Pressure Drop for Laminar Flow

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

1 formulas

10 exam-style examples

~47 min

All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The equation for Q in terms of the pressure drop ∆Pf is the Hagen-Poiseuille equation. This relation is valid only for flow in the

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow

Find the gauge pressure 22.5 ft below the surface of a fluid with specific gravity 7.80.

Given

$h = 22.5 ft$
$SG = 7.80$
$\gamma_water = 62.4 lb/ft^{3}$

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure 1 — schematic for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow

Step-by-step solution

Specific weight
$\gamma = SG \times \gamma_w = 7.80(62.4) = 486.7 lb/ft^{3}$
Hydrostatic — p = γh
Substituting
$p = 486.7(22.5) = 10,951 psf$
Convert
$p = 10,951/144 = 76.05 psi$

Answer:

p ≈ 10,951 psf (76.05 psi)

Why the other options are there

1,404 psf (specific gravity ignored)
1,576,973 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 2

Reynolds number and flow regime — Pressure Drop for Laminar Flow

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 5.00 ft/s in a 0.25 ft pipe. Compute Re and classify the flow.

Given

$v = 5.00 ft/s$
$D = 0.25 ft$
$\nu = 1.08e-5 ft^{2}/s$

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$Re = vD/\nu$
Substituting
$Re = 5.00(0.25)/1.08e-5$
Evaluate
$Re = 115,741$
Classify — turbulent

Answer:

Re ≈ 115,741 → turbulent

Why the other options are there

9,645 (diameter left in inches)
1,851,852 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 3

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (2)

Find the gauge pressure 7.5 ft below the surface of a fluid with specific gravity 12.70.

Given

$h = 7.5 ft$
$SG = 12.70$
$\gamma_water = 62.4 lb/ft^{3}$

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure 3 — schematic for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (2)

Step-by-step solution

Specific weight
$\gamma = SG \times \gamma_w = 12.70(62.4) = 792.5 lb/ft^{3}$
Hydrostatic — p = γh
Substituting
$p = 792.5(7.5) = 5,944 psf$
Convert
$p = 5,944/144 = 41.28 psi$

Answer:

p ≈ 5,944 psf (41.28 psi)

Why the other options are there

468.0 psf (specific gravity ignored)
855,878 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 4

Reynolds number and flow regime — Pressure Drop for Laminar Flow (2)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 5.25 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

$v = 5.25 ft/s$
$D = 0.50 ft$
$\nu = 1.08e-5 ft^{2}/s$

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$Re = vD/\nu$
Substituting
$Re = 5.25(0.50)/1.08e-5$
Evaluate
$Re = 243,056$
Classify — turbulent

Answer:

Re ≈ 243,056 → turbulent

Why the other options are there

20,255 (diameter left in inches)
972,222 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 5

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (3)

Find the gauge pressure 4.0 ft below the surface of a fluid with specific gravity 10.40.

Given

$h = 4.0 ft$
$SG = 10.40$
$\gamma_water = 62.4 lb/ft^{3}$

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure 5 — schematic for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (3)

Step-by-step solution

Specific weight
$\gamma = SG \times \gamma_w = 10.40(62.4) = 649.0 lb/ft^{3}$
Hydrostatic — p = γh
Substituting
$p = 649.0(4.0) = 2,596 psf$
Convert
$p = 2,596/144 = 18.03 psi$

Answer:

p ≈ 2,596 psf (18.03 psi)

Why the other options are there

249.6 psf (specific gravity ignored)
373,801 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 6

Reynolds number and flow regime — Pressure Drop for Laminar Flow (3)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 1.50 ft/s in a 1.25 ft pipe. Compute Re and classify the flow.

Given

$v = 1.50 ft/s$
$D = 1.25 ft$
$\nu = 1.08e-5 ft^{2}/s$

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$Re = vD/\nu$
Substituting
$Re = 1.50(1.25)/1.08e-5$
Evaluate
$Re = 173,611$
Classify — turbulent

Answer:

Re ≈ 173,611 → turbulent

Why the other options are there

14,468 (diameter left in inches)
111,111 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 7

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (4)

Find the gauge pressure 20.5 ft below the surface of a fluid with specific gravity 0.90.

Given

$h = 20.5 ft$
$SG = 0.90$
$\gamma_water = 62.4 lb/ft^{3}$

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure 7 — schematic for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (4)

Step-by-step solution

Specific weight
$\gamma = SG \times \gamma_w = 0.90(62.4) = 56.16 lb/ft^{3}$
Hydrostatic — p = γh
Substituting
$p = 56.16(20.5) = 1,151 psf$
Convert
$p = 1,151/144 = 8.00 psi$

Answer:

p ≈ 1,151 psf (8.00 psi)

Why the other options are there

1,279 psf (specific gravity ignored)
165,784 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 8

Reynolds number and flow regime — Pressure Drop for Laminar Flow (4)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 4.50 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

$v = 4.50 ft/s$
$D = 0.50 ft$
$\nu = 1.08e-5 ft^{2}/s$

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$Re = vD/\nu$
Substituting
$Re = 4.50(0.50)/1.08e-5$
Evaluate
$Re = 208,333$
Classify — turbulent

Answer:

Re ≈ 208,333 → turbulent

Why the other options are there

17,361 (diameter left in inches)
833,333 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 9

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (5)

Find the gauge pressure 4.0 ft below the surface of a fluid with specific gravity 0.90.

Given

$h = 4.0 ft$
$SG = 0.90$
$\gamma_water = 62.4 lb/ft^{3}$

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure 9 — schematic for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (5)

Step-by-step solution

Specific weight
$\gamma = SG \times \gamma_w = 0.90(62.4) = 56.16 lb/ft^{3}$
Hydrostatic — p = γh
Substituting
$p = 56.16(4.0) = 224.6 psf$
Convert
$p = 224.6/144 = 1.56 psi$

Answer:

p ≈ 224.6 psf (1.56 psi)

Why the other options are there

249.6 psf (specific gravity ignored)
32,348 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 10

Reynolds number and flow regime — Pressure Drop for Laminar Flow (5)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 3.75 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

$v = 3.75 ft/s$
$D = 0.50 ft$
$\nu = 1.08e-5 ft^{2}/s$

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$Re = vD/\nu$
Substituting
$Re = 3.75(0.50)/1.08e-5$
Evaluate
$Re = 173,611$
Classify — turbulent

Answer:

Re ≈ 173,611 → turbulent

Why the other options are there

14,468 (diameter left in inches)
694,444 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow