Week 8 Cheatsheet — Fluid Properties, Pressure, Buoyancy

How this week breaks down

Three pillars, all built on a single force balance.

Topic	What you do
Fluid properties	Compute $\rho$, $\gamma$, $SG$. Apply Newton’s law of viscosity over a thin gap.
Hydrostatic pressure	Force balance on a column $\Rightarrow$ $\Delta p=\rho g\Delta h$. Walk paths through manometers.
Buoyancy	$F_B=\rho gV$ with displaced-fluid density. Sink / float / neutral by comparing to weight.

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1 — Fluid properties

Definitions and conversions

Quantity	Symbol	Formula	SI units
Density	$\rho$	$m/V$	$kg/{\mathrm{m}}^3$
Specific weight	$\gamma$	$\rho g$	$N/{\mathrm{m}}^3$
Specific gravity	$SG$	$\rho /{\rho }_{\text{water}}$	dimensionless
Dynamic viscosity	$\mu$	(constitutive)	$\mathrm{Pa}\cdot \mathrm{s}$
Kinematic viscosity	$\nu$	$\mu /\rho$	${\mathrm{m}}^2/s$

Reference values

Fluid / quantity	Value
${\rho }_{\text{water}}$	$1000kg/{\mathrm{m}}^3$
${\rho }_{\text{air}}$	$\approx 1.3kg/{\mathrm{m}}^3$
${\rho }_{\text{mercury}}$	$13550\text{–}13600kg/{\mathrm{m}}^3$ ($SG=13.55$–$13.6$)
$g$	$9.8m/{\mathrm{s}}^2$ (lecture) / $9.81$ (some tut. answers)
Standard atmosphere	$101.325\mathrm{kPa}$

Viscosity recipe (thin-gap bearing)

1. Velocity gradient $\frac{\Delta u}{\Delta y}$ across the gap (linear profile assumed).
2. Shear stress $\tau =\mu \frac{\Delta u}{\Delta y}$.
3. Drag force on the moving plate $V=\tau A$.
4. Power dissipated $P=V\cdot u$ (force $\times$ velocity).

For a concentric-cylinder bearing, $\Delta u=\omega r_2$ and $A=2\pi r_2\ell$.

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2 — Hydrostatic pressure and manometers

Core results

Concept	Formula
Pressure	$P=F/A$ in $\mathrm{Pa}=N/{\mathrm{m}}^2$
Pascal’s law	$P$ isotropic at a point
Hydrostatic increment	$\Delta p=\rho g\Delta h$
Linear depth law	$p(z)=\rho gz+p_0$
Absolute pressure	$p_{\text{abs}}=p_{\text{atm}}+p_{\text{gauge}}$
Gauge pressure	$p_{\text{gauge}}=\rho gz$

Two-point rule

Situation	Pressure relation
Same fluid, same depth, continuous path	$p_A=p_B$
Crossing a fluid–fluid interface	Walk the path, $\rho g\Delta h$ per segment
Going down	$p$ increases (add $\rho g\Delta h$)
Going up	$p$ decreases (subtract $\rho g\Delta h$)

Manometer types

Device	Use
Barometer	Vertical column with vacuum on top, reads atmosphere
Piezometer	Open vertical tap into a pressurised line
U-tube manometer	Heavy manometer fluid (often Hg) for larger pressures
Differential manometer	Pressure difference between two containers
Inclined-tube manometer	Amplify resolution; remember $\Delta h=L\sin \theta$

Path-walking recipe

1. Start at a known pressure (atmosphere, a labelled point).
2. Move segment-by-segment toward the unknown.
3. Down $\to$ add $\rho g\Delta h$.
4. Up $\to$ subtract $\rho g\Delta h$.
5. Use the local fluid’s density for each segment.
6. Air columns are usually negligible — only a few Pa per $\mathrm{m}$.

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3 — Buoyancy (Archimedes)

Result	Formula
Buoyant force	$F_B={\rho }_{\text{fluid}}g{V}_{\text{displaced}}$
Fully submerged	$V_{\text{displaced}}=V_{\text{object}}$
Floating	$V_{\text{displaced}}=V_{\text{submerged}}$ only
Net force submerged	$F_{\text{net}}=W-F_B$

Sink / float / neutral

Comparison	Behaviour
${\rho }_{\text{object}}>{\rho }_{\text{fluid}}$	sinks; needs cable tension $F_T=W-F_B$
${\rho }_{\text{object}}<{\rho }_{\text{fluid}}$	floats; submerged fraction $={\rho }_{\text{object}}/{\rho }_{\text{fluid}}$
${\rho }_{\text{object}}={\rho }_{\text{fluid}}$	neutral — hovers anywhere

Hydrometer-style problems use the same idea: total mass of the device equals mass of displaced fluid: $m={\rho }_{\text{fluid}}{V}_{\text{disp}}$.

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Worked snippets (one per topic)

Problem	Setup	Result
Viscous drag plate ($\mu =0.89\times 10^{-3}$, gap $5\mathrm{mm}$, $A=0.5{\mathrm{m}}^2$, $u=0.25m/s$)	$\tau =\mu \Delta u/\Delta y$, then $V=\tau A$, then $m=V/g$	$\tau =0.0445\mathrm{Pa}$, $V=0.0223\mathrm{N}$, $m\approx 2.3\mathrm{g}$
Pool with trapezoidal cross-section ($4\times 8$, depths $1\mathrm{m}$ and $3\mathrm{m}$)	$V=$ trap area $\times$ length, $m=\rho V$	$V=64{\mathrm{m}}^3$, $m=64000\mathrm{kg}$
Oil ($SG=0.9$, $0.5\mathrm{m}$) over water ($1.2\mathrm{m}$) — gauge pressure top and bottom of water	$p_A={\rho }_{\text{oil}}g(0.5)$; $p_B=p_A+{\rho }_{w}g(1.2)$	$p_A=4410\mathrm{Pa}$, $p_B=16170\mathrm{Pa}$
Submerged concrete block ($V=0.048{\mathrm{m}}^3$, ${\rho }_c=2300$, ${\rho }_{\text{sea}}=1025$)	$F_T=({\rho }_c-{\rho }_{\text{sea}})Vg$	$F_T=600\mathrm{N}$ (vs $1083\mathrm{N}$ in air)

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Common mistakes

1. Sign of $\Delta h$. Down adds, up subtracts. Pick a direction, stay consistent.
2. Same-fluid-same-depth-same-pressure is only valid on a continuous path inside one fluid. Crossing an interface breaks the shortcut — walk the path.
3. Plugging $SG$ into $\rho gh$. $SG$ is dimensionless. Convert to $\rho =SG\cdot {\rho }_{\text{water}}$ first.
4. Confusing absolute vs gauge. Negative gauge pressure is allowed (vacuum). Read which one the question wants.
5. Air columns. Almost always negligible (a few Pa over a metre) — but drop them only when explicitly safe to.
6. Viscosity gap geometry. $\Delta u$ is the velocity difference across the gap, $\Delta y$ is the gap thickness. Not the plate length or thickness.
7. Buoyancy uses displaced fluid density — not the object’s. For a floating body, use the submerged volume.
8. Inclined manometers. Convert inclined length to vertical drop with $\sin \theta$ before $\rho gh$.
9. $g$ value mismatch. Lecture uses $9.8$, some tutorial answers use $9.81$. Stay consistent within a problem.

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Key formulas

$$\rho =\frac{m}{V},\gamma =\rho g,SG=\frac{\rho }{{\rho }_{\text{water}}},\nu =\frac{\mu }{\rho }$$ $$\tau =\mu \frac{du}{dy},V_{\text{shear}}=\tau A$$ $$\Delta p=\rho g\Delta h,p_{\text{abs}}=p_{\text{atm}}+p_{\text{gauge}}$$ $$F_B={\rho }_{\text{fluid}}g{V}_{\text{displaced}}$$

For derivations, the why, and a worked exam-style example, see the in-depth note.