Basic equations of fluid flow

The principles of physics most useful in the applications of the fluid mechanics are mass- balance, or continuity ; the linear- and angular- momentum-balance equation, and the mechanical-energy balance. Mass balance Momentum balance Mechanical energy balance

One-dimensional flow In discussing fluid flow it is helpful to visualize, in the fluid stream, fluid paths called streamlines. Flow along a streamline is therefore one- dimensional, and a single term for velocity is all that is needed. A stream tube is a tube of large or small cross section and of any convenient cross-section shape that is entirely bounded by streamline.

Measures of Flow If the tube has a differential cross-sectional area dS, the velocity through the tube can also be denoted by the single term u. The mass flow through the differential area is a b

If the fluid is being heated or cooled, the fluid density also varies, but usually the variation is small and can be neglected. The mass flow rate through the entire cross section is ρ is constant across the cross section

In general, the local velocity u varies across the cross section. The average velocity V of the entire stream flowing through cross-sectional area S is defined by V also can be considered to be the flux of volume, m 3 /m 2 ·s q is the volumetric flow rate

q is the volumetric flow rate, m 3 /s G the mass velocity, kg/m 2 · s mass current density or mass flux

Mass balance in a flowing fluid Consider the flow through a conduit of cross-sectional area S a at the entrance and area S b at the exit. a b rate of mass flow in – rate of mass flow out= rate of mass accumulation

Continuity equation at steady state the mass flow in =the mass flow out then continuity equation becomes  Where is mass flow rate, kg/s  q is volumetric flow rate, m 3 /s  V is average velocity, m/s  S is cross-section area, m 2   is density of fluid, kg/m 3

The equation is also called the equation of continuity.  For an incompressible fluids  So the volumetric flow rate is

Flow is through channels of circular cross section a and b  from which

Crude oil, density 887 kg/m 3, flows through the piping shown in Fig Pipe A is 50-mm inside diameter, pipe B is 75-mm inside diameter, and each of pipes C is 38-mm inside diameter. An equal quantity of liquid flows through each of the pipes C. The flow through pipe A is 6.65 m 3 /h. Calculate (a) the mass flow rate in each pipe, (b) the average linear velocity in each pipe, and (c) the mass velocity in each pipe. Example 1.2

Solution: q a =6.65m 3 /h= m 3 /s, ρ=887 kg/m 3, D a =50mm=0.05 m, D b =75mm=0.075 m, D c =38mm=0.038m (a) the mass flow rate in each pipe the mass flow in =the mass flow out =0.5× ×887=0.8192kg/s = ×887 = kg/s

(b) the average linear velocity in each pipe For incompressible fluid

(c) the mass velocity in each pipe

Overall Energy Balance for Steady-state Flow System Consider a volume element of a stream tube in steady potential flow, as shown bellow s S+ΔS Δ L p p+Δp p´ (Rate of energy input)-(Rate of energy output)= (Rate of energy accumulation) (1.3-8)

The Energy E present within a system can be classified in three ways. 1) Potential energy zg of a unit mass of a fluid is the energy present because of the position of the mass in a gravitational field, where z is the relative height from a reference plane. The unit for zg are J/kg or m 2 /s 2 2) Kinetic energy u 2 /2 of a unit mass of a fluid is the energy present because of the translational or rotational motion of the mass, where u is the velocity in m/s relative to the boundary of the system at a given point. The units are J/kg or m 2 /s 2

3) Internal energy U of a unit mass of a fluid is all of the other energy present, such as rotational and vibrational energy in chemical bonds. The units are J/kg or m 2 /s 2 The total energy of the fluid per unit mass is The net energy transported into the system by the flowing stream is

In addition, 4) Heat transfer rate heat absorbed from surroundings 5) Work rate the work of all kinds obtained from surroundings Eq(1.3-8) be written: (1.3-10)

includes the pressure-volume work per unit mass fluid (pν) and other form of work So (1.3-11)

In most cases (except for precise work) the u is replaced by average velocity V At steady state, the accumulation term of Eq. (1.3-11) is zero. And the control volume has but one entrance and one exit. Thus (1.3-18)

Divided by gives: Eq.(1.3-19)is the Energy balance for a steady-state, steady-flow process between on entrance and one exit. Where Q=the heat absorbed per unit mass of fluid from surroundings W e =the obtained from the surroundings per unit mass of fluid (1.3-19)

Overall Mechanical Energy Balance for Steady-state Flow System (1.3-20) According to the first law of thermodynamics For the case of steady state flow (1.3-21) is the sum of all friction loss per unit mass So (1.3-22)

(1.3-23) Using the above Eqs, We obtain As, the overall mechanical-energy balance equation is obtained: (1.3-24)

Using the above equation, the Eq. (1.3-24) becomes (1.3-25) For incompressive fluid ， ρ is a constant

Bernoulli Equation Deduction of Mechanical Energy Balance Eq. In the special case where no mechanical energy is added W e =0 and for no friction loss h f =0, the Eq.(1.3-25) becomes (1.3-26)

Example 1.4 Brine, specific gravity =1.15, is draining from the bottom of a large open tank through a 50-mm pipe. The drainpipe ends at a point 5 m below the surface of the brine in the tank. Considering a streamline starting at the surface of the brine in the tank and passing through the center of the drain line to the point of discharge, and assuming that friction along the streamline is negligible, calculate the velocity of flow along the streamline at the point of discharge from the pipe. Solution: ρ=1.15 ×10 3 kg/m 3, D b =50mm=0.05 m take plane b as reference plane, so a=5m. P a =P b =0 (gauge), D a >>D b, so V a ≈0. W e =0, h f =0 V b =9.90 m/s

Pump work in Bernoulli Eq. W e a pump is used in a flow system to increase the mechanical energy of the flowing fluid. In an actual pump not only are all the sources of fluid friction active, but mechanical friction occurs as well. (1.3-27) h fp : the total friction in the pump per unit mass of fluid. η: a pump efficiency Where η<1. Eq.(1.3-25) corrected for pump work is (1.3-28)

Example 1.5 In the equipment shown in Fig.1.14, a pump draws a solution of specific gravity 1.84 from a storage tank through a 75-mm inside diameter steel pipe. The efficiency of the pump is 60 percent. The velocity in the suction line is m/s. The pump discharges through a 50-mm inside diameter pipe to an overhand tank. The end of the discharge pipe is 15.2 m above the level of the solution in the feed tank. Friction losses in the entire piping system are 29.9 J/kg. What pressure must the pump develop? What is the power delivered to the fluid by the pump? Solution: ρ=1.84 ×10 3 kg/m 3, D 1 =75mm=0.075m, D b =50mm=0.05m, η=60%, V 1 =0.914 m/s, h f =29.9J/kg, take a as reference plane, so Z a =0, Z b =15.2m, P a =P b =0 (gage), V a =0

Solution: ρ=1.84 ×10 3 kg/m 3, D 1 =75mm=0.075m, D b =50mm=0.05m, η=60%, V 1 =0.914 m/s, h f =29.9J/kg, take a as reference plane, so Z a =0, Z b =15.2m, V a =0, P a =P b =0 (gage), What pressure must the pump develop? Write the bernoulli equation between the suction 1 and discharge 2 pipe of the pump Z 1 ≈ Z 2, V 2 =V b =2.06m/s, h f ’=0 2. What is the power delivered to the fluid by the pump?

Fluid Friction h f represents all the friction generated per unit mass of fluid 1)h f represents the loss of mechanical energy at all points between stations a and b 2)The sign of h f is always positive. It is zero in potential flow. 3)H f is included by the skin friction h fs and the form friction h ff.

Macroscopic momentum balance Deduction of Bernoulli Equation An overall momentum balance can be written for control volume shown in figure. Assuming that the flow is steady and flows in the x direction. ab MaMa MbMb

The sum of forces acting in the x direction equals the difference between the momentum leaving with the fluid per unit time and that brought in per unit time by the fluid or

The momentum flow rate M of a fluid stream having a mass flow rate m and all moving at a velocity V equals mV

It is true if the velocity V is an average velocity at the cross section. If V varies from point to point in the cross section of stream, however, the total momentum flow does not equal the product of mass flow rate and average velocity In general it is somewhat greater than this.

Several force components acting on the fluid in the direction of the velocity component may appear: 1. Pressure change in the direction of flow; 2. Shear stress at the boundary between the fluid stream and the conduit, or external forces acting on the solid wall; 3. The appropriate component of the force of gravity.

For one-dimensional flow, a typical situation is represented by the equation (1.3-36)  Where p a, p b =inlet and outlet pressure, respectively F w =net force of wall of channel on fluid F g =component of force of gravity

Layer flow with free surface In one form of layer flow the liquid layer has a free surface and flows under the force of gravity over an inclined or vertical surface. If such flow is in steady state, with fully developed velocity gradients, the thickness of the layer is constant.

 Consider a layer of a Newtonian liquid flowing in steady flow at constant rate and thickness over a flat plate, as shown in Fig.1.16 l r g τ Φ

The shear stress can be ignored at the free liquid surface. If the flow is laminar and the liquid surface is flat and free from ripples. The fluid motion can be analyzed mathematically

The possible forces acting on the control volume in a direction parallel to the flow are 1)the pressure forces at the ends 2)the shear forces on the upper and lower faces 3)the component of the force of gravity in the direction of flow.

Since the pressure on the outer surface is atmospheric, the pressures on the control volume at the ends of the volume are equal and oppositely directed. Also, by assumption, the shear on the upper surface of the element is neglected.

The two forces remaining are:  the shear force on the lower surface of the control volume  the component of the gravity in the direction of flow

From this equation, noting that A=bL and F g =ρrLbg

So  Since the flow is laminar,

Rearranging and integrating between limits gives  And Where δ is the total thickness of the liquid layer

 The pressure forces normal to the cross section of the tube at the inlet and outlet of the tube  Inlet : p a S a =pS (4-18)  Outlet: p b S b =(p+Δp)(S+ ΔS)

The external forces acting on the surface of element in the direction of flow

 The component of gravity acting along the axis (4-20) Substitution from Eqs(4-18) to (2-20) in Eq(4-16) gives (4-21)  Dividing Eq(4-21) by ρSΔL (4-22)

 Now,fine the limits of all terms in Eq(4- 22) as ΔL → 0.  Then and the ratios of increments all become the corresponding differential coefficient, so that, in the limit, (4-23)

Then mass flow rate is  Substituting in Eq(4-23) gives  And (4-24) Equation(4-24) is the point form of the Bernoulli Equation without friction.

For static fluid, the term d(u 2 /2) is zero, Eq(4-24) becomes identical with Eq(2-3) for a stationary fluid.

Bernoulli Equation deduced from energy balance  Assumption:  1 incompressible flow  2 ideal fluid  3 mechanical energy such as kinetic,  Potential and pressure energy  Kinetic energy 1/2 mu 2  Potential energy mgZ  Pressure energy PV

 For section a-a and b-b, energy balance  Energy in = energy out  1/2mu a 2 +mgZ a +P a V a = 1/2mu b 2 +mgZ b +P b V b  Or 1/2u a 2 +gZ a +P a V a /m=1/2u b 2 +gZ b +P b V b /m  since =m/V  the above equation turns out to be  1/2u a 2 +gZ a +P a /=1/2u b 2 +gZ b +P b /

 Between two definite point in the tube, say station a and b, Eq(4-25) can be integrated, since ρ is constant, the  Bernoulli Equation is given as follows (4-26)

unit  Each term in the equation is a scalar and the dimensions of energy per unit mass The unit for Eq(4-26) can be derived (4-27)  To represent the mechanical work done by forces ， external to stream ， on the fluid in pushing it into the tube or the work recovered from the fluid leaving the tube 。

 The term gZ and u 2 /2 are potential and kinetic energy ， respectively.  For section 1-1 and 2-2, Equation(4-26) is  divided by g, gives (4-28)

The unit for each term is J/(kg.m/s 2 )=m In the same way the equation (4-26) is multiplied by ρg (4-29)

Discussion of Bernoulli equation  The principle of conservation of energy permits the extension of the equation to potential flow taking place in curved stream tube of variable cross section. The equation can be modified for use in boundary layer flow.

It is essential to choose upstream and downstream station. Station a and b are chosen on the basis of convenience and are usually taken at locations where the most information about pressure, velocity, and heights is available.

Bernoulli equation: correction for effects of solid boundaries To extend the Bernoulli equation to cover these practical situations, two modifications are needed: 1) correction of the kinetic-energy term 2) correction of the equation for the existence of fluid friction

Kinetic-energy correction factor kinetic-energy can be calculated from average velocity by using αu 2 /2. laminar flow: α=2.0 highly turbulent flow: α is about 1.05 It is usual to take α to 1 in the calculation

Correction of Bernoulli equation for fluid friction Fluid friction can be defined as any conversion of mechanical- energy into heat in a flowing stream.  For incompressible fluid, the Bernoulli equation is corrected for friction by adding a term to the right-hand side of Eq.(4-26) (4-30)

Pump work in Bernoulli equation The work done per unit mass of liquid passing through an pump. Pump flash

 Equation (4-30) corrected for pump work is

Power gained by a fluid through the pump  P e =WSu Where P e and W are the power and work gained by the fluid through the pump u and  are velocity and density of fluid. S is cross-section area. P e unit is J/s, watt.

Example  As shown in Fig, water (1000kg/m 3 ) in the pipe is ( )  A static; B flowing upward; C flowing downward; D uncertain  If the diameters of sections 1-1 and 2-2 are 0.05 and 0.1 m, respectively, and the velocity u 2 in section 2-2 is 1m/s, what is the frictional loss h f between two sections?

The only way to learn and use the Bernoulli Equation is work and practice the problem. Before we start, there are some steps to follow.  1 Draw the picture and determine the system you choose.  2 Select the suitable sections which should be vertical to flow direction and the flow between two sections must be continuous.

 3 The reference (datum) plane section is Parallel to the ground. Z is the distance between reference Plane and section center.  4 Units of each term should be consistent, for instance, in SI.  In addition, the pressures of two Sections should be absolute or gauge pressure or vacuum.

 As shown in Fig., water flows from the bottom to the top through a contracted pipe with d 1 ＝ 300mm, d 2 ＝ 150mm. Pressures measured at large and small tubes are 0.2MPa and 0.16MPa （ gauge ）,respectively. Vertical distance between pressure measurement points is 1500mm. If the frictional loss can be neglected, what is flow rate of water, in m 3 /h ？ (water density 1000 kg/m 3 )

 Solution: For sections (pressure measurement points) 1 — 1 and 2 － 2, the Bernoulli equation is  Take section 1-1 as reference plane, we have z 1 =0, z 2 =1.5m  From continuity equation:

 So  the velocity in section 2-2 is  u 2 =7.3ms -1  flow rate of water V, in m 3 /h as follows