Question 1 of 7: Question 1 (Part A, Question A1): Drag and Power for a Flat Plate Towed Edgewise

Question 1 (Part A, Question A1): Drag and Power for a Flat Plate Towed Edgewise (14 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

In arrangement A side a = 5 m is normal to the flow (it is the leading edge, as drawn in Fig. A1), so the stream runs only \(L = b = 1\ \text{m}\) along the plate over a width \(w = a = 5\ \text{m}\). In arrangement B side b = 1 m is the leading edge, so \(L = a = 5\ \text{m}\) and \(w = b = 1\ \text{m}\). The plate is thin, so both faces carry a boundary layer and the wetted area is \(2ab = 10\ \text{m}^2\) in either case.

Find. The towing force and the towing power for each arrangement in each fluid, which arrangement costs less power, and by what percentage.

Approach. Integrate the aid-sheet local skin-friction law along each face, splitting the integral at the transition point \(x_{tr}\) where \(Re_x = 5\times10^5\), to obtain a mean coefficient \(C_F\); then \(D = 2\left(\tfrac{1}{2}\rho U_o^2 wL\,C_F\right)\) and \(P = D\,U_o\).

1. Locate the transition point in each fluid. Transition sits where \(Re_x = U_o x/\nu = Re_{cr}\), so \[x_{tr} = \frac{\nu\,Re_{cr}}{U_o}.\] In water \(x_{tr} = (10^{-6})(5\times10^{5})/15 = 0.0333\ \text{m}\); in air \(x_{tr} = (15\times10^{-6})(5\times10^{5})/15 = 0.500\ \text{m}\). This one number decides the whole question: in water transition happens within the first 3 cm of either plate, while in air it takes half a metre — which is half of the 1 m plate in arrangement B.
2. Integrate the wall shear over one face. The drag on one face of width \(w\) is \(D_1=\int_0^L \tau_w(x)\,w\,dx\) with \(\tau_w = \tfrac{1}{2}\rho U_o^2 C_{fx}\). Defining the mean coefficient \(C_F = \frac{1}{L}\int_0^L C_{fx}\,dx\) and substituting the two aid-sheet laws, \[\begin{aligned}\int_0^{x_{tr}}\!\frac{0.67}{Re_x^{1/2}}\,dx&=\frac{1.34\,x_{tr}}{Re_{cr}^{1/2}},\\\int_{x_{tr}}^{L}\!\frac{0.0266}{Re_x^{1/7}}\,dx&=\frac{7}{6}(0.0266)\!\left[\frac{L}{Re_L^{1/7}}-\frac{x_{tr}}{Re_{cr}^{1/7}}\right].\end{aligned}\] Both integrals are elementary because \(Re_x\) is linear in \(x\); the \(7/6\) factor is the exponent bookkeeping of \(x^{-1/7}\).
3. Evaluate the mean coefficient for each case. Adding the two contributions and dividing by \(L\) gives \[\boxed{C_F = \frac{1}{L}\left[\frac{1.34\,x_{tr}}{\sqrt{Re_{cr}}}+\frac{7}{6}(0.0266)\left(\frac{L}{Re_L^{1/7}}-\frac{x_{tr}}{Re_{cr}^{1/7}}\right)\right]}\] which yields the four values tabulated below. Note that arrangement B in water has \(Re_L = 7.5\times10^{7}\) — essentially an all-turbulent plate — whereas arrangement A in air is 50 per cent laminar.
4. Convert to force and power. With both faces wetted, \[D = 2\cdot\tfrac{1}{2}\rho U_o^2\,(wL)\,C_F=\rho U_o^{2}\,(wL)\,C_F,\qquad P = D\,U_o .\] Because \(wL = 5\ \text{m}^2\) and \(U_o = 15\ \text{m/s}\) are the same in every case, the force ratio between arrangements is the \(C_F\) ratio.

The two fluids give opposite answers, and that is the point of the question. In water the laminar run is negligible in both arrangements, so the only thing that matters is that a longer plate has a thicker, slower-growing turbulent layer and therefore a smaller mean coefficient: arrangement B wins. In air the laminar run is a full half of the short plate, and the laminar contribution is so much cheaper than the turbulent one that arrangement A wins despite its larger turbulent-layer growth rate. The plot below shows \(C_F(L)\) for both fluids with the two operating points marked.

5. Percentage power reduction, water. Arrangement B is the efficient one: \[\begin{aligned}\text{saving}&=\frac{P_A-P_B}{P_A}=\frac{47.81-38.95}{47.81}=0.1854\\&\Rightarrow\ \boxed{\text{arrangement B uses }18.5\text{ per cent less power}}\end{aligned}\]
6. Percentage power reduction, air. Here arrangement A is the efficient one: \[\begin{aligned}\text{saving}&=\frac{P_B-P_A}{P_B}=\frac{63.58-58.30}{63.58}=0.0830\\&\Rightarrow\ \boxed{\text{arrangement A uses }8.3\text{ per cent less power}}\end{aligned}\]