Fluid Flow Systems lecture notes

Fluid Flow Systems lecture notes Csaba H˝os September 27, 2010

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Contents 1 Some basic hydraulic principles and hydraulic 1.1 Continuity equation . . . . . . . . . . . . . . . 1.2 Bernoulli’s equation . . . . . . . . . . . . . . . 1.3 Energy equation for compressible flow . . . . . 1.4 Frictional head loss in pipes . . . . . . . . . . . 1.5 Head-discharge curves and operating point . . . 1.6 Parallel and series connection . . . . . . . . . . 2 Steady state hydraulic simulation 2.1 Formulating the equations . . . . 2.1.1 Nodes . . . . . . . . . . . 2.1.2 Branches . . . . . . . . . 2.1.3 Example network . . . . . 2.2 Solving the equations . . . . . . .

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3 Steady flow in open channels 3.1 Wave celerity . . . . . . . . . . . . . . 3.2 Flow in an open channel . . . . . . . . 3.3 Critical depth in a rectangular channel 3.4 Calculating the losses . . . . . . . . . 3.5 The flow in a circular section . . . . .

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machinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 3 3 4 4 5 6

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7 7 7 7 8 9

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11 11 11 12 13 13

4 Homeworks 15 4.1 Homework 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

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1 1.1

Some basic hydraulic principles and hydraulic machinery Continuity equation

In the absence of nuclear reactions, matter can neither be created or destroyed. This is the principle of energy conservation and gives the continuity equation. Its general form is ∂ρ + div(ρv) = 0, ∂t

(1)

where divv = ∇v = ∂Fx /∂x + ∂Fy /∂y + ∂Fz /∂z. If the flow is steady (∂ . . . /∂t = 0) and one-dimensional, we have div(ρv) = 0.

(2)

Moreover, in many engineering applications the density can be considered to be constant, leading to (3) div(v) = 0. The above forms are so-called differential forms of the continuity equation. However, one can derive the so-called integral forms. For example, for the steady-state case, we have Z Z ρv ⊥ dA = 0. (4) ρvdA = A

A

Note that the surface is defined by its normal unit vector dA and one has to compute the scalar product vdA. One can resolve the velocity to a component parallel to and another perpendicular to the surface as v = v ⊥ + v k . Thus vdA = v ⊥ dA. In many engineering applications, there is an inflow A1 and an outflow A2 , between which we have rigid walls, e.g. pumps, compressors, pipes, etc. Let us denote the average perpendicular velocities and the densities at the inlet A1 and outlet A2 by v1 , ρ1 and v2 , ρ2 respectively. Than, we have ρ1 v1 A1 = ρ2 v2 A2

1.2

(5)

Bernoulli’s equation

In the case of steady frictionless flow, the energy of the fluid along a streamline remains constant. Mostly we deal with incompressible fluids, for which the energy content per unit volume is Energy per unit volume =

mgh + 21 mv 2 + pV ρ = p + v 2 + ρgh = constant. V 2 (6) 3

Considering two points of a streamline (the flow is from 1 to 2), we have1 ρ ρ p1 + v12 + ρgh1 = p2 + v22 + ρgh2 . (7) 2 2

1.3

Energy equation for compressible flow

Without derivation, we simply state that the energy equation for frictionless, stationary flow of a compressible ideal gas without heat transfer takes the following form: v2 + cp T = constant, 2

(8)

where cp [J/kgK] is the specific heat capacity taken at constant pressure and T [K] is the absolute (!) temperature.

1.4

Frictional head loss in pipes

In hydraulic machinery, instead of pressure p[P a], usually the term head p is used: H[m] = ρg . In real moving fluids, energy is dissipated due to friction, as well as turbulence. Note that as the hydraulic power is P = ρgHQ, but - because of the continuity equation - the flow rate is constant, the energy loss manifests itself in head (pressure) loss. Head loss is divided into two main categories, ”major losses” associated with energy loss per length of pipe, and ”minor losses” associated with bends, fittings, valves, etc. The most common equation used to calculate major head losses is the DarcyWeisbach equation: h′f = λ

L 8Q2 L v2 =λ , D 2g D D5 π2

(9)

where the friction coefficient λ (sometimes denoted by f ) depends on the Reynolds number (Re = vD/ν, ν[m2 /s] = µ/ρ being the kinematic viscosity of the fluid) and the relative roughness e/D (e[m] being the roughness projections and D the inner diameter of the pipe). Based on Nikuradse’s experiments, we have different regimes based on the Reynolds number. • For laminar flow Re < 2300, we have λ =

64 Re .

• For transitional flow 2300 < Re < 4000, the value of λ is uncertain and falls into the range of 0.03 . . . 0.08 for commercial pipes. √ • For turbulent flow in smooth pipes, we have √1λ = 1.95 lg(Re λ)−0.55. However, this equation need iteration for computing the actual value of λ. Instead, in the range √ of 4000 < Re < 105 , the Blasius’s formula is usually used: λ = 0.316/ 4 Re. 1

Note that the density is not indexed as density is assumed to be constant!

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• For turblent flow in rough pipes, Karman-Prandtl equation may be
e used: √1λ = −2 lg 3.7D .

• The Colebrook-White both the smooth and rough  equation covers  e 1.88 1 regime: √λ = −2 lg Re√λ + 3.7D .

For relatively short pipe systems, with a relatively large number of bends and fittings, minor losses can easily exceed major losses. These losses are usually taken into account with the help of the loss factor ζ in the form of ρ h′ = ζ v 2 . 2

(10)

In design, minor losses (ζ) are usually estimated from tables using coefficients or a simpler and less accurate reduction of minor losses to equivalent length of pipe (giving the length of a straight pipe with the same head loss). Such relationships can be found in [2], [1] or [3].

1.5

Head-discharge curves and operating point

Let us consider a single pipe with several elbows, fittings, etc. that ends up in a reservoir, see Figure 1. The head h = (p1 −p0 )/(ρg) needed to convey Q flow rate consists of two parts: geodetic height difference hb + h − h1 and the losses of the pipe h′f : Hs(ystem) = hb + h − h1 + {z } | static head

X

|

ζ+

X L  v2 λ = A + BQ2 D 2g {z }

(11)

friction head

The typical performance curve of a pump is a falling curve starting off with Hmax. at zero flow rate and arriving at Qmax at zero head. Often it can be approximated with a second-order polynomial as Hp(ump) = aQ2 +bQ+c, with a < 0. Note that the coefficients a, b and c are valid only for a given driving motor revolution number. An important issue is the so-called affinity law giving relationships between performance curves at different revolution numbers:  2 n n H Q , (12) = , and = Q1 n1 H1 n1 where Q1 and H1 represent a point on the performance curve at n1 . Thus if the performance curve at n1 is given by Hp(ump) = aQ2 + bQ + c, the performance curve at an arbitrary n revolution number is n Hp = aQ + b Q + c n1 2

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n n1

2

.

(13)

p0

ζe p1

ζe h

h1 ζe

ζe

hb

ζv

Figure 1: Simple pumping system.

1.6

Parallel and series connection

When two (or more) elements (either pumps or other elements) are arranged in serial, their resulting performance curve is obtained by adding their heads at same flow rate. For example, centrifugal pumps in series are used to overcome larger system head loss than one pump can handle alone. For two identical pumps in series the head will be twice the head of a single pump at the same flow rate. When two or more elements are arranged in parallel their resulting performance curve is obtained by adding their flow rates at the same head. E.g. centrifugal pumps in parallel are used to overcome larger volume flows than one pump can handle alone. For two identical pumps in parallel the flowrate will double compared to a single pump if head is kept constant. H

H

2Hmax Hmax

Hmax

H

H1 + H2 H2

Q Qmax 2Qmax pipe 1 pipe 2

Qmax H

Q pipe 1 pipe 2

H1 Q

Q Q1 Q2 Q1 + Q2

Figure 2: Two identical pumps (above) and (different) pipes (below) connected in series (left column) and parallel (right).

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2

Steady state hydraulic simulation

Let us consider a hydraulic system consisting of pumps, reservoirs, the connecting pipes and other minor elements such as valves, filters, elbows, etc. In this section we develop a mathematical technique to compute the unknown pipe discharges Q and nodal heads H in a hydraulic system of arbitrary size and complexity. We assume that the geometry of the system (i.e. pipe data, elevations, pump performance curves, etc.) are known. Such a system can be modelled as a graph, having n nodes and m edges (pumps, reservoirs, pipes, etc.). The pressures are computed at the nodes, while flow rates are related to edges.

2.1 2.1.1

Formulating the equations Nodes

At the nodes we prescribe continuity as X ρi Qi = ρd Qd ,

(14)

i

where Qi is the signed flow rate in the ith pipe and Qd is the demand flow rate at the node (i.e. consumption). By convention, if the flow is towards the node, Q is positive, if the flow is outwards the node, the flow rate is negative. In many engineering situations, the density change is negligible, in such cases, we have X Qi = Qd . (15) i

2.1.2

Branches

The branches connect the pressure of the starting node (ps ) and the ending node (pe ) with an equation of the form pb − pe = f (Q).

(16)

Pipes For a simple, straight pipe of constant diameter, we have ps − pe = he − hs + λ

L ρ Q|Q|. D 2A2

(17)

Pumps Let us denote the performance curve of the pump by H(Q), and denote the pressure side by p and the suction side by s. Than, we have ! vp2 vs2 (18) − + hp − hs pp − ps = ρgH(Q) − ρg 2 2 7

Reservoirs By definition, the reservoir is connected to a starting node and the ending node is ’virtual’ with p0 pressure. If the bottom level is Hb , the actual water is level is Hw , and we neglect the losses in the connecting pipes, we have ps = p0 + ρg(Hb + Hw − hs ).

(19)

Minor losses (elbows, filters, etc. These elements are simply characterized by a loss coefficient ζ: ps − pe = ζ 2.1.3

ρ Q|Q| . 2 A2ref

(20)

Example network iv d iii

iii

5 d iv

4

6

3

1

i

2

ii

d ii

Figure 3: Example network Let dx and hx denote the consumption and geodetic height of the xth node. The nodal continuity equations are 0 = −Q1 + Q2

dii = Q2 − Q3 − Q4

diii = Q4 − Q5

div = Q5 + Q3 − Q6

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The branch equations are pump reservoir:

pi = p0 + ρg (Hb,1 + Hw,1 − hi )  2  Q2 Q22 pump: pii − pi = ρgH(Q2 ) − ρg − + hii − hi 2A2p 2A2s L3 ρ Q3 |Q3 | pipe 3: pii − piv = hiv − hii + λ D3 2A23 L4 ρ pipe 4: pii − piii = hiii − hii + λ Q4 |Q4 | D4 2A24 L5 ρ Q5 |Q5 | pipe 5: piii − piv = hiv − hiii + λ D5 2A25

upper reservoir:

piv = p0 + ρg (Hb,6 + Hw,6 − hiv )

Thus we have 10 equations for the ten unknowns (pi...iv and Q1...6 ). However, due to the branch equations, this is a system of nonlinear algebraic equations that can be solved numerically.

2.2

Solving the equations

We want to solve a general nonlinear algebraic equation of a single variable x of the form f (x) = 0. After providing a previous guess, say xn , Newton’s technique - see Figure 4 - tries to find a better solution xn+1 along the tangent line: f (xn ) = f ′ (xn ) xn − xn+1

→

xn+1 = xn −

f (xn ) . f ′ (xn+1 )

(21)

f (xn ) f (x)

xn+1

xn

x

Figure 4: Newton’s technique. Example: Solve the equation 0 = x3 − 2 sin x with initial guess x0 = 2

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no. of step 0 1 2 3 4 5

x 2.00000 1.51829 1.29765 1.24019 1.23620 1.23618

f(x) 6.18141 1.50275 0.25924 0.01582 0.00007 0.00000

f’(x) 12.83229 6.81068 4.51214 3.96501 3.92782 3.92765

Systems of equations Let us assume now that we have to solve several (maybe hundreds of) nonlinear equations simultaneously. Let x denote the vector of unknowns and f (x) the equations to be solved: 0 = f (x).

(22)

Without actually proving it, we just show the multidimensional Newton’s technique. If the previous step was xn , the improved solution can be obtained by (23) xn+1 = xn − J −1 f (xn ), where J denotes the Jacobian of system (22): Jij =

∂fi . ∂xj

(24)

Example: Solve the equations f1 = x31 −2 sin x2 = 0 and f2 = x1 x22 −1 =

0. The Jacobian is:

J=

∂f1 ∂x1 ∂f2 ∂x1

∂f1 ∂x2 ∂f2 ∂x2

!

=



10

 3x21 −2 cos(x2 ) x22 2x1 x2 ;

(25)

3 3.1

Steady flow in open channels Wave celerity

Consider the water wave depicted in Fig. 5. We employ the small amplitude theory requiring that both a/L and a/d are small. Using this assumption and solving the equation of motion for small amplitude waves yields the following expression for the wave celerity: s   d gL c= tanh 2π (26) 2π L The shallow water wave requires that d/L < 0.05. In this case, we have s   p d gL csw = lim tanh 2π = gd. L→∞ 2π L (27) On the other hand, if d/L > 0.5 i.e. a deep water wave, we have r gL . (28) cdw = 2π

L a H d

Figure 5: Wave

Todo: extend with the derivations.

3.2

Flow in an open channel

The flow of a fluid in a conduit may either be open channel or pipe flow. Open channel flow has a free surface, whereas pipe flow has none. This free surface is subject to atmospheric pressure, as depicted in Figure 6 Let us consider the energy balance of the fluid: y+

v2 + yb + h′ = const., 2g

(29)

where y(x) [m] is the water height, relative to the bottom yb (x) [m] kpest. v(x) [m/s] denotes the mean velocity of the lfuid, h′ (x) [m] stands for the head loss. The flow rate through the channel is Q = A(x, y(x)) v(x) = konst, with A(x, y(x)) being the wetted area. By differentiating (29) w.r.t. x we obtain   dy d v2 dh′ 0= + = 0, (30) −i+ dx dx 2g dx where the slope of the channel was considered to be constant: i = d yf /dx = (he − hv )/L. The second term is       ∂A ∂A dy 1 Q2 Q2 d d v(x)2 + =− 3 = . (31) dx 2g 2g dx A(x, y(x))2 gA ∂x ∂y dx 11

v12 2g

h+ h1

τf z1

h′

v2 2g

v22 2g

fluid surface g sin θ bottom g θ L

B

h2 h z2

Figure 6: Open-channel flow

In what follows, we consider prismatic channels, i.e. the channel profile does not change in the x direction. The quantity B is defined as B(y) =

∂A , ∂y

(32)

with which (30) turns into ′

i − dh dy dx = Q2 B dx 1 − A3 g

(33)

Let us define the Froude number (similar to the Mach number in gas dynamics) as the ratio of the flow velocity (v) and the wave velocity (c): F r = v/c. Note that for a rectangular channel (A = By), we have ′

′

′

i − dh i − dh i − dh dy dy dx dx dx = = = = Q2 Q2 B dx dx 1 − F r2 1 − A3 g 1 − AAg2

(34)

B

3.3

Critical depth in a rectangular channel

Let us fix the flow rate Q and calculate the energy-like quantity y +v 2 /2g as a function of the water depth h: e=y+

Q2 , 2gB 2 y 2

(35)

which has an minimum value where de Q2 0= =1−2 dy 2gB 2 yc3 12

→

yc =

s 3

Q2 , gB 2

(36)

Type of Conduit Cement, neat surface Cement, mortar Wood, planed, untreated Concrete, unfinished Brick, glazed Cast iron, new

Minimum 0.010 0.011 0.010 0.014 0.011 0.013

Maximum 0.013 0.015 0.014 0.020 0.015 0.017

Table 1: Values of n in Manning’s formula

and the corresponding mean velocity is v r r u u Q3 √ Q √ 3 Qg 3 Qg 3 = = 3 vc y c g → vc = y c g =t vc = = Q2 3 Byc B B B gB 2

(37)

Thus the specific energy is a minimum at the critical depth yc , at which the velocity is the wave celerity. For a fixed discharge, for all energy values, there are two possible depths √ • supercritical flow y < yc , v > gy, F r > 1 √ • subcritical flow y > yy , v < gy, F r < 1.

3.4

Calculating the losses

The friction term in (29) is usually computed with the Chezy formula dh′ v2 = 2 , dx C Rh

(38)

where Rh is the hydraulic radius, the ratio of the wetted area and perimeter (Rh = A/P ). C is the Chezy constant, which can be calculate with the Manning formula: 1

C = Rh6 /n.

(39)

Here n is an empirically derived coefficient, which is dependent on many factors, including surface roughness and sinuosity, see Table 1. Note that there is a natural connection between the friction factor in filled pipes λ and n: s v2 v 2 n2 λ d h′ 1/6 =λ = (40) ⇒ n = Rh 4 dx 2gD 8g R3 h

3.5

The flow in a circular section

For a pipe with radius R, we have θ = arccos (1 − y/R), A = R2 (θ − sin θ cos θ), P = Dθ and B = 2R sin θ. 13

The flow rate in the case of normal flow is p Q = AC iRh , (41)

θ R−y

which, compared to the fully filled section is √ 1/6 Q A(y) Rh n−1 iR p h = 2 Qf ull R π (R/2)1/6 n−1 iR/2 (42)

R y

Figure 7: Circular cross section

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=

(θ − sin θ cos θ)) 3 2

πθ 3

.

(43)

Note that this function is maximum if θ = 2.63905, i.e. y/R = 1.87636 (y/D = 0.938). Moreover, this number is independent of both n and the slope. 1.4 1.2

X: 1.879 Y: 1.076

Q/Q

full

1 0.8 0.6 0.4 0.2 0 0

0.5

1 y/R

1.5

2

Figure 8: The flow in a circular cross section

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4

Homeworks

4.1

Homework 1

We have two pumps, whose performance curves are given by Hp,1 = 50 − 1000Q2

and

Hp,2 = 70 − 2000Q2 ,

(44)

and two pipes: L1 = 100m, D1 = 0.3m, λ1 = 0.02 and L2 = 200m, D2 = 0.25m, λ2 = 0.015 The pump and pipe connections are listed below. E = pump or pipe. The homework code is two letters, e.g. DB, where the first letter stands for the pipe, the second for the pipes. For example, code DB means 2× pump 1 in parallel (the first letter is D), with pipe 1 and pipe 2 connected in series (the second letter is B). Group A B C D E F

Element 1 E1 E1 E2 E1 E1 E2

Element 2 E1 E2 E2 E1 E2 E2

Connection Series Series Series Parallel Parallel Parallel

1. Using graphical technique, compute the operating point of the system, giving the flow rate and head of all four elements. 2. Re-compute the results with hydraulic system simulation.

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References [1] P.R. Bhave and R. Gupta. Analysis of Water Distribution Systems. Alpha Science, 2006. [2] I.E. Idelchik. Handbook of Hydraulic Resistance. Jaico Publishing House, 2008. [3] D.S. Miller. Internal Flow Systems. BHRA Fluid Engineering, 1978.

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