Question

The figure shows two fluids held by a hinged gate. The atmospheric pressure is \(P_a\) = 100 kPa. The moment per unit width about the base of the hinge is ............... kNm/m. (Rounded off to one decimal place) 
Take the acceleration due to gravity to be g = 9.8 m/s². 

 

Hint:

For submerged surfaces under multiple fluid layers, it's often easiest to break down the pressure diagram into simple shapes (rectangles and triangles). Calculate the force and moment for each shape separately and then sum them up. Remember that the force from a rectangular pressure distribution acts at the midpoint, while the force from a triangular distribution acts at one-third of the height from the base.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the moment about a hinge at the bottom of a vertical gate due to hydrostatic pressure from two immiscible fluids. The total moment is the sum of the moments created by the forces from each fluid section. The hydrostatic pressure increases linearly with depth, and the force due to this pressure acts at a specific point called the center of pressure. Since atmospheric pressure acts on both sides, its effect on the net moment can be neglected, and we can work with gauge pressures.
Step 2: Key Formula or Approach:
1. Hydrostatic Force (\(F\)): For a rectangular area, \(F = P_c A\), where \(P_c = \rho g h_c\) is the gauge pressure at the centroid of the area \(A\).
2. Center of Pressure (\(y_p\)): For a rectangular area submerged vertically from the surface, the depth of the center of pressure is \(y_p = \frac{2}{3}h\), where \(h\) is the height of the rectangle.
3. Moment (\(M\)): \(M = F \times d\), where \(d\) is the lever arm from the point of action of the force to the hinge.
Step 3: Detailed Explanation or Calculation:
We will calculate the moment per unit width, so let the width \(w = 1\) m. The gate has a total height of 3 m. The hinge is at the bottom. Part 1: Top Fluid (\(\rho_1 = 1000\) kg/m³, height \(h_1 = 1\) m)
This section of the gate is a rectangle of height 1 m and width 1 m. The pressure distribution is triangular.
- Area of the top section, \(A_1 = h_1 \times w = 1 \times 1 = 1\) m².
- Hydrostatic force, \(F_1 = (\rho_1 g h_{c1}) A_1\), where \(h_{c1} = h_1/2 = 0.5\) m.
\[ F_1 = (1000 \times 9.8 \times 0.5) \times 1 = 4900 \text{ N/m} \] - This force acts at the center of pressure for this section, which is at a depth of \(\frac{2}{3}h_1 = \frac{2}{3}\) m from the free surface.
- The distance of this force from the hinge (lever arm), \(d_1 = (2+1) - \frac{2}{3} = 3 - \frac{2}{3} = \frac{7}{3}\) m.
- Moment from the top fluid, \(M_1 = F_1 \times d_1 = 4900 \times \frac{7}{3} \approx 11433.3\) Nm/m.
Part 2: Bottom Fluid (\(\rho_2 = 2000\) kg/m³, height \(h_2 = 2\) m)
The pressure on this section is trapezoidal. It consists of a uniform pressure from the top fluid (\(P_{top} = \rho_1 g h_1\)) and a triangular pressure from the bottom fluid itself. We can split the force calculation into two parts.
- Rectangular Part (due to pressure from top fluid):
- Pressure at the interface, \(P_{interface} = \rho_1 g h_1 = 1000 \times 9.8 \times 1 = 9800\) Pa.
- Force, \(F_{2,rect} = P_{interface} \times A_2 = 9800 \times (2 \times 1) = 19600\) N/m.
- This force acts at the centroid of the bottom section, which is 1 m below the interface, or 2 m from the top surface.
- Lever arm, \(d_{2,rect} = 3 - 2 = 1\) m.
- Moment, \(M_{2,rect} = 19600 \times 1 = 19600\) Nm/m.
- Triangular Part (due to bottom fluid's own weight):
- Force, \(F_{2,tri} = (\rho_2 g h_{c2}) A_2\), where \(h_{c2} = h_2/2 = 1\) m (relative to the interface).
\[ F_{2,tri} = (2000 \times 9.8 \times 1) \times (2 \times 1) = 39200 \text{ N/m} \] - This force acts at \(\frac{2}{3}h_2 = \frac{2}{3} \times 2 = \frac{4}{3}\) m below the interface. Its total depth from the free surface is \(1 + \frac{4}{3} = \frac{7}{3}\) m.
- Lever arm, \(d_{2,tri} = 3 - \frac{7}{3} = \frac{2}{3}\) m.
- Moment, \(M_{2,tri} = 39200 \times \frac{2}{3} \approx 26133.3\) Nm/m.
Total Moment:
\[ M_{total} = M_1 + M_{2,rect} + M_{2,tri} \] \[ M_{total} = 11433.3 + 19600 + 26133.3 = 57166.6 \text{ Nm/m} \] Convert to kNm/m:
\[ M_{total} = 57.1666 \text{ kNm/m} \] Step 4: Final Answer:
Rounding off to one decimal place, the moment per unit width is 57.2 kNm/m.
Step 5: Why This is Correct:
The solution correctly breaks down the complex pressure distribution into manageable parts: a triangular distribution for the top fluid and a trapezoidal distribution (split into rectangular and triangular components) for the bottom fluid. The forces and their respective centers of pressure are calculated accurately, leading to the correct moments about the hinge. The sum of these moments gives the total moment, and the final value of 57.2 kNm/m is within the provided answer range of 55.9 to 58.5.