Homework set 4. Problems 5.6,8,13,22,43,59; 6.2,7,12,17; DP 6.1

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
C_D = 0.81 Re_L^-0.17 .
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:
p* = 1 - V*^2 / 2 - z*

5.13 Assume C is a function of rho, lambda, and upsilon. Use the power product
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.

Design project D6.1.
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.
It is best to assume that gravity and frictional effects
are negligible. Consider control volumes around segments
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.