a) Q = 1.3044 m^3 / s
b) h2 = 0.37 m
c) Pi = Pe = Patm = 101.325 KPa
Explanation:
Given:-
- The constant width of the rectangular channel, b = 0.06 m
- The density of air, ρa = 1.23 kg/m^3
- The density of water, ρw = 1000 kg / m^3
- The height of the channel at inlet and exit, hi = he = 0.04 m
- The height of the channel at point 2 = h2
- The height of the channel at point 3 - Throat , ht = 0.02 m
- The change height of the water in barometer at throat, ΔHt = 0.1 m
- The change height of the water in barometer at point 2, ΔH2 = 0.05 m
- The flow rate = Q
Solution:-
- The flow rate ( Q ) of air through the venturi remains constant because the air is assumed to be incompressible i.e ( constant density ). We have steady state conditions for the flow of air.
- So from continuity equation of mass flow rate of air we have:
m ( flow ) = ρa*An*Vn = Constant
Where,
Ai : The area of the channel at nth point
Vi : The velocity of air at nth point.
- Since, the density of air remains constant throughout then we can say that flow rate ( Q ) remains constant as per continuity equation:
Q = m ( flow ) / ρa
Hence,
Q = Ai*Vi = A2*V2 = At*Vt = Ae*Ve
- We know that free jet conditions apply at the exit i.e the exit air is exposed to atmospheric pressure P_atm.
- We will apply the bernoulli's principle between the points of throat and exit.
Assuming no changes in elevation between two points and the effect of friction forces on the fluid ( air ) are negligible.
Pt + 0.5*ρa*Vt^2 = Pe + 0.5*ρa*Ve^2
- To determine the gauge pressure at the throat area ( Pt ) we can make use of the barometer principle.
- There is an atmospheric pressure acting on the water contained in the barometric tube ( throat area ). We see there is a rise of water by ( ΔHt ).
- The rise in water occurs due to the pressure difference i.e the pressure inside the tube ( Pt ) and the pressure acting on the water free surface i.e ( Patm ).
- The change in static pressure leads to a change in head of the fluid.
Therefore from Barometer principle, we have:
Patm - Pt-abs = pw*g*ΔHt
101,325 - Pt-abs = 1000*9.81*0.1
Pt-abs = 101,325 - 981
Pt-abs = 100,344 Pa Absolute pressure
- We will convert the absolute pressure into gauge pressure by the following relation:
Pt = Pt-abs - Patm
Pt = 100,344 - 101,325
Pt = -981 Pa ... Gauge pressure
- Now we will use the continuity equation for points of throat area and exit.
At*Vt = Ae*Ve
b*ht*Vt = b*he*Ve
Ve = ( ht / he ) * Vt
Ve = ( 0.02 / 0.04 ) * Vt
Ve = 0.5*Vt
- Now substitute the pressure at throat area ( Pt ) and the exit velocity ( Ve ) into the bernoulli's equation expressed before:
Pt + 0.5*ρa*Vt^2 = 0 + 0.5*ρa*( 0.5*Vt )^2
-981 = 0.5*ρa*( 0.25*Vt^2 - Vt^2 )
-981 = - 0.1875*ρa*Vt^2
Vt^2 = 981 / ( 0.1875*1.23 )
Vt = √4253.65853
Vt = 65.22 m/s
- The flow rate ( Q ) of air in the venturi is as follows:
Q = At*Vt
Q = ( 0.02 )*( 65.22 )
Q = 1.3044 m^3 / s Answer part a
- We will apply the bernoulli's principle between the points of throat and point 2.
Assuming no changes in elevation between two points and the effect of friction forces on the fluid ( air ) are negligible.
Pt + 0.5*ρa*Vt^2 = P2 + 0.5*ρa*V2^2
- To determine the gauge pressure at point 2 ( P2 ) we can make use of the barometer principle.
Therefore from Barometer principle, we have:
Patm - P2-abs = pw*g*ΔH2
101,325 - P2-abs = 1000*9.81*0.05
P2-abs = 101,325 - 490.5
Pt-abs = 100834.5 Pa Absolute pressure
- We will convert the absolute pressure into gauge pressure by the following relation:
P2 = P2-abs - Patm
Pt = 100,344 - 100834.5
Pt = -490.5 Pa ... Gauge pressure
- Now substitute the pressure at point 2 ( P2 ) bernoulli's equation expressed before:
Pt + 0.5*ρa*Vt^2 = P2 + 0.5*ρa*( V2 )^2
( Pt - P2 ) + 0.5*ρa*Vt^2 = 0.5*ρa*( V2 )^2
2*( Pt - P2 ) / ρa + Vt^2 = V2^2
2*( -981 + 490.5 ) / 1.23 + 65.22^2 = V2^2
-981/1.23 + 4253.6484 = V2^2
V2 = √3456.08742
V2 = 58.79 m/s
- The flow rate ( Q ) of air in the venturi remains constant is as follows:
Q = A2*V2
Q = b*h2*V2
h2 = Q / b*V2
h2 = 1.3044 / ( 0.06*58.79)
h2 = 0.37 m Answer part b
- We will apply the bernoulli's principle between the points of inlet and exit.
Assuming no changes in elevation between two points and the effect of friction forces on the fluid ( air ) are negligible.
Pi + 0.5*ρa*Vi^2 = Pe + 0.5*ρa*Ve^2
- Now we will use the continuity equation for points of inlet area and exit.
Ai*Vi = Ae*Ve
b*hi*Vi = b*he*Ve
Vi = ( he / hi ) * Ve
Vi = ( 0.04 / 0.04 ) * 0.5*Vt
Vi = Ve = 0.5*Vt = 0.5*65.22 = 32.61 m/s
- Now substitute the velocity at inlet in bernoulli's equation expressed before:
Pi + 0.5*ρa*Vi^2 = 0 + 0.5*ρa*( Ve )^2
Since, Vi = Ve then:
Pi = Pe = 0 ( gauge pressure ).
Pi = Pe = Patm = 101.325 KPa
Comment: If the viscous effects are considered then the Pressure at the inlet must be higher than the exit pressure to do work against the viscous forces to drive the fluid through the venturi assuming the conditions at every other point remains same.
