Help with a question please.
Discussion
Could someone help with this? I'm sure we're missing something simple, but, anyway....
The ends of a U-tubed manometer are connected to two different pipes each containing water. Pipe A is at a higher pressure than pipe B, and pipe A is 0.75m higher than pipe B. The U-tube manometer has a difference in mercury column height of 0.58m. The left hand limb mercury level is the lowest due to the relative higher pressure in pipe A and is 1.5m below the central axis of pipe A.
Assuming that acceleration due to gravity is 9.81m/s^2, and the densities of water and mercury are 1,000kg/m^3 and 13,600kg/m^3 respectively, calculate the difference in pressure between the two water pipes.
Could you show me the method used for your answer, so I understand how the value was calculated?
Thanks in advance.
The ends of a U-tubed manometer are connected to two different pipes each containing water. Pipe A is at a higher pressure than pipe B, and pipe A is 0.75m higher than pipe B. The U-tube manometer has a difference in mercury column height of 0.58m. The left hand limb mercury level is the lowest due to the relative higher pressure in pipe A and is 1.5m below the central axis of pipe A.
Assuming that acceleration due to gravity is 9.81m/s^2, and the densities of water and mercury are 1,000kg/m^3 and 13,600kg/m^3 respectively, calculate the difference in pressure between the two water pipes.
Could you show me the method used for your answer, so I understand how the value was calculated?
Thanks in advance.
https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&...
I believe the example you want is second from the bottom. Just work out the heights, use the correct density at the right points and that should see you to the answer.
Been a long while since I have done anything like this so I will have a go too if I get chance at lunch and see what I get!
I believe the example you want is second from the bottom. Just work out the heights, use the correct density at the right points and that should see you to the answer.
Been a long while since I have done anything like this so I will have a go too if I get chance at lunch and see what I get!
Done it just now. I got a difference in pressure of 64333.98 Pa.
So Pipe A is the start, with pressure p(A). Where the manometer joins, this is pressure p(1). p(A) = p(1).
The pressure then increases to point 2, pressure p(2). It increases by density(water)*g*h(1). h(1) in this case is 1.5 m which is the height of pipe A central axis, to the height on the mercury in the left hand side of the manometer.
pressure at point 2 is the same as point 3 which is the same height as point 2 but on the right hand side of the manometer.
from point 3, the pressure decreases to point 4, pressure p(4). Point 4 is where the mercury ends in the right hand side. Pressure here is reduced by density(mercury)*g*h(2). Here h(2) is 0.58 m or the difference in mercury height.
from point 4 to point 5, pressure p(5) where the manometer and pipe meet the pressure reduces again. This time it reduces by density(water)*g*h(3). In this case, h(3) is 0.17m (1.5 - 0.58 - 0.75). Pressure at point 5 is the same as in pipe B.
Your equation should look like:
p(A) + density(water)*g*h(1) - density(mercury)*g*h(2) - density(water)*g*h(3) = p(B).
Pressure at A is higher so re-arrange to get p(A) - p(B) =
= density(mercury)*g*h(2) + density(water)*g*h(3) - density(water)*g*h(1)
= 13600*9.81*0.58 + 1000*9.81*0.17 - 1000*9.81*1.5 = 64333.98 Pa.
64.33 kPa or about 0.63 of a bar.
So Pipe A is the start, with pressure p(A). Where the manometer joins, this is pressure p(1). p(A) = p(1).
The pressure then increases to point 2, pressure p(2). It increases by density(water)*g*h(1). h(1) in this case is 1.5 m which is the height of pipe A central axis, to the height on the mercury in the left hand side of the manometer.
pressure at point 2 is the same as point 3 which is the same height as point 2 but on the right hand side of the manometer.
from point 3, the pressure decreases to point 4, pressure p(4). Point 4 is where the mercury ends in the right hand side. Pressure here is reduced by density(mercury)*g*h(2). Here h(2) is 0.58 m or the difference in mercury height.
from point 4 to point 5, pressure p(5) where the manometer and pipe meet the pressure reduces again. This time it reduces by density(water)*g*h(3). In this case, h(3) is 0.17m (1.5 - 0.58 - 0.75). Pressure at point 5 is the same as in pipe B.
Your equation should look like:
p(A) + density(water)*g*h(1) - density(mercury)*g*h(2) - density(water)*g*h(3) = p(B).
Pressure at A is higher so re-arrange to get p(A) - p(B) =
= density(mercury)*g*h(2) + density(water)*g*h(3) - density(water)*g*h(1)
= 13600*9.81*0.58 + 1000*9.81*0.17 - 1000*9.81*1.5 = 64333.98 Pa.
64.33 kPa or about 0.63 of a bar.
Edited by Otispunkmeyer on Monday 2nd November 10:29
Edited by Otispunkmeyer on Monday 2nd November 10:31
Gassing Station | Science! | Top of Page | What's New | My Stuff