# Fluid mechanics problem watch

1. 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.

Thanks
2. (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.

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.
3. (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.
Also i do not know the velocity of the mass going into the jet.
4. (Original post by Hello1992)
Thank you.
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)
5. (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
6. What answer did you get? without checking what I've done, I've got the jet velocity to be 47.2m/s.
7. 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
8. (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
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
I understand everything you say except how you got 0.0189Vjet^2 from A1V1=A2V2

Sorry for my lack of knowledge,
10. (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 just got the answer. I was being stupid.
Thanks again so much!
11. No worries, just glad I got it right and wasn't leading you down th wrong path!

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 25, 2011
Today on TSR

### He broke up with me because of long distance

Now I'm moving to his city

### University open days

Wed, 25 Jul '18
2. University of Buckingham
Wed, 25 Jul '18
3. Bournemouth University