Homework set 5. Problems 6.18,24,43,78,87,141

6.18 Water at20 deg C flows in a 9 cm diameter pipe under fully developed conditions.
The centerline velocity is 10 m/s. Compute a. Q , b. V, c. tau_wall , and d. delta p for a 100 m pipe
length.

Soln. Look up rho = 998 kg/m^3 and mu = 0.001 kg/ m s . Check Re = 900,000 >> 2100.
Use the law of the wall to find the friction velocity u* = (tau/rho)^1/2 . That is:
u_center / u* = (1/kappa) ln (Ru*/nu) + B
where kappa is von Karman's constant 0.42, u_center is 10 m/s, R is 0.045 m, and B is 5.
This gives u* = 0.35 m/s. Thus tau_wall is 122 Pa. Estimate V from u_center to be 8.5 m/s.
(or use 6.59). Thus Q = AV = 0.054 m^3/s. A force balance for steady flow gives the expression:
delta p = 2 tau_wall delta L / R = 542,000 Pa.

6.24 Flow from A to B or B to A? For glycerin look up rho = 1260 kg/m^3 and mu = 1.49
kg/m s . Check that Re = 30 << 2100 to confirm laminar flow. Assuming the flow is from
A to B gives Bernoulli's equation as:
pA/rho g + zA + V^2 / 2 = pB/rho g + zB + V^2 / 2 + hf
2.1x101350/1260x9.81 + 0 = 3,7x101350/1260x9.81 + 12 + hf
17.2 m + 0 = 30.3 m + 12 m + hf, so hf = - -25.1 m.
But hf must always be positive, so the assumption is wrong, and the flow is from B to A.

6.43 Water flows through a 3 in diameter horizontal wrought iron pipe for a mile at 250 gal/min.
Estimate the head loss and the pressure drop in the pipe.

Soln.: Look up rho = 1.94 slug/ft^3 and mu = 2.09 E-5 slug/ft s. Convert 250 gal/min to 0.557
ft^3/s. For wrought iron take eps = 0.00015 ft. Compute V = 11.35 ft/s and Re = 263000.
Find eps/d = 0.0006 from the pipe diam. d. Look up f = 0.0189 on the Moody chart, or use
6.64. Determine hf = f (L/d) (V^2 / 2 g) = 800 ft. Thus delta p = rho g hf = 1.94x32.2x800 = 49900 lbf/ft^2.