A problem about fluid mechanics

In summary: That's the velocity head. To find the velocity, take the square root of both sides.ok now finding the pressure head of water 531552(pressure as found in cgs unit) /[1 (rho of water) *980] =542 and equating this with v^2/2g i m getting 1032 .the answer is not this as...That's the velocity head. To find the velocity, take the square root of both sides.In summary, the problem involves finding the speed at which the velocity head of water is equal to a pressure head of 40 cm of Hg. The equation used is v^2/2g = p/rho*g, where p is the pressure,
  • #1
harini07
79
2
1. The problem statement, all variables and given/known d
At what speed, the velocity head of water is equal to pressure head of 40 cm of Hg?

(1) 10.32 m/s
(2) 2.8 m/s
(3) 5.6 m/s

Homework Equations


v^2/2g = p/rho*g (velocity head= pressure head).

The Attempt at a Solution


since pressure head is given as 40 cm/s, i did it like v^2 = 40*2*g which i got after taking square root and all as 28.8 cm/s. i couldn't find my answer matching with any of the given options how to do this?
 
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  • #2
Where did you bring in the fact that the pressure is given in terms of Hg?
Also, please show all your working.
 
  • #3
it was given in the ques that the pressure is in Hg. and i have already shown my working in "my attempt at the solution" .
 
  • #4
haruspex said:
Where did you bring in the fact that the pressure is given in terms of Hg?
Also, please show all your working.
i have already shown my working in "my attempt at the solution" section. that's all i did..i equated both the equations and got an answer which is irrelevant to the options!
 
  • #5
harini07 said:
it was given in the ques that the pressure is in Hg. and i have already shown my working in "my attempt at the solution" .
harini07 said:
v^2 = 40*2*g
How do you get that from the relevant equation you quoted? What happened to p and rho?
 
  • #6
haruspex said:
How do you get that from the relevant equation you quoted? What happened to p and rho?
so as per the question, isn't the pressure head = 40 cm of Hg which means p/ rho*2 =40 and we know v^2/2g= presssure head = 40 cm of Hg. this is how i proceeded..is that right?
 
  • #7
harini07 said:
pressure head = 40 cm of Hg which means p/ rho*2 =40
I assume you meant p/ (rho*g) =40, but there are two things you need to be careful about there.
1. The 40 is in cm - always keep track of units;
2. The rho is the density of Hg, whereas in this equation:
harini07 said:
v^2/2g = p/rho*g (velocity head= pressure head).
the rho is the density of the flowing fluid.
 
  • #8
haruspex said:
I assume you meant p/ (rho*g) =40, but there are two things you need to be careful about there.
1. The 40 is in cm - always keep track of units;
2. The rho is the density of Hg, whereas in this equation:

the rho is the density of the flowing fluid.
so if rho is the density of flowing fluid (water) is 1 g/cm^3 right? but i m still confused whether the given value of 40 cm of Hg is the value of pressure or that of pressure head. 'cause as you told if i have to take care of rho as the density of fluid in flow (1 gm/cm^3) and p=40 cm of Hg which means 40/(1*980) =0.041 and then equating it with v^2/2g, i am not getting the result :/
 
  • #9
harini07 said:
i m still confused whether the given value of 40 cm of Hg is the value of pressure or that of pressure head
"40cm" is neither here. If the problem were about flowing Hg it could be the pressure head.
"40cm of Hg" is the pressure.
You have p/ (rhoHg*g) =40cm. Look up what rhoHg is and calculate the pressure p.
 
  • #10
haruspex said:
"40cm" is neither here. If the problem were about flowing Hg it could be the pressure head.
"40cm of Hg" is the pressure.
You have p/ (rhoHg*g) =40cm. Look up what rhoHg is and calculate the pressure p.
why should i calculate pressure there since i have been asked to find velocity only.can't i simply equate v^2/2g=40 cm of Hg ?
 
  • #11
harini07 said:
why should i calculate pressure there since i have been asked to find velocity only.can't i simply equate v^2/2g=40 cm of Hg ?
Sure, but you need the pressure head expressed in terms of such-and-such a height of the flowing fluid, i.e. water. The pressure head is NOT 40cm.
 
  • #12
haruspex said:
Sure, but you need the pressure head expressed in terms of such-and-such a height of the flowing fluid, i.e. water. The pressure head is NOT 40cm.
so what is the "40 cm of Hg" given in the question?
 
  • #13
harini07 said:
so what is the "40 cm of Hg" given in the question?
It is the gauge pressure at a depth of 40cm in a column of Hg. What depth of water would produce the same pressure?
 
  • #14
haruspex said:
It is the gauge pressure at a depth of 40cm in a column of Hg. What depth of water would produce the same pressure?
*40 cm of Hg column! so P=13.56(rho)*40*980(g) right?
 
  • #15
harini07 said:
*40 cm of Hg column! so P=13.56(rho)*40*980(g) right?
Assuming consistent units for P, yes.
 
  • #16
haruspex said:
Assuming consistent units for P, yes.
so i m getting 531552 in the cgs units. should i equate this with v^2/2g ? the answer is exceeding the relevant options beyond limits :/
 
  • #17
harini07 said:
i m getting 531552 in the cgs units
That's as a pressure in cgs units, right?
harini07 said:
should i equate this with v^2/2g ?
No, that formula needs a distance, i.e. the pressure head of water that results in the above pressure.
 
  • #18
haruspex said:
That's as a pressure in cgs units, right?

No, that formula needs a distance, i.e. the pressure head of water that results in the above pressure.
how to find that? i am baffled :/ so i have found the pressure value in cgs unit and now how to convert it as such distance value?
 
  • #19
harini07 said:
how to find that? i am baffled :/ so i have found the pressure value in cgs unit and now how to convert it as such distance value?
You have found the value of the pressure, p. To find what that is as pressure head of water, you can use p/(ρg) again, but now using the density of water.
 
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  • #20
haruspex said:
You have found the value of the pressure, p. To find what that is as pressure head of water, you can use p/(ρg) again, but now using the density of water.
ok now finding the pressure head of water 531552(pressure as found in cgs unit) /[1 (rho of water) *980] =542 and equating this with v^2/2g i m getting 1032 .the answer is not this as per the options right?
 
  • #21
harini07 said:
ok now finding the pressure head of water 531552(pressure as found in cgs unit) /[1 (rho of water) *980] =542 and equating this with v^2/2g i m getting 1032 .the answer is not this as per the options right?
sorry wait voila...the answer in metre.i found in cm/sec! so it would be 10.32 m/s. i got it.. :D thanks for helping me to solve this..sorry i took more time..*a slow student *
 
  • #22
harini07 said:
sorry wait voila...the answer in metre.i found in cm/sec! so it would be 10.32 m/s. i got it.. :D thanks for helping me to solve this..sorry i took more time..*a slow student *
It might be useful to go back to your original equations, in a bit more detail
v2/(2g)=p/(ρwaterg)=(ρHghHgg)/(ρwaterg)=ρHghHg/ρwater
So all you needed to do was multiply the 40cm by the density ratio,
 
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1. What is fluid mechanics?

Fluid mechanics is the study of how fluids (liquids and gases) behave and interact with their surroundings. It involves understanding the properties of fluids, such as density and viscosity, and how these properties affect the motion of fluids.

2. What are some real-world applications of fluid mechanics?

Fluid mechanics has many practical applications, including designing efficient engines and turbines, understanding weather patterns, and developing medical devices such as ventilators. It is also used in industries such as aviation, automotive, and chemical engineering.

3. What is the difference between a laminar and turbulent flow?

In a laminar flow, the fluid particles move in a smooth, orderly fashion, while in a turbulent flow, the fluid particles move in an irregular, chaotic manner. Laminar flows are typically observed at low velocities and in highly viscous fluids, while turbulent flows are observed at higher velocities and in less viscous fluids.

4. How is Bernoulli's principle related to fluid mechanics?

Bernoulli's principle states that as the speed of a fluid increases, its pressure decreases. This principle is a fundamental concept in fluid mechanics and is often used to explain the lift force on an airplane wing or the flow of water through a pipe.

5. What is the significance of Reynolds number in fluid mechanics?

The Reynolds number is a dimensionless quantity that describes the ratio of inertial forces to viscous forces in a fluid. It is used to determine whether a flow is laminar or turbulent, and it helps predict the behavior of fluids in different situations, such as around objects or in pipes.

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