Question

A long body with elliptical cross section is held perpendicular to a 2D uniform steady flow field of horizontal velocity \(U_\infty\) as shown in the following figure. The heights of the control volume (bounded by the dashed lines) at the inlet and outlet are \(2h\) and \(4h\), respectively. The profile of the horizontal velocity far downstream is given by \(U(y) = \frac{U_\infty y}{2h}\). The density of the fluid is \(\rho\). The magnitude of the drag force per unit length acting on the body is 

 

• A. \(\frac{2\rho U_\infty^2 h}{3}\)
• B. \(\frac{\rho U_\infty^2 h}{3}\)
• C. \(\frac{\rho U_\infty^2 h}{2}\)
• D. \(\frac{2\sqrt{2}\rho U_\infty^2 h}{3}\)

Hint:

When applying control volume analysis, always verify mass conservation first, especially if the problem statement seems ambiguous or contains conflicting information. It can help you identify the correct parameters or assumptions to use.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The drag force on the body can be determined by applying the integral form of the momentum equation to a control volume surrounding the body. The net force on the control volume (including the reaction force from the body) equals the net rate of momentum leaving the control volume.
Step 2: Key Formula or Approach:
The x-momentum equation for a steady flow is: \[ \sum F_x = (\text{Momentum flux out}) - (\text{Momentum flux in}) \] Here, \(\sum F_x = -F_D\), where \(F_D\) is the drag force of the fluid on the body. We assume pressure forces on the control volume boundaries cancel out. \[ -F_D = \dot{M}_{out} - \dot{M}_{in} \] where \(\dot{M} = \int_A \rho u^2 \, dA\).
We must also check for conservation of mass: \(\dot{m}_{in} = \dot{m}_{out}\), where \(\dot{m} = \int_A \rho u \, dA\).
Step 3: Detailed Explanation or Calculation:
Let's consider a unit length (width) for the 2D flow.
The problem has conflicting information between the text (\(U(y) = \frac{U_\infty y}{2h}\)) and the diagram (\(U(y) = \frac{U_\infty y}{h}\)). We will first check which one satisfies mass conservation. The origin y=0 is the centerline. Due to symmetry, we assume the formula applies to \(y \geq 0\) and the profile is mirrored for \(y<0\), i.e., \(U(y) = \frac{U_\infty |y|}{k . h}\).
Mass Conservation Check:
Inlet mass flux: \(\dot{m}_{in} = \rho \times A_{in} \times U_{in} = \rho \times (2h \times 1) \times U_\infty = 2\rho U_\infty h\).
Outlet mass flux: \(\dot{m}_{out} = \int_{A_{out}} \rho U(y) \, dA = 2 \int_{0}^{2h} \rho U(y) (1 . dy)\) (due to symmetry).
Using the formula from the text, \(U(y) = \frac{U_\infty y}{2h}\) for \(y \geq 0\): \[ \dot{m}_{out} = 2\rho \int_{0}^{2h} \frac{U_\infty y}{2h} dy = \frac{\rho U_\infty}{h} \left[\frac{y^2}{2}\right]_0^{2h} = \frac{\rho U_\infty}{h} \frac{(2h)^2}{2} = 2\rho U_\infty h \] Since \(\dot{m}_{in} = \dot{m}_{out}\), the formula from the text is correct. The diagram's formula is a typo.
Momentum Flux Calculation:
Inlet momentum flux: \[ \dot{M}_{in} = \int_{A_{in}} \rho u^2 \, dA = \rho U_\infty^2 \times A_{in} = \rho U_\infty^2 (2h \times 1) = 2\rho U_\infty^2 h \] Outlet momentum flux: \[ \dot{M}_{out} = \int_{A_{out}} \rho U(y)^2 \, dA = 2 \int_{0}^{2h} \rho \left(\frac{U_\infty y}{2h}\right)^2 (1 . dy) \] \[ \dot{M}_{out} = 2\rho \frac{U_\infty^2}{4h^2} \int_{0}^{2h} y^2 dy = \frac{\rho U_\infty^2}{2h^2} \left[\frac{y^3}{3}\right]_0^{2h} = \frac{\rho U_\infty^2}{2h^2} \frac{(2h)^3}{3} = \frac{\rho U_\infty^2}{2h^2} \frac{8h^3}{3} = \frac{4}{3}\rho U_\infty^2 h \] Drag Force Calculation:
Using the momentum equation: \[ -F_D = \dot{M}_{out} - \dot{M}_{in} \] \[ -F_D = \frac{4}{3}\rho U_\infty^2 h - 2\rho U_\infty^2 h = \left(\frac{4}{3} - 2\right) \rho U_\infty^2 h = -\frac{2}{3}\rho U_\infty^2 h \] \[ F_D = \frac{2}{3}\rho U_\infty^2 h \] Step 4: Final Answer:
The magnitude of the drag force per unit length is \(\frac{2\rho U_\infty^2 h}{3}\).
Step 5: Why This is Correct:
The calculated drag force matches option (A). The key to solving this problem was identifying the correct velocity profile by enforcing the principle of mass conservation for the control volume.