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Fluid Properties: Concepts & Notes

Hydraulics and Fluid Properties

Hydraulics is the branch of mechanics that studies the behavior of water and other liquids at rest or in motion. The usual branches are hydrostatics, hydrokinetics, and hydrodynamics.

The term fluid is used to denote matter in a liquid or gaseous phase.

Density (Unit Mass), ρ$$\rho=\frac{\text{mass}}{\text{volume}}=\frac{m}{V}$$
Unit Weight, γ$$\gamma=\frac{\text{weight}}{\text{volume}}=\frac{W}{V}=\rho g$$
Specific Gravity, SG$$SG=\frac{\rho_{fluid}}{\rho_{water}}$$
Dry Air / Common Gas$$\rho=\frac{p}{RT}$$
Ideal Gases$$\rho=\frac{pn}{RT}$$

Where: p = absolute pressure; n = number of moles (in chemistry); R = gas constant = 287.058 J/(kg·K) for dry air = 1,716 lb·ft/(slug·°R); and T = absolute temperature.

For board problems, the most frequent first move is converting a pressure into an equivalent head or converting a fluid head into pressure.

$$p = \gamma h$$ $$h = \frac{p}{\gamma}$$ $$\gamma = SG\,\gamma_w$$

Bulk Modulus, Compressibility, and Capillarity

These properties describe how fluids respond to pressure change and surface interaction.

$$E_v = \frac{\Delta p}{\Delta V/V}$$ $$\beta = \frac{1}{E_v}$$ $$c = \sqrt{\frac{E_v}{\rho}}$$ $$h = \frac{2\sigma \cos\theta}{\rho g r}$$
Use consistent SI units. In most CE board computations, $\gamma_w = 9.81 \text{ kN/m}^3$, $1 \text{ atm} \approx 101.3 \text{ kPa}$, and mercury has $SG \approx 13.6$.

Pressure Classifications

Pressure may be reported as absolute pressure, gage pressure, or vacuum pressure. Always identify which one is being used before substituting values.

$$p_{abs} = p_{atm} + p_{gage}$$ $$p_{gage} = p_{abs} - p_{atm}$$ $$p_{abs} = p_{atm} - p_{vac}$$ $$p_{vac} = p_{atm} - p_{abs}$$

A positive gage pressure means the pressure is above atmospheric pressure. A vacuum reading means the pressure is below atmospheric pressure.

Temperature, Relative Density, and Reference Values

$$\mathrm{K}={}^{\circ}\mathrm{C}+273$$ $${}^{\circ}\mathrm{R}={}^{\circ}\mathrm{F}+460$$ $$1\ \mathrm{atm}=101.325\ \mathrm{kPa}=14.7\ \mathrm{psi}=760\ \mathrm{mmHg}$$

Relative Density (RD)

$$RD=\frac{\text{density of substance}}{\text{density of pure water at standard temperature and pressure}}$$ $$RD=\frac{\text{density of substance }(\mathrm{kg/m^3})}{1000\ \mathrm{kg/m^3}}\quad\text{at }4^{\circ}\mathrm{C}$$

Density and specific weight of water at standard condition:

Density, ρwSpecific Weight, γw
1000 kg/m3
= 1.94 slug/ft3
= 1.0 gram/cm3		9810 N/m3
= 62.4 lb/ft3

Specific Gravity of Common Fluid
Note: The specific gravity of each substance is at 20°C and 1 atm unless stated otherwise.

SubstanceSpecific Gravity
Water (at 4°C)1.0
Seawater1.025
Gasoline0.68
Ethyl Alcohol0.790
Mercury13.60
Air0.001204

Surface Tension and Capillary Action

Surface tension, σ, is the force per unit length at a liquid-vapor or liquid-liquid interface resulting from the imbalance in attractive forces among like liquid molecules at the interface.

$$\sigma=\frac{\text{force along the interface}}{\text{length of the interface}}=\frac{\text{energy required to increase the surface area of the fluid}}{\text{unit area}}$$

See images:

Surface tension of water at any temperature

Capillary action, h. Liquids have both cohesion and adhesion, which are forms of molecular attraction. Capillarity, the rise or fall of liquid in small-diameter tubes, is due to this attraction.

$$\text{total surface tension}=\text{weight of capillary action}$$ $$h=\frac{4\sigma\cos\theta}{\gamma d}$$
List of Common Contact Angle, θ
LiquidSolidContact Angle, θ
WaterSoda-lime glass0°
Carbon tetrachlorideLead glass0°
Acetic acid (vinegar)Fused quartz0°
WaterParaffin wax107°
WaterSilver90°
MercurySoda-lime glass140°

Viscosity and Newton's Law of Viscosity

Viscosity is a property of a fluid that quantifies the ratio of shear stress to the rate of deformation (strain rate) of a fluid particle. It is a measure of resistance to gradual deformation.

$$\tau=\frac{F}{A}=\mu\frac{U}{y}$$

Where μ = coefficient of dynamic viscosity; τ = shear stress; and U/y = strain rate.

Fluid Properties: Problem Sets

Problem: Pressure Head from Gas Tank Pressure

The pressure in a gas tank is 2.75 atmospheres absolute. Compute the pressure head in meters of water.

Pressure head is the height of a fluid column that produces a given pressure, from $p = \gamma_w h$, so $h = \dfrac{p}{\gamma_w}$. A handy shortcut is to express one atmosphere directly as a column of water:

$$1 \text{ atm} = \frac{101.325}{9.81} \approx 10.33 \text{ m of water}$$

Absolute pressure head. Using the full 2.75 atm absolute:

$$h_{abs} = \frac{p_{abs}}{\gamma_w} = \frac{2.75(101.325)}{9.81}$$ $$h_{abs} = 2.75(10.33)$$ $$\boxed{h_{abs} = 28.4 \text{ m of water}}$$

Meaning: the gas pressure of 2.75 atmospheres is equivalent to the pressure at the bottom of a 28.4 m tall column of water.

Gage pressure head. If instead the head above atmospheric pressure is required, subtract one atmosphere first:

$$p_{gage} = p_{abs} - p_{atm} = (2.75 - 1)(101.325) = 177.32 \text{ kPa}$$ $$h_{gage} = \frac{p_{gage}}{\gamma_w} = \frac{177.32}{9.81} = (2.75 - 1)(10.33)$$ $$\boxed{h_{gage} = 18.1 \text{ m of water}}$$
If the problem states 2.75 atm absolute and asks for the pressure head directly, the answer is the absolute head, 28.4 m of water. If it asks for the head of the gage pressure (the amount above atmospheric), use 18.1 m of water. Read the question carefully to decide which datum applies.

Problem: Atmospheric Pressure as Water Column

If water were used to measure atmospheric pressure, determine the height of water column in meters.

$$p = \gamma_w h$$ $$101.3 = 9.81h$$ $$\boxed{h = 10.32 \text{ m}}$$

Problem: Variable Unit Weight

The unit weight of a liquid is variable and given by $\gamma = 10 + 0.5h$, where $\gamma$ is in $\text{kN/m}^3$ and $h$ is the depth from the free surface in meters. Determine the gage pressure at a depth of 5 m.

$$dp = \gamma\,dh$$ $$p = \int_0^5 (10 + 0.5h)\,dh$$ $$p = \left[10h + 0.25h^2\right]_0^5$$ $$p = 10(5) + 0.25(5)^2$$ $$\boxed{p = 56.25 \text{ kPa}}$$

Problem: Vacuum Reading in Mercury

A gage on the suction side of a pump shows a vacuum of 250 mm of mercury. Compute the pressure head in meters of water and the gage pressure in kPa.

$$h_w = -0.250(13.6) = -3.40 \text{ m of water}$$ $$p_{gage} = \gamma_w h_w = 9.81(-3.40) = -33.35 \text{ kPa}$$ $$p_{atm} = 101.356\left(\frac{725}{760}\right) = 96.69 \text{ kPa}$$ $$p_{abs} = p_{atm} + p_{gage} = 96.69 - 33.35$$ $$\boxed{p_{abs} = 63.33 \text{ kPa}}$$

Pressure head is -3.40 m of water and gage pressure is -33.35 kPa.

Problem: Pressure at the Bottom of a Layered Tank

A closed tank contains three stratified liquids stacked vertically. At the top is 0.60 m of oil with SG = 0.80. Below it is 0.90 m of fresh water. At the bottom is 0.30 m of a heavy liquid with SG = 1.60. The pressure at the very top (oil surface) is 20 kPa gage. Find the gage pressure in kPa at the bottom of the tank.

$$p_{bottom} = p_{top} + \gamma_{oil}h_{oil} + \gamma_w h_w + \gamma_{heavy}h_{heavy}$$ $$p_{bottom} = 20 + 9.81(0.80)(0.60) + 9.81(1.00)(0.90) + 9.81(1.60)(0.30)$$ $$p_{bottom} = 20 + 4.709 + 8.829 + 4.709$$ $$p_{bottom} = 38.25 \text{ kPa}$$

Answer: The gage pressure at the bottom of the tank is 38.25 kPa.

Problem: Specific Gravity by Submerged Weighing

An unknown solid object weighs 450 N in air. When fully submerged in fresh water, it weighs 300 N. When submerged in an unknown liquid, it weighs 270 N. Determine: (a) the volume and density of the object, and (b) the specific gravity of the unknown liquid.

$$BF_{water} = 450 - 300 = 150 \text{ N}$$ $$V = \frac{BF}{\gamma_w} = \frac{150}{9810} = 0.01529 \text{ m}^3$$ $$\rho_{obj} = \frac{W}{gV} = \frac{450}{9.81(0.01529)} = 3000 \text{ kg/m}^3 \Rightarrow SG_{obj} = 3.0$$
$$BF_{unknown} = 450 - 270 = 180 \text{ N}$$ $$\gamma_{unknown} = \frac{BF}{V} = \frac{180}{0.01529} = 11774 \text{ N/m}^3$$ $$SG_{unknown} = \frac{11774}{9810} = 1.20$$

Answer: The object has volume 0.01529 m³, density 3000 kg/m³ (SG = 3.0). The unknown liquid has SG = 1.20.

Problem: Capillary Rise in a Glass Tube

A clean glass tube with inside diameter 1.5 mm is inserted vertically into water at 20°C. The surface tension of water is σ = 0.0728 N/m and the contact angle with glass is 0°. Determine the capillary rise in millimeters. Would a tube of diameter 3.0 mm result in twice less rise?

$$h = \frac{2\sigma\cos\theta}{\rho g r}$$ $$r = \frac{0.0015}{2} = 0.00075 \text{ m}$$ $$h = \frac{2(0.0728)\cos 0°}{1000(9.81)(0.00075)} = \frac{0.1456}{7.358} = 19.8 \text{ mm}$$

For the 3.0 mm tube, $r = 0.0015$ m:

$$h = \frac{2(0.0728)}{1000(9.81)(0.0015)} = 9.9 \text{ mm}$$

Answer: Water rises 19.8 mm in the 1.5 mm tube and 9.9 mm in the 3.0 mm tube. Yes — doubling the diameter exactly halves the capillary rise because $h \propto 1/r$.

Problem: Speed of Sound Using Bulk Modulus

The bulk modulus of elasticity of sea water is $E_v = 2.34 \text{ GPa}$ and its mass density is 1025 kg/m³. Compute: (a) the speed of sound in sea water, and (b) the time for a pressure pulse to travel 1.80 km along a submarine pipeline.

$$c = \sqrt{\frac{E_v}{\rho}} = \sqrt{\frac{2.34 \times 10^9}{1025}} = \sqrt{2{,}283{,}000} = 1511 \text{ m/s}$$ $$t = \frac{L}{c} = \frac{1800}{1511} = 1.19 \text{ s}$$

Answer: Sound travels at 1511 m/s in sea water, and the pulse reaches 1.80 km in 1.19 s. This celerity value directly enters the water hammer pressure formula $\Delta p = \rho c \Delta V$.

Problem: Average Unit Weight and SG of a Mixed Volume

A pipeline carries a mixture of 0.80 m³ of oil (SG = 0.85) and 0.20 m³ of fresh water. The two fluids are stratified (not emulsified). Determine the total weight of liquid in the pipe and the average specific gravity based on the combined volume.

$$W_{oil} = \gamma_w \cdot SG_{oil} \cdot V_{oil} = 9.81(0.85)(0.80) = 6.671 \text{ kN}$$ $$W_{water} = 9.81(1.00)(0.20) = 1.962 \text{ kN}$$ $$W_{total} = 8.633 \text{ kN}$$ $$V_{total} = 1.00 \text{ m}^3$$ $$SG_{avg} = \frac{W_{total}}{\gamma_w V_{total}} = \frac{8.633}{9.81} = 0.88$$

Answer: Total weight is 8.633 kN. The average specific gravity of the liquid-filled pipeline is 0.88.

Water Vapor Density at Absolute Pressure

Calculate the density of water vapor at 350 kPa abs and 20°C if its gas constant is 0.462 kPa-m3/(kg-°K).

For a common gas, use the ideal-gas density relation with absolute pressure and absolute temperature.

$$T=20+273=293\ \mathrm{K}$$ $$\rho=\frac{p}{RT}=\frac{350}{(0.462)(293)}$$ $$\boxed{\rho=2.59\ \mathrm{kg/m^3}}$$

Pressure of Confined Oxygen

Four hundred fifty kilograms of oxygen is confined in a 110 L tank at -75°C. What is the pressure if the gas constant is 4.115 kJ/(kg-°K)?

Use the ideal-gas equation, with 1 kJ = 1 kPa-m3.

$$V=110\ \mathrm{L}=0.110\ \mathrm{m^3},\qquad T=-75+273=198\ \mathrm{K}$$ $$pV=mRT$$ $$p=\frac{mRT}{V}=\frac{(450)(4.115)(198)}{0.110}$$ $$p=3.33\times10^6\ \mathrm{kPa}$$ $$\boxed{p=3.33\ \mathrm{GPa}}$$

Capillary Correction in a Glass Tube

Distilled water at 20°C stands in a glass tube of 6.0 mm diameter at a height of 18.0 mm. What is the true static height?

For clean water and glass, take the contact angle as 0° and the surface tension at 20°C as 0.0728 N/m.

$$h_c=\frac{4\sigma\cos\theta}{\gamma d}$$ $$h_c=\frac{4(0.0728)\cos0^{\circ}}{(9810)(0.006)}$$ $$\boxed{h_c=0.00495\ \mathrm{m}=4.96\ \mathrm{mm}}$$

The capillary correction is 4.96 mm. If the 18.0 mm observed reading is to be corrected to the level without capillary rise, the corrected height is 18.0 - 4.96 = 13.04 mm.

Pressure Difference Inside an Air Bubble

By how much does the pressure inside a 5-mm-diameter air bubble in 15°C water exceed the pressure in the surrounding water?

An air bubble in water has one liquid-gas interface, so its excess pressure is 4σ/d. At 15°C, use σ = 0.07335 N/m.

$$p_{in}-p_{out}=\frac{4\sigma}{d}=\frac{4(0.07335)}{0.005}$$ $$\boxed{p_{in}-p_{out}=58.68\ \mathrm{Pa}}$$

Shear Stress Between Parallel Plates

SAE 30 oil (μ = 9.20 × 10-3 lb-s/ft2) at 20°C is sheared between two parallel plates 0.005 in apart, with the lower plate fixed and the upper plate moving at 13 ft/s. Compute the shear stress in the oil.

$$y=\frac{0.005}{12}=4.1667\times10^{-4}\ \mathrm{ft}$$ $$\tau=\mu\frac{U}{y}=(9.20\times10^{-3})\frac{13}{4.1667\times10^{-4}}$$ $$\boxed{\tau=287.04\ \mathrm{lb/ft^2}}$$

Terminal Velocity of a Slab on an Oil Film

An 18-kg slab slides down a 15° inclined plane on a 3-mm-thick film of SAE 10 oil (μ = 8.14 × 10-2 N-s/m2) at 20°C; the contact area is 0.3 m2. Find the terminal velocity of the slab.

At terminal velocity, the downslope component of weight is balanced by the viscous resistance.

$$mg\sin\theta=\tau A=\mu\frac{U}{y}A$$ $$U=\frac{mg\sin\theta\,y}{\mu A}$$ $$U=\frac{(18)(9.81)\sin15^{\circ}(0.003)}{(8.14\times10^{-2})(0.3)}$$ $$\boxed{U=5.615\ \mathrm{m/s}}$$

Force Required to Move a Piston Through an Oil Film

A piston shown is moving through a cylinder at a speed of 19 ft/s. The film of oil separating the piston from the cylinder has a viscosity of 0.020 lb-s/ft2. What force is required to maintain this motion?

See images:

Piston moving through a cylinder with an oil film

From the figure, the cylinder diameter is 5.000 in, the piston diameter is 4.990 in, and the piston length is 3 in. The uniform radial oil-film thickness is half the diametral clearance.

$$y=\frac{5.000-4.990}{2}=0.005\ \mathrm{in}=4.1667\times10^{-4}\ \mathrm{ft}$$ $$A=\pi DL=\pi\left(\frac{4.990}{12}\right)\left(\frac{3}{12}\right)=0.32666\ \mathrm{ft^2}$$ $$F=\tau A=\mu\frac{U}{y}A$$ $$F=(0.020)\frac{19}{4.1667\times10^{-4}}(0.32666)$$ $$\boxed{F=297.85\ \mathrm{lb}}$$

Oil Resistance on a Shaft in a Bearing Sleeve

A shaft 70.0 mm in diameter is being pushed at a speed of 400 mm/s through a bearing sleeve 70.2 mm in diameter and 250 mm long. The clearance, assumed uniform, is filled with oil at 20°C with ν = 0.005 m2/s and SG = 0.90. Use 1000 kg/m3 as the density of water.

(a) Find the force exerted by the oil on the shaft.

(b) Determine the resisting torque exerted by the oil if the shaft is fixed axially and rotated inside the sleeve at 2000 rpm.

First obtain the oil density, dynamic viscosity, radial clearance, and shearing area.

$$\rho=SG(1000)=900\ \mathrm{kg/m^3}$$ $$\mu=\rho\nu=(900)(0.005)=4.50\ \mathrm{Pa\cdot s}$$ $$y=\frac{70.2-70.0}{2}=0.10\ \mathrm{mm}=1.0\times10^{-4}\ \mathrm{m}$$ $$A=\pi DL=\pi(0.070)(0.250)=0.05498\ \mathrm{m^2}$$

For axial translation:

$$F=\mu\frac{U}{y}A=(4.50)\frac{0.400}{1.0\times10^{-4}}(0.05498)$$ $$\boxed{F=989.6\ \mathrm{N}}$$

For rotation, use the shaft surface speed and multiply the viscous force by the shaft radius.

$$\omega=2000\left(\frac{2\pi}{60}\right)=209.44\ \mathrm{rad/s}$$ $$U=\omega r=(209.44)(0.035)=7.330\ \mathrm{m/s}$$ $$T=\left(\mu\frac{U}{y}A\right)r$$ $$T=\left[(4.50)\frac{7.330}{1.0\times10^{-4}}(0.05498)\right](0.035)$$ $$\boxed{T=634.7\ \mathrm{N\cdot m}}$$
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