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Before you turn this problem in, make sure everything runs as expected. First, restart the kernel (in the menubar, select Kernel\(\rightarrow\)Restart) and then run all cells (in the menubar, select Cell\(\rightarrow\)Run All).

Make sure you fill in any place that says YOUR CODE HERE or “YOUR ANSWER HERE”, as well as your name and collaborators below:

NAME = ""
COLLABORATORS = ""

This figure shows a schematic of a pipeline that delivers constant temperature water from point 1 to point 2.

The general mechanical energy balance on this system results in:

\(-\frac{1}{2} \nu^2 + g \Delta z + \frac{g_C \Delta P}{\rho} + 2\frac{f_F L \nu^2}{D} = 0\)

variable

value

ν

flow velocity (ft/s)

g

acceleration of gravity, 32.174 ft/s2

Δ z

z2 - z1

gc

conversion factor, 32.174 ft lb_m / lb_f s^2

Δ P

p2 - p1 lbf / ft2

fF

Fanning friction factor (see eq)

L

length of pipe (ft)

D

inner diameter of pipe (ft), 7.981 inches

Our goal is to compute the flow rate of 60 °F water through a 1000 ft long pipe. The water is pumped uphill 300 ft (i.e. z2 - z1 = 300 ft). The pressure at p1 is 150 psig, and the pressure at 2 is atmospheric pressure, so \(\Delta P = -150\) psig.

The density (ρ) of water is temperature dependent. With T in °F, the density (in lbm/ft3) is given by:

\(\rho(T) = 62.122 + 0.0122 T - (1.54e-4) T^2 + (2.65e-7) T^3 - (2.24e-10) T^4\)

The Fanning friction factor depends on the Reynold’s number: \(Re = \frac{\nu \rho D}{\mu}\). In this equation, \(\mu\) is the viscosity (in lbm/ft/s), and it is also dependent of temperature:

\(\ln \mu = -11.0318 + \frac{1057.51}{T + 214.624}\)

When \(Re < 2100\) the Fanning friction factor is defined as \(f_F = 16 / Re\), but when \(Re > 2100\) it has this more complex expression:

\(f_F = \frac{1}{16 (\log (\frac{\epsilon / D}{3.7} - \frac{5.02}{Re} \log (\frac{\epsilon / D}{3.7} + \frac{14.5}{Re})))^2}\)

In this expression, \(\epsilon = 0.00015\) ft, and represents the surface roughness of the pipe.

Given this information,

  1. Estimate the velocity of water in the pipe (hint: it is 11.61 ft/s)

  2. Compute the mass flow of water (in lbm/min) in the pipe.

Some notes:

  1. In numpy, the ln(x) is computed with np.log(x).

  2. In numpy, the log(x) is computed with np.log10(x).

  3. 1 psig = 144 lbf/ ft2

  4. The velocity is 11.61 ft / s, and the mass flow is 15094 lb_m / min.

When you are done, download a PDF and turn it in on Canvas. Make sure to save your notebook, then run this cell and click on the download link.

%run ~/f23-06623/f23.py
%pdf
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