Fluid Mechanics Oral Exam Notes
Week 1 & 2
Question 2.34
Question:
A closed tank is filled with water and is 1.5 m long. The pressure gauge on the tank reads 49
kPa. Determine: (a) the height h in the open water column, (b) the gauge pressure acting on
the bottom tank surface AB, and (c) the absolute pressure of the air in the top of the tank if
the local atmospheric pressure is 101.3 kPa.
Assumptions:
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- The pressure at the air–water interface equals the gauge pressure (49 kPa)
- The fluid is incompressible and static
- Water density is 1000 kg/m³
- Gravitational acceleration is 9.8 m/s²
- The tank and tube are in pressure equilibrium
Spoken-style Answer:
(a) I use pressure balance across the bottom of the U-tube. The pressure at the bottom of
the tank is 49 kPa plus 0.6 m of water depth, giving 54.88 kPa. I equate that to rho g h to find
the open column height. That gives h = 54,880 / (1000 × 9.8) = 5.6 meters.
(b) To find the pressure at the bottom AB, I add 1.2 m of water depth to the 49 kPa gauge
pressure. That gives 49,000 + 1000 × 9.8 × 1.2 = 60.76 kPa or 60.8 kPa.
(c) For absolute pressure, I add gauge and atmospheric pressure: 49 + 101.3 = 150.3 kPa.
Key Concepts:
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- Hydrostatic pressure balance
- Gauge vs absolute pressure
- Continuity of pressure in connected fluids at the same elevation
Question 2.40
Question:
Two pipes are connected by a manometer as shown. Determine the pressure difference P_A
minus P_B between the pipes.
Assumptions:
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- Fluids are static and incompressible
- Gauge fluid has SG = 2.6, so its density is 2600 kg/m³
- Water density is 1000 kg/m³
- Gravitational acceleration is 9.8 m/s²
- Manometer is ideal with no capillarity or flow
Spoken-style Answer:
To find P_A minus P_B, I move from A to B through the manometer, adding or subtracting
hydrostatic pressure changes. Going down 1.1 meters of water, then down 0.6 meters of
dense fluid, and finally up 0.8 meters of water. So, P_A minus P_B equals -9.8×10³ times (1.1
+ 0.8) plus 2.6×10³ times 9.8 times 0.6. That gives -3316 Pascals or -3.3 kilopascals.
Key Concepts:
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- Manometer pressure balancing
- Hydrostatic pressure adds when going down, subtracts when going up
- Use SG to determine fluid density
Week 3
Question 2.47
Question:
Determine the elevation difference Δh between the water levels in the two open tanks,
where a U-tube contains a fluid with specific gravity 0.90 and one leg has a 0.4 m height
difference.
Assumptions:
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- Tanks are open to atmosphere (p1 = p2 = 0 gauge)
- Fluid is static and incompressible
- Hydrostatic pressure applies
- Specific gravity relates to water: SG = 0.90 means fluid is 90% as dense as water
HD-Level Oral Answer:
To determine the elevation difference, I apply pressure continuity between the two tanks.
Both are open to atmosphere, so surface pressures cancel out. Starting from the left tank, I
go down 1 meter in water, up 0.4 meters in the SG = 0.9 fluid, and up Δh meters in water on
the right. Using hydrostatics, I write: γ_water × 1 = γ_fluid × 0.4 + γ_water × Δh. Replacing
γ_fluid with 0.9 γ_water, and simplifying, I get: 1 = 0.9 × 0.4 + Δh → Δh = 0.04 meters. So the
water on the right side is 4 centimeters lower than the left.
Key Concepts:
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- Hydrostatic pressure balance
- Specific gravity to density conversion
- Same depth = same pressure in static fluids
Question 2.122
Question:
A quarter-cylinder gate holds water to a depth of 4 m. Determine the weight per meter of
length of the gate if it pivots when the water just reaches 4 m.
Assumptions:
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- Water is incompressible and at rest
- Gate is homogeneous and weight is evenly distributed
- Forces act at centroids and centres of pressure
- No friction at the hinge
HD-Level Oral Answer:
To determine the weight of the gate per meter length, I consider the hydrostatic forces
acting on the curved surface. I calculate the vertical force by taking the weight of the
imaginary volume of fluid directly above the gate, including the quarter cylinder. Next, I
calculate the horizontal force using pressure at the centroid of the projected vertical area.
Using the moment equation about the hinge, I balance the moment from the gate’s own
weight against the moments from hydrostatic vertical and horizontal forces. From this, I
solve for the gate’s weight and find that it must be 64.4 kilonewtons per meter of length to
keep the gate just on the verge of opening at 4 meters of water depth.
Key Concepts:
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- Hydrostatic vertical force from imaginary fluid above the gate
- Horizontal force from pressure on projected area
- Moment balance about a hinge
- Composite centroid and centre of pressure calculation
Question 2.121
Question:
A conical plug in the bottom of a pressurised tank is held in place by fluid and air pressure.
The air pressure is 40 kPa, and the liquid has a specific weight of 27 kN/m³. Find the
magnitude, direction, and line of action of the total force on the cone.
Assumptions:
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- Air pressure acts uniformly at the liquid surface and transfers downwards
- Fluid is static and incompressible
- Cone is symmetrical, so resultant force acts along the cone's central axis
HD-Level Oral Answer:
To find the total force on the cone, I add the pressure force from the air, which acts on the
horizontal circular top of the cone, and the weight of the liquid directly above it. I subtract
the internal pressure pushing up inside the cone from the liquid below. The cone geometry
gives me a base area and volume based on the 60-degree angle, allowing me to calculate the
force from each contribution. Adding them up gives a resultant force of 117 kilonewtons
acting vertically down through the axis of the cone.
Key Concepts:
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- Force = pressure × area + fluid weight
- Use of cone geometry (tan relationships)
- Resultant acts vertically through cone centerline due to symmetry
Question 4.2
Question:
Given the velocity field V = (3y + 2)i + (x – 8)j + 5zk, determine the fluid speed at the origin
and along the y-axis (where x = z = 0).
Assumptions:
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- Steady 3D velocity field
- Fluid is ideal and continuous
- Velocity components are functions of position
HD-Level Oral Answer:
At the origin, I substitute x = y = z = 0 into the velocity field. That gives u = 2, v = -8, and w =
0. The speed is the square root of the sum of squares: speed = square root of 2 squared plus
8 squared, which is 8.25 meters per second. Along the y-axis, I set x and z to 0 and simplify
the expressions. The general formula for speed becomes square root of (3y + 2) squared
plus 64. So the speed depends on y and follows the expression: square root of (9y² + 12y +
68).
Key Concepts:
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- Vector magnitude of velocity field
- Evaluation of functions at specific points
- Understanding 3D flow properties
Question 4.4
Question:
Given a 2D velocity field where u = 1 + y and v = 1, determine the equation of the streamline
passing through the origin and sketch it.
Assumptions:
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- 2D, steady flow
- Streamlines are tangent to the velocity field at all points
HD-Level Oral Answer:
To find the streamline, I use the differential equation dy/dx = v/u. Substituting u and v gives
dy/dx = 1 / (1 + y). Separating variables gives (1 + y) dy = dx. Integrating both sides, I get y
plus half y squared equals x plus a constant. To find the specific streamline through the
origin, I use the point (0, 0), which gives the constant as zero. So the streamline equation is x
= y + 0.5 y squared.
Key Concepts:
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- Streamline equation from dy/dx = v/u
- Integration and applying initial conditions
- Streamlines represent paths of fluid particles
Question 4.15
Question:
A swirling velocity field is given by u = -Ky / (x² + y²), v = Kx / (x² + y²). Show that (a) the
velocity magnitude is inversely proportional to the distance from the origin, and (b) that the
streamlines are circles.
Assumptions:
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- Flow is steady and planar
- K is a constant
- Swirl is symmetric about the origin
HD-Level Oral Answer:
For part (a), I calculate the magnitude of velocity using square root of u squared plus v
squared. Substituting the expressions and simplifying gives V = K divided by the square root
of x squared plus y squared, which means velocity is inversely proportional to radial
distance. For part (b), I find the streamline by taking dy/dx = v/u. Substituting gives dy/dx =
-x/y. Separating variables and integrating, I get y dy = -x dx, which integrates to x squared
plus y squared equals a constant. So streamlines are circular.
Key Concepts:
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- Polar velocity behavior
- Velocity magnitude from vector components
- Circular streamline derivation
Week 4
Question 4.1
Question:
Given a 2D velocity field where u = x² + y and v = 4x − y², find the acceleration field at any
point.
Assumptions:
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- Flow is steady and 2D
- Fluid properties like density are constant
- Velocity components are functions of position
- Material derivative applies: includes convective acceleration
HD-Level Oral Answer:
To find the acceleration field, I take the material derivative of velocity, which gives total
acceleration. Since the flow is steady, I ignore the time derivatives. I calculate the xcomponent of acceleration as u times du/dx plus v times du/dy. Similarly, for the ycomponent, I take u times dv/dx plus v times dv/dy. After computing partial derivatives and
substituting, I get the acceleration vector as: [u·2x + v] in the x-direction and [u·4 − 2y·v] in
the y-direction.
Key Concepts:
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- Material derivative = local + convective acceleration
- Use chain rule with partial derivatives
- Steady flow → local derivative is zero