Homework set 1.
Problems 1.3,12,20,37,45,58,73,75; 2.5,7,17,29,44,86,109
1.3. Find number of molecules in 1 mm^3 of liquid water at 20 deg
C. At what pressure
will liquid water become rarefied and deviate from the continuum concept?
Solution: Since the molecular wt. of water is about 18 and Avagodro's number
molecules/18g water there are about 3x10^-23 grams/molecule. A mm^3 weighs
grams, so it contains about 3x10^19 molecules. The separation between molecules
a mm times 1/(3x10^19)^.333 , or 3x10^-10 m. Kolmogorov scales in water
smaller than 1 mm for flows in the laboratory or ocean, giving a comfortable
range of length scales for the continuum hypothesis. Use the ideal gas
law to find the density
of water vapor at 20 deg C and p = 2337 Pa (Table A.5).
1.12 Check the units of B in the expression u = B delta p/ r )(r_o^2 -
r^2) by inserting L/T
for the left hand side. This must equal [B] times [p] times L^2 times /
[ mu ] = [B] x M/LT^2
/ M/LT = [B] L^2 / T. Thus B [=] L^-1. See section 6.4 for the correct
1.20 Derive the substantive derivative, as done in class.
1.37 Shear stress is mu V / h for the flow between plates. For glycerin,
mu is 1.5 Pa s
from Table 1.4 giving 2800 Pa. The Reynolds number rho V h / mu is 25.
Thus the flow
will be laminar, as assumed.
1.45 The force balance for the block in the x direction is Wsin theta =
tau A , where theta is
the angle of the plate to the horizontal, tau is the shear stress, A is
the plate area, and W is the
weight of the block. Substitute Newton's law of viscosity to find V_terminal
= h W sin theta / mu A.
1.58 Apply the formula to compute mu from the data and plot delta p versus
mu. For the small values of delta p mu is constant (about 0.04 kg/m s ),
corresponding to laminar flow, but larger values give larger mu, showing
the effective viscosities of turbulent flows are larger than those for laminar
1.73 Cavitation occurs when the pressure falls below the vapor pressure
of the water, which is 2337 Pa from A.5. The cavitation number Ca = [p_a
- p_v] / [(rho V^2)(1/2)] is critical at a value of 0.27. Solve for a critical
velocity of 29.7 m/s.
1.75 At mach number 1 the airplane moves at a sound speed of 318 m/s.
From eq. 1.42 the sound speed of air a_air = (kRT)^1/2 , giving T = 252
K. From A.6 the altitude is 5500 m.
2.5 Standard atmospheric pressure as head h = p / rho g . Find specific
weights rho g in Table 2.1. Substitute in the expression to find h = 43
ft for ethanol, 10.35 m for water, 30 in. for mercury, and 6510 mm for carbon
2.7 This problem was done as an example in lecture, to give 1121 atm pressure
at the bottom of the Mariana Trench.
2.17 Integrate the hydrostatic equation dp/dz = - rho g to find p_B = p_A
- rho_water g (z_B - z_A) = 2000 - 62.4 x 1 = 1938 lbf/ft^2. etc.
2.29 Derive the given expression by substituting p = C rho^k in the hydrostatic
equation. Separate variables and integrate to get C rho^(k-1) / (k-1) =
-gz/k + const. Find the const. at z = 0 and divide this equation by it.
Substitute the p expression to eliminate rho. Substitute C rho_o^(k-1)
= RT_o to get the desired form, using the ideal gas law. From the standard
atm table A.6 at z = 10,000 ft. the pressure is 26416 Pa versus 23809 from
the derived expression.
2.44 Integrate the hydrostatics equation from pt. 1 to pt. 2 to get p_1
+ rho g (5 sin 45 + h + 6/12) - 846 (6/12) = p_2 , so p_1 - p_2 = frict.
loss - gravity head = 392 - 221 = 171 lbf/ft^2 . Note that for the engineering
units used, rho g for water is actually rho g/g_c = 62.4 lbf/ft^3 . The
manometer reads only the frictional pressure loss because equal vertical
legs of water in the pipe and manometer legs cancel.
2.86 The horizontal force is zero by symmetry. The vertical force is the
difference between the weights of a vertical cylinder minus a hemisphere
of water. F_V = 7841 - 1046 lbf = 6800 lbf.
2.109 If the submerged volume is V_o with SG = 1, and A = pi D^2 / 4 ,
then W = rho g V_o = SG rho g (V_o -Ah), or h = W(SG-1)/[SG rho g ( pi D^2
/ 4 )].