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Pipeline with pump

CE 382

Dr. Huidae Cho

Department of Civil and Environmental Engineering...New Mexico State University

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1 Learning objectives

Apply the extended Bernoulli equation including pump head
Distinguish pump head from pressure rise
Derive the system head curve
Interpret pump curves physically
Determine operating point analytically and graphically
Compute pump power and efficiency
Assess physical realism of solutions

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2 Energy equation with a pump

Between sections 1 and 2:

\[ \frac{p_1}{\gamma} + z_1 + \frac{V_1^2}{2g} - h_L + h_p = \frac{p_2}{\gamma} + z_2 + \frac{V_2^2}{2g} \]

where

$h_L$ = total head loss (major + minor)
$h_p$ = pump head (energy added per unit weight)

Key points:

Pump head is energy per unit weight
Units are meters (or feet)
Pump head is not pressure
Pump head is not power
A pump increases total mechanical energy of the fluid

Physical interpretation:

If no pump exists, total head decreases in the direction of flow because of losses. A pump reverses this trend by adding energy.

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3 Static head vs. dynamic head

Static head:

\[ H_{static} = \Delta z \]

Dynamic head (frictional losses):

\[ H_{dynamic} = h_L \]

Total required head:

\[ H_s = H_{static} + H_{dynamic} \]

Important:

Static head does not depend on discharge. Dynamic head increases with discharge.

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4 Head loss model

\[ h_L = h_f + \sum h_m \]

Major loss:

\[ h_f = f \frac{L}{D} \frac{V^2}{2g} \]

Minor loss:

\[ h_m = K \frac{V^2}{2g} \]

Since

\[ V = \frac{4Q}{\pi D^2} \]

Substitute into head loss expression:

\[ h_L \propto Q^2 \]

Thus system curve becomes:

\[ H_s = \Delta z + b Q^2 \]

where

\[ b = \left( f \frac{L}{D} + \sum K \right) \frac{8}{g \pi^2 D^4} \]

Key insight:

System curve is quadratic in discharge.

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5 Pump characteristic curve

Real centrifugal pumps behave approximately as:

\[ H_p = H_0 - a Q^2 \]

where

$H_0$ = shutoff head ($Q = 0$)
Head decreases as discharge increases

Physical explanation:

As discharge increases, internal losses inside the pump increase, reducing available head.

At $Q = 0$:

Maximum head Zero hydraulic power

At maximum discharge:

Head approaches zero

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6 Operating point

Operating point occurs when:

\[ H_p = H_s \] \[ H_0 - a Q^2 = \Delta z + b Q^2 \]

Solve for Q:

\[ Q = \sqrt{\frac{H_0 - \Delta z}{a + b}} \]

Condition for real solution:

\[ H_0 > \Delta z \]

If not:

No operating point exists. Pump cannot overcome elevation rise.

Graphical interpretation:

Intersection of pump curve and system curve.

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7 Effect of system changes

If pipe roughness increases:

b increases
System curve becomes steeper
Operating discharge decreases

If valve is partially closed:

Additional K added
b increases
Operating point shifts left

If elevation increases:

System curve shifts upward
Discharge decreases

If a larger pump is installed:

Pump curve shifts upward
Discharge increases

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8 Pump power

Hydraulic power delivered to fluid:

\[ P_{hyd} = \gamma Q H_p \]

Input power:

\[ P_{in} = \frac{\gamma Q H_p}{\eta} \]

Important:

At Q = 0:

\[ P_{hyd} = 0 \]

Even though head is maximum.

Power depends on both head and discharge.

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9 Efficiency considerations

Pump efficiency varies with discharge.

There is typically a best efficiency point (BEP).

Operating too far from BEP:

Reduces efficiency
Increases vibration
Shortens pump life

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10 Cavitation introduction

Net positive suction head measures how much pressure head is available at the pump inlet above vapor pressure.

Net positive suction head available:

\[ \text{NPSH}_a = \frac{p_{abs}}{\gamma} + z - \frac{p_v}{\gamma} - h_{loss} \]

\[ \text{NPSH}_a < \text{NPSH}_r \]

where $\text{NPSH}_r$ is net positive suction head required, cavitation occurs.

$\text{NPSH}_r$ is the minimum pressure safety margin the pump needs to avoid cavitation.

Physical meaning:

Local pressure drops below vapor pressure.

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11 Flow regime check

\[ Re = \frac{V D}{\nu} \]

Interpretation:

Laminar if Re < 2000
Transitional if 2000 < Re < 4000
Turbulent if Re > 4000

Municipal pipelines are typically turbulent.

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12 Conceptual questions

Is pump head the same as pressure increase? Explain.
Why is pump head expressed in meters instead of Pascals?
Can a pump produce head at zero discharge?
Why does pump head decrease with increasing discharge?
Can the system curve slope downward?
What happens if H_0 < Δz?

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13 Analytical problems

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13.1 Exercise 1: Deriving the system curve

Given:

$L$
$D$
$f$
$\Delta z$

Tasks:

Starting from the extended Bernoulli equation, derive: \[ H_s = \Delta z + b Q^2 \]
Derive expression for $b$.

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13.2 Exercise 2: Operating point derivation

Pump:

\[ H_p = H_0 - a Q^2 \]

System:

\[ H_s = \Delta z + b Q^2 \]

Tasks:

Derive formula for $Q$.
State condition for no solution.
Interpret physically.

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14 Computational problems

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14.1 Exercise 3: Basic pump system

Water at 20°C:

L = 250 m
D = 0.20 m
f = 0.022
Elevation rise = 18 m

Pump curve:

\[ H_p = 40 - 600 Q^2 \]

Find:

Operating discharge
Pump head
Velocity
Reynolds number
State whether Re is realistic for water systems

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14.2 Exercise 4: Including minor losses

Add:

Two elbows K = 0.9
One gate valve K = 0.15

Find:

New discharge
Percent reduction

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14.3 Exercise 5: Pump power

Using Exercise 3:

Compute hydraulic power
If efficiency is 78%, compute input power
Estimate hourly electricity cost at 0.12 dollars per kWh

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15 Design problems

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15.1 Exercise 6: Pump selection

System:

$\Delta z$ = 15 m
$b$ = 350

Pump A:

\[ H_p = 30 - 300 Q^2 \]

Pump B:

\[ H_p = 45 - 700 Q^2 \]

Pump C:

\[ H_p = 60 - 1200 Q^2 \]

Tasks:

Compute discharge for each pump
Which delivers largest discharge?
Which is likely most efficient?
Which would you select and why?

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16 Challenge: Series and parallel pumps

What happens to pump curve in series?
What happens in parallel?
Which configuration increases discharge?
Which increases head?

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