A sports scientist has constructed a prototype vuvuzela in order to achieve the optimal audio frequency for filling soccer stadiums with noise while mitigating potential hearing loss for players and fellow spectators. Air is blown in to the inlet of the tubular instrument, and shortly after, passes through a piezometer which measures the static pressure (gauge) of the air. A stagnation tube is located further downstream, measuring the total pressure (gauge) at a stagnation point in the flow. In addition, an inclinometer measures the vuvuzela\'s angle of inclination towards the vertical (see Figure below). After months of tinkering the scientist has achieved the perfect frequency using the following conditions: a = 0.25 m Atmospheric Pressure =101 kPa b = 0.4 m Piezometer Reading = 16 mm H_2O c = 0.3 m Stagnation Tube Reading = 24 mm H_2O dx = 2 y theta = 45 degree y = z Air Temperature = 20 degree C At what velocity must air be blown in to the inlet to achieve the perfect frequency? If the piezometer and stagnation tube were replaced by a pitot-static probe (a gauge designed o measure the difference between total and static pressure) placed at the original location of e stagnation tube, what would its reading have been under the same ideal conditions (in P)?
solution:
1)here applying bernoulli\'s equation at distance of 0,a,(a+b) means at inlet1,piezometer 2and at stagnation tube 3we get
p1/density*g+v1^2/2*g+z1=p2/density*g+v2^2/2*g+z2=p3/density*g+v3^2/2*g+z3=constant
here piezometer measyre static head and stagnation tube measure static plus dynamic head
where z1=0
z2=asin45=.176776 m
z3=(a+b)sin45=.459619 m
p1=0
hence putting value we get
v1^2/2*g=.016+v2^2/2*g+.176776=.024+.459619=.483619
hence V2=2.3887 m/s
V1=3.080 m/s
hence velocity at inlet is v1=3.080 m/s
2)if we replace piezomter and stagnation tube by pitot static probe,the it will represent head as difference of dynamic and static head
as from geometry
radius y=z
means area rae equal at 2 and 3 point
means V2=V3=2.3887 m/s
hence dynamic head is=v3^2/2*g=.2908 m
by applying bernaullies at point 2 and 3
p2/density*g+v2^2/2*g+z2=p3/density*g+v3^2/2*g+z3=.483619
p3/density*g=-.2668 m
4)here pitot static tube will give difference between dynamic and static head for same condition
H=v3^2/2*g-p3/density*g=.2908-(-.266819)=.557638 m of water or 557.638 mm of water
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