Homework set 1. (*unassigned)

Problems 1.2,7,*3,*12,*20,37,45,*58,59,73,*75,78; 2.5,7,17,26,*29,44,*86,91,109. Prepare your own solution to the assigned problems. Discussion is allowed but not copying.

1.2 The earth's atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0.6 kg/m^3 (see Table A-6). Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth.

Solution: The earth's radius is 6377 km. The total mass of the atmosphere is then the integral of the density \rho time the volume of the atmosphere in each layer. M_t = \rho_ave x 4 \pi x thickness_ave = .6 x 4 x 3.14 x 6.377e6 x 20e3 = 6.1e18 kg. A proton mass is 1.67e-27 kg, so an air "molecule" has mass 28.97 times larger or 4.8e-26 kg. Thus there are about 1.3e44 "molecules" of air in the earth's atmosphere. Note that you can work the problem by setting the mass of the atmosphere times the acceleration of gravity equal to the surface pressure times the surface area to get the same mass as above.

*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 is 6x10^23
molecules/18g water there are about 3x10^-23 grams/molecule. A mm^3 weighs about 10^-3
grams, so it contains about 3x10^19 molecules. The separation between molecules is thus
a mm times 1/(3x10^19)^.333 , or 3x10^-10 m. Kolmogorov scales in water rarely are
smaller than 1 mm for flows in the laboratory or ocean, giving a comfortable ten million
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).

1.7 A small village draws 1.5 acre-foot of water per day from its reservoir. Convert this water usage into (a) gallons per minute; and (b) liters per second.

Solution: Look up the necessary conversion factors in the textbook (inside front cover). An acre has 4.0469e3 m^2 and a foot is 0.3048 m, so an acre-foot is 1.233e3 m^3. A US gallon is 3.7854e-3 m^3, and a day has 24x60=1.44e3 minutes, so (a) is 1.5x1.233e3/3.7854e-3 x 1.44e3 = 339 gallons/min. (b) requires you to know that there are 10^3 L/m^3 and 60 seconds/min, giving 21 L/s. Write out the expressions in full cancelling the units of the conversion factors to be sure you have the right result. Check for reasonableness. How big is the village?

*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 expression.

*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 flows.

1.59 A solid cylinder of diameter D, length L, density \rho_s falls due to gravity inside a tube of diameter D_o. The clearance, D_o - D << D, is filled with a film of viscous fluid. Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20 deg C for a steel cylinder with D=2 cm, D_o = 2.04 cm, and L = 15 cm. Neglect the effect of any air in the tube.

Solution: Terminal velocity is reached when the drag on the cylinder equals the weight of the cylinder. The drag is the area times the viscous stress. The weight is the gravitational acceleration times the mass. The mass is the density of steel times the volume of the cylinder. The volume of the cylinder is the area times the length. The density of steel is 7850 kg/m^3. The viscosity of SAE 30 oil is 0.29 kg/m s from Table 1.4. Thus, the viscous force on the cylinder is \mu 2V/(D_o- D) \pi DL and the weight is g \rho_s \pi D^2 L/4. Solve for V = \rho_s g D (D_o- D)/9 \mu = 0.265 m/s (ans).

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.

1.78 Sir Isaac Newton measured sound speed by timing the difference between seeing a connon's puff of smoke and hearing its boom. If the cannon is on a mountain 5.2 miles away, estimate the air temperature in deg C if the time difference is (a) 24.2 s; (b) 25.1 s.

Solution: Cannons produce shock waves that travel faster than sound waves at first, so you must neglect this supersonic portion of the trajectory of the sound to the ear. The sound speed is V_S = (1.4xRxT)^1/2 = distance/time. For air R = 287 m^2/s^2 K (8314/28.9). Solve for (a) T = 298 K, 25 C; (b) T = 277 K = 4 C.

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 tetrachloride.

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.

2.26 Investigate the effect of doubling the lapse rate on atmospheric pressure. Compare Table A.6 with a lapse rate twice as high, B_2 = 0.0130 K/m. Find the altitude at which the difference in pressure is (a) 1%; and (b) 5%. Comment.

Solution: Evaluate Eq. 2.27 for the two different values of B and compare the values to pressures of Table A.6. The error is small up to about 5 km. (a) 1% at z=2578 m, and (b) 5% at z = 5409 m.

*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.

2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and is filled with water and attached to the floor by 6 equally spaced bolts. What is the force in each bolt required to hold the dome down?

Solution: The total force required of the bolts is the weight of the missing water from a cylinder 2 m in diameter and 6 m high, minus the weight of the dome and pipe. Divide this by 6. The missing water weight is 574088 N, so the total force is 544088 N, giving the force per bolt of 90700 N.

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 )].