Turbulent Flow and Pipe Losses Questions and Answers

Answer: B

The correct option is I is False and II is True.

Statement II: the pressures at N, E and S are equal and all three pipes start at J, so the pressure drop from J to each end is the same. II is true.

Statement I: with the same friction factor and cross-section, the Darcy-Weisbach loss in each pipe is \(\Delta p \propto L\,Q^{2}\). Equal pressure drops therefore require

\[ \begin{aligned} L\,q_{N}^{2} &= 10L\,q_{E}^{2} \Rightarrow q_{E} = \frac{q_{N}}{\sqrt{10}} \\ Q &= q_{N} + q_{S} + q_{E} = q_{E}(2\sqrt{10} + 1) \\ q_{E} &= \frac{Q}{7.32} \end{aligned} \]

So the flow to E is about Q/7.3, not Q/21. The value Q/21 is what a loss proportional to Q (laminar flow) would give, \(q_{E} = q_{N}/10\) and \(Q = 2.1q_{N}\), but a constant friction factor makes the loss vary with \(Q^{2}\). That rules out the options with I true.

Answer: A

The channel is running full, so the area is 4 x 2 = 8 m\(^2\) and the wetted perimeter is 4 + 2 + 2 = 8 m (the free surface is not part of it). The hydraulic mean depth is \(m = A/P = 8/8 = 1\) m. By Chezy's formula:

The velocities of 2, 3 and 4 m/s would result from using too large a hydraulic mean depth, for example taking the whole perimeter as only the bed width.

Answer: D

Running full, the cross-sectional area is \(4 \times 2 = 8\) m\(^2\) and the wetted perimeter is \(4 + 2 + 2 = 8\) m, so the hydraulic mean depth is \(m = 1\) m. Chezy's formula gives the velocity

The options 6, 10 and 12 m\(^3\)/s come from a wrong area or a wrong velocity, such as using the whole perimeter of 12 m.

Answer: B

Only statement ii goes with turbulent flow. In turbulent flow the liquid moves in a zigzag, random way, with eddies and mixing. Orderly flow, in which layers of the liquid slide smoothly past one another, is laminar flow, so statement i describes laminar flow and not turbulent flow. The options that include statement i are therefore wrong. The usual way to tell the two apart is that laminar flow is smooth and orderly, while turbulent flow is irregular, with eddies, which is the picture in statement ii.

Answer: A

The ratio is \(\left(\frac{d_{2}}{d_{1}}\right)^{(5-n)}\).

The Darcy-Weisbach equation gives \(\Delta P = f\frac{L}{d}\frac{\rho V^{2}}{2}\). With the same flow rate in both pipes, \(V \propto 1/d^{2}\) and \(Re \propto Vd \propto 1/d\), so \(f \propto Re^{-n} \propto d^{n}\):

\[ \begin{aligned} \Delta P &\propto d^{n} \cdot \frac{1}{d} \cdot \frac{1}{d^{4}} = d^{\,n-5} \\ \frac{\Delta P_{1}}{\Delta P_{2}} &= \left(\frac{d_{1}}{d_{2}}\right)^{n-5} = \left(\frac{d_{2}}{d_{1}}\right)^{5-n} \end{aligned} \]

The exponent 5 alone is the constant-friction-factor result, (5 + n) has the Reynolds-number effect in the wrong direction, and (3 - n) mishandles the velocity dependence.

Answer: A

The head loss is about 116.18 m.

Find the velocity, then apply the Darcy-Weisbach equation:

\[ \begin{aligned} A &= \tfrac{\pi}{4}(0.2)^{2} = 0.031416\ \text{m}^{2} \\ v &= \frac{0.2}{0.031416} = 6.366\ \text{m/s} \end{aligned} \]

\[ \begin{aligned} h_{f} &= \frac{fLv^{2}}{2gD} \\ &= \frac{0.0225 \times 500 \times 6.366^{2}}{2 \times 9.81 \times 0.2} \\ &= \frac{455.9}{3.924} = 116.2\ \text{m} \end{aligned} \]

232.36 is twice the correct value, the result of dropping the 2 in 2g. 0.116 is the right digits with the decimal point moved three places, and 18.22 does not come from the Darcy-Weisbach equation with these data.

Answer: A

The correct statement is \(\tau_{A} \gt \tau_{B}\).

At the pipe entrance the boundary layer is very thin, so the velocity gradient at the wall, and with it the wall shear stress \(\tau = \mu (du/dy)_{wall}\), is large. As the boundary layer thickens along the developing zone the wall gradient falls, so the shear stress decreases with distance: section A, upstream of B, has the larger value.

Once the flow is fully developed the velocity profile no longer changes along the pipe, so the wall shear stress is constant: \(\tau_{C} = \tau_{D}\), which rules out \(\tau_{C} \gt \tau_{D}\). The fully developed value is the lowest along the pipe, so \(\tau_{C} \gt \tau_{B}\) is also wrong, as is \(\tau_{B} \gt \tau_{A}\).

Answer: C

The hydraulic diameter is 2R.

Hydraulic diameter is defined as four times the flow area divided by the wetted perimeter:

\[ D_h = \frac{4A}{P} = \frac{4\pi R^{2}}{2\pi R} = 2R \]

For a full circular pipe it is simply the pipe diameter, which is why the definition carries the factor 4. 0.5R is the hydraulic radius \(A/P\), R is the pipe radius, and 4R double-counts the factor of 2.

Answer: C

The correct pair is \(\Delta h = h_{P} + h_{Q} + h_{R}\) and \(V_{P}d_{P}^{2} = V_{Q}d_{Q}^{2} = V_{R}d_{R}^{2} + V_{S}d_{S}^{2}\).

P and Q are in series, so their head losses add and they carry the same flow. R and S are in parallel between stations 3 and 4, so each has the same head loss, \(h_{R} = h_{S}\), and the loss across the pair is counted once. Hence \(\Delta h = h_{P} + h_{Q} + h_{R}\).

The flow rate \(Q = \frac{\pi}{4}d^{2}V\) is the same in P and Q, and it divides between R and S, so \(V_{P}d_{P}^{2} = V_{Q}d_{Q}^{2} = V_{R}d_{R}^{2} + V_{S}d_{S}^{2}\).

Adding \(h_{S}\) as well double-counts the parallel branch, and writing \(V_{R}d_{R}^{2} = V_{S}d_{S}^{2}\) equal to the total flow treats the parallel branches as if each carried all of it.

Answer: A

The flow is laminar for both (i), and (ii).

For a circular pipe the Reynolds number can be written directly in terms of the mass flow rate:

\[ Re = \frac{\rho VD}{\mu} = \frac{4\dot{m}}{\pi D\mu} \]

\[ \begin{aligned} Re_{i} &= \frac{4(0.25)}{\pi(0.5)(10^{-3})} \approx 637 \\ Re_{ii} &= \frac{4(0.8)}{\pi(0.5)(10^{-3})} \approx 2037 \end{aligned} \]

Both are below the usual transition value of about 2300 for pipe flow, so both flows are laminar. The large 50 cm diameter keeps the velocity, and so the Reynolds number, low. The options calling either case turbulent overestimate Re.