A tank of cross sectional area A1 discharges water through an orifice of area A2 at the bottom of the tank at a steady rate. If the density of the water is sigma find the mass flow rate of water from the tank. note a pipe of area A2 and length H/2b is attached to the orifice what's is the increase in the flow rate find the pressure at point C for the two cases.
Answers
Guys all parts are there.
Case 1: P1 + 0.5rhov1^2 = P2 + 0.5rho(A1/A2)^2 * v1^2
Case 2: P3 + 0.5rhov3^2 = P2 + 0.5rho(A1/A2)^2 * v3^2
To find,
The pressures at point C for the two cases.
Given,
Cross-sectional area A1
Orifice of area A2 at the bottom of the tank and
A pipe of area A2 and length H/2b
Solution,
The mass flow rate of water from the tank can be calculated using the continuity equation, which states that the mass flow rate is equal to the product of the density, cross-sectional area, and velocity of the fluid:
m_dot = rho * A1 * v1 = rho * A2 * v2,
where m_dot is the mass flow rate, rho is the density of water, A1 is the cross-sectional area of the tank, v1 is the velocity of the water in the tank, A2 is the area of the orifice, and v2 is the velocity of the water flowing through the orifice.
Since the water is discharging at a steady rate, the velocities v1 and v2 are constant. Therefore, we can write:
m_dot = rho * A1 * v1 = rho * A2 * v2
v2 = (A1/A2) * v1
The pressure at point C can be found using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a horizontal flow:
P1 + 0.5rhov1^2 + rhogh1 = P2 + 0.5rhov2^2 + rhogh2,
where P1 and P2 are the pressures at points 1 and 2, v1 and v2 are the velocities at points 1 and 2, h1 and h2 are the heights of the water surface above points 1 and 2, and g is the acceleration due to gravity.
In the first case, without the pipe attached to the orifice, the velocity of the water flowing through the orifice is v2 = (A1/A2) * v1. Since the height of the water surface is constant, we have h1 = h2. Therefore, Bernoulli's equation reduces to:
P1 + 0.5rhov1^2 = P2 + 0.5rho(A1/A2)^2 * v1^2
The pressure at point C in this case is P2, which can be found by rearranging the equation:
P2 = P1 + 0.5rhov1^2 - 0.5rho(A1/A2)^2 * v1^2
In the second case, with the pipe attached to the orifice, the water flows through the pipe before discharging into the atmosphere. Let the height of the water in the pipe be h3. Applying Bernoulli's equation between points 1 and 3, we get:
P1 + 0.5rhov1^2 + rhogh1 = P3 + 0.5rhov3^2 + rhogh3,
where v3 is the velocity of water in the pipe. Since the pipe is long and narrow, we can assume that the velocity of the water in the pipe is much smaller than the velocity of the water flowing through the orifice, i.e., v3 << v2. Therefore, we can neglect the term 0.5rhov3^2 in the above equation. Also, since the height of the water surface is constant, we have h1 = h3. Therefore, the equation reduces to:
P1 + 0.5rhov1^2 = P3
Applying Bernoulli's equation between points 3 and 2, we get:
P3 + 0.5rhov3^2 = P2 + 0.5rho(A1/A2)^2 * v3^2
Since v3 << v2, we can neglect the term 0.5rho(
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