Question

Water is flowing with a free stream velocity of 0.25 m/s around a submerged flat plate of 2 m length (in the direction of flow) and 1 m width. The local shear stress at a distance \(x\) from the leading edge of the plate is given by \[ \tau = \frac{0.332\rho u^2}{\sqrt{Re_x}} \] where \(\rho = 1000\) kg/m³ is the density of the water, \(u\) is the free stream velocity and \(Re_x\) is the Reynolds number at \(x\). Assume that the flow is laminar, and the kinematic viscosity of water is \(10^{-6}\) m²/s. The drag force (in Newton) acting on one side of the plate lies between

• A. 0 and 0.05
• B. 0.05 and 0.10
• C. 0.10 and 0.15
• D. 0.15 and 0.20

Hint:

Notice that the total drag force on a flat plate in laminar flow is proportional to \(L^{1/2}\) and \(u^{3/2}\). The expression \(F_D = 0.664 \rho u^2 w \sqrt{\frac{\nu L}{u}}\) which simplifies to the one used, is a standard result worth remembering for flat plate boundary layer problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The total drag force on a flat plate is obtained by integrating the local shear stress (\(\tau\)) over the entire surface area of the plate. The local shear stress itself depends on the position \(x\) along the plate through the local Reynolds number \(Re_x\).
Step 2: Key Formula or Approach:
1. Express shear stress \(\tau\) as a function of \(x\).
2. Integrate \(\tau\) over the area of the plate to find the drag force \(F_D\).
Drag Force: \( F_D = \int_A \tau \, dA = \int_0^L \tau(x) (w \, dx) \)
Reynolds Number: \( Re_x = \frac{ux}{\nu} \)
Step 3: Detailed Explanation or Calculation:
Given values:
Free stream velocity, \(u = 0.25\) m/s
Length of plate, \(L = 2\) m
Width of plate, \(w = 1\) m
Density of water, \(\rho = 1000\) kg/m³
Kinematic viscosity, \(\nu = 10^{-6}\) m²/s
First, substitute the formula for \(Re_x\) into the expression for \(\tau\): \[ \tau(x) = \frac{0.332 \rho u^2}{\sqrt{\frac{ux}{\nu}}} = 0.332 \rho u^2 \sqrt{\frac{\nu}{ux}} = 0.332 \rho u^{3/2} \nu^{1/2} x^{-1/2} \] Now, calculate the total drag force \(F_D\) by integrating \(\tau(x)\) over the length of the plate: \[ F_D = \int_0^L \tau(x) w \, dx = w \int_0^L (0.332 \rho u^{3/2} \nu^{1/2} x^{-1/2}) \, dx \] \[ F_D = 0.332 w \rho u^{3/2} \nu^{1/2} \int_0^L x^{-1/2} \, dx \] The integral is: \[ \int_0^L x^{-1/2} \, dx = [2x^{1/2}]_0^L = 2\sqrt{L} - 0 = 2\sqrt{L} \] So, the drag force formula becomes: \[ F_D = 0.332 w \rho u^{3/2} \nu^{1/2} (2\sqrt{L}) = 0.664 w \rho u^{3/2} \nu^{1/2} \sqrt{L} \] Now, substitute the given numerical values: \[ u^{3/2} = (0.25)^{3/2} = ((0.5)^2)^{3/2} = (0.5)^3 = 0.125 \] \[ \nu^{1/2} = (10^{-6})^{1/2} = 10^{-3} \] \[ \sqrt{L} = \sqrt{2} \approx 1.414 \] \[ F_D = 0.664 \times (1) \times (1000) \times (0.125) \times (10^{-3}) \times (\sqrt{2}) \] \[ F_D = 0.664 \times 1000 \times 0.125 \times 0.001 \times \sqrt{2} \] \[ F_D = 0.664 \times 0.125 \times \sqrt{2} \] \[ F_D = 0.083 \times 1.4142 \approx 0.11738 \] N Step 4: Final Answer:
The calculated drag force is approximately 0.117 N.
Step 5: Why This is Correct:
The value 0.117 N lies between 0.10 N and 0.15 N. Thus, option (C) is the correct answer.