1. Fluid kinematics Chapter 3
Description and visualization of fluid’s motion (velocity, acceleration).

2. Fluid Flow
Up till now, we have pretty much focused on fluids at rest. Now let's look at fluids in motion
It is important that you understand that an IDEAL FLUID:
Is non viscous (meaning there is NO internal friction)
Is incompressible (meaning its Density is constant)
Its motion is steady and NON – TURBULENT

3. Fluid Flow
The ideal fluid model simplifies fluid-flow analysis

4. Fluid Flow Laminar & Turbulent Flows
The criterion which determines whether flow is laminar or turbulent is the quantity (ρvd/μ), known as the Reynolds number (Re).
below a critical value of Re = 2000, flow will normally be laminar (viscous), otherwise it is turbulent.

5. Principles of Fluid Flow 1/2
The continuity equation results from conservation of mass:
mass in = mass out
ρQ (in) = ρQ (out)
Continuity equation:
A1V1 = A2V2
Area  speed in region 1 = area  speed in region 2
For liquid, same density

9. Example
The speed of blood in the aorta is 50 cm/s and this vessel has a radius of 1.0 cm. If the capillaries have a total cross sectional area of 3000 cm2, what is the speed of the blood in them?
0.052 cm/s

10. Examples Example. 1 Discharge in a 25cm pipe is 0.03 m3/s. What is the
average velocity?
Example. 2
A pipe whose diameter is
8 cm transports air with
a temp. of 20oC and
pressure of 200 kPa
abs. At 20 m/s, what is
the mass flow rate?
PV=nRT , R=8.31 J/mol.K, MW=29

11. Example

13. Example

14. a) b)

15. c)

16. EXAMPLE : Water Flow through a Garden Hose Nozzle
A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit (Fig. 5–12). If it takes 50 s to fill the bucket with water, determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.

17. (b) The cross-sectional area of the nozzle exit is
The volume flow rate through the hose and the nozzle is constant. Then the average velocity of water at the nozzle exit becomes

18. Example
Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit opening is 8 mm2. Determine the mass flow rate of the steam and the exit velocity.
Assumptions:
1 )steady flow,
2)ΔKE=ΔPE=0,
3)Saturation conditions

19. Solution
a) the mass flow rate of the steam and the exit velocity

20. The Energy Balance (First Law of Thermodynamics)

21. Conservation of energy - 1st law of thermodynamics
“when heat or work are transferred to a control mass this will result in change of energy stored in it”
Q heat added to the system
W work done by the system

22. The Energy Balance - Reminder: The general balance equation is:
Rate of Rate of Rate of Rate of Rate of
Creation – Destruction + Flow in – Flow out = Accumulation
In nature energy can be neither created nor destroyed. With respect to a control volume (CV):
Rate of energy flow into CV
Rate of energy efflux from CV
Rate of accumulation of energy into CV - =
(1)
Units [Energy/time]

23. The Energy Balance Rate of Energy Flow into CV:
Rate of Energy Flow out of CV:
Rate of Energy Accumulation:
e [=J/kg] is the specific energy: e (internal) + e (kinetic)+ e (potential=gravity)
e = u + V2/ 2+ g z

24. The Energy Balance Substituting in equation (1) :
Q and W are positive when transferred from surroundings to system.
MODERN Convention
Note: 2nd edition of textbook uses old convention: Q is positive when transferred from surroundings to system. W is positive when transferred from system to surroundings

25. The Energy Balance The work term includes: Shaft work, Wshaft
The energy balance reduces to:
(2)
The work term includes:
Shaft work, Wshaft
Work due to pressure (injection work)

26. pressure

27. The Energy Balance The net work is:
Where specific volume u = 1/  [m3/kg]
Jule (j)= N.m
From (2):
(3)
This is the Energy equation

28. The Energy Balance We may use the enthalpy as: h = u + P/ρ
For multiple inlets – outlets the energy equation becomes:
(4)
We may use the enthalpy as: h = u + P/ρ

29. Simplification of Energy Equation
For single inlet (1) – outlet (2) and steady-state conditions
(ie no accumulation of energy and mass) ,
The energy equation becomes:
(5)

30. The Bernoulli Equation
• It is an approximate relation between pressure, velocity and elevation
• It is valid in regions of steady, incompressible flow where net frictional forces are negligible
• Viscous effects are negligible compared to inertial, gravitational and pressure effects.
• Applicable to inviscid regions of flow (flow regions outside of boundary layers)
• Steady flow (no change with time at a specified location)

31. BERNOULLI'S PRINCIPLE
An increase in the speed of fluid flow results in a decrease in the pressure. (In an ideal fluid.)

33. p large
p large
p small
v small
v large
v small

35. for any point along a flow tube or streamline
Bernoulli’s Equation
for any point along a flow tube or streamline
p + ½  v2 +  g y = constant
Dimensions
p [Pa] = [N.m-2] = [N.m.m-3] = [J.m-3]
½  v2 [kg.m-3.m2.s-2] = [kg.m-1.s-2] = [N.m.m-3] = [J.m-3]
g h [kg.m-3 m.s-2. m] = [kg.m.s-2.m.m-3] = [N.m.m-3] = [J.m-3]
Each term has the dimensions of energy / volume or energy density.
½  v KE of bulk motion of fluid
g h GPE for location of fluid
p pressure energy density arising from internal forces within
moving fluid (similar to energy stored in a spring)

40. Ideal fluid
Real fluid

41. Modifications of Bernoulli Equation
In practice, the total energy of a streamline does not remain constant. Energy is ‘lost’ through friction, and external energy may be either :
added by means of a pump or
extracted by a turbine.
Consider a streamline between two points 1 and 2. If the energy head lost through friction is denoted by Hf and the external energy head added (say by a pump) is or extracted (by a turbine) HE, then Bernoulli's equation may be rewritten as :
± HE = H2 + Hf (3.11) or
HE = energy head added/loss due to external source such as pump/turbines
This equation is really a restatement of the First Law of Thermodynamics for an incompressible fluid.
(3.12)

42. Applications of the Bernoulli Equation
EXAMPLE: Spraying Water into the Air
Water is flowing from a garden hose. A child places his thumb to cover most of the hose outlet, causing a thin jet of high-speed water to emerge. The pressure in the hose just upstream of his thumb is 400 kPa. If the hose is held upward, what is the maximum height that the jet could achieve?

43. Applications of the Bernoulli Equation
Solution:
Water from a hose attached to the water main is sprayed into the air.
The maximum height the water jet can rise is to be determined.
Assumptions:
1) The flow exiting into the air is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable).
2 ) The surface tension effects are negligible.
3) The friction between the water and air is negligible.
4) The irreversibilities that occur at the outlet of the hose due to abrupt contraction are not taken into account.
Properties :We take the density of water to be 1000 kg/m3.

44. Applications of the Bernoulli Equation
Analysis :
This problem involves the conversion of flow, kinetic, and potential energies to each other without involving any pumps, turbines, and wasteful components with large frictional losses, and thus it is suitable for the use of the Bernoulli equation.
The water height will be maximum under the stated assumptions.
The velocity inside the hose is relatively low (V12 << Vj2 , and thus V1 =0 compared to Vj) and we take the elevation just below the hose outlet as the reference level (z1 = 0).
At the top of the water trajectory V2 = 0, and atmospheric pressure pertains.
Then the Bernoulli equation along a streamline from 1 to 2 simplifies to

45. Example
Water circulates throughout the house in a hot-water heating system. If the water is pumped at a speed of 0.50 m/s through a 4.0 cm diameter pipe in the basement under a pressure of 3.0 atm, what will be the flow speed and pressure in a 2.6 cm-diameter pipe on the second floor 5.0 m above?
1 atm = 1x105 Pa
1.183 m/s
2.5x105 Pa(N/m2) or 2.5 atm

46. Applications of the Bernoulli Equation

47. Applications of the Bernoulli Equation

48. Applications of the Bernoulli Equation