A problem about fluid mechanics

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Homework Help Overview

The problem involves fluid mechanics, specifically relating to the velocity head and pressure head of water in a system where the pressure is given in terms of mercury (Hg). The original poster attempts to find the speed at which the velocity head of water equals a pressure head of 40 cm of Hg.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between pressure head and velocity head, questioning how to correctly interpret the given pressure in terms of Hg. There are attempts to equate pressure head with velocity head using various equations, while also addressing unit conversions and the significance of fluid densities.

Discussion Status

The discussion is ongoing, with participants providing guidance on the need to express pressure in terms of the flowing fluid (water) and clarifying the role of density in the calculations. There is a recognition of the complexity involved in converting pressure values to pressure head, and participants are exploring different interpretations of the problem.

Contextual Notes

There is confusion regarding whether the given value of 40 cm of Hg represents pressure or pressure head, and participants are encouraged to keep track of units and fluid densities throughout their calculations.

harini07
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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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Where did you bring in the fact that the pressure is given in terms of Hg?
Also, please show all your working.
 
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" .
 
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!
 
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?
 
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?
 
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.
 
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 :/
 
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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