A tank has a circular orifice 20mm diameter in the vertical side near the bottom. It contains water to a depth of Im above the orifice with oil of relative density 0.8 for a depth of Im above the water. Acting on the upper surface of the oil-is an air pressure on 20kN/m². The jet of water issuing from the orifice travels a horizontal distance of 1.5m from the orifice while falling a vertical distance of 0.156m. If the coefficient of contraction of the orifice is 0.65, estimate. find:(b) The value of the coefficient of velocity, d (ii) The actual discharge through the orifice.

To solve this problem, we need to use the Torricelli's theorem, Bernoulli's equation, and the definitions of the coefficients of contraction (Cc) and velocity (Cv).

1. First, we need to calculate the total pressure head (H) at the orifice. This is the sum of the pressure head due to the air pressure, the pressure head due to the oil, and the pressure head due to the water.

   The pressure head due to the air pressure is given by P/ρg, where P is the air pressure, ρ is the density of the fluid (water in this case), and g is the acceleration due to gravity. Substituting the given values, we get 20,000/(1000*9.81) = 2.04m.

   The pressure head due to the oil is the height of the oil column times the relative density of the oil, which is 1*0.8 = 0.8m.

   The pressure head due to the water is simply the height of the water column, which is 1m.

   Adding these up, we get H = 2.04 + 0.8 + 1 = 3.84m.

2. According to Torricelli's theorem, the theoretical velocity (Vt) of the jet at the orifice is given by sqrt(2gH), where g is the acceleration due to gravity and H is the total pressure head at the orifice. Substituting the given values, we get Vt = sqrt(29.813.84) = 8.72m/s.

3. The actual velocity (Va) of the jet at the orifice can be calculated from the horizontal and vertical distances travelled by the jet. The horizontal velocity (Vx) is the horizontal distance divided by the time taken to fall the vertical distance, and the vertical velocity (Vy) is given by sqrt(2gD), where D is the vertical distance. The time taken to fall the vertical distance is sqrt(2D/g). Substituting the given values, we get Vx = 1.5/sqrt(20.156/9.81) = 8.5m/s and Vy = sqrt(29.81*0.156) = 1.75m/s. The actual velocity is the vector sum of these, which is sqrt(Vx^2 + Vy^2) = sqrt(8.5^2 + 1.75^2) = 8.68m/s.

4. The coefficient of velocity (Cv) is the ratio of the actual velocity to the theoretical velocity, which is Va/Vt = 8.68/8.72 = 0.995.

5. The actual discharge (Qa) through the orifice is given by the product of the area of the orifice (A), the actual velocity (Va), and the coefficient of contraction (Cc). The area of the orifice is πd²/4, where d is the diameter of the orifice. Substituting the given values, we get Qa = 0.65π(0.02)^2/4*8.68 = 0.00017m³/s or 0.17 litres/s.