A problem about fluid mechanics

[harini07]

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?

 

[haruspex]

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

 

[harini07]

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]

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!

 

[haruspex]

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" .

v^2 = 40*2*g

How do you get that from the relevant equation you quoted? What happened to p and rho?

 

[harini07]

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?

 

[haruspex]

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:

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

the rho is the density of the flowing fluid.

 

[harini07]

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 :/

 

[haruspex]

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/ (rho_Hg*g) =40cm. Look up what rho_Hg is and calculate the pressure p.

 

[harini07]

"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/ (rho_Hg*g) =40cm. Look up what rho_Hg 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 ?

 

[haruspex]

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.

 

[harini07]

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?

 

[haruspex]

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?

 

[harini07]

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?

 

[haruspex]

*40 cm of Hg column! so P=13.56(rho)*40*980(g) right?

Assuming consistent units for P, yes.

 

[harini07]

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 :/

 

[haruspex]

i m getting 531552 in the cgs units

That's as a pressure in cgs units, right?

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.

 

[harini07]

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?

 

[haruspex]

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.

 

[harini07]

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?

 

[harini07]

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 *

 

[haruspex]

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/(ρ_waterg)=(ρ_Hgh_Hgg)/(ρ_waterg)=ρ_Hgh_Hg/ρ_water
So all you needed to do was multiply the 40cm by the density ratio,