Pipe Bends, Enlargements, and Contractions

Example 1

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend — Pipe Bends, Enlargements, and Contractions

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 1,035 kg/m^3; flow rate (Q) = 0.2100 m^3/s; inlet velocity (V_1) = 2.3000 m/s; outlet velocity (V_2) = 5.6000 m/s; bend angle (theta) = 75.0000 deg; inlet pressure (p_1) = 255,500 Pa; inlet area (A_1) = 0.0300 m^2; outlet pressure (p_2) = 71,000 Pa; outlet area (A_2) = 0.0850 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,035 kg/m^3$
$flow rate (Q) = 0.2100 m^3/s$
$inlet velocity (V_1) = 2.3000 m/s$
$outlet velocity (V_2) = 5.6000 m/s$
$bend angle (theta) = 75.0000 deg$
$inlet pressure (p_1) = 255,500 Pa$
$inlet area (A_1) = 0.0300 m^2$
$outlet pressure (p_2) = 71,000 Pa$
$outlet area (A_2) = 0.0850 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 1 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend — Pipe Bends, Enlargements, and Contractions

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,035 kg/m^3, flow rate (Q) = 0.2100 m^3/s, inlet velocity (V_1) = 2.3000 m/s, outlet velocity (V_2) = 5.6000 m/s, bend angle (theta) = 75.0000 deg, inlet pressure (p_1) = 255,500 Pa, inlet area (A_1) = 0.0300 m^2, outlet pressure (p_2) = 71,000 Pa, outlet area (A_2) = 0.0850 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = 5918\ \text{N}$
Step 6 — Check: returning F_x = 5,918 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = 5918\ \text{N}

Why the other options are there

11,836 — kept a factor of two that cancels in the correct rearrangement.
2,959 — dropped that same factor in the other direction.
6,510 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 2

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 2) — Pipe Bends, Enlargements, and Contractions (2)

the reaction force at a pipe enlargement transition Given fluid density (rho) = 960.0 kg/m^3; flow rate (Q) = 0.0900 m^3/s; inlet velocity (V_1) = 5.7000 m/s; outlet velocity (V_2) = 2.0000 m/s; bend angle (theta) = 5.0000 deg; inlet pressure (p_1) = 131,500 Pa; inlet area (A_1) = 0.1250 m^2; outlet pressure (p_2) = 274,500 Pa; outlet area (A_2) = 0.0150 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 960.0 kg/m^3$
$flow rate (Q) = 0.0900 m^3/s$
$inlet velocity (V_1) = 5.7000 m/s$
$outlet velocity (V_2) = 2.0000 m/s$
$bend angle (theta) = 5.0000 deg$
$inlet pressure (p_1) = 131,500 Pa$
$inlet area (A_1) = 0.1250 m^2$
$outlet pressure (p_2) = 274,500 Pa$
$outlet area (A_2) = 0.0150 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 2 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 2) — Pipe Bends, Enlargements, and Contractions (2)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 960.0 kg/m^3, flow rate (Q) = 0.0900 m^3/s, inlet velocity (V_1) = 5.7000 m/s, outlet velocity (V_2) = 2.0000 m/s, bend angle (theta) = 5.0000 deg, inlet pressure (p_1) = 131,500 Pa, inlet area (A_1) = 0.1250 m^2, outlet pressure (p_2) = 274,500 Pa, outlet area (A_2) = 0.0150 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = 12015\ \text{N}$
Step 6 — Check: returning F_x = 12,015 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = 12015\ \text{N}

Why the other options are there

24,031 — kept a factor of two that cancels in the correct rearrangement.
6,008 — dropped that same factor in the other direction.
13,217 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 3

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 3) — Pipe Bends, Enlargements, and Contractions (3)

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 1,040 kg/m^3; flow rate (Q) = 0.3000 m^3/s; inlet velocity (V_1) = 2.1000 m/s; outlet velocity (V_2) = 3.1000 m/s; bend angle (theta) = 5.0000 deg; inlet pressure (p_1) = 263,000 Pa; inlet area (A_1) = 0.0300 m^2; outlet pressure (p_2) = 152,500 Pa; outlet area (A_2) = 0.1450 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,040 kg/m^3$
$flow rate (Q) = 0.3000 m^3/s$
$inlet velocity (V_1) = 2.1000 m/s$
$outlet velocity (V_2) = 3.1000 m/s$
$bend angle (theta) = 5.0000 deg$
$inlet pressure (p_1) = 263,000 Pa$
$inlet area (A_1) = 0.0300 m^2$
$outlet pressure (p_2) = 152,500 Pa$
$outlet area (A_2) = 0.1450 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 3 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 3) — Pipe Bends, Enlargements, and Contractions (3)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,040 kg/m^3, flow rate (Q) = 0.3000 m^3/s, inlet velocity (V_1) = 2.1000 m/s, outlet velocity (V_2) = 3.1000 m/s, bend angle (theta) = 5.0000 deg, inlet pressure (p_1) = 263,000 Pa, inlet area (A_1) = 0.0300 m^2, outlet pressure (p_2) = 152,500 Pa, outlet area (A_2) = 0.1450 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = -13830\ \text{N}$
Step 6 — Check: returning F_x = -13,830 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = -13830\ \text{N}

Why the other options are there

-27,660 — kept a factor of two that cancels in the correct rearrangement.
-6,915 — dropped that same factor in the other direction.
-15,213 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 4

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 4) — Pipe Bends, Enlargements, and Contractions (4)

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 1,020 kg/m^3; flow rate (Q) = 0.3300 m^3/s; inlet velocity (V_1) = 4.7000 m/s; outlet velocity (V_2) = 4.3000 m/s; bend angle (theta) = 50.0000 deg; inlet pressure (p_1) = 150,500 Pa; inlet area (A_1) = 0.0400 m^2; outlet pressure (p_2) = 169,500 Pa; outlet area (A_2) = 0.1400 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,020 kg/m^3$
$flow rate (Q) = 0.3300 m^3/s$
$inlet velocity (V_1) = 4.7000 m/s$
$outlet velocity (V_2) = 4.3000 m/s$
$bend angle (theta) = 50.0000 deg$
$inlet pressure (p_1) = 150,500 Pa$
$inlet area (A_1) = 0.0400 m^2$
$outlet pressure (p_2) = 169,500 Pa$
$outlet area (A_2) = 0.1400 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 4 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 4) — Pipe Bends, Enlargements, and Contractions (4)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,020 kg/m^3, flow rate (Q) = 0.3300 m^3/s, inlet velocity (V_1) = 4.7000 m/s, outlet velocity (V_2) = 4.3000 m/s, bend angle (theta) = 50.0000 deg, inlet pressure (p_1) = 150,500 Pa, inlet area (A_1) = 0.0400 m^2, outlet pressure (p_2) = 169,500 Pa, outlet area (A_2) = 0.1400 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = -9885\ \text{N}$
Step 6 — Check: returning F_x = -9,885 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = -9885\ \text{N}

Why the other options are there

-19,770 — kept a factor of two that cancels in the correct rearrangement.
-4,943 — dropped that same factor in the other direction.
-10,874 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 5

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 5) — Pipe Bends, Enlargements, and Contractions (5)

the reaction force at a pipe enlargement transition Given fluid density (rho) = 990.0 kg/m^3; flow rate (Q) = 0.3600 m^3/s; inlet velocity (V_1) = 2.4000 m/s; outlet velocity (V_2) = 1.8000 m/s; bend angle (theta) = 0.0000 deg; inlet pressure (p_1) = 126,500 Pa; inlet area (A_1) = 0.0950 m^2; outlet pressure (p_2) = 77,000 Pa; outlet area (A_2) = 0.0650 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 990.0 kg/m^3$
$flow rate (Q) = 0.3600 m^3/s$
$inlet velocity (V_1) = 2.4000 m/s$
$outlet velocity (V_2) = 1.8000 m/s$
$bend angle (theta) = 0.0000 deg$
$inlet pressure (p_1) = 126,500 Pa$
$inlet area (A_1) = 0.0950 m^2$
$outlet pressure (p_2) = 77,000 Pa$
$outlet area (A_2) = 0.0650 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 5 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 5) — Pipe Bends, Enlargements, and Contractions (5)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 990.0 kg/m^3, flow rate (Q) = 0.3600 m^3/s, inlet velocity (V_1) = 2.4000 m/s, outlet velocity (V_2) = 1.8000 m/s, bend angle (theta) = 0.0000 deg, inlet pressure (p_1) = 126,500 Pa, inlet area (A_1) = 0.0950 m^2, outlet pressure (p_2) = 77,000 Pa, outlet area (A_2) = 0.0650 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = 6799\ \text{N}$
Step 6 — Check: returning F_x = 6,799 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = 6799\ \text{N}

Why the other options are there

13,597 — kept a factor of two that cancels in the correct rearrangement.
3,399 — dropped that same factor in the other direction.
7,479 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 6

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 6) — Pipe Bends, Enlargements, and Contractions (6)

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 1,025 kg/m^3; flow rate (Q) = 0.0800 m^3/s; inlet velocity (V_1) = 3.5000 m/s; outlet velocity (V_2) = 1.6000 m/s; bend angle (theta) = 85.0000 deg; inlet pressure (p_1) = 107,500 Pa; inlet area (A_1) = 0.1450 m^2; outlet pressure (p_2) = 245,000 Pa; outlet area (A_2) = 0.0850 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,025 kg/m^3$
$flow rate (Q) = 0.0800 m^3/s$
$inlet velocity (V_1) = 3.5000 m/s$
$outlet velocity (V_2) = 1.6000 m/s$
$bend angle (theta) = 85.0000 deg$
$inlet pressure (p_1) = 107,500 Pa$
$inlet area (A_1) = 0.1450 m^2$
$outlet pressure (p_2) = 245,000 Pa$
$outlet area (A_2) = 0.0850 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 6 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 6) — Pipe Bends, Enlargements, and Contractions (6)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,025 kg/m^3, flow rate (Q) = 0.0800 m^3/s, inlet velocity (V_1) = 3.5000 m/s, outlet velocity (V_2) = 1.6000 m/s, bend angle (theta) = 85.0000 deg, inlet pressure (p_1) = 107,500 Pa, inlet area (A_1) = 0.1450 m^2, outlet pressure (p_2) = 245,000 Pa, outlet area (A_2) = 0.0850 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = 13497\ \text{N}$
Step 6 — Check: returning F_x = 13,497 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = 13497\ \text{N}

Why the other options are there

26,994 — kept a factor of two that cancels in the correct rearrangement.
6,748 — dropped that same factor in the other direction.
14,847 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 7

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 7) — Pipe Bends, Enlargements, and Contractions (7)

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 995.0 kg/m^3; flow rate (Q) = 0.0600 m^3/s; inlet velocity (V_1) = 3.0000 m/s; outlet velocity (V_2) = 3.2000 m/s; bend angle (theta) = 40.0000 deg; inlet pressure (p_1) = 122,500 Pa; inlet area (A_1) = 0.0300 m^2; outlet pressure (p_2) = 238,000 Pa; outlet area (A_2) = 0.0800 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 995.0 kg/m^3$
$flow rate (Q) = 0.0600 m^3/s$
$inlet velocity (V_1) = 3.0000 m/s$
$outlet velocity (V_2) = 3.2000 m/s$
$bend angle (theta) = 40.0000 deg$
$inlet pressure (p_1) = 122,500 Pa$
$inlet area (A_1) = 0.0300 m^2$
$outlet pressure (p_2) = 238,000 Pa$
$outlet area (A_2) = 0.0800 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 7 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 7) — Pipe Bends, Enlargements, and Contractions (7)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 995.0 kg/m^3, flow rate (Q) = 0.0600 m^3/s, inlet velocity (V_1) = 3.0000 m/s, outlet velocity (V_2) = 3.2000 m/s, bend angle (theta) = 40.0000 deg, inlet pressure (p_1) = 122,500 Pa, inlet area (A_1) = 0.0300 m^2, outlet pressure (p_2) = 238,000 Pa, outlet area (A_2) = 0.0800 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = -10943\ \text{N}$
Step 6 — Check: returning F_x = -10,943 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = -10943\ \text{N}

Why the other options are there

-21,886 — kept a factor of two that cancels in the correct rearrangement.
-5,472 — dropped that same factor in the other direction.
-12,038 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 8

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 8) — Pipe Bends, Enlargements, and Contractions (8)

the reaction force at a pipe enlargement transition Given fluid density (rho) = 1,040 kg/m^3; flow rate (Q) = 0.3200 m^3/s; inlet velocity (V_1) = 4.4000 m/s; outlet velocity (V_2) = 1.4000 m/s; bend angle (theta) = 15.0000 deg; inlet pressure (p_1) = 66,000 Pa; inlet area (A_1) = 0.1550 m^2; outlet pressure (p_2) = 221,000 Pa; outlet area (A_2) = 0.1850 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,040 kg/m^3$
$flow rate (Q) = 0.3200 m^3/s$
$inlet velocity (V_1) = 4.4000 m/s$
$outlet velocity (V_2) = 1.4000 m/s$
$bend angle (theta) = 15.0000 deg$
$inlet pressure (p_1) = 66,000 Pa$
$inlet area (A_1) = 0.1550 m^2$
$outlet pressure (p_2) = 221,000 Pa$
$outlet area (A_2) = 0.1850 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 8 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 8) — Pipe Bends, Enlargements, and Contractions (8)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,040 kg/m^3, flow rate (Q) = 0.3200 m^3/s, inlet velocity (V_1) = 4.4000 m/s, outlet velocity (V_2) = 1.4000 m/s, bend angle (theta) = 15.0000 deg, inlet pressure (p_1) = 66,000 Pa, inlet area (A_1) = 0.1550 m^2, outlet pressure (p_2) = 221,000 Pa, outlet area (A_2) = 0.1850 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = -30276\ \text{N}$
Step 6 — Check: returning F_x = -30,276 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = -30276\ \text{N}

Why the other options are there

-60,552 — kept a factor of two that cancels in the correct rearrangement.
-15,138 — dropped that same factor in the other direction.
-33,304 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 9

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 9) — Pipe Bends, Enlargements, and Contractions (9)

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 1,005 kg/m^3; flow rate (Q) = 0.3300 m^3/s; inlet velocity (V_1) = 3.0000 m/s; outlet velocity (V_2) = 1.9000 m/s; bend angle (theta) = 55.0000 deg; inlet pressure (p_1) = 138,500 Pa; inlet area (A_1) = 0.0500 m^2; outlet pressure (p_2) = 300,000 Pa; outlet area (A_2) = 0.0700 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 1,005 kg/m^3$
$flow rate (Q) = 0.3300 m^3/s$
$inlet velocity (V_1) = 3.0000 m/s$
$outlet velocity (V_2) = 1.9000 m/s$
$bend angle (theta) = 55.0000 deg$
$inlet pressure (p_1) = 138,500 Pa$
$inlet area (A_1) = 0.0500 m^2$
$outlet pressure (p_2) = 300,000 Pa$
$outlet area (A_2) = 0.0700 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 9 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 9) — Pipe Bends, Enlargements, and Contractions (9)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 1,005 kg/m^3, flow rate (Q) = 0.3300 m^3/s, inlet velocity (V_1) = 3.0000 m/s, outlet velocity (V_2) = 1.9000 m/s, bend angle (theta) = 55.0000 deg, inlet pressure (p_1) = 138,500 Pa, inlet area (A_1) = 0.0500 m^2, outlet pressure (p_2) = 300,000 Pa, outlet area (A_2) = 0.0700 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = -5754\ \text{N}$
Step 6 — Check: returning F_x = -5,754 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = -5754\ \text{N}

Why the other options are there

-11,507 — kept a factor of two that cancels in the correct rearrangement.
-2,877 — dropped that same factor in the other direction.
-6,329 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Example 10

Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 10) — Pipe Bends, Enlargements, and Contractions (10)

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 965.0 kg/m^3; flow rate (Q) = 0.3700 m^3/s; inlet velocity (V_1) = 4.0000 m/s; outlet velocity (V_2) = 5.4000 m/s; bend angle (theta) = 75.0000 deg; inlet pressure (p_1) = 67,500 Pa; inlet area (A_1) = 0.0950 m^2; outlet pressure (p_2) = 119,500 Pa; outlet area (A_2) = 0.0950 m^2, determine the resultant force on bend (F_x) in N.

Given

$fluid density (rho) = 965.0 kg/m^3$
$flow rate (Q) = 0.3700 m^3/s$
$inlet velocity (V_1) = 4.0000 m/s$
$outlet velocity (V_2) = 5.4000 m/s$
$bend angle (theta) = 75.0000 deg$
$inlet pressure (p_1) = 67,500 Pa$
$inlet area (A_1) = 0.0950 m^2$
$outlet pressure (p_2) = 119,500 Pa$
$outlet area (A_2) = 0.0950 m^2$

Find

resultant force on bend (F_x), in N

Start with the thinking

The governing relation printed in this handbook section is Pipe bends, enlargements, and contractions (momentum force).
Everything except F_x is given, so isolate F_x symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Pipe bends, enlargements, and contractions require a momentum balance to find the anchoring force on the fitting.

Figure 10 — schematic for Pipe bends, enlargements, and contractions (momentum force) — solve for resultant force on bend (case 10) — Pipe Bends, Enlargements, and Contractions (10)

Step-by-step solution

Step 1 — State the governing relation:
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.
Step 3 — List the givens: fluid density (rho) = 965.0 kg/m^3, flow rate (Q) = 0.3700 m^3/s, inlet velocity (V_1) = 4.0000 m/s, outlet velocity (V_2) = 5.4000 m/s, bend angle (theta) = 75.0000 deg, inlet pressure (p_1) = 67,500 Pa, inlet area (A_1) = 0.0950 m^2, outlet pressure (p_2) = 119,500 Pa, outlet area (A_2) = 0.0950 m^2.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$F_{x} = 2545\ \text{N}$
Step 6 — Check: returning F_x = 2,545 N to
$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$
reproduces the given quantities, and both sides carry the same units.

Answer:

F_{x} = 2545\ \text{N}

Why the other options are there

5,090 — kept a factor of two that cancels in the correct rearrangement.
1,273 — dropped that same factor in the other direction.
2,800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Pipe Bends, Enlargements, and Contractions

Pipe Bends, Enlargements, and Contractions

Handbook notes for this section

Core formulas for this FE topic

Worked exam-style examples

Pipe Bends, Enlargements, and Contractions

Handbook notes for this section

Core formulas for this FE topic

P1A1−P2A2cos⁡α−Fx=Qρ(v2cos⁡α−v1)P_{1}A_{1} - P_{2}A_{2}\cos \alpha - Fx = Q\rho (v_{2}\cos \alpha - v_{1})P1​A1​−P2​A2​cosα−Fx=Qρ(v2​cosα−v1​)

Fy−W−P2A2sin⁡α=Qρ(v2sin⁡α−0)Fy - W - P_{2}A_{2}\sin \alpha = Q\rho (v_{2}\sin \alpha - 0)Fy−W−P2​A2​sinα=Qρ(v2​sinα−0)

F=forceexertedbythebendonthefluid(theforceexertedbythefluidonthebendisequalinmagnitudeandF = force exerted by the bend on the fluid (the force exerted by the fluid on the bend is equal in magnitude andF=forceexertedbythebendonthefluid(theforceexertedbythefluidonthebendisequalinmagnitudeand

oppositeinsign),FxandFyarethex−componentandy−componentoftheforceF=Fx2+Fy2andopposite in sign), Fx and Fy are the x-component and y-component of the force F = Fx_{2} + Fy_{2} andoppositeinsign),FxandFyarethex−componentandy−componentoftheforceF=Fx2​+Fy2​and

i=tan⁡1dFni = \tan 1 d F ni=tan1dFn

P=internalpressureinthepipelineP = internal pressure in the pipe lineP=internalpressureinthepipeline

A=cross−sectionalareaofthepipelineA = cross-sectional area of the pipe lineA=cross−sectionalareaofthepipeline

W=weightofthefluidW = weight of the fluidW=weightofthefluid

v=velocityofthefluidflowv = velocity of the fluid flowv=velocityofthefluidflow

α=anglethepipebendmakeswiththehorizontal\alpha = angle the pipe bend makes with the horizontalα=anglethepipebendmakeswiththehorizontal

ρ=densityofthefluid\rho = density of the fluidρ=densityofthefluid

Q=fluidvolumetricflowrateQ = fluid volumetric flowrateQ=fluidvolumetricflowrate

Worked exam-style examples