- #1

- 613

- 3

Please see attached.

I already got the answer verified for part a which is v1=1.5 m/s and mass flow rate=23.1 kg/s

For part b, I'd like someone to check my answer and answer a question.

Sum of forces in the x direction ƩFx=m'out * V2 - m'in * V1

m'=mass flow rate

V=velocity

For the left hand side of the equation gives:

ƩFx=-Rx + P1*A1 - P2*A2*cos(50)

Can someone explain why the force due to the pressure @ 2 causes a negative force in the x direction?

=-3880+P1*pi*(.07^2) - 199*pi(.035^2)cos50

I think my professor said that we need to use the gauge pressure in this conservation of momentum equation, thus P2=300-101=199kPa

Fort the right hand side of the equation:

density*V2*A2*cos(50)*V2 - density*V1*A1*V1=1000*6*PI*(.035^2)*COS(50)*6-1000*1.5*pi*(.07^2)*1.5

P1=256 kPa (gage pressure)

I already got the answer verified for part a which is v1=1.5 m/s and mass flow rate=23.1 kg/s

For part b, I'd like someone to check my answer and answer a question.

Sum of forces in the x direction ƩFx=m'out * V2 - m'in * V1

m'=mass flow rate

V=velocity

For the left hand side of the equation gives:

ƩFx=-Rx + P1*A1 - P2*A2*cos(50)

Can someone explain why the force due to the pressure @ 2 causes a negative force in the x direction?

=-3880+P1*pi*(.07^2) - 199*pi(.035^2)cos50

I think my professor said that we need to use the gauge pressure in this conservation of momentum equation, thus P2=300-101=199kPa

Fort the right hand side of the equation:

density*V2*A2*cos(50)*V2 - density*V1*A1*V1=1000*6*PI*(.035^2)*COS(50)*6-1000*1.5*pi*(.07^2)*1.5

P1=256 kPa (gage pressure)