Things to grab for this session (in priority order) Pencil Henderson, Perry, and Young text (Principles of Process Engineering) Calculator Eraser Scratch paper Units conversion chart Tables of fluid properties Moody diagram Pump affinity laws

Core Ag Engineering Principles – Session 1Bernoulli’s EquationPump Applications

Core Principles

Conservation of massConservation of energy

Assumption/Conditions Hydrodynamics (the fluid is moving)

Incompressible fluid (liquids and gases at low pressures) Therefore changes in fluid density are not

considered

Conservation of Mass If the rate of flow is constant at any

point and there is no accumulation or depletion of fluid within the system, the principle of conservation of mass (where mass flow rate is in kg/s) requires:

i21 m...mm

For incompressible fluids – density remains constant and the equation becomes:

Q...VAVA 2211

Q is volumetric flow rate in m3/sA is cross-sectional area of pipe (m2) andV is the velocity of the fluid in m/s

Example Water is flowing in a 15 cm ID pipe at a

velocity of 0.3 m/s. The pipe enlarges to an inside diameter of 30 cm. What is the velocity in the larger section, the volumetric flow rate, and the mass flow rate?

ExampleD1 = 0.15 m D2 = 0.3 m

V1 = 0.3 m/s V2 = ?

How do we find V2?

ExampleD1 = 0.15 m D2 = 0.3 m

V1 = 0.3 m/s V2 = ?

How do we find V2?

We know A1V1 = A2V2

Answer

V2 = 0.075 m/s

4π(0.3m)

(0.3m/s)4

π(0.15m)

A

VAV 2

2

2

112

What is the volumetric flow rate?

Volumetric flow rate = Q

s

m0.0053

)s

m(0.3

4

π(0.15m)

VAVAQ

3

2

2211

What is the mass flow rate in the larger section of pipe?

Mass flow rate =

m

s

kg5.3

)m

kg(1000

s

m0.0053

Qρm

3

3

Questions?

Bernoulli’s Theorem (conservation of energy) Since energy is neither created nor

destroyed within the fluid system, the total energy of the fluid at one point in the system must equal the total energy at any other point plus any transfers of energy into or out of the system.

Bernoulli’s Theorem

γ

γ

W = work done to the fluidh = elevation of point 1 (m or ft)P1 = pressure (Pa or psi)

= specific weight of fluidv = velocity of fluidF = friction loss in the system

2g

v

γ

PhFW

2g

v

γ

Ph

222

2

211

1

Bernoulli’s Theorem Special Conditions(situations where we can simplify the equation)

Special Condition 1 When system is open to the

atmosphere, then P=0 if reference pressure is atmospheric (gauge pressure)

Either one P or both P’s can be zero depending on system configuration

Special Condition 2 When one V refers to a storage tank and

the other V refers to a pipe, then V of tank <<<< V pipe and assumed zero

Special Condition 3 If no pump or fan is between the two

points chosen, W=0

Example

Find the total energy (ft) at B; assume flow is frictionless

A

B

C

125’

75’

25’

Preliminary Thinking Why is total energy in units of ft? Is that

a correct measurement of energy?

What are the typical units of energy?

How do we start the problem?

Preliminary Thinking Feet is a measure of pressure; it can be

converted to more traditional pressure units.

Typical units of energy: energy = work = Nm or ft-lb. If we multiply feet or meter by the specific weight of the fluid, we obtain units of pressure.

How do we start the problem?

ExampleTotal EnergyA = Total EnergyB

2g

v

γ

PhFW

2g

v

γ

Ph

2BB

B

2AA

A

Total EnergyB

hA =

ExampleTotal EnergyA = Total EnergyB

2g

v

γ

PhFW

2g

v

γ

Ph

2BB

B

2AA

A

Total EnergyB

hA = 125’ = Total EnergyB

ExampleFind the velocity at point C.

ExampleFind the velocity at point C.

s

ft80.2v

)sft

2(32.2

v25'125'

γ

P

2g

vh125'

2

2

2C

C2C

C

0

Try it yourself:

pump

9’

1’

x’

1’

Water is pumped at the rate of 3 cfs through piping system shown. If the energy of the water leaving the pump is equivalent to the discharge pressure of 150 psig, to what elevation can the tank be raised? Assume the head loss due to friction is 10 feet.

Answer 150 psig *62.4 lb/ft3 * 144 in2 / ft2 = 346

ft

1ft + 346 ft + 0 + 0 – 10 = 11 + x

X = 326’

Bernoulli’s EquationAdding on – how do we calculate F (instead of having it given to us or assuming it is negligible like in the previous problems)

Bernoulli’s Theorem

γ

γ

h = elevation of point 1 (m or ft)P1 = pressure (Pa or psi)

= specific weight of fluidv = velocity of fluidF = friction loss in the system

2g

v

γ

PhFW

2g

v

γ

Ph

222

2

211

1

Determining F for Piping Systems

Step 1Determine

Reynolds numberDynamic

viscosity unitsDiameter of pipeVelocityDensity of fluid

μ

ρVDRe

ExampleMilk at 20.2C is to be lifted 3.6 m

through 10 m of sanitary pipe (2 cm ID pipe) that contains two Type A elbows. Milk in the lower reservoir enters the pipe through a type A entrance at the rate of 0.3 m3/min. Calculate Re.

Step 1: Calculate Re number

μ

ρVDRe

Calculate v = ?

Calculate v2 / 2g, because we’ll need this a lot

12.9m)

sm

2(9.81

)sm

(15.9

2g

v

s

m15.9

4π(0.02m)

60s1min

minm

0.3

A

Qv

2

22

2

3

What is viscosity? What is density?

Viscosity = 2.13 x 10-3 Pa · sρ = 1030 kg/m3

So Re = 154,000

2

23

3

s

kgmN

mNs

102.13

)mkg

)(1030sm

0.02m(15.9Re

Reynolds numbers:< 2130 Laminar> 4000 Turbulent

Affects what?

Reynolds numbers:To calculate the f in Darcy’s

equation for friction loss in pipe; need Re

Laminar: f = 64 / ReTurbulent: Colebrook equation or

Moody diagram

Total F in piping sytem

F = Fpipe + Fexpansion +

Fexpansion+ Ffittings

Darcy’s Formula

2g

v

D

LfF

2

pipe

Where do you use relative roughness?

Relative roughness is a function of the pipe material; for turbulent flow it is a value needed to use the Moody diagram (ε/D) along with the Reynolds number

ExampleFind f if the relative roughness is 0.046 mm, pipe diameter is 5 cm, and the Reynolds number is 17312

Solution ε / D = 0.000046 m / 0.05 m = 0.00092 Re = 1.7 x 104

Re > 4000; turbulent flow – use Moody diagram

Find ε/D , move to left until hit dark black line – slide up line until intersect with Re #

Answer f = 0.0285

Energy Loss due to Fittings and Sudden Contractions

g2

vKF

2

Energy Loss due to Sudden Enlargement

2g

)V(VF

221

ExampleMilk at 20.2C is to be lifted 3.6 m

through 10 m of sanitary pipe (2 cm ID pipe) that contains two Type A elbows. Milk in the lower reservoir enters the pipe through a type A entrance at the rate of 0.3 m3/min. Calculate F.

Step 1:

Step 1

Re = 154,000

2

23

3

s

kgmN

mNs

102.13

)mkg

)(1030sm

0.02m(15.9Re

f = ?

Fpipe =

0.026f

:sMoody'

101.5Re

0.00230.02m

0.000046m

0.02m

0.046mm

D

ε

5

2g

v

D

LfF

2

pipe

167.5m

12.9m0.02m

10m0.026

2g

v

D

LfF

2

pipe

167.5m

12.9m0.02m

10m0.026

2g

v

D

LfF

2

pipe

Ffittings =

Fexpansion =

Fcontraction=

6.45m2g

v0.5F

F

F

2

contr

exp

fittings

6.45m2g

v0.5F

12.9m2g

v

2g

)v(vF

12.9m2g

v0.5)(0.5F

2

contr

21

221

exp

2

fittings

Ftotal = 199.7 m

Try it yourself Find F for milk at 20.2 C flowing at 0.075

m3/min in sanitary tubing with a 4 cm ID through 20 m of pipe, with one type A elbow and one type A entrance. The milk flows from one reservoir into another.

Pump Applications

How is W determined for a pump? Compute all the terms in the Bernoulli

equation except W

Solve for W algebraically

Why is W determined for a pump?

W is used to compute the size of pump needed

PowerThe power output

of a pump is calculated by:

W = work from pump (ft or m)Q = volumetric flow rate (ft3/s or m3/s)ρ = densityg = gravity

Remember:

Po = the power delivered to the fluid (sometimes referred to as hydraulic power)

Pin = Po/pump efficiency

(sometimes referred to as brake horsepower)

To calculate Power(out)

What we know What we need

To calculate Power(out)

What we know W g

What we need Q

Flow rate is variableDepends on “back pressure”

Intersection of system characteristic curve and the pump curve

System Characteristic Curves

A system characteristic curve is calculated by solving Bernoulli’s theorem for many different Q’s and solving for W’s

This curve tells us the power needed to be supplied to move the fluid at that Q through that system

Example system characteristic curve

Pump Performance Curves

Given by the manufacturer – plots total head against Q: volumetric discharge rate

Note: these curves are good for ONLY one speed, and one impeller diameter – to change speeds or diameters we need to use pump laws

Pump Performance Curve

Efficiency

Total head

Power

N = 1760 rpmD = 15 cm

Pump Operating Point Pump operating point is found by the

intersection of pump performance curve and system characteristic curve

What volumetric flow rate will this pump discharge on this system?

Performance of centrifugal pumps while pumping water is used as standard for comparing pumps

To compare pumps at any other speed than that at which tests were conducted or to compare performance curves for geometrically similar pumps (for ex. different impeller diameters or different speeds) – use pump affinity laws (or pump laws)

Pump Affinity Laws (p. 106) Q1/Q2=(N1/N2)(D1/D2)3

W1/W2=(N1/N2)2(D1/D2)2

Po1/Po2=(N1/N2)3(D1/D2)5(ρ1/ρ2)

NOTE: For changing ONLY one property at a time

Size a pump that is geometrically similar to the pump given in the performance curve below, for the same system. Find D and N to achieve Q= 0.005 m3/s against a head of 19.8 m?

0.01 m3/s

Example

N = 1760 rpmD = 17.8 cm

(Watt)

Procedure Step 1: Find the operating point of the

original pump on this system (so you’ll have to plot your system characteristic curve (calculated) onto your pump curve (given) or vice-versa.

N = 1760 rpmD = 17.8 cm

P(W)

0.01 m3/sN = 1760 rpmD = 17.8 cm

P(W)

882.9 W

What is the operating point of first pump?

N1 = 1760

D1 = 17.8 cm

Q1 = 0.01 m3/s Q2 = 0.005 m3/s

W1 = W2 = 19.8 m

How do we convert Po to W?

How do we convert Po to W? W = Po/Qg W=(882.9 Nm/s)/(0.01 m3/s *1000 kg/m3

* 9.81 m/s2) W= 9 m

Find D that gives both new W and new Q (middle of p. 109 – can’t use p. 106 for

two conditions changing) D2=D1(Q2/Q1)1/2(W1/W2)1/4

D2=

D2=

Find D that gives both new W and new Q (middle of p. 109) D2=D1(Q2/Q1)1/2(W1/W2)1/4

D2=0.178m(.005 m3/s/0.01m3/s)1/2(9 m/19.8 m)1/4

D2=0.141 m

Find N that corresponds to new Q and D (p. 109 equ) N2=N1(Q2/Q1)(D1/D2)3

N2=

Find N that corresponds to new Q and D N2=N1(Q2/Q1)(D1/D2)3

N2= 1770 rpm

Try it yourself

If the system used in the previous example was changed by removing a length of pipe and an elbow – what changes would that require you to make?

Would N1 change? D1? Q1? W1? P1? Which direction (greater or smaller)

would “they” move if they change?

Answers Removing pipe + elbow would reduce F

and therefore reduce W (increase Q). System curve would move to the right.

N would likely change. D- no. Q- yes. P – depends on the shape of the power curve but likely it will change.

N would increase (N2=N1(Q2/Q1)(D1/D2)3) Q – increase; P – depends.