no motion)
= 112.75m
3. The tank on figure has a 4cm diameter plug which will pop out if the hydrostatic force on it reaches 25N. What will be the reading h on the manometer when this happens?
Water 0
H
50
F =25N
̅
Plug
h 2cm X
X
D=4cm
Hg
Solution: Hydrostatic force ( )
̅
̅ ̅ = 2.028m ̅
= 2.043m
Writing Manometric equation at XX ( ) 13.6x9810h=9810(2.043+0.02) h = 0.15m
4. Prove that the rise of water equals the fall in case of forced vortex motion in cylindrical tank at side and center of tank. Solution: R
A
B
hr hf
N
M C
O
D
h P
Q
R = Radius of cylinder MN = Water level at absolute equilibrium (original water level) After rotation, AOB is the profile of the liquid surface. hr = Rise of liquid at end hf = Fall of liquid at the end Volume of liquid before rotation = ( ) Volume of liquid rotation = Volume of cylinder ABQP- Volume of paraboloid AOB (
)
(
)
Volume of liquid before rotation = Volume of liquid after rotation (
)
(
)
(
)
Hence, rise of liquid level at the end = fall of liquid level at the axis of rotation 5. The stream function for the 2-D flow of a liquid is given by . In the range of values of x and y between 0 and 5 plot the stream lines and equipotential lines passing through co-ordinates (1,1), (1,2), (2,2). Also determine the velocity in magnitude and direction at the point (1,2). Solution:
(a) (b) Integrating eq. (a) w.r.t. x (c) Differentiating eq. (c) w.r.t. y (d) Equating eq. (b) and (d) Integrating w.r.t. y Hence Velocity at (1,2) u = -2x = -2x1 =-2 v = 2y =2x2 =4
Velocity = √ Direction
√(
)
= 4.47 unit = -63.40
Plotting of streamlines (constant line) and equipotential line (constant For point (1,1), (1,2), (2,2), is 2, 4, 8 > =2 =4 =8 x y x y x y 1 1.0 1 2.0 1 4.0 2 0.5 2 1.0 2 2.0 3 0.3 3 0.7 3 1.3 4 0.3 4 0.5 4 1.0 5 0.2 5 0.4 5 0.8 For point (1,1), (1,2), (2,2), is 0, -3, 0 ( ) > =0 =-3 x y x y 1 1.0 1 2.0 2 2.0 2 2.6 3 3.0 3 3.5 4 4.0 4 4.4 5 5.0 5 5.3
Plotting of streamlines (si) and equipotential line (phi)
line)
6. Write down and explain the terms of Euler’s equation and Navier-Stoke’s equation of motion of fluid. Solution: Euler’s equation
= density of fluid = change in pressure with space = change in velocity with space = change in elevation with space Navier-Stoke’s equation in 1D First term= local acceleration, second term = convective acceleration, third term = pressure force per unit mass, fourth term= gravity force per unit mass, fifth term = shear force per unit mass
7. Derive an expression for the time required to empty a conical tank without inflow. Solution: R1 x
dh H1
h
H2
H0
O R0
Let A = cross-sectional area of conical tank, a = cross-sectional area of orifice, H1 = Initial level of liquid, H2 = Final level of liquid, R = radius of tank at H1, R0 = radius of tank at the bottom, T = time to fall from H1 to H2. Let at any instant of time, the liquid surface is at height h above the orifice and let dh is the decrease of liquid surface in an interval of time dt.
From similar triangles (
)
Time of emptying tank from H1 to H2 ∫
√
∫
√
(A also varies with h)
∫
√ √
√
[|
(
∫
[
(
)
)
| √
]
∫ (
|
where
)
|
(
[ (
|
| ]
)
)
(
)
(
)]
This is the expression to compute T. H0 is found from similar triangles.
8. The inlet pipe of a pump rises at an angle of 450 to the horizontal and is 10cm in diameter. For design reasons, it is undesirable that the pressure at the inlet to the pump should fall below 25 KN/m2. The inlet pipe is 6m long, the lower end being just below the surface of the water in the sump. Find the maximum discharge that the pump may deliver. 10cm dia 2
6m5 m
Inlet 0
45 sump
Solution:
1
Pump
Taking datum at 1 Z1 = 0m Z2 = 6sin45 = 4.24m V1 =0m/s P1 = Patm = 0 (in terms of gauge) Gauge pressure at the inlet is negative. The given pressure 25Kpa is absolute pressure. Pabs=Patm+Pgauge Pgauge = Pabs-Patm = 25-101.3 = -76.3 Kpa = P2 Applying Bernoulli’s equation at section 1 and 2
V2 = 8.33 m/s Q = A2V2 =
=0.065m3/s = 65 lps
9. When a jet of fluid strikes a moving vertical plate, show that the maximum efficiency of the system is 8/27. Solution:
u
V
Jet of water
Consider a jet moving with velocity V strikes a flat vertical plate, which is moving with velocity u. The velocity with which the jet strikes the plate will be the relative velocity V-u. Force exerted by the jet in X-direction (Fx) = mass of water striking/sec x (V1x – V2x) ] ( )[( ) ( ) Work done per sec by the jet on the plate = ( Efficiency ( ) = For maximum efficiency,
(
) )
(
)
(
(
)
)
(V-u)(v-3u) = 0 V=u (work done =0) V=3u (
)
( (
) )
10. Explain with neat sketches Boundary layer concept along a thin plate. Solution: Turbulent Boundary layer
U Laminar Boundary layer
Laminar zone
Laminar sublayer
Transition Turbulent zone zone
Consider the flow of fluid, having free stream velocity U, over a smooth thin plate which is flat and parallel to the direction of free stream of fluid. At the leading edge of the fluid, the thickness of the boundary layer is zero. On the downstream, for the fluid in contact with the boundary, the velocity of flow is zero and at some distance from the boundary the velocity is u. Hence a velocity gradient is set up which retards the motion of fluid due to the shear resistance. Near the leading edge, the fluid is retarded in thin layer. At subsequent point downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded. Laminar boundary layer As the boundary layer develops, up to some distance from the leading edge, the flow in the boundary layer is laminar irrespective of whether the incoming flow is laminar or turbulent. This is known as laminar boundary layer. When the boundary layer is thin, the velocity gradient is large. As the boundary layer thickens, velocity gradient reduces and shear stress decreases. Eventually it is too small to drag the slow fluid along. Up to this point the flow is laminar. Viscous force is dominant in this case. Transition zone
As the boundary layer grows to a certain thickness, instability occurs, leading to transition from laminar to turbulent boundary layer. This length is known as transition zone. Turbulent boundary layer After the transition zone, the boundary layer is turbulent and continues to grow in thickness. This layer is known as turbulent boundary layer. If the viscous forces were the only action, the fluid would come to a rest. Viscous shear stresses have held the fluid particles in a constant motion within layers. Eventually they become too small to hold the flow in layers and the fluid starts to rotate. The fluid motion rapidly becomes turbulent. Momentum transfer occurs between fast moving main flow and slow moving near wall flow. Thus the fluid by the wall is kept in motion. The net effect is an increase in momentum in the boundary layer.
11. A jet plane having a wing area of 30 m2 and weighing 28 KN flies at 900 km/hr. The engine develops 8000 KW and has a mechanical efficiency of 70%. Determine the lift and drag coefficient for the wind taking density of air as 1.2 kg/m3. Solution: Weight of plane (W) = 28KN Wing area (A) = 30m2 Speed (V) = 900 km/hr =
= 250m/s
Power = 8000KW Efficiency = 70% Power used to overcome drag resistance (P) = 0.7x8000 =5600KW Density of air (ρ) = 1.2 kg/m3 Coefficient of drag (CD) = ? Coefficient of lift (CL) = ? P = FD x V 5600x1000 = FD x 250 FD = 22400 N
CD = 0.019 Lift force (FL) = Weight of plane = 28000N
CL = 0.025
12. Show that the resistance F to the motion of a sphere of diameter D moving with a uniform velocity V through a real fluid of density and viscosity is given by (
)
Solution: ( ) Total number of variables = 5 No. of fundamental dimensions = 3 No. of π terms = 5-3= 2 ( ) (I) Choose V, D and ρ as repeating variables. First π term (II) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c1+1 =0> c1 = -1 a1+b1-3c1+1 = 0 -a1-2 = 0> a1 = -2 -2+b1+3+1=0 b1 = -2 Substituting the values of a1, b1 and c1 in II
Second π term (III) Writing dimensions ( ) ( ) ( ) Equating the powers of M, L and T c2+1 =0> c2 = -1 a2+b2-3c2-1 = 0 -a2-1 = 0> a2 = -1 -1+b2+3-1 =0 b2 =-1 Substituting the values of a2, b2 and c2 in III
Substituting values of π1 andπ2 in I (
) (
) (
)
2070 Ashadh (Back exam) 1. What is continuum concept in Fluid Mechanics? Explain the cavitation phenomena. Solution: In Fluid Mechanics, a fluid is considered as a continuous substance. This concept is called continuum concept. In this concept, molecular structure of the fluid is not considered and the separation between molecules is neglected. The fluid properties such as velocity and pressure are a continuous function of space and time. The fluid properties can be considered to be constant at any point in space, which is average of large number of molecules surrounding that point within a characteristic distance. Using continuum concept, the mathematical equations relating the physical laws can be derived easily as we don’t need to consider the motion of individual molecule. This concept is not valid if the mean free path of molecules is greater than the characteristic dimension of fluid considered for analysis. The ratio of mean free path length to the characteristic length is known as Knudesn number (Kn). The continuum hypothesis is valid for Kn<0.01. When flow of liquid passes through a region having pressure less than vapor pressure, there will be local boiling and a cloud of vapor bubbles will form. This phenomenon is known as cavitation. The bubbles of low pressure zone move towards the high pressure zone and collapse under that pressure. If this occurs in contact with a solid surface, serious damage can result. Cavitation can affect the performance of hydraulic machinery such as propellers, turbines and pumps and the impact of collapsing bubbles can cause local erosion of metal surface.
2. Compute the pressure of water flowing through a pipe shown in the figure. 32Kpa
Air Water
Water
60cm
Pipe
8cm 10cm 8cm θ
Fluid of Sp. gr. = 2.4
Vertical height for fluid pressure in right limb = (10/16)x8 = 5cm Vertical height for water pressure in right limb = (10/16)x8 = 5cm Starting from left
Ppipe = 36218 pa = 36.218 Kpa 3. Gate AB in figure below is a homogeneous mass of 180kg, 1.2m wide into the paper, resting on smooth bottom B. For what depth h will the force at point B be zero? Solution: Patm
D C
Glycerin 3 Sp wt= 12360 N/m 2m
Water h
1m
A 0
60 B
Wt. of gate AB (W) = 180x9.81 = 1765.8N Width of gate = 1.2m
ycp(gly) A Fgly
AC = 1-1sin60 = 1.13m
F
G
AD =h-1sin60 =h-0.87 B
Fwater W
Hydrostatic force
Zero
̅̅̅̅̅ = 12360x(1.2X1)x(1.13+0.87/2) = 23212 N ̅ = 9810x(1.2x1)x(h-0.87+0.87/2) = 11772(h-0.435) N Center of pressure (
)
̅
̅̅̅̅̅ ̅
(
̅̅̅̅̅̅̅
(
)
(
= 1.605m
) ) (
)
(
)
(
)
ycp
AG = (1.605-1.13)/sin60 = 0.55m )
AF = [ycp-(h-0.87)]/sin60 = [( =[
(
)
]
(
(
)
(
)]
)
Taking moment about A (
)(
)
h =2.4m 4. A very tall 10cm diameter vase contains 1178 cm3 of water. When spun steadily to achieve rigid body rotation, a 4cm diameter dry spot appears at the bottom of the vase. What is the rotation rate, in rev/min, for this condition? Solution: r1 =0.05m A
B
h Z1
C
D
Z2
P
r2 = 0.02m
Radius of vase (r1) = 0.05m Radius of dry spot at bottom (r2) = 0.02m Volume of water = 1178 cm3 = 1178x10-6 m3 N=?
h = z1-z2 =
Volume of water left=Volume of cylinder- Volume of ABDC (
) (
)=
Volume of water left=1178x10-6 m3 Equating = 57.2 rad/s = 546rpm 5. The x and y components of fluid velocity in a 2-D flow field are u = x and v = -y respectively. Determine the stream function and plot the streamline for . Solution: u = x, v = -y (a) (b) Integrating a with respect to x (c) c is a constant which is independent of x, but may be a function of y. Differentiating c with respect to y (d) From b and d c=0 Substituting c in eq. c
Si = 1 x 1 2 3 4 5
y -1 -0.5 -0.3 -0.25 -0.2
si =2 x 1 2 3 4 5
si = 3 y -2 -1 -0.67 -0.5 -0.4
x 1 2 3 4 5
y -3 -1.5 -1 -0.75 -0.6
Plot of stream lines 6. From Euler’s equation, derive the Bernoulli’s equation. Solution: Integrating ∫
∫
Dividing by g
or,
∫
7. Derive an expression which calculates time required to completely emptying the hemispherical tank which was full at the beginning. Solution: R
x R dh H1
h H2
Let A = cross-sectional area of hemispherical tank, a = cross-sectional area of orifice, H1 = Initial level of liquid, H2 = Final level of liquid, R = radius of tank at H1, T = time to fall from H1 to H2. Let at any instant of time, the liquid surface is at height h above the orifice and let dh is the decrease of liquid surface in an interval of time dt. ( )
Time of emptying tank from H1 to H2 ∫
√
∫
(A also varies with h)
∫
√
[
[ (
)
∫(
√ √
(
∫
√
√
√
|
) |
|
)
(
| ] )]
For completely emptying the tank, H2 = 0 √
[
]
In this condition H1 =R √
[
]
√
8. A turbine is set 43m below water level of a reservoir and is fed by a 60cm diameter pipe as shown in the figure below. A short pipe of 45cm diameter discharges the water at a rate of 0.9m 3/s from the turbine to the atmosphere. If the total loss of head is 3m of oil of sp. gr. 0.9 and the turbine efficiency is 85%, find the power output of the turbine. [6]
43m
60cm dia.
Turbine
Solution: Diameter of bigger pipe (d1) = 60cm Diameter of smaller pipe (d2) = 45cm = 0.45m Discharge (Q) = 0.9 m3/s Velocity at point 1 (V1) = 0m/s Velocity at point 2 (V2) =Q/A2 = 0.3/(
)= 5.66 m/s
Pressure at 1 (P1) = 0 pa (atm) Pressure at 2 (P2) = 0 Pa Datum head at 1 (Z2) = 43m Datum head at 2 (Z1) = 0m Efficiency ( ) = 0.85 Head loss = 3m of oil of sg 0.9 Head loss in m of water (hL) = 3x0.9 = 2.7m Energy output of machine = ? ht = Head extracted by turbine Applying Bernoulli’s equation at 1 (water surface) and 2 (exit point)
45cm dia.
ht = 38.m Energy output = (
) = 0.85x(9.81x0.9x38.7) = 290KW
9. A flat plate is struck normally by a jet of water 50mm diameter with a velocity of 18 m/s. Calculate (a) the force on the plate when it is stationary (b) the force on the plate when it moves in the same direction as the jet with a velocity of 6 m/s Solution: Diameter of jet (d) = 50mm =0.05m Area of jet (A) =
= 0.001963m2
Velocity of jet (V) = 18 m/s a. Force exerted by the jet on the plate ( ) = 636N b. Velocity of plate (u) = 6m/s Force exerted by the jet on the plate when the plate is moving ( ( ) = 283N
)
(
)
10. When a jet of fluid strikes series of flat vanes, show that the maximum efficiency of the system is 1/2. Solution:
V
V = absolute velocity of jet, u = absolute velocity of plate Force exerted by the jet in X-direction (Fx) = mass of water striking/sec x (V1x – V2x) [( ] ) ( ) (Since entire fluid mass issuing from the jet strikes the plate, mass of water striking/sec = ( ) Work done per sec by the jet on the plate = Fxu =
)
(
Efficiency ( ) =
)
(
)
For maximum efficiency, (
(
)
)
V=2u (
)
( (
) )
11. The laminar boundary layer profile in a case is approximated by a cubic parabola as, ( )
( ) , where u is the velocity at a distance y from the surface and
.
Calculate the displacement thickness and momentum thickness and workout the shear stress at the surface. Solution: ( )
( )
Displacement thickness ( ) = ? Momentum thickness ( ) = ? Shear stress at surface ( ) = ? ∫ ( |
)
∫ (
)
| (
∫
)
∫ (
)(
∫ (
) )
∫ (
)
|
| (
Shear stress =
) (
)
At y = Shear stress =
(
)
12. A kite has an effective area of 0.6 m2 and mass 0.4 kg. It experiences a drag of 15 N in a wind speed of 40 km/hr. Determine (a) the tension in the chord if it makes an angle 45 0 with the horizontal (b) lift coefficient for the kite. Consider the density of air 1.2kg/m3. Solution: Lift Kite 0
45
Drag
String T
W
Solution: Weight of kite (W) = 0.4x9.81=3.924N Area of kite (A) = 0.6 m2 Velocity (V) = 40 km/hr =
=11.11m/s
Drag (FD) = 15N Density of air ( ) = 1.2kg/m3 Tension (T) = ? Coefficient of lift (CL) = ? Forces in X-dir FD = Tcos45 15=Tcos45 T = 21.2N Force in Y-dir FL = Tsin45+W = 21.2sin45+3.924=18.88N
CL = 0.425
13. The force F on a circular cylinder depends on the free stream velocity V, the diameter of the cylinder D, density of the fluid , viscosity of the fluid and time t. By using Buckingham’s π theorem, show that (
). Take , V and D as repeating variables.
Solution: ( ) Total number of variables = 6 No. of fundamental dimensions = 3
No. of π terms = 6-3= 3 ( ) (I) , V and D are repeating variables. First π term (II) ( ) ( ) ( ) Equating the powers of M, L and T a1+1 =0 -3a1+b1+c1+1 = 0 -b1- 2 = 0 a1 = -1, b1 = -2 3-2+c1+1=0 c1 = -2 Substituting the values of a1, b1 and c1 in II
Second π term (III) ( ) ( ) ( ) Equating the powers of M, L and T a2+1 =0 -3a2+b2+c2-1 = 0 -b2- 1 = 0 a2 = -1, b2 = -1 3-1+c2-1=0 c1 = -1 Substituting the values of a2, b2 and c2 in III
Third π term (IV) ( ) ( ) ( ) Equating the powers of M, L and T a3 =0 -3a3+b3+c3 = 0
-b3+ 1 = 0 b3 =1 0+1+c3=0 c3 = -1 Substituting the values of a3, b3 and c3 in IV
(
)
(
) (
) (
)
