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Fluid mechanics problem

Hi, I have a fluids problem.

A design contest features a sumbarine that will travel at a steady speed of Vsub=1m/s om 15degC of water.
The sub is powered by a water jet. This jet is created by drawing water from an inlet of diameter 25mm, passing this watter through a pump and then accelerating the water through a nozzle of diameter 5mm to a speed of Vjet.
The hydrodynamice drag force can be calculated using F=CdAp(density*(Vsub^2)/2) where the Cd=0.3 and Ap=0.28m^2.
Specify an acceptable value of Vjet.

Please help.
Thanks
Reply 1
Original post by Hello1992
Hi, I have a fluids problem.

A design contest features a sumbarine that will travel at a steady speed of Vsub=1m/s om 15degC of water.
The sub is powered by a water jet. This jet is created by drawing water from an inlet of diameter 25mm, passing this watter through a pump and then accelerating the water through a nozzle of diameter 5mm to a speed of Vjet.
The hydrodynamice drag force can be calculated using F=CdAp(density*(Vsub^2)/2) where the Cd=0.3 and Ap=0.28m^2.
Specify an acceptable value of Vjet.

Please help.
Thanks



Just balance forces. THrust generated by the jet = mass flow times change in velocity. THe inlet and outlet velocities are linked by mass flow in = mass flow out and with density constant the only unknown when you balance will be the exit velocity of the jet. Mass flow = density x area of inlet x inlet velocity. (ie replace inlet velocity in the rate of momentum equation at the start by exit velocity x exit area / inlet area.
(edited 13 years ago)
Reply 2
Original post by wdywuk
Just balance forces. THrust generated by the jet = mass flow times change in velocity. THe inlet and outlet velocities are linked by mass flow in = mass flow out and with density constant the only unknown when you balance will be the exit velocity of the jet. Mass flow = density x area of inlet x inlet velocity. (ie replace inlet velocity in the rate of momentum equation at the start by exit velocity x exit area / inlet area.


Thank you.
How about the hydrostatic equation?
Also i do not know the velocity of the mass going into the jet.
(edited 13 years ago)
Reply 3
Original post by Hello1992
Thank you.
How about the hydrostatic equation?
Also i do not know the velocity of the mass going into the jet.


Don't need the hydrostatic equation

Anytime you see the inlet velocity in a formula replace it by exit velocity * (exit area) / (inlet area)
Reply 4
Original post by wdywuk
Don't need the hydrostatic equation

Anytime you see the inlet velocity in a formula replace it by exit velocity * (exit area) / (inlet area)


Hi, sorry but i keep gettint the wrong answer.
Can you show me how?
Sorry for the poorness of my fluids knowledge :frown:
Reply 5
What answer did you get? without checking what I've done, I've got the jet velocity to be 47.2m/s.
Reply 6
I got 86.6.
you got it right.
when you say thrust = mass flow x change in velocity, what did you use as change in velocity?
Sorry just tryong to get my head around it.
Thanks
Reply 7
Original post by Hello1992
I got 86.6.
you got it right.
when you say thrust = mass flow x change in velocity, what did you use as change in velocity?
Sorry just tryong to get my head around it.
Thanks


Change in velocity = Vjet - Vinlet and again,
replace Vinlet by Vjet*(A-outlet/A-inlet) (You can quickly get this by realising that the by continuity the mass flow in = mass flow out)

As density of the water is constant then A1V1 = A2V2 (1- inlet, 2 - outlet)

If I remember correctly the drag force on the sub worked out to be about 42N
On the other side, the thrust worked out to be something like 0.0189 Vjet^2
Reply 8
Original post by wdywuk
Change in velocity = Vjet - Vinlet and again,
replace Vinlet by Vjet*(A-outlet/A-inlet) (You can quickly get this by realising that the by continuity the mass flow in = mass flow out)

As density of the water is constant then A1V1 = A2V2 (1- inlet, 2 - outlet)

If I remember correctly the drag force on the sub worked out to be about 42N
On the other side, the thrust worked out to be something like 0.0189 Vjet^2


I understand everything you say except how you got 0.0189Vjet^2 from A1V1=A2V2

Sorry for my lack of knowledge,
Reply 9
Original post by wdywuk
Change in velocity = Vjet - Vinlet and again,
replace Vinlet by Vjet*(A-outlet/A-inlet) (You can quickly get this by realising that the by continuity the mass flow in = mass flow out)

As density of the water is constant then A1V1 = A2V2 (1- inlet, 2 - outlet)

If I remember correctly the drag force on the sub worked out to be about 42N
On the other side, the thrust worked out to be something like 0.0189 Vjet^2


Thank you for your help.
I just got the answer. I was being stupid.
Thanks again so much!
Reply 10
No worries, just glad I got it right and wasn't leading you down th wrong path!

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