5.6 SAE 10 oil at 20 deg C has rho of 870 kg/m^3 and mu of 0.104 kg/m s .

Compute C_D = F/(rho V^2 D^2) values from the data and plot C_D versus

Re_L = rho V / mu on a log-log paper to determine the relation

The new velocity is 11 m/s, and for glycerin at 20 deg C take rho = 1260 kg/m^3

and mu = 1.49 kg/m s. The new Re_L is744. Substitute in the empirical relation

to find C_D = 0.263. F = 257 N.

5.8 The Bernoulli equation in the form p = p_o - rho V^2 / 2 - rho g z can

be made dimensionless with p_o , and rho. Thus p* = p/p_o , V* = V/(p_o rho )^1/2,

z* = rho g z / p_o . The dimensionless Bernoulli equation is therefore:

method to find C = constant x ( upsilon / rho lambda )^1/2. Thus, if upsilon is

doubled holding the other variables constant the wave speed C increases by 41%.

5.22 Use the Buckingham pi method to show that

F / rho V^2 L^2 = function ( V^2 / g L , rho V L / mu ).

5.43 Make the given salt conservation equation dimensionless using the

variables u* = u/U, x* = x/L , t* = U t / L etc. Note that S is already

dimensionless. The dimensionless equation is

dS/dt* + u* dS/dx* + v* dS/dy* + w* dS/dz* = (k/UL) del*^S ,

where d indicate partial differentiation. k is the diffusivity of

salt with units L^2 / t . UL/k = (UL/ nu) x (nu / k ) = Re Sc ,

where Re is the Reynolds number and Sc is the Schmidt number.

5.59 You must assume the prototype blimp and model blimp are

dynamically similar, which means they will have the same Re

and C_D. Plug in the given values to find the model velocity is

12 m/s from the Re equation. Similarly find the prototype force

to be 730 N by equating C_D values. The prototype (real) blimp

power is the force times the velocity, giving 4400 watts.

6.2 The air and water boundary layers are assumed to be dynamically

similar, so the Re values are the same at the transition points of interest.

For air at 20 deg C take rho = 1.2 kg/m^3 and mu = 1.8 x 10^-5 kg/m s.

For water rho is 998 kg/m^3 and mu = 0.001 kg/m s. Transition

occurs at about Re_x = 2.8 E6 . Since the wing moves at 20 m/s in both

air and water, x will be = 140000 mu / rho for both. Substitute the different

mu and rho values to find x for air is 2.1 m and x for water is 0.14 m.

6.7 For cola "water" assume rho = 998 kg/m^3 and mu = 0.001 kg/m s. Assume a transition

Reynolds number for the tube of 2000-2300. From the expression Re_crit = 4 rho Q_crit /

pi mu D find Q_crit, and a filling time of about 26 s. Then find a temperature from

Table A-1 that will cause the viscosity to increase the filling time to one minute. This occurs

at a temperature of about 66 deg C.

6.12 Substituting the Reynolds decomposition expressions for the velocity components

gives three Reynolds stresses per direction for a total of nine (it is a second order tensor).

The tensor is symmetric, so only six components are unique.

6.17 The logarithmic turbulent velocity profile for the constant stress boundary layer

was found by von Karman by setting the turbulent stress tau_t equal to an eddy viscosity

epsilon times the mean velocity gradient du/dy. tau_t = tau_wall = rho u*^2 , where u* is

the friction velocity. epsilon is rho nu , where nu is the eddy diffusivity proportional to

y times du/dy times y. The proportionality constant is k .

Thus rho u*^2 = rho k^2 y^2 |du/dy|^2 . Solve for du/dy = u* / ky . Integrate

du = (u*/k) dy / y to find u = (u* / k) ln y + const. k is von Karman's constant 0.42.

The hydroponic garden perforated pipe watering

system should be designed to have nearly uniform flow

through the holes along the pipe. The flow rate is not

specified, so an obvious easy solution is to make the holes

so small that there will be only a small change in the pressure

75 kPa along the pipe due to pipe frictional losses and

Bernoulli effects. Please don't do this. You are free to

decide what drill bit sizes are available to the "typical"

machine shop. From the sketch on p 386 the author has in

mind that the watering flow is high enough for the pressure

for the first holes near the pump to be significantly reduced,

so these holes must be larger to keep their watering rate the

same (nearly) as the later holes. Keep in mind that time is

money, and the profit margin for hydroponic gardens is very

low...don't spend much time on this problem trying to

perfectly optimize the solution.

are negligible.

containing individual holes and apply the Reynolds Transport

Theorems for mass and momentum, and Bernoulli's equation

to determine the pressure variation along the pipe and the

velocity of the jets emerging from the pipe holes. The mass

balance gives a linear decrease in the velocity along the pipe

since Q/50 volume per unit time is removed by each of the

50 holes. The momentum balance does not give the pressure,

but does give the longitudinal force on the pipe walls, in case

you needed this information. Bernoulli's equation shows

that the pressure increases downstream because the velocity

decreases. The velocity through each jet therefore increases

along the pipe (vj = (2 pj / rho )^1/2 , where vj is the jet

velocity, rho is the fluid density, and pj is the pressure for

the jth jet). The volume flow rate for each jet should be Q/50

which is vj x pi Dj^2 / 4 . Thus the jet hole diameter Dj may

be calculated from the pressure pj along the pipe, and your

assumed value of Q. Make sample calculations as you proceed

to be sure your computer or calculator program is working.

The new PowerMacs have Excel 5, which can easily do the

calculations and plots you need (339 EBII).

It is possible to apply Bernoulli's equation from the entrance

of the pipe to the jth segment. Thus,

pj = rho [ po/rho + (1/2)(Q/A)^2 (j/50)[2 - j/50]]

for j=1 to 50.