 #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=300101=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)*61000*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=300101=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)*61000*1.5*pi*(.07^2)*1.5
P1=256 kPa (gage pressure)
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