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Jet Propulsion

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics
8 formulas
10 exam-style examples
~60 min
All Fluid Mechanics lectures

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
Jet propulsion thrust — solve for thrust — Jet Propulsion

jet propulsion thrust of a water jet boat Given fluid density (rho) = 1,044 kg/m^3; jet flow rate (Q) = 1.5400 m^3/s; jet exit velocity (V_j) = 284.0 m/s; vehicle velocity (V_0) = 24.0000 m/s, determine the thrust (F) in N.

Given

  • fluiddensity(rho)=1,044kg/m3fluid density (rho) = 1,044 kg/m^3
  • jetflowrate(Q)=1.5400m3/sjet flow rate (Q) = 1.5400 m^3/s
  • jetexitvelocity(Vj)=284.0m/sjet exit velocity (V_j) = 284.0 m/s
  • vehiclevelocity(V0)=24.0000m/svehicle velocity (V_0) = 24.0000 m/s

Find

thrust (F), in N

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except F is given, so isolate F 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 1,044 kg/m^3, jet flow rate (Q) = 1.5400 m^3/s, jet exit velocity (V_j) = 284.0 m/s, vehicle velocity (V_0) = 24.0000 m/s.

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

  5. Step 5 — Evaluate:

    F=418018 NF = 418018\ \text{N}
  6. Step 6 — Check: returning F = 418,018 N to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=418018 NF = 418018\ \text{N}

Why the other options are there

  • 836,035 — kept a factor of two that cancels in the correct rearrangement.
  • 209,009 — dropped that same factor in the other direction.
  • 459,819 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 2
Jet propulsion thrust — solve for jet flow rate — Jet Propulsion (2)

jet propulsion of a fireboat monitor nozzle Given fluid density (rho) = 800.0 kg/m^3; jet exit velocity (V_j) = 265.0 m/s; vehicle velocity (V_0) = 17.0000 m/s; thrust (F) = 134,460 N, determine the jet flow rate (Q) in m^3/s.

Given

  • fluiddensity(rho)=800.0kg/m3fluid density (rho) = 800.0 kg/m^3
  • jetexitvelocity(Vj)=265.0m/sjet exit velocity (V_j) = 265.0 m/s
  • vehiclevelocity(V0)=17.0000m/svehicle velocity (V_0) = 17.0000 m/s
  • thrust(F)=134,460Nthrust (F) = 134,460 N

Find

jet flow rate (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except Q is given, so isolate Q 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 800.0 kg/m^3, jet exit velocity (V_j) = 265.0 m/s, vehicle velocity (V_0) = 17.0000 m/s, thrust (F) = 134,460 N.

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

  5. Step 5 — Evaluate:

    Q = 0.6777\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 0.6777 m^3/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 0.6777\ \text{m^3/s}

Why the other options are there

  • 1.3554 — kept a factor of two that cancels in the correct rearrangement.
  • 0.3389 — dropped that same factor in the other direction.
  • 0.7455 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 3
Jet propulsion thrust — solve for jet exit velocity — Jet Propulsion (3)

jet propulsion thrust from an aircraft engine exhaust Given fluid density (rho) = 515.0 kg/m^3; jet flow rate (Q) = 0.6300 m^3/s; vehicle velocity (V_0) = 11.0000 m/s; thrust (F) = 274,730 N, determine the jet exit velocity (V_j) in m/s.

Given

  • fluiddensity(rho)=515.0kg/m3fluid density (rho) = 515.0 kg/m^3
  • jetflowrate(Q)=0.6300m3/sjet flow rate (Q) = 0.6300 m^3/s
  • vehiclevelocity(V0)=11.0000m/svehicle velocity (V_0) = 11.0000 m/s
  • thrust(F)=274,730Nthrust (F) = 274,730 N

Find

jet exit velocity (V_j), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except V_j is given, so isolate V_j 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that V_j stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 515.0 kg/m^3, jet flow rate (Q) = 0.6300 m^3/s, vehicle velocity (V_0) = 11.0000 m/s, thrust (F) = 274,730 N.

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

  5. Step 5 — Evaluate:

    Vj=857.8 m/sV_{j} = 857.8\ \text{m/s}
  6. Step 6 — Check: returning V_j = 857.8 m/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Vj=857.8 m/sV_{j} = 857.8\ \text{m/s}

Why the other options are there

  • 1,716 — kept a factor of two that cancels in the correct rearrangement.
  • 428.9 — dropped that same factor in the other direction.
  • 943.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 4
Jet propulsion thrust — solve for thrust (case 2) — Jet Propulsion (4)

jet propulsion thrust of a water jet boat Given fluid density (rho) = 835.0 kg/m^3; jet flow rate (Q) = 1.1100 m^3/s; jet exit velocity (V_j) = 163.0 m/s; vehicle velocity (V_0) = 32.0000 m/s, determine the thrust (F) in N.

Given

  • fluiddensity(rho)=835.0kg/m3fluid density (rho) = 835.0 kg/m^3
  • jetflowrate(Q)=1.1100m3/sjet flow rate (Q) = 1.1100 m^3/s
  • jetexitvelocity(Vj)=163.0m/sjet exit velocity (V_j) = 163.0 m/s
  • vehiclevelocity(V0)=32.0000m/svehicle velocity (V_0) = 32.0000 m/s

Find

thrust (F), in N

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except F is given, so isolate F 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 835.0 kg/m^3, jet flow rate (Q) = 1.1100 m^3/s, jet exit velocity (V_j) = 163.0 m/s, vehicle velocity (V_0) = 32.0000 m/s.

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

  5. Step 5 — Evaluate:

    F=121417 NF = 121417\ \text{N}
  6. Step 6 — Check: returning F = 121,417 N to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=121417 NF = 121417\ \text{N}

Why the other options are there

  • 242,835 — kept a factor of two that cancels in the correct rearrangement.
  • 60,709 — dropped that same factor in the other direction.
  • 133,559 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 5
Jet propulsion thrust — solve for jet flow rate (case 2) — Jet Propulsion (5)

jet propulsion of a fireboat monitor nozzle Given fluid density (rho) = 154.0 kg/m^3; jet exit velocity (V_j) = 129.0 m/s; vehicle velocity (V_0) = 45.0000 m/s; thrust (F) = 406,440 N, determine the jet flow rate (Q) in m^3/s.

Given

  • fluiddensity(rho)=154.0kg/m3fluid density (rho) = 154.0 kg/m^3
  • jetexitvelocity(Vj)=129.0m/sjet exit velocity (V_j) = 129.0 m/s
  • vehiclevelocity(V0)=45.0000m/svehicle velocity (V_0) = 45.0000 m/s
  • thrust(F)=406,440Nthrust (F) = 406,440 N

Find

jet flow rate (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except Q is given, so isolate Q 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 154.0 kg/m^3, jet exit velocity (V_j) = 129.0 m/s, vehicle velocity (V_0) = 45.0000 m/s, thrust (F) = 406,440 N.

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

  5. Step 5 — Evaluate:

    Q = 31.4193\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 31.4193 m^3/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 31.4193\ \text{m^3/s}

Why the other options are there

  • 62.8386 — kept a factor of two that cancels in the correct rearrangement.
  • 15.7096 — dropped that same factor in the other direction.
  • 34.5612 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 6
Jet propulsion thrust — solve for jet exit velocity (case 2) — Jet Propulsion (6)

jet propulsion thrust from an aircraft engine exhaust Given fluid density (rho) = 1,003 kg/m^3; jet flow rate (Q) = 0.6000 m^3/s; vehicle velocity (V_0) = 46.0000 m/s; thrust (F) = 480,300 N, determine the jet exit velocity (V_j) in m/s.

Given

  • fluiddensity(rho)=1,003kg/m3fluid density (rho) = 1,003 kg/m^3
  • jetflowrate(Q)=0.6000m3/sjet flow rate (Q) = 0.6000 m^3/s
  • vehiclevelocity(V0)=46.0000m/svehicle velocity (V_0) = 46.0000 m/s
  • thrust(F)=480,300Nthrust (F) = 480,300 N

Find

jet exit velocity (V_j), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except V_j is given, so isolate V_j 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that V_j stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 1,003 kg/m^3, jet flow rate (Q) = 0.6000 m^3/s, vehicle velocity (V_0) = 46.0000 m/s, thrust (F) = 480,300 N.

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

  5. Step 5 — Evaluate:

    Vj=844.1 m/sV_{j} = 844.1\ \text{m/s}
  6. Step 6 — Check: returning V_j = 844.1 m/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Vj=844.1 m/sV_{j} = 844.1\ \text{m/s}

Why the other options are there

  • 1,688 — kept a factor of two that cancels in the correct rearrangement.
  • 422.1 — dropped that same factor in the other direction.
  • 928.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 7
Jet propulsion thrust — solve for thrust (case 3) — Jet Propulsion (7)

jet propulsion thrust of a water jet boat Given fluid density (rho) = 867.0 kg/m^3; jet flow rate (Q) = 0.3300 m^3/s; jet exit velocity (V_j) = 129.0 m/s; vehicle velocity (V_0) = 13.0000 m/s, determine the thrust (F) in N.

Given

  • fluiddensity(rho)=867.0kg/m3fluid density (rho) = 867.0 kg/m^3
  • jetflowrate(Q)=0.3300m3/sjet flow rate (Q) = 0.3300 m^3/s
  • jetexitvelocity(Vj)=129.0m/sjet exit velocity (V_j) = 129.0 m/s
  • vehiclevelocity(V0)=13.0000m/svehicle velocity (V_0) = 13.0000 m/s

Find

thrust (F), in N

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except F is given, so isolate F 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 867.0 kg/m^3, jet flow rate (Q) = 0.3300 m^3/s, jet exit velocity (V_j) = 129.0 m/s, vehicle velocity (V_0) = 13.0000 m/s.

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

  5. Step 5 — Evaluate:

    F=33189 NF = 33189\ \text{N}
  6. Step 6 — Check: returning F = 33,189 N to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=33189 NF = 33189\ \text{N}

Why the other options are there

  • 66,378 — kept a factor of two that cancels in the correct rearrangement.
  • 16,594 — dropped that same factor in the other direction.
  • 36,508 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 8
Jet propulsion thrust — solve for jet flow rate (case 3) — Jet Propulsion (8)

jet propulsion of a fireboat monitor nozzle Given fluid density (rho) = 839.0 kg/m^3; jet exit velocity (V_j) = 201.0 m/s; vehicle velocity (V_0) = 42.0000 m/s; thrust (F) = 445,170 N, determine the jet flow rate (Q) in m^3/s.

Given

  • fluiddensity(rho)=839.0kg/m3fluid density (rho) = 839.0 kg/m^3
  • jetexitvelocity(Vj)=201.0m/sjet exit velocity (V_j) = 201.0 m/s
  • vehiclevelocity(V0)=42.0000m/svehicle velocity (V_0) = 42.0000 m/s
  • thrust(F)=445,170Nthrust (F) = 445,170 N

Find

jet flow rate (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except Q is given, so isolate Q 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 839.0 kg/m^3, jet exit velocity (V_j) = 201.0 m/s, vehicle velocity (V_0) = 42.0000 m/s, thrust (F) = 445,170 N.

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

  5. Step 5 — Evaluate:

    Q = 3.3371\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 3.3371 m^3/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 3.3371\ \text{m^3/s}

Why the other options are there

  • 6.6742 — kept a factor of two that cancels in the correct rearrangement.
  • 1.6685 — dropped that same factor in the other direction.
  • 3.6708 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 9
Jet propulsion thrust — solve for jet exit velocity (case 3) — Jet Propulsion (9)

jet propulsion thrust from an aircraft engine exhaust Given fluid density (rho) = 911.0 kg/m^3; jet flow rate (Q) = 1.8100 m^3/s; vehicle velocity (V_0) = 19.0000 m/s; thrust (F) = 142,390 N, determine the jet exit velocity (V_j) in m/s.

Given

  • fluiddensity(rho)=911.0kg/m3fluid density (rho) = 911.0 kg/m^3
  • jetflowrate(Q)=1.8100m3/sjet flow rate (Q) = 1.8100 m^3/s
  • vehiclevelocity(V0)=19.0000m/svehicle velocity (V_0) = 19.0000 m/s
  • thrust(F)=142,390Nthrust (F) = 142,390 N

Find

jet exit velocity (V_j), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except V_j is given, so isolate V_j 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that V_j stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 911.0 kg/m^3, jet flow rate (Q) = 1.8100 m^3/s, vehicle velocity (V_0) = 19.0000 m/s, thrust (F) = 142,390 N.

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

  5. Step 5 — Evaluate:

    Vj=105.4 m/sV_{j} = 105.4\ \text{m/s}
  6. Step 6 — Check: returning V_j = 105.4 m/s to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
Vj=105.4 m/sV_{j} = 105.4\ \text{m/s}

Why the other options are there

  • 210.7 — kept a factor of two that cancels in the correct rearrangement.
  • 52.6770 — dropped that same factor in the other direction.
  • 115.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

Example 10
Jet propulsion thrust — solve for thrust (case 4) — Jet Propulsion (10)

jet propulsion thrust of a water jet boat Given fluid density (rho) = 1,024 kg/m^3; jet flow rate (Q) = 0.1300 m^3/s; jet exit velocity (V_j) = 220.0 m/s; vehicle velocity (V_0) = 3.0000 m/s, determine the thrust (F) in N.

Given

  • fluiddensity(rho)=1,024kg/m3fluid density (rho) = 1,024 kg/m^3
  • jetflowrate(Q)=0.1300m3/sjet flow rate (Q) = 0.1300 m^3/s
  • jetexitvelocity(Vj)=220.0m/sjet exit velocity (V_j) = 220.0 m/s
  • vehiclevelocity(V0)=3.0000m/svehicle velocity (V_0) = 3.0000 m/s

Find

thrust (F), in N

Start with the thinking

  • The governing relation printed in this handbook section is Jet propulsion thrust.
  • Everything except F is given, so isolate F 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.
  • Jet propulsion thrust results from momentum change of fluid ejected relative to the vehicle.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3 — List the givens: fluid density (rho) = 1,024 kg/m^3, jet flow rate (Q) = 0.1300 m^3/s, jet exit velocity (V_j) = 220.0 m/s, vehicle velocity (V_0) = 3.0000 m/s.

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

  5. Step 5 — Evaluate:

    F=28887 NF = 28887\ \text{N}
  6. Step 6 — Check: returning F = 28,887 N to

    F=ρQ(Vj−V0)F = \rho Q (V_j - V_0)

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=28887 NF = 28887\ \text{N}

Why the other options are there

  • 57,774 — kept a factor of two that cancels in the correct rearrangement.
  • 14,444 — dropped that same factor in the other direction.
  • 31,776 — rounded an intermediate value before the final step.

Reference: FE Handbook — Jet Propulsion

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