What Does Head in Fluid Dynamics Refer To?

[foo9008]

Homework Statement

the water flow in the pipe in upward direction . the pipe with internal diameter 60mm carries water with velocity = 3m/s under head of 30cm . Determine the resultant force on the bend of 70 degree and its direction.

Homework Equations

The Attempt at a Solution

my question is what does the head refer to ? from where to where ? red pen refer to from datum to the inlet , blue pen means from inlet to outlet. which is correct , blue or red?

second is is P1 = P2 ? ( assuming no friction loss)

 

[BvU]

OK, for the head you can take the center of the flow lines, so where you wrote the 'circumcircled' F1 and F2

And the pressures are not identical: $\Delta p = \rho g \Delta h$ (Archimedes, Bernoulli)

[edit] Oops, I suppose 'head of 30 cm' refers to the pressure at F1. So at F1 the pressure is $p = \rho g \times 0.3 {\rm \ m}$

 

[foo9008]

OK, for the head you can take the center of the flow lines, so where you wrote the 'circumcircled' F1 and F2

And the pressures are not identical: $\Delta p = \rho g \Delta h$ (Archimedes, Bernoulli)

[edit] Oops, I suppose 'head of 30 cm' refers to the pressure at F1. So at F1 the pressure is $p = \rho g \times 0.3 {\rm \ m}$

if inlet diameter = outlet diameter , the pressure at both inlet and outlet also not the same ? assuming no friction loss

 

[foo9008]

OK, for the head you can take the center of the flow lines, so where you wrote the 'circumcircled' F1 and F2

And the pressures are not identical: $\Delta p = \rho g \Delta h$ (Archimedes, Bernoulli)

[edit] Oops, I suppose 'head of 30 cm' refers to the pressure at F1. So at F1 the pressure is $p = \rho g \times 0.3 {\rm \ m}$

do you mean the red line is correct?

 

[Chestermiller]

The head H is defined as $H=\frac{p}{\rho g}+z$, where z is the elevation above a fixed datum. In this case, the datum for z = 0 is to be taken as the elevation of the pipe inlet (red location). So, at this location, H = 0.3 m. At the outlet, the pressure may be different and, of course, the elevation is different. This datum applies to the entire pipe and bend.

The head H can be interpreted physically as the height to which fluid in a static column would rise above the datum elevation (if the column were attached to the pipe) at any given location along the pipe.

 

[foo9008]

The head H is defined as $H=\frac{p}{\rho g}+z$, where z is the elevation above a fixed datum. In this case, the datum for z = 0 is to be taken as the elevation of the pipe inlet (red location). So, at this location, H = 0.3 m. At the outlet, the pressure may be different and, of course, the elevation is different. This datum applies to the entire pipe and bend.

The head H can be interpreted physically as the height to which fluid in a static column would rise above the datum elevation (if the column were attached to the pipe) at any given location along the pipe.

so , the 0.3 refers to the distance of inlet and outlet ? ( blue line ) ?

 

[Chestermiller]

so , the 0.3 refers to the distance of inlet and outlet ? ( blue line ) ?

No.
The 0.3m refers to how high water would rise vertically within a tube that is attached to the pipe at its inlet cross section. There is no tube actually attached there, but, if there were, this is how high the fluid would rise.

 

[foo9008]

No.
The 0.3m refers to how high water would rise vertically within a tube that is attached to the pipe at its inlet cross section. There is no tube actually attached there, but, if there were, this is how high the fluid would rise.

ok, so the Fx = rho(g)(v2-v1) - P1A1 +P2A2cos(theta) ?
Fx= 1000(0.848)(3cos70 -3) - (0.3)(9.81 x1000)(A1) , but how to find P2 ?

 

[Chestermiller]

ok, so the Fx = rho(g)(v2-v1) - P1A1 +P2A2cos(theta) ?
Fx= 1000(0.848)(3cos70 -3) - (0.3)(9.81 x1000)(A1) , but how to find P2 ?

I'm not checking your equation but, if it is a section of pipe in which there is just a bend in the pipe, the assumption is that the exit cross section area is equal to the inlet cross section area. What does the Bernoulli equation tell you about the relationship between P1 and P2? (The equation you are using to find the force is not the Bernoulli equation; it is the macroscopic momentum balance equation). What does the Bernoulli equation tell you about H1 and H2?

 

[foo9008]

I'm not checking your equation but, if it is a section of pipe in which there is just a bend in the pipe, the assumption is that the exit cross section area is equal to the inlet cross section area. What does the Bernoulli equation tell you about the relationship between P1 and P2? (The equation you are using to find the force is not the Bernoulli equation; it is the macroscopic momentum balance equation). What does the Bernoulli equation tell you about H1 and H2?

so P 1= P 2?

since bernoullis equation =
P1 / (rho)(g) + z1 + (V1)^2 / 2g = P2 / (rho)(g) + z2 + (V2)^2 / 2g

since Q=Av = constant , cancelling off z and v , P 1= P2 ?

 

[Chestermiller]

so P 1= P 2?

since bernoullis equation =
P1 / (rho)(g) + z1 + (V1)^2 / 2g = P2 / (rho)(g) + z2 + (V2)^2 / 2g

since Q=Av = constant , cancelling off z and v , P 1= P2 ?

No. In this problem, z1 and z2 are not equal. The inlet and outlet are at different elevations. If you rewrite Bernoulli's equation in terms of head H, then
$$H_1+\frac{v_1^2}{2g}=H_2+\frac{v_2^2}{2g}$$
What does this tell you about H1 and H2?

 

[foo9008]

No. In this problem, z1 and z2 are not equal. The inlet and outlet are at different elevations. If you rewrite Bernoulli's equation in terms of head H, then
$$H_1+\frac{v_1^2}{2g}=H_2+\frac{v_2^2}{2g}$$
What does this tell you about H1 and H2?

Like this? Sorry, I am not in front of computer now

 

[Chestermiller]

Like this? Sorry, I am not in front of computer now

Huh?? What are you asking?

 

[foo9008]

Huh?? What are you asking?

H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??
like this ?

 

[foo9008]

Huh?? What are you asking?

i asked my senior , he told me that this is a closed pipe , so the Q =constant , as area for inlet and outlet same velocity = constant (v1=v2) , this act like the hydarulic jack , so the pressure transferred is constant throughout the pipe ... i doubt is the statement correct ? can someone explain ?

 

[Chestermiller]

i asked my senior , he told me that this is a closed pipe , so the Q =constant , as area for inlet and outlet same velocity = constant (v1=v2) , this act like the hydarulic jack , so the pressure transferred is constant throughout the pipe ... i doubt is the statement correct ? can someone explain ?

If that's what he said, it's not correct. Suppose that Q = 0. Certainly your analysis should be able to handle this specific case. If Q = 0, the system is in hydrostatic equilibirum.

Multiple Choice Question:
Which of the following answers are correct for a system in hydrostatic equilibrium (like a hydraulic jack)?

(a) the pressure is uniform throughout
(b) the pressure is constant horizontally
(c) the pressure varies vertically
(d) the head is uniform throughout

Chet

 

[foo9008]

If that's what he said, it's not correct. Suppose that Q = 0. Certainly your analysis should be able to handle this specific case. If Q = 0, the system is in hydrostatic equilibirum.

Multiple Choice Question:
Which of the following answers are correct for a system in hydrostatic equilibrium (like a hydraulic jack)?

(a) the pressure is uniform throughout
(b) the pressure is constant horizontally
(c) the pressure varies vertically
(d) the head is uniform throughout

Chet

for MCQ , my ans is (a) the pressure is uniform throughout
so , the conclusion is P1 are not the same with P2 ?
relationship between P1 and P2 must b e
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??

 

[Chestermiller]

for MCQ , my ans is (a) the pressure is uniform throughout
so , the conclusion is P1 are not the same with P2 ?
relationship between P1 and P2 must b e
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??

(b), (c), and (d) are all correct. (a) is incorrect.

 

[foo9008]

(b), (c), and (d) are all correct. (a) is incorrect.

in hydraulic jack , why the pressure isn't uniform throughout ? the hydarulic jack has the same pressure at both ends , enabling the small force applied to carry big force , right ?

 

[Chestermiller]

in hydraulic jack , why the pressure isn't uniform throughout ? the hydarulic jack has the same pressure at both ends , enabling the small force applied to carry big force , right ?

In a system at hydrostatic equilibrium, does the pressure increase with depth or doesn't it? Why should the pressure within a hydraulic jack be uniform?

 

[foo9008]

In a system at hydrostatic equilibrium, does the pressure increase with depth or doesn't it? Why should the pressure within a hydraulic jack be uniform?

i mean the pressure is constant at both ends

 

[foo9008]

In a system at hydrostatic equilibrium, does the pressure increase with depth or doesn't it? Why should the pressure within a hydraulic jack be uniform?

maybe i heard it wrongly . i forgt he gt said 'hydraulic jacks' or not , but most importantly , he stressed that , the Q is constant (not equal to 0 ) , so v1 = v2 , leading to P1 = P2 (consequence of bernoulli's principle)

 

[Chestermiller]

i mean the pressure is constant at both ends

Only if the two ends are at the same elevation. Otherwise, there is a difference in the pressures. In analyzing a hydraulic jack, we usually neglect the hydrostatic pressure variations in the vertical direction because they are small compared to the applied pressure. In your problem, this is not the case. Otherwise, they wouldn't show the inlet and outlet at different elevations. Of course, at hydrostatic equilibrium, the hydraulic head is independent of vertical location, and is thus uniform.

 

[Chestermiller]

maybe i heard it wrongly . i forgt he gt said 'hydraulic jacks' or not , but most importantly , he stressed that , the Q is constant (not equal to 0 ) , so v1 = v2 , leading to P1 = P2 (consequence of bernoulli's principle)

Bernoulli's principle does not predict this. You left out the z terms.

 

[foo9008]

Bernoulli's principle does not predict this. You left out the z terms.

ok , so
relationship between P1 and P2 is
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??
am i correct ?

 

[Chestermiller]

ok , so
relationship between P1 and P2 is
H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
P1- P2 = (V1 ^2 ) - (V2 ^2) ??
am i correct ?

No. $$H_1-H_2=\frac{(P_1-P_2)}{\rho g}+(z_1-z_2)$$
You need to get the algebra correct and you need to include the elevations.
Also,
H1 - H2 =[ ( (V2 ^2 ) - (V1 ^2) ) / 2g ]

 

[Chestermiller]

For this problem, I think that they expect you to neglect the change in elevation between inlet and outlet. Why? Because they don't even tell you what these elevations are. They also expect you to neglect the weight of the water in the pipe bend in determining the vertical force on the bend, because they don't even give you enough information to determine the weight. So, if the effect of the change in elevation is negligible, you can neglect the elevation terms, and you can take the inlet and outlet pressures as being essentially equal.

chet