Water at room temperature flows with a constant speed of 8 m/s through a nozzle with a square cross section, as shown in the figure. Water enters the nozzle at point A and exits the nozzle at point B. The lengths of the sides of the square cross section at A and B are 64 cm and 44 cm, respectively.
(a) What is the volume flow rate at the exit?
(b) What is the acceleration at the exit? The length of the nozzle is 2 m.
(c) If the volume flow rate through the nozzle is increased to 7 m3/s, what is the acceleration of the fluid at the exit?
I got the correct answer for a which is 3.27 m^3/s, help me on part b and c please.
(b) and (c) are poorly worded questions. Once the water reaches the exit, acceleration stops. Prior to that, while water is still in the channel,
V*A = b^2*V = constant
= volume flow rate, Q
b is the side length of the square water channel
2 ln b + ln V = 0
(1/V)dV/dt = (1/V)*acceleration
= -2 (db/dt)(1/b)
= -2 (db/dx)*(1/b)*V
acceleration = -V^2*(1/b)*db/dx
To get an answer, you need to know the distance over which the side length b decreases from 0.64 to 044 m.
To solve part (b), we can use the principle of conservation of mass. The volume flow rate is given by the equation:
Q = A * v
where Q is the volume flow rate, A is the cross-sectional area, and v is the velocity of the fluid.
At the entrance (point A), the side length of the square cross-section is 64 cm, which is equal to 0.64 m. Therefore, the cross-sectional area at point A is:
A_A = (0.64 m)^2 = 0.4096 m^2
At the exit (point B), the side length of the square cross-section is 44 cm, which is equal to 0.44 m. Therefore, the cross-sectional area at point B is:
A_B = (0.44 m)^2 = 0.1936 m^2
We know that the volume flow rate is constant throughout the nozzle, so Q_A = Q_B.
At point A, we have:
Q_A = A_A * v_A
At point B, we have:
Q_B = A_B * v_B
Since Q_A = Q_B, we can write:
A_A * v_A = A_B * v_B
Substituting the values we have:
0.4096 m^2 * v_A = 0.1936 m^2 * 8 m/s
Solving for v_A:
v_A = (0.1936 m^2 * 8 m/s) / 0.4096 m^2
v_A = 3.78 m/s (approximately)
To find the acceleration at the exit (point B), we can use the equation:
a = (v_B - v_A) / t
Where a is the acceleration, v_B is the velocity at the exit, v_A is the velocity at the entrance, and t is the time taken to travel through the nozzle.
The length of the nozzle is given as 2 m. So, the time taken to travel through the nozzle is:
t = 2 m / v_A
Substituting the value of v_A, we get:
t = 2 m / 3.78 m/s
t = 0.53 s (approximately)
Substituting the values of v_B = 8 m/s and v_A = 3.78 m/s into the acceleration equation, we get:
a = (8 m/s - 3.78 m/s) / 0.53 s
a = 4.22 m/s^2 (approximately)
For part (c), if the volume flow rate through the nozzle is increased to 7 m^3/s, we can use the same approach as above to find the acceleration at the exit (point B).
First, find the velocity at the entrance (point A) using the equation:
Q_A = A_A * v_A
7 m^3/s = 0.4096 m^2 * v_A
v_A = 17.09 m/s (approximately)
Then, find the time taken to travel through the nozzle:
t = 2 m / v_A
t = 2 m / 17.09 m/s
t = 0.117 s (approximately)
Finally, find the acceleration at the exit (point B) using the equation:
a = (v_B - v_A) / t
Substituting the values v_B = 8 m/s, v_A = 17.09 m/s, and t = 0.117 s into the acceleration equation, we get:
a = (8 m/s - 17.09 m/s) / 0.117 s
a = -96.67 m/s^2 (approximately)
Note that the negative sign indicates that the fluid is decelerating or slowing down at the exit.
To solve part (b) and (c) of the problem, we need to use the principles of fluid dynamics and Bernoulli's equation. Let's break down the steps:
(b) To find the acceleration at the exit, we need to apply Bernoulli's equation, which relates pressure, velocity, and height of a fluid. However, before we can apply Bernoulli's equation, we need to determine the velocity at the exit.
1. Convert the lengths of the sides of the square cross-section at points A and B to meters:
- Length at A = 64 cm = 0.64 m
- Length at B = 44 cm = 0.44 m
2. Use the equation of continuity, which states that the volume flow rate at any point in a steady flow is constant:
- A1V1 = A2V2, where A1 and V1 are the cross-sectional area and velocity at point A, and A2 and V2 are the cross-sectional area and velocity at point B.
- Since the cross-section is square, the area is equal to the length of a side squared.
- A1 = (0.64 m)^2, A2 = (0.44 m)^2
3. From the problem statement, the velocity at point A is given as 8 m/s (constant speed). So, we can calculate the velocity at point B using the equation of continuity:
- V2 = (A1V1) / A2
4. Now that we have the velocity at the exit (V2), we can apply Bernoulli's equation:
- P1 + 1/2ρV1^2 + ρgh1 = P2 + 1/2ρV2^2 + ρgh2
- P1 is the pressure at point A (which we can assume to be atmospheric pressure), P2 is the pressure at point B (which we want to find), ρ is the density of water, g is the acceleration due to gravity (9.8 m/s^2), and h1 and h2 are the heights at points A and B (which we can assume to be the same).
5. Since the heights h1 and h2 are the same, they cancel out, and we can rewrite Bernoulli's equation as:
- P2 = P1 + 1/2ρ(V1^2 - V2^2)
6. Calculate the pressure difference P2 - P1 to find the pressure at point B.
Now, for part (c), we are given a new volume flow rate (7 m^3/s). We need to find the corresponding acceleration at the exit.
1. Repeat steps 1 and 2 from part (b) to find the new velocities at points A and B using the new volume flow rate.
2. Apply Bernoulli's equation as before, using the new velocities, to find the pressure at point B.
3. Calculate the pressure difference P2 - P1 to determine the pressure at point B.
4. Once you have the pressure at point B, you can calculate the acceleration at the exit by using Newton's second law for fluid flow:
- F = m * a, where F is the net force acting on the fluid, m is the mass flow rate, and a is the acceleration.
- The net force is the pressure difference (P2 - P1) multiplied by the cross-sectional area at the exit.
5. Rearrange the formula to solve for the acceleration:
- a = (P2 - P1) * A2 / ρ
Now you have the steps to solve parts (b) and (c) of the problem using the principles of fluid dynamics. Apply these steps with the values provided to find the solutions.