Homework help,engineers?

S

smooticus

Active member
So in order for me to make it thanksgiving i have to finish this homework that has me stumped.
Heres the first. I have a 2L bottle that can hold 8atm of pressure. Im adding enough CO2 to meet that max. I need to estimate the velocity of the gas leaving the bottle if the opening iss 25mm.
the second
Bournoulli equation=(p2-p1)/P+(u2-u1)/2+g(Z2-Z1)=0p2=psip1=atmP=lbm/ft^3u2=ft/hru1=cm/sZ1=ft(i do not know what the dimensions of Z2 are)

a. using the fundamental dimensions (M,L,t) show the the equation is dimensionally consistent (i beleive that this problem is asking me to show how the dimensions should cancel out, but i cant get them to im not completely sure what is i am doing wrong)
 
What happened to the rest of your values? Or are you just given pressure in the bottle is 8atm. Because then it's pretty easy to solve for the hole size.
 
Agreed, it was one of those courses that you just crammed into your temporary memory for the exams then purged quickly after to make room for more useful knowledge.
 
I dont understand the question, i need to solve for the velocity. I just said fuck it and "estimated" a flowrate.
And for the second i didnt realise that the g variable was gravitational acceleration. So i think im good.
And ya fuck fluids. Its not even for an actual fluid mechanics class its for this stupid engineering class that just throws crazy shit at us and tells us to learn it. Im a freshmen so im still a year from fluid mechanics.
 
For question #2, your bernoulli equation is wrong, your speed need to be squared. You will have:

g(z2-z1)=(p2-p1)/P+((u2)^2-(u1)^2)/2

p2=psi === N/m^2 === M*t^-2*L^-1

p1=atm === N/m^2 === M*t^-2*L^-1

P=lbm/ft^3 ========= M*L^-3

(u2=ft/hr)^2 === (L*t^-1)^2===L^2*t^-2

(u1=cm/s)^2 === (L*t^-1)^2===L^2*t^-2

Z1=ft =============== L

The last shit you dont know are g === m/s^2 === L*t^-2

and P === kg/m^3 === M*L^-3

Put all this in yout bernoulli equation...

(L*t^-2)(z2-z1)=(M*t^-2*L^-1)/(M*L^-3)+(L^2*t^-2)

(L*t^-2)(L)=(L^2*t^-2)

L=L weeeeeeeeeeeeeeeeetttttttt!!!!

For question 1.. this question suck

 
Im not sure how to do it, i see 2 ways.. using bernoulli equation...

p1=1 atm = 101,325 kPa (P outside the can, it may be something else)

p2=8atm = 810,6 kPa (P inside the can)

v1= ???

v2=0 (Speed of the fluid inside the can initally... 0)

P= 1562 kg/m^3 (value I found on internet, you may have other numbers in your book)

(p1-p2)/P=(v2^2-v1^2)/2

(709,275 kPa/1562 kg/m^3)*2=v1^2

v1=30,1357m/s

Or you can use the ''Conservation de la quantitée de mouvement'' (ehh not sure how to call it in english... conservation of quantity of velocity??) if its not bernoulli, its this but yeah I dont want to start doing integrations... but if the awnser's not good with bernoulli, look for this.

Hope it helps man, keep it up!!
 
ok. for the first part of your equation you need to make sure all of the units are consitant. Then just plug the knowns into the bernoulli equation.

basically,

Z1,Z2=0

u1=0

p2=atmospheric pressure

p1=8atm

I cant remember if you need densitys but you can find that from a table if you need it.

now you have 1 unknown which is u2. Plug and solve.

for the next part: dimensions. The dimension of Z2 will be meters(or inches or feet) but basically it has the fundamental dimension of distance or length. assuming you wrote your equation correctly the dimensions will cancel.

****Note***

subtracting dimensional quantities does not cancel out dimensions
 
It looks a lot harder than it actually is. You just have to remember all the formulas, and some of them are really fucked up.

I dont wanna do fluid at 10am
 
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