Water flows at 0.25 m³/s from a 300 mm pipe (p₁ = 250 kPa) into a 200 mm pipe at the same elevation. Neglecting losses, what is p₂?

Start with the thinking

• Continuity gives both velocities; Bernoulli converts the velocity change to pressure.
• Same elevation means the z terms cancel.

Step-by-step solution

1. Areas

   $$A_{1} = \pi(0.30)^{2}/4 = 0.0707 m^{2} and A_{2} = \pi(0.20)^{2}/4 = 0.0314 m^{2}$$

2. Velocities

   $$V_{1} = 0.25/0.0707 = 3.54 m/s and V_{2} = 0.25/0.0314 = 7.96 m/s$$

3. Bernoulli

   $$p_{2} = p_{1} + \rho(V_{1}^{2} - V_{2}^{2})/2$$

4. Velocity terms

   $$(3.54^{2} - 7.96^{2})/2 = (12.5 - 63.4)/2 = -25.4 m^{2}/s^{2}$$

5. Result

   $$p_{2} = 250,000 + 1,000(-25.4) = 224,600 Pa = 225 kPa$$

Why the other options are there

• 275 kPa (sign of the velocity term reversed)
• 250 kPa (velocity change ignored)

Reference: FE Reference Handbook — Fluid Mechanics — Energy equation

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 1,045 kg/m^3; flow rate (Q) = 0.0600 m^3/s; inlet velocity (V_1) = 4.8000 m/s; outlet velocity (V_2) = 1.5000 m/s; bend angle (theta) = 30.0000 deg; inlet pressure (p_1) = 251,000 Pa; inlet area (A_1) = 0.1300 m^2; outlet pressure (p_2) = 220,000 Pa; outlet area (A_2) = 0.1850 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 1,045 kg/m^3, flow rate (Q) = 0.0600 m^3/s, inlet velocity (V_1) = 4.8000 m/s, outlet velocity (V_2) = 1.5000 m/s, bend angle (theta) = 30.0000 deg, inlet pressure (p_1) = 251,000 Pa, inlet area (A_1) = 0.1300 m^2, outlet pressure (p_2) = 220,000 Pa, outlet area (A_2) = 0.1850 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = -2837\ \text{N}$$

6. Step 6 — Check: returning F_x = -2,837 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.

Why the other options are there

• -5,673 — kept a factor of two that cancels in the correct rearrangement.
• -1,418 — dropped that same factor in the other direction.
• -3,120 — rounded an intermediate value before the final step.

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

the reaction force at a pipe enlargement transition Given fluid density (rho) = 985.0 kg/m^3; flow rate (Q) = 0.2500 m^3/s; inlet velocity (V_1) = 2.2000 m/s; outlet velocity (V_2) = 3.0000 m/s; bend angle (theta) = 70.0000 deg; inlet pressure (p_1) = 240,500 Pa; inlet area (A_1) = 0.0450 m^2; outlet pressure (p_2) = 239,000 Pa; outlet area (A_2) = 0.0850 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 985.0 kg/m^3, flow rate (Q) = 0.2500 m^3/s, inlet velocity (V_1) = 2.2000 m/s, outlet velocity (V_2) = 3.0000 m/s, bend angle (theta) = 70.0000 deg, inlet pressure (p_1) = 240,500 Pa, inlet area (A_1) = 0.0450 m^2, outlet pressure (p_2) = 239,000 Pa, outlet area (A_2) = 0.0850 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = 3585\ \text{N}$$

6. Step 6 — Check: returning F_x = 3,585 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.

Why the other options are there

• 7,171 — kept a factor of two that cancels in the correct rearrangement.
• 1,793 — dropped that same factor in the other direction.
• 3,944 — rounded an intermediate value before the final step.

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

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 975.0 kg/m^3; flow rate (Q) = 0.3200 m^3/s; inlet velocity (V_1) = 2.2000 m/s; outlet velocity (V_2) = 5.8000 m/s; bend angle (theta) = 75.0000 deg; inlet pressure (p_1) = 290,500 Pa; inlet area (A_1) = 0.1150 m^2; outlet pressure (p_2) = 117,000 Pa; outlet area (A_2) = 0.1550 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 975.0 kg/m^3, flow rate (Q) = 0.3200 m^3/s, inlet velocity (V_1) = 2.2000 m/s, outlet velocity (V_2) = 5.8000 m/s, bend angle (theta) = 75.0000 deg, inlet pressure (p_1) = 290,500 Pa, inlet area (A_1) = 0.1150 m^2, outlet pressure (p_2) = 117,000 Pa, outlet area (A_2) = 0.1550 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = 28496\ \text{N}$$

6. Step 6 — Check: returning F_x = 28,496 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.

Why the other options are there

• 56,992 — kept a factor of two that cancels in the correct rearrangement.
• 14,248 — dropped that same factor in the other direction.
• 31,345 — rounded an intermediate value before the final step.

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

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 1,045 kg/m^3; flow rate (Q) = 0.2300 m^3/s; inlet velocity (V_1) = 5.8000 m/s; outlet velocity (V_2) = 4.9000 m/s; bend angle (theta) = 80.0000 deg; inlet pressure (p_1) = 219,000 Pa; inlet area (A_1) = 0.0250 m^2; outlet pressure (p_2) = 251,500 Pa; outlet area (A_2) = 0.0300 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 1,045 kg/m^3, flow rate (Q) = 0.2300 m^3/s, inlet velocity (V_1) = 5.8000 m/s, outlet velocity (V_2) = 4.9000 m/s, bend angle (theta) = 80.0000 deg, inlet pressure (p_1) = 219,000 Pa, inlet area (A_1) = 0.0250 m^2, outlet pressure (p_2) = 251,500 Pa, outlet area (A_2) = 0.0300 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = 2975\ \text{N}$$

6. Step 6 — Check: returning F_x = 2,975 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.

Why the other options are there

• 5,951 — kept a factor of two that cancels in the correct rearrangement.
• 1,488 — dropped that same factor in the other direction.
• 3,273 — rounded an intermediate value before the final step.

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

the reaction force at a pipe enlargement transition Given fluid density (rho) = 990.0 kg/m^3; flow rate (Q) = 0.2200 m^3/s; inlet velocity (V_1) = 4.0000 m/s; outlet velocity (V_2) = 5.5000 m/s; bend angle (theta) = 5.0000 deg; inlet pressure (p_1) = 227,000 Pa; inlet area (A_1) = 0.0200 m^2; outlet pressure (p_2) = 63,000 Pa; outlet area (A_2) = 0.1050 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 990.0 kg/m^3, flow rate (Q) = 0.2200 m^3/s, inlet velocity (V_1) = 4.0000 m/s, outlet velocity (V_2) = 5.5000 m/s, bend angle (theta) = 5.0000 deg, inlet pressure (p_1) = 227,000 Pa, inlet area (A_1) = 0.0200 m^2, outlet pressure (p_2) = 63,000 Pa, outlet area (A_2) = 0.1050 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = -1728\ \text{N}$$

6. Step 6 — Check: returning F_x = -1,728 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.

Why the other options are there

• -3,455 — kept a factor of two that cancels in the correct rearrangement.
• -863.8 — dropped that same factor in the other direction.
• -1,900 — rounded an intermediate value before the final step.

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

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 1,040 kg/m^3; flow rate (Q) = 0.0500 m^3/s; inlet velocity (V_1) = 3.4000 m/s; outlet velocity (V_2) = 4.9000 m/s; bend angle (theta) = 45.0000 deg; inlet pressure (p_1) = 52,500 Pa; inlet area (A_1) = 0.1050 m^2; outlet pressure (p_2) = 181,500 Pa; outlet area (A_2) = 0.1850 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 1,040 kg/m^3, flow rate (Q) = 0.0500 m^3/s, inlet velocity (V_1) = 3.4000 m/s, outlet velocity (V_2) = 4.9000 m/s, bend angle (theta) = 45.0000 deg, inlet pressure (p_1) = 52,500 Pa, inlet area (A_1) = 0.1050 m^2, outlet pressure (p_2) = 181,500 Pa, outlet area (A_2) = 0.1850 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = -18227\ \text{N}$$

6. Step 6 — Check: returning F_x = -18,227 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.

Why the other options are there

• -36,454 — kept a factor of two that cancels in the correct rearrangement.
• -9,114 — dropped that same factor in the other direction.
• -20,050 — rounded an intermediate value before the final step.

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

the thrust force on a pipe bend in a pressurized pipeline Given fluid density (rho) = 1,005 kg/m^3; flow rate (Q) = 0.0700 m^3/s; inlet velocity (V_1) = 5.0000 m/s; outlet velocity (V_2) = 3.6000 m/s; bend angle (theta) = 80.0000 deg; inlet pressure (p_1) = 115,500 Pa; inlet area (A_1) = 0.0250 m^2; outlet pressure (p_2) = 157,000 Pa; outlet area (A_2) = 0.1150 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 1,005 kg/m^3, flow rate (Q) = 0.0700 m^3/s, inlet velocity (V_1) = 5.0000 m/s, outlet velocity (V_2) = 3.6000 m/s, bend angle (theta) = 80.0000 deg, inlet pressure (p_1) = 115,500 Pa, inlet area (A_1) = 0.0250 m^2, outlet pressure (p_2) = 157,000 Pa, outlet area (A_2) = 0.1150 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = -555.5\ \text{N}$$

6. Step 6 — Check: returning F_x = -555.5 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.

Why the other options are there

• -1,111 — kept a factor of two that cancels in the correct rearrangement.
• -277.7 — dropped that same factor in the other direction.
• -611.0 — rounded an intermediate value before the final step.

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

the reaction force at a pipe enlargement transition Given fluid density (rho) = 960.0 kg/m^3; flow rate (Q) = 0.1400 m^3/s; inlet velocity (V_1) = 5.4000 m/s; outlet velocity (V_2) = 5.5000 m/s; bend angle (theta) = 70.0000 deg; inlet pressure (p_1) = 183,000 Pa; inlet area (A_1) = 0.1750 m^2; outlet pressure (p_2) = 59,500 Pa; outlet area (A_2) = 0.1500 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 960.0 kg/m^3, flow rate (Q) = 0.1400 m^3/s, inlet velocity (V_1) = 5.4000 m/s, outlet velocity (V_2) = 5.5000 m/s, bend angle (theta) = 70.0000 deg, inlet pressure (p_1) = 183,000 Pa, inlet area (A_1) = 0.1750 m^2, outlet pressure (p_2) = 59,500 Pa, outlet area (A_2) = 0.1500 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = 28500\ \text{N}$$

6. Step 6 — Check: returning F_x = 28,500 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.

Why the other options are there

• 56,999 — kept a factor of two that cancels in the correct rearrangement.
• 14,250 — dropped that same factor in the other direction.
• 31,349 — rounded an intermediate value before the final step.

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

the anchoring force needed at a pipe contraction fitting Given fluid density (rho) = 1,005 kg/m^3; flow rate (Q) = 0.2100 m^3/s; inlet velocity (V_1) = 2.1000 m/s; outlet velocity (V_2) = 1.3000 m/s; bend angle (theta) = 75.0000 deg; inlet pressure (p_1) = 146,500 Pa; inlet area (A_1) = 0.0800 m^2; outlet pressure (p_2) = 198,000 Pa; outlet area (A_2) = 0.2000 m^2, determine the 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.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$F_x = \rho Q (V_2 \cos\theta - V_1) + p_1 A_1 - p_2 A_2 \cos\theta$$

2. Step 2 — Rearrange the relation so that F_x stands alone on the left-hand side.

3. Step 3 — List the givens: fluid density (rho) = 1,005 kg/m^3, flow rate (Q) = 0.2100 m^3/s, inlet velocity (V_1) = 2.1000 m/s, outlet velocity (V_2) = 1.3000 m/s, bend angle (theta) = 75.0000 deg, inlet pressure (p_1) = 146,500 Pa, inlet area (A_1) = 0.0800 m^2, outlet pressure (p_2) = 198,000 Pa, outlet area (A_2) = 0.2000 m^2.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F_{x} = 1099\ \text{N}$$

6. Step 6 — Check: returning F_x = 1,099 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.

Why the other options are there

• 2,197 — kept a factor of two that cancels in the correct rearrangement.
• 549.3 — dropped that same factor in the other direction.
• 1,208 — rounded an intermediate value before the final step.

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