Homework set 2. Problems 6.24, 43, 78, 87, 141.

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.

6.78. Use Bernoulli's equation for the streamline
starting at the top of the open reservoir, ending at the top
of the closed reservoir, where the pressure is known.
Calculate h_f for the pipe connecting the tanks using
Bernoulli's equation and the known pressure and heights,
to give h_f = 5.43 m. Find the velocity from the
Moody chart by iteration. Assume turbulent flow and guess
a value of f, using the roughness ratio for commercial steel,
where e_w = 0.046 mm. Compute an estimate of V from
h_f = f (L/d) (V^2 /2g). Find the Reynolds number and
a better guess for f from the Moody chart for the commercial
steel wall roughness. Keep it up until f converges to 0.0205.
6.85. The pump with the h_p versus Q performance
curve shown in Fig. P6.80 delivers 0.7 m^3/s of methanol
through 95 m of cast iron pipe. What is the pipe diameter?
Look up the density and viscosity of methanol. Look up the
roughness length for cast iron (0.26 mm). From Bernoulli's
equation the head loss h_f = h_p for a constant diameter,
horizontal pipe (note: h_p = - h_shaft). Express the equation
h_f = f (L/d) (V^2 /2g) in terms of Q, to get
h_f = f (L/d) (V^2 /2g) = 8fLQ^2/pi^2 g d^5.
Find the constants a and b in the "parabolic" pump performance
curve h_p = a + bQ^2 (h_p = 80 -20Q^2). Substitute and solve
for d = 0.559 f^1/5. Guess f = 0.02 and compute d. From
Re = 4 rho Q/ pi mu d find Re = 4.61x10^6. From the roughness
ratio and the Moody chart find a better estimate of f. Iterate to
d = 0.255 m.

Example 6.87. A commercial steel annulus 40 ft long, with radii a =
1 inch and b = 1/2 inch, connects two reservoirs which
differ in surface height by 20 ft. Compute the flow rate in
ft^3 / s through the annulus if the fluid is water at 20 deg C.
Solution: For water take rho = 1.94 slug/ft^3 and
mu = 2.09x10^-5 slug/ft s. For commercial steel, take e_w
= 0.00015 ft. Compute the hydraulic diameter of the
annulus. Dh = 4A/P, where A is the wetted area and P is the
wetted perimeter. A = pi (a^2 - b^2) , and P = 2 pi (a+b),
so Dh = 2(a-b) = 1 inch. Hence,
hf = 20 ft = f (L/Dh) V^2 / 2g = f (40 / 1/12 ) (V^2 /
2x32.2) ;
or
fV^2 = 2.683.
Such an equation must be solved by "cut and try" methods;
that is, guess a value for f, find Re and thus V from the
Moody chart, and compute a new f from the equation above.
Keep doing this until the values of f and V converge.
To use the Moody chart we need the roughness ratio.
For Dh = 1 in and e_w = 0.00015 ft we find e_w/D =
0.00015/(1/12) = 0.0018. Guess f = 0.023, so V = 10.8 ft /
s from the equation. The Reynolds number is then
1.94x10.8x(1/12)/2.09x10^-5 = 83550. Use this to find a
better value of f = 0.0249, Vbetter = 10.4 ft/s , etc. This
converges to f = 0.0250, V = 10.37 ft/s. Thus Q = pi (a^2 -
b^2)V = 0.17 ft^3 / s. ans.

Example 6.141. Gasoline at 20 deg C flows at 105 m^3 / h in a 10
cm diameter pipe. We wish to meter the flow with a thin
plate orifice and a differential pressure transducer which
reads best at about 55 kPa. What is the proper beta ratio for
the orifice?
Solution: For the gasoline rho = 680 kg/m^3, and
mu = 2.92x10^-4 kg/m s. The pipe velocity is V1 = Q/A1 =
(105/3600) / (pi/4)(0.1)^2 = 3.71 m/s. Re_D =
680x3.71x0.1/2.92E-4 = 865,000.
From Fig. 6.37, read C_d = 0.61. Therefore the
throat velocity V_throat = 3.71/beta^2 = C_d
(2x55000/680(1-beta^4))^1/2 ; or beta^2/ (1-beta^4)^1/2 =
0.4778. Solve for beta = 0.66. ans. Now check back with
Fig. 6.37 to see this is consistent, so no further iteration is
needed.