Fluid Mechanics Practice

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1.1Relate Density, Specific Weight, and Viscosity

Exam I | Problem 1.1 |

A fluid has density $1000\ \text{kg/m}^3$ and dynamic viscosity $1.0\times 10^{-3}\ \text{Pa}\cdot\text{s}$.

Find:

its specific weight $\gamma$
its specific volume $v$
its kinematic viscosity $\nu$

Also state whether this fluid is usually modeled as incompressible and Newtonian in ordinary pipe flow.

Solution

Use the definitions:

\gamma = \rho g = 1000(9.81) = 9810\ \text{N/m}^3

v = \frac{1}{\rho} = \frac{1}{1000} = 1.0\times 10^{-3}\ \text{m}^3/\text{kg}

\nu = \frac{\mu}{\rho} = \frac{1.0\times 10^{-3}}{1000} = 1.0\times 10^{-6}\ \text{m}^2/\text{s}

This fluid is commonly modeled as incompressible and Newtonian in ordinary pipe flow.

1.2Classify Flow by Reynolds Number

Exam I | Problem 1.2 |

Oil flows through a circular pipe with

\rho = 850\ \text{kg/m}^3,\quad \mu = 0.17\ \text{Pa}\cdot\text{s},\quad V = 0.40\ \text{m/s},\quad D = 0.020\ \text{m}.

Find the Reynolds number and classify the flow as laminar, transitional, or turbulent.

Solution

Compute the Reynolds number:

\mathrm{Re} = \frac{\rho V D}{\mu} = \frac{850(0.40)(0.020)}{0.17} = 40

Since $\mathrm{Re} \ll 2300$, the flow is laminar.

1.3Find Pressure at a Depth

Exam I | Problem 1.3 |

What is the gauge pressure $6$ m below the free surface in fresh water?

Use $\rho = 1000\ \text{kg/m}^3$ and $g = 9.81\ \text{m/s}^2$.

Solution

For a fluid at rest, pressure increases with depth:

p = \rho g h

p = 1000(9.81)(6) = 58860\ \text{Pa}

p \approx 58.9\ \text{kPa}

1.4Convert Gauge Pressure to Pressure Head

Exam I | Problem 1.4 |

Pressure Head

A gauge reads $150\ \text{kPa}$ in water.

What pressure head does this correspond to?

Solution

Pressure head is

\frac{p}{\gamma}

For water, $\gamma = \rho g = 1000(9.81) = 9810\ \text{N/m}^3$.

Thus

\frac{p}{\gamma} = \frac{150000}{9810} \approx 15.3\ \text{m}

So the pressure head is about $15.3$ m of water.

1.5Compute Buoyant Force from Displaced Volume

Exam I | Problem 1.5 |

A submerged object displaces $0.030\ \text{m}^3$ of water.

What buoyant force acts on it?

Solution

By Archimedes' principle, the buoyant force equals the weight of the displaced fluid:

F_B = \rho g V_{\text{disp}}

F_B = 1000(9.81)(0.030) = 294.3\ \text{N}

1.6Use Continuity in a Converging Pipe

Exam I | Problem 1.6 |

Water flows through a $60$ mm pipe at $3.0\ \text{m/s}$ and then enters a $30$ mm nozzle.

Find the volumetric flow rate and the nozzle exit speed.

Solution

First compute the flow rate at the larger section:

A_1 = \frac{\pi}{4}(0.060)^2

Q = A_1V_1 = \frac{\pi}{4}(0.060)^2(3.0) \approx 8.48\times 10^{-3}\ \text{m}^3/\text{s}

Now use continuity at the nozzle:

A_2 = \frac{\pi}{4}(0.030)^2

V_2 = \frac{Q}{A_2} = \frac{8.48\times 10^{-3}}{\pi(0.030)^2/4} = 12\ \text{m/s}

1.7Apply Bernoulli at the Same Elevation

Exam I | Problem 1.7 |

In a horizontal water line, the speed increases from $2\ \text{m/s}$ to $6\ \text{m/s}$.

What is the pressure drop between the two points?

Solution

For steady, incompressible, inviscid flow at the same elevation,

\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g}

p_1 - p_2 = \frac{1}{2}\rho\left(V_2^2 - V_1^2\right)

Substitute the values:

p_1 - p_2 = \frac{1}{2}(1000)(6^2 - 2^2)

p_1 - p_2 = 500(36 - 4) = 16000\ \text{Pa}

So the pressure drops by $16\ \text{kPa}$.

1.8Find the Darcy Friction Factor for Laminar Flow

Exam I | Problem 1.8 |

A pipe flow has Reynolds number $1600$.

Assuming the flow is laminar, what is the Darcy friction factor?

Solution

For laminar pipe flow,

f = \frac{64}{\mathrm{Re}}

Therefore

f = \frac{64}{1600} = 0.04

1.9Compute Stagnation Pressure

Exam I | Problem 1.9 |

Air flows with static pressure $101\ \text{kPa}$, density $1.2\ \text{kg/m}^3$, and speed $30\ \text{m/s}$.

What is the stagnation pressure?

Solution

Use the stagnation pressure relation:

p_0 = p + \frac{1}{2}\rho V^2

Substitute the values:

p_0 = 101000 + \frac{1}{2}(1.2)(30^2)

p_0 = 101000 + 540 = 101540\ \text{Pa}

p_0 \approx 101.5\ \text{kPa}

1.10Check for Compressibility from Mach Number

Exam I | Problem 1.10 |

Air moves at $120\ \text{m/s}$ and the local speed of sound is $340\ \text{m/s}$.

Find the Mach number and say whether compressibility effects may matter.

Solution

Compute the Mach number:

\mathrm{Ma} = \frac{V}{a} = \frac{120}{340} \approx 0.35

Since this is greater than about $0.3$, compressibility effects may matter.

Difficulty

Integrated

2.1Read a Two-Fluid Manometer

Exam II | Problem 2.1 |

A tank gas is connected to a U-tube manometer. On the tank side, $5$ cm of water sits above mercury. The mercury level on the open side is $15$ cm higher than the mercury-water interface on the tank side. The open side is at atmospheric pressure.

Find the gas gauge pressure in the tank.

Solution

Start at the open surface, where the pressure is atmospheric.

Move down $0.15$ m in mercury:

p = p_{\text{atm}} + \rho_{\text{Hg}} g(0.15)

Then move up $0.05$ m in water to the tank gas:

p_{\text{tank}} = p_{\text{atm}} + \rho_{\text{Hg}} g(0.15) - \rho_w g(0.05)

So the gauge pressure is

p_g = \rho_{\text{Hg}} g(0.15) - \rho_w g(0.05)

Using $\rho_{\text{Hg}} = 13600\ \text{kg/m}^3$ and $\rho_w = 1000\ \text{kg/m}^3$:

$$ p_g = 13600(9.81)(0.15) - 1000(9.81)(0.05) $$

p_g \approx 19.5\ \text{kPa}

2.2Resultant Force on a Submerged Gate

Exam II | Problem 2.2 |

A vertical rectangular gate is $2$ m high and $1$ m wide. Its top edge is $3$ m below the free surface in water.

Find the resultant hydrostatic force and the depth of the center of pressure below the free surface.

Solution

The centroid of the gate is at a depth

h_c = 3 + 1 = 4\ \text{m}

The area is

A = 2(1) = 2\ \text{m}^2

So the resultant force is

F_R = \rho g h_c A = 1000(9.81)(4)(2) = 78480\ \text{N}

F_R \approx 78.5\ \text{kN}

For the center of pressure,

h_p = h_c + \frac{I_G}{h_cA}

For a rectangle,

I_G = \frac{bh^3}{12} = \frac{1(2^3)}{12} = \frac{8}{12} = 0.667\ \text{m}^4

Thus

h_p = 4 + \frac{0.667}{(4)(2)} \approx 4.08\ \text{m}

2.3Match Mass Flow Rates Across a Nozzle

Exam II | Problem 2.3 |

Water enters a $60$ mm pipe at $3.0\ \text{m/s}$ and leaves through a $30$ mm nozzle.

Find the nozzle exit speed and the mass flow rate.

Solution

Compute the inlet area:

A_1 = \frac{\pi}{4}(0.060)^2

Then the flow rate is

Q = A_1V_1 = \frac{\pi}{4}(0.060)^2(3.0) \approx 8.48\times 10^{-3}\ \text{m}^3/\text{s}

At the nozzle,

A_2 = \frac{\pi}{4}(0.030)^2

V_2 = \frac{Q}{A_2} = 12\ \text{m/s}

The mass flow rate is

\dot{m} = \rho Q = 1000(8.48\times 10^{-3}) \approx 8.48\ \text{kg/s}

2.4Find Pump Head in an Extended Bernoulli Balance

Exam II | Problem 2.4 |

Water is pumped from one large open reservoir to another large open reservoir that is $8$ m higher.

The system loses $3$ m of head along the way.

What pump head is required if the free-surface velocities are negligible?

Solution

Between the two reservoir surfaces, the pressure terms cancel and the velocities are negligible. The pump must supply the elevation gain plus the losses:

h_p = 8 + 3 = 11\ \text{m}

2.5Compute the Force on a 90-Degree Elbow

Exam II | Problem 2.5 |

A $0.10$ m diameter pipe carries water at $4.0\ \text{m/s}$ through a 90-degree elbow.

Neglect pressure forces and weight.

What is the magnitude of the force the fluid exerts on the elbow?

Solution

First find the mass flow rate:

A = \frac{\pi}{4}(0.10)^2

\dot{m} = \rho AV = 1000\left(\frac{\pi}{4}(0.10)^2\right)(4.0) \approx 31.4\ \text{kg/s}

The inlet velocity is along one axis and the outlet velocity is along a perpendicular axis, both with speed $4.0\ \text{m/s}$.

So the momentum change has components of magnitude

\dot{m}V = 31.4(4.0) = 125.6\ \text{N}

The resultant force magnitude is

F = \sqrt{125.6^2 + 125.6^2} = 125.6\sqrt{2} \approx 177.6\ \text{N}

So the fluid exerts a force of about $178\ \text{N}$ on the elbow.

2.6Scale a Free-Surface Model

Exam II | Problem 2.6 |

A spillway model is built at $1:36$ scale using the same fluid as the prototype.

If the prototype surface speed is $6\ \text{m/s}$, what model speed gives Froude similarity?

Solution

For free-surface flow, match the Froude number:

\mathrm{Fr} = \frac{V}{\sqrt{gL}}

\frac{V_m}{V_p} = \sqrt{\frac{L_m}{L_p}} = \sqrt{\frac{1}{36}} = \frac{1}{6}

Therefore

V_m = 6\left(\frac{1}{6}\right) = 1\ \text{m/s}

Exact Reynolds similarity is not possible at this scale with the same fluid, so Froude similarity is the governing choice.

2.7Combine Major and Minor Losses

Exam II | Problem 2.7 |

Water flows through a $40$ m long pipe of diameter $50$ mm at $2.0\ \text{m/s}$.

The Darcy friction factor is $0.03$, and the total minor-loss coefficient is $2.0$.

Find the total head loss and the equivalent pressure drop.

Solution

First compute the velocity head:

\frac{V^2}{2g} = \frac{2.0^2}{2(9.81)} \approx 0.204\ \text{m}

Major loss:

h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.03\left(\frac{40}{0.05}\right)(0.204)

h_f \approx 4.89\ \text{m}

Minor loss:

h_m = K\frac{V^2}{2g} = 2.0(0.204) \approx 0.41\ \text{m}

Total head loss:

h_L = h_f + h_m \approx 5.30\ \text{m}

Equivalent pressure drop:

\Delta p = \rho g h_L = 1000(9.81)(5.30) \approx 5.20\times 10^4\ \text{Pa}

So the pressure drop is about $52\ \text{kPa}$.

2.8Calculate Drag from a Coefficient

Exam II | Problem 2.8 |

A bluff body in air has reference area $2.0\ \text{m}^2$ and drag coefficient $C_D = 1.2$.

The air density is $1.2\ \text{kg/m}^3$ and the speed is $25\ \text{m/s}$.

What drag force acts on the body?

Solution

Use the drag-coefficient relation:

C_D = \frac{D}{\tfrac{1}{2}\rho V^2 A}

D = \frac{1}{2}\rho V^2 A C_D

Substitute the values:

D = \frac{1}{2}(1.2)(25^2)(2.0)(1.2)

D = 900\ \text{N}

Difficulty

Applied

3.1Size the Inlet Pressure for a Rising Pipe

Final | Problem 3.1 |

Water flows at $0.015\ \text{m}^3/\text{s}$ through an $80$ m long, $0.10$ m diameter pipe that rises $5$ m and discharges to atmosphere at the same diameter.

Take the Darcy friction factor as $0.02$ and neglect minor losses.

What inlet gauge pressure is needed?

Solution

First find the pipe speed:

A = \frac{\pi}{4}(0.10)^2

V = \frac{Q}{A} = \frac{0.015}{\pi(0.10)^2/4} \approx 1.91\ \text{m/s}

The velocity head is

\frac{V^2}{2g} \approx \frac{1.91^2}{2(9.81)} \approx 0.186\ \text{m}

The friction loss is

h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02\left(\frac{80}{0.10}\right)(0.186)

h_f \approx 2.97\ \text{m}

Because the pipe diameter is the same at inlet and outlet, the velocity heads cancel. The inlet gauge pressure must supply the elevation rise plus the friction loss:

\frac{p_{in}}{\gamma} = 5 + 2.97 = 7.97\ \text{m}

Thus

p_{in} = 1000(9.81)(7.97) \approx 7.82\times 10^4\ \text{Pa}

So the required inlet gauge pressure is about $78.2\ \text{kPa}$.

3.2Find the Force of a Jet on a Flat Plate

Final | Problem 3.2 |

A $50$ mm diameter water jet moves at $20\ \text{m/s}$ and is brought to rest in the jet direction by a flat plate.

What average force does the jet exert on the plate?

Solution

The jet area is

A = \frac{\pi}{4}(0.050)^2

The mass flow rate is

\dot{m} = \rho AV = 1000\left(\frac{\pi}{4}(0.050)^2\right)(20) \approx 39.3\ \text{kg/s}

The change in velocity in the jet direction is from $20\ \text{m/s}$ to $0$, so the force magnitude is

F = \dot{m}V \approx 39.3(20) \approx 785\ \text{N}

So the jet exerts about $785\ \text{N}$ on the plate.

3.3Choose the Right Similarity Condition for a Spillway Model

Final | Problem 3.3 |

A spillway model is built at $1:36$ scale in the same fluid as the prototype.

The prototype surface speed is $6\ \text{m/s}$.

Which similarity condition should be matched, and what model speed follows from that choice?

Solution

For a free-surface flow, the key similarity condition is Froude similarity because gravity effects control the wave pattern and surface profile.

\frac{V_m}{V_p} = \sqrt{\frac{L_m}{L_p}} = \sqrt{\frac{1}{36}} = \frac{1}{6}

Therefore

V_m = 6\left(\frac{1}{6}\right) = 1\ \text{m/s}

Reynolds similarity is not matched exactly at this scale, but Froude similarity is the dominant choice.

3.4Check Whether a Converging Nozzle Chokes

Final | Problem 3.4 |

Air flows from a large reservoir with stagnation pressure $500\ \text{kPa}$ and stagnation temperature $300\ \text{K}$ through a converging nozzle.

The back pressure is $200\ \text{kPa}$.

Determine whether the nozzle is choked. If it is, find the throat pressure and throat temperature.

Solution

For air, take $\gamma = 1.4$.

The critical pressure ratio for choking is

\left(\frac{p^*}{p_0}\right) = \left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)} = \left(\frac{2}{2.4}\right)^{3.5} \approx 0.528

The back-pressure ratio is

\frac{p_b}{p_0} = \frac{200}{500} = 0.40

Since $0.40 < 0.528$, the nozzle is choked.

At the throat, $\mathrm{Ma}=1$, so

p^* = 0.528(500\ \text{kPa}) \approx 264\ \text{kPa}

Also,

\frac{T_0}{T^*} = 1 + \frac{\gamma-1}{2}(1^2) = 1.2

T^* = \frac{300}{1.2} = 250\ \text{K}

3.5Estimate the Minimum Flight Speed for Level Lift

Final | Problem 3.5 |

A glider must support a weight of $5400\ \text{N}$ in level flight.

Its wing area is $16\ \text{m}^2$, the air density is $1.2\ \text{kg/m}^3$, and the lift coefficient is $0.8$.

Estimate the minimum speed for level flight.

Solution

For level flight, lift equals weight:

L = \frac{1}{2}\rho V^2 A C_L = W

Solve for $V$:

V = \sqrt{\frac{2W}{\rho A C_L}}

Substitute the values:

V = \sqrt{\frac{2(5400)}{(1.2)(16)(0.8)}}

V = \sqrt{703.125} \approx 26.5\ \text{m/s}

Difficulty

Challenge

4.1Balance a Multi-Fluid Manometer

Final | Problem 4.1 |

An open mercury surface is connected to a tank through a multi-fluid column. Starting at the open surface, move down $0.12$ m in mercury, then up $0.40$ m in water, then up $0.18$ m in oil with density $850\ \text{kg/m}^3$ to reach the tank gas.

What is the tank gauge pressure?

Solution

Follow the pressure changes along the path:

p_g = \rho_{\text{Hg}}g(0.12) - \rho_w g(0.40) - \rho_o g(0.18)

Using $\rho_{\text{Hg}} = 13600\ \text{kg/m}^3$, $\rho_w = 1000\ \text{kg/m}^3$, and $\rho_o = 850\ \text{kg/m}^3$:

$$ p_g = 13600(9.81)(0.12) - 1000(9.81)(0.40) - 850(9.81)(0.18) $$

p_g \approx 10.6\ \text{kPa}

4.2Size the Pump for a Reservoir-to-Reservoir Line

Final | Problem 4.2 |

Water is pumped from one large open reservoir to another large open reservoir that is $12$ m higher.

The line is $120$ m long, the pipe diameter is $0.08$ m, the Darcy friction factor is $0.025$, and the flow rate is $0.010\ \text{m}^3/\text{s}$.

Two 90-degree elbows have loss coefficients of $0.9$ each.

What pump head is required?

Solution

First find the pipe speed:

A = \frac{\pi}{4}(0.08)^2

V = \frac{Q}{A} = \frac{0.010}{\pi(0.08)^2/4} \approx 1.99\ \text{m/s}

Then compute the velocity head:

\frac{V^2}{2g} \approx \frac{1.99^2}{2(9.81)} \approx 0.202\ \text{m}

Major loss:

h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.025\left(\frac{120}{0.08}\right)(0.202) \approx 7.56\ \text{m}

Minor loss:

h_m = K\frac{V^2}{2g} = (0.9+0.9)(0.202) \approx 0.36\ \text{m}

Between the two reservoir surfaces, the pump must supply the elevation rise plus the losses:

h_p = 12 + 7.56 + 0.36 \approx 19.9\ \text{m}

4.3Match Reynolds Number in a Wind-Tunnel Model

Final | Problem 4.3 |

A car is tested in a wind tunnel with a $1:5$ scale model.

The prototype speed is $30\ \text{m/s}$, and the same air is used in both cases.

What model speed is needed to match Reynolds number?

Solution

For Reynolds similarity,

\mathrm{Re} \propto VL

$$ V_mL_m = V_pL_p $$

With a $1:5$ model, $L_m = L_p/5$, so

V_m = 5V_p = 5(30) = 150\ \text{m/s}

That speed is often impractical, which is why exact dynamic similarity is not always achievable in a wind tunnel.

4.4Use Critical Pressure Ratio to Test for Choking

Final | Problem 4.4 |

A converging nozzle receives air from a large reservoir at $p_0 = 500\ \text{kPa}$ and $T_0 = 300\ \text{K}$.

What is the largest back pressure that still allows choking? If the nozzle is choked, what are the throat pressure and throat temperature?

Solution

For air, take $\gamma = 1.4$.

The critical pressure ratio is

\frac{p^*}{p_0} = \left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)} \approx 0.528

So the largest back pressure that still allows choking is

p_{b,\text{crit}} = 0.528(500\ \text{kPa}) \approx 264\ \text{kPa}

At choking, the throat Mach number is $1$ and

p^* \approx 264\ \text{kPa}

Also,

\frac{T_0}{T^*} = 1 + \frac{\gamma-1}{2}(1^2) = 1.2

T^* = \frac{300}{1.2} = 250\ \text{K}

Fluid Mechanics Practice

Selected problems

Quick reference

Direct

1.1Relate Density, Specific Weight, and Viscosity

1.2Classify Flow by Reynolds Number

1.3Find Pressure at a Depth

1.4Convert Gauge Pressure to Pressure Head

1.5Compute Buoyant Force from Displaced Volume

1.6Use Continuity in a Converging Pipe

1.7Apply Bernoulli at the Same Elevation

1.8Find the Darcy Friction Factor for Laminar Flow

1.9Compute Stagnation Pressure

1.10Check for Compressibility from Mach Number

Integrated

2.1Read a Two-Fluid Manometer

2.2Resultant Force on a Submerged Gate

2.3Match Mass Flow Rates Across a Nozzle

2.4Find Pump Head in an Extended Bernoulli Balance

2.5Compute the Force on a 90-Degree Elbow

2.6Scale a Free-Surface Model

2.7Combine Major and Minor Losses

2.8Calculate Drag from a Coefficient

Applied

3.1Size the Inlet Pressure for a Rising Pipe

3.2Find the Force of a Jet on a Flat Plate

3.3Choose the Right Similarity Condition for a Spillway Model

3.4Check Whether a Converging Nozzle Chokes

3.5Estimate the Minimum Flight Speed for Level Lift

Challenge

4.1Balance a Multi-Fluid Manometer

4.2Size the Pump for a Reservoir-to-Reservoir Line

4.3Match Reynolds Number in a Wind-Tunnel Model

4.4Use Critical Pressure Ratio to Test for Choking