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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
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?
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
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
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 = 125’ = Total EnergyB
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.
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
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.
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
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
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.
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.
How is W determined for a pump? Compute all the terms in the Bernoulli
equation except W
Solve for W algebraically
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)
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
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 Operating Point Pump operating point is found by the
intersection of pump performance curve and system characteristic curve
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
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.
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
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
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?
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