Hi
Probably there are errors in my description below but I believe you can see what I'm driving at. In simple words, I want to know what happens at microscopic levels which reduces the pressure in region B of the tube when speed of flow gets doubled. Please help me with it. Thank you very much for your time and help.
Suppose the tube has rectangular cross section like this.
Let's assume that the central region, B, can hold five liters of water and both regions, A and C, can hold 10 liters of water. Suppose that water is entering the tube at the rate of 10 liters per second which means every second 10 liters of water enters A region and at the same time 10 liters leave the C region. This would simply mean that water in the region B has to have double the speed compared to other two regions. Let's say in regions A and C speed of water is 50 m/s and therefore speed in the region B is 100 m/s.
At microscopic level liquid pressure is result of collisions with the walls of tube. The more forceful collisions are, the more pressure will be exerted. I believe the liquid pressure has also to do something with the frequency of collisions. For instance, if the number of collisions of individual molecules with a certain wall of the tube becomes halved, then I believe the pressure would also get halved. Please correct me if I'm wrong.
Although the speed of the water in B region is twice (which was assumed to be 100 m/s), I don't think it would affect the individual velocity of molecules (let's say average velocity of water molecules is 350 m/s). Only the "y" component of molecules' velocity contribute to the pressure.
Suppose that if the tube were of uniform cross section as that of region B and water were flowing at rate of 50 m/s and volume of water flowing per second through region B was 5 liters (please note that region B was assumed to hold only 5 liters of water), then the number of collisions of water molecules every second per wall of region B was 500. But in our case the tube is not of uniform cross section and speed of water flow in region B is 100 m/s which results in half the number of collisions per second, i.e. 250. But twice the speed in region B means that twice the volume of water flows per second, i.e. 10 liters. This twice the volume would keep the number of collisions per second per wall of region B constant, i.e. 500. Right?
If the number of collisions per wall of region B per second does not change when the speed of the flow through region B gets doubled, then I think pressure should not change either; it should remain constant.
Regards
PG
Probably there are errors in my description below but I believe you can see what I'm driving at. In simple words, I want to know what happens at microscopic levels which reduces the pressure in region B of the tube when speed of flow gets doubled. Please help me with it. Thank you very much for your time and help.
Suppose the tube has rectangular cross section like this.
Let's assume that the central region, B, can hold five liters of water and both regions, A and C, can hold 10 liters of water. Suppose that water is entering the tube at the rate of 10 liters per second which means every second 10 liters of water enters A region and at the same time 10 liters leave the C region. This would simply mean that water in the region B has to have double the speed compared to other two regions. Let's say in regions A and C speed of water is 50 m/s and therefore speed in the region B is 100 m/s.
At microscopic level liquid pressure is result of collisions with the walls of tube. The more forceful collisions are, the more pressure will be exerted. I believe the liquid pressure has also to do something with the frequency of collisions. For instance, if the number of collisions of individual molecules with a certain wall of the tube becomes halved, then I believe the pressure would also get halved. Please correct me if I'm wrong.
Although the speed of the water in B region is twice (which was assumed to be 100 m/s), I don't think it would affect the individual velocity of molecules (let's say average velocity of water molecules is 350 m/s). Only the "y" component of molecules' velocity contribute to the pressure.
Suppose that if the tube were of uniform cross section as that of region B and water were flowing at rate of 50 m/s and volume of water flowing per second through region B was 5 liters (please note that region B was assumed to hold only 5 liters of water), then the number of collisions of water molecules every second per wall of region B was 500. But in our case the tube is not of uniform cross section and speed of water flow in region B is 100 m/s which results in half the number of collisions per second, i.e. 250. But twice the speed in region B means that twice the volume of water flows per second, i.e. 10 liters. This twice the volume would keep the number of collisions per second per wall of region B constant, i.e. 500. Right?
If the number of collisions per wall of region B per second does not change when the speed of the flow through region B gets doubled, then I think pressure should not change either; it should remain constant.
Regards
PG
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