这是一份BMEN90036的作业。首先，Question 1 [30 marks] To this point we have assumed that the height in tanks remains constant, which is appropriate for the case when we want to know the instantaneous flow rate for fluid moving between tanks. This is more complicated for analysing systems over longer timescales, however, when we want to know how the overall system evolves when the level in tanks change. We want to fill a 0.5m diameter tank of a bioprocessing slurry (ρ = 1020 kg/m^3, μ = 0.0015 kg/ms) initially with a height of 2m to 6m using a pump operating with 82% mechanical efficiency and a brake power of 1500W if it is moving fluid from an immediately adjoining 1.5m diameter tank, initially filled to 2m. The pump and any entrance/exits to the tanks are located on the floor between the two tanks (at a height of 0m). The pump curve for this pump can be approximated by the equation hp = 20.2 – 0.02Q^2, where Q is the flow rate in m^3/h and hp has units of meters. Both tanks are vented to atmosphere. Because these tanks are so close together, you can ignore minor losses and friction effects from pipe tubing. (Hint: you may not require all the values provided in the problem statement to calculate the answers to each of the following) A. Potentially using a MATLAB script, calculate the amount of time to fill Tank 2 to the desired height. [20 marks] B. What is the maximum (absolute) height to which the fluid could be pumped in Tank 2 (provided Tank 2 were sufficiently tall)? [5 marks] C. Following on from your answer to part (A), take the case where there is now a pipe (5 cm diameter, surface roughness of 0.03 mm) connecting Tank 1 to the pump, which we can assume has an NPSHR of 2.8m. The fluid has a vapour pressure of 4.2kPa. Ignore minor losses associated with entry and exit of pipes. Find the longest this pipe can be before cavitation occurs at any point in the pumping process for the situation described, where the height in Tank 2 is to be raised to 6m high. For simplicity, you can assume the pump head from the pump curve for the tank height at which cavitation is most likely to occur, though you should state whether this assumption will yield an overestimate or underestimate of the actual pipe length. [5 marks]
Question
这是一份BMEN90036的作业。首先，Question 1 [30 marks] To this point we have assumed that the height in tanks remains constant, which is appropriate for the case when we want to know the instantaneous flow rate for fluid moving between tanks. This is more complicated for analysing systems over longer timescales, however, when we want to know how the overall system evolves when the level in tanks change. We want to fill a 0.5m diameter tank of a bioprocessing slurry (ρ = 1020 kg/m^3, μ = 0.0015 kg/ms) initially with a height of 2m to 6m using a pump operating with 82% mechanical efficiency and a brake power of 1500W if it is moving fluid from an immediately adjoining 1.5m diameter tank, initially filled to 2m. The pump and any entrance/exits to the tanks are located on the floor between the two tanks (at a height of 0m). The pump curve for this pump can be approximated by the equation hp = 20.2 – 0.02Q^2, where Q is the flow rate in m^3/h and hp has units of meters. Both tanks are vented to atmosphere. Because these tanks are so close together, you can ignore minor losses and friction effects from pipe tubing. (Hint: you may not require all the values provided in the problem statement to calculate the answers to each of the following) A. Potentially using a MATLAB script, calculate the amount of time to fill Tank 2 to the desired height. [20 marks] B. What is the maximum (absolute) height to which the fluid could be pumped in Tank 2 (provided Tank 2 were sufficiently tall)? [5 marks] C. Following on from your answer to part (A), take the case where there is now a pipe (5 cm diameter, surface roughness of 0.03 mm) connecting Tank 1 to the pump, which we can assume has an NPSHR of 2.8m. The fluid has a vapour pressure of 4.2kPa. Ignore minor losses associated with entry and exit of pipes. Find the longest this pipe can be before cavitation occurs at any point in the pumping process for the situation described, where the height in Tank 2 is to be raised to 6m high. For simplicity, you can assume the pump head from the pump curve for the tank height at which cavitation is most likely to occur, though you should state whether this assumption will yield an overestimate or underestimate of the actual pipe length. [5 marks]
Solution
This is a complex problem that involves fluid dynamics and pump mechanics. Here's how you can approach each part:
A. To calculate the time to fill Tank 2 to the desired height, you need to first calculate the volume of fluid that needs to be added to Tank 2. The volume of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height. The radius of Tank 2 is 0.5m/2 = 0.25m, and the height that needs to be filled is 6m - 2m = 4m. So, the volume to be filled is V = π*(0.25m)²*4m.
Next, you need to find the flow rate Q at which the pump operates. This can be found from the pump curve equation given, hp = 20.2 – 0.02Q². You can solve this equation for Q using the given brake power and mechanical efficiency of the pump.
Finally, the time to fill the tank can be calculated by dividing the volume to be filled by the flow rate.
B. The maximum height to which the fluid could be pumped can be found by setting the pump head hp to zero in the pump curve equation and solving for Q. This gives the maximum flow rate at which the pump can operate. The maximum height can then be found by using the formula for the energy equation of a pump, h = (P/ρg) + (v²/2g) + z, where P is the pressure, ρ is the density, g is the acceleration due to gravity, v is the velocity, and z is the height. Since the tanks are vented to atmosphere, P = 0, and since we're looking for the maximum height, v = 0. So, the maximum height is just z = h.
C. Cavitation occurs when the pressure in the fluid drops below its vapor pressure. The Net Positive Suction Head Required (NPSHR) is given as 2.8m. The longest pipe length before cavitation occurs can be found by using the formula for the energy equation of a pump and setting it equal to the NPSHR. This will give an equation in terms of the pipe length, which can be solved to find the maximum length. Note that this is a simplification and assumes that the pump head from the pump curve for the tank height at which cavitation is most likely to occur. Whether this yields an overestimate or underestimate of the actual pipe length depends on the specific characteristics of the pump and fluid.
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