Fig. 1. Definition of viscosity (see text).
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Viscosity is a measure of the ease with which a fluid flows. Resistance to fluid flow is caused by friction in the fluid.
Fig. 1. Definition of viscosity (see text).
Suppose a layer of fluid of area A is at rest in contact with a solid surface in the xy-plane (see Fig. 1). Suppose another layer also of area A at height z above the xy-plane has velocity u in the x-direction. Then the velocity gradient in the fluid between the planes is u/z. Friction in the fluid produces a force F opposing the motion of the upper plane given by
F = eta.Au/z,
where eta is a constant for the fluid called the coefficient of viscosity.
The coefficient of viscosity of a gas is independent of pressure, but it increases as the temperature rises. The coefficient of viscosity of a liquid decreases as the temperature rises. Table 1 gives the approximate viscosities of air and liquid water at several different temperatures.
| Substance | Temperature | Viscosity |
|---|---|---|
| Air | 27°C | 2.0×10-5Ns/m2 |
| 77°C | 2.1×10-5Ns/m2 | |
| 127°C | 2.3×10-5Ns/m2 | |
| Liquid Water | 15°C | 1.1×10-3Ns/m2 |
| 30°C | 0.8×10-3Ns/m2 | |
| 50°C | 0.6×10-3Ns/m2 |
Calculate the friction force on a square with side 10cm in water at 30°C between two layers of water 1.0mm apart moving with a relative velocity 2.0cm/s.
Poiseuille's formula for the steady laminar flow of a liquid with viscosity eta through a pipe with radius r and length L is:
volume per second = k.etax.ry(p/L)z,
where k is a dimensionless constant and p is the pressure difference across the length L. Use dimensional analysis to find x, y and z. (The constant k, which cannot be found by dimensional analysis, is pi/8.)
Imagine a fluid flowing steadily through a pipe with different cross section areas at different distances along its length (see Fig. 2). Let rho1 be the density of the fluid, let v1 be the velocity of the fluid, and let A1 be the cross section area of the pipe at point 1. Let rho2, v2 and A2 be the corresponding quantities at point 2.
Fig. 2. Steady flow of a fluid through a pipe.
Since the flow is steady (independent of time) and the mass of the fluid is conserved (no fluid is created or destroyed), the mass of the fluid passing point 1 per unit time is the same as the mass of the fluid passing point 2 per unit time. Therefore
rho1v1A1 = rho2v2A2.
This is called the continuity equation.
If the fluid is incompressible, which is true for liquids but not always true for gases, then rho1 = rho2 and the continuity equation becomes
v1A1 = v2A2.
The velocity is low where the pipe has a large cross section area and high where the pipe has a small cross section area.
Water flows at 2.4m/s through a pipe with internal diameter 1.6cm. The internal diameter of the nozzle at the end of the pipe is only 1.2cm. What is the speed at which the water emerges from the nozzle?
We shall apply the principle of conservation of energy to an incompressible non-viscous fluid flowing between points 1 and 2 in a pipe as shown in Fig. 2.
Let P1 be the pressure in the fluid at point 1, and let P2 be the pressure in the fluid at point 2. Let y1 be the height of point 1 above a standard level, and let y2 be the height of point 2 above the same standard level. Suppose the diameter of the pipe is small compared with the difference (y2 - y1) between the heights.
The rate at which work is done on the fluid entering the pipe at point 1 is P1A1v1. The rate at which work is done by the fluid leaving the pipe is P2A2v2. Therefore the net rate at which work is done on the fluid in the pipe between points 1 and 2 is
dW/dt = P1A1v1 - P2A2v2.
The volume rate of flow of the fluid is dV/dt = A1v1 = A2v2. Therefore
dW/dt = (P1 - P2)dV/dt.
Since the kinetic energy per unit volume of the fluid is (½)rho.v12 at point 1 and (½)rho.v22 at point 2, it follows that the rate of change of kinetic energy of the fluid in the pipe between points 1 and 2 is
(½)rho(v22 - v12)dV/dt.
No work is done against friction in the fluid because we have assumed that the fluid is non-viscous. Therefore, by the principle of conservation of energy, the rate at which work is done on the fluid is equal to the sum of the rate of change of kinetic energy and the rate of change of potential energy. In other words
(P1 - P2)dV/dt = (½)rho(v22 - v12)dV/dt + rho.g(y2 - y1)dV/dt.
Therefore
P1 + (½)rho.v12 + rho.g.y1 = P2 + (½)rho.v22 + rho.g.y2.
This is called Bernoulli's equation.
A large tank of water has a small hole in its side at a distance y below the surface of the water. Water emerges from the hole with velocity v. (a) Prove that v2 = 2gy. (b) Calculate the velocity of water emerging from a hole 1.0 m below the surface of the water.
The wing of an airplane has area 80m2. Air flows over the top of the wing at 200m/s and under the wing at 180m/s. What is the net force on the wing due to the Bernoulli effect?
Heat conduction is the flow of thermal energy through a body from points at a high temperature to points at a low temperature. Figure 3 shows a metal bar of length x and cross section area A. One end of the bar is at a low temperature T1, and the other end is at a high temperature T2. Heat Q flows by conduction from the high temperature end to the low temperature end. The sides of the bar are thermally insulated so that no heat flows through them.
Fig. 3. Definiton of heat conductivity (see text).
The rate of heat transfer dQ/dt is proportional to the cross section area A of the bar, and to the temperature gradient dT/dx = (T2 - T1)/x. We write
dQ/dt = -kA.dT/dx,
where k is a constant called the thermal conductivity of the material of the bar. The minus sign indicates that the direction of the flow of heat is opposite to the direction of the temperature gradient.
Metals have high thermal conductivities and non-metals have low thermal conductivities. Fluids also have thermal conductivities, which are low for liquids and very low for gases. However, heat is often transferred more quickly through fluids by the fluid motion in a process called convection. Table 2 shows the thermal conductivities of some common substances.
| Substance | Thermal Conductivity |
|---|---|
| Copper | 400 W/mK |
| Iron | 80 W/mK |
| Glass | 0.9 W/mK |
| Wood | 0.15 W/mK |
| Water | 0.6 W/mK |
| Air | 0.024 W/mK |
A sheet of glass of area 2.5 m2 is 4 mm thick. There is a temperature difference of 10°C across the sheet. Calculate the rate of heat transfer from one side of the glass to the other side.
The walls of a cold storage room consist of iron sheet 6 mm thick and plastic foam 10 cm thick on the inside. The temperature of the outside metal surface is 35°C. The temperature of the inside plastic foam surface is -10°C. (a) What is the temperature of the junction between the two materials? (b) What is the heat transfer rate through the walls? [The thermal conductivity of plastic foam is practically the same as the thermal conductivity of air.]
A vector field is a vector valued function of position in space. Let V = (u, v) be a vector field representing the velocity of a fluid at a point (x, y) in two dimensions, for example the horizontal velocity of water on the surface of a lake. Let the values of the partial derivatives of the velocity be
ux = partial du/dx, uy = partial du/dy,
vx = partial dv/dx, vy = partial dv/dy.
Then the divergence and curl of the velocity field are defined by the equations
div V = ux + vy, curl V = vx - uy.
The physical meanings of the divergence and the curl can be visualized as follows.
If div V is positive, then water must be rising up to the surface from below. If div V is negative, then water must be sinking down from the surface.
If curl V is zero, then a small object floating on the water will not rotate, although it may move with the flow of the water. If curl V is positive, then the object will rotate counterclockwise. If curl V is negative, then the object will rotate clockwise.
The arrows in Fig. 4 represent the directions and magnitudes of the velocities of the water on the surface of a lake for different patterns of flow.
(a) Try to see (without calculation) whether the divergence and curl in each pattern is positive, zero, or negative.
(b) Check your guess by calculating div V and curl V using the formulas for the velocity vectors given below the figure.
Fig. 4. Examples of vector fields representing the velocity of water on the surface of a lake.
Pattern 1: u = y, v = 0. Pattern 2: u = x, v = y. Pattern 3: u = -x + y, v = -x - y.
Suppose the velocity vector on the half-plane x > 0 is given by V = (-x, y). Sketch the flow, and calculate the divergence and curl of the velocity.
The wind near the ground is horizontal and the speed u depends on height h in accordance with the formula u = A ln (h/h0), where A and h0 are constants. (a) Show that the divergence of the wind is zero, and the curl is inversely proportional to height. (b) What are the units of measure of the curl?
By R. H. B. Exell, 2002. King Mongkut's University of Technology Thonburi.
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