5-9. By applying a force F, a saline solution is ejected from the 15-mm-diameter syringe through a 0.6-mm-diameter needle. Determine the average velocity of the solution through the needle as a function of the force Fapplied to the plunger. Plot this velocity (vertical axis) as a function of the force for 0 s Fs 20 N. Give values for increments of AF = 5 N. Takep 1050 kg/m. 15 mm

Elements Of Electromagnetics
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5-9. By applying a force F, a saline solution is ejected from
the 15-mm-diameter syringe through a 0.6-mm-diameter
needle. Determine the average velocity of the solution
through the needle as a function of the force Fapplied to the
plunger. Plot this velocity (vertical axis) as a function of the
force for 0s Fs 20 N. Give values for increments of
AF = 5 N. Take p = 1050 kg/m.
15 mm
SOLUTION
The saline solution can be considered as an ideal fluid (incompressible and inviscid).
Also, the flow is steady. Therefore, Bernoulli's equation is applicable. Applying this
equation between a point in the syringe and the other at the tip of the needle of
which both points are on the central streamline,
P. V?
Pa
gz
Hello, professor,
solved in a detailed
+ gza
Since V<<<<V, the term
to the atmosphere, pa = 0. Here, P, =
is negligible. Since the tip of the needle is exposed
= 5.659(10) F and the
way, the same
solution, steps, but it
is a detailed solution
7(0.0075 m)
datum will coincide with the control streamline. Then
A,
5.659(10') F
1050 kg/m
V2
+ 0 +0=0 + + 0
for mathematics.
V = (3.2831 VF) m/s
Va = (3.283 VF) m/s where Fis in N
Ans.
The plot of V vs Fis shown in Fig. a.
Using the continuity equation,
-V,A, + V A = 0
-v,[7(0.0075 m)] + (3.2831 VF){=[0.3(103)m]²}
= 0
V, = 0.00525 VF
Transcribed Image Text:5-9. By applying a force F, a saline solution is ejected from the 15-mm-diameter syringe through a 0.6-mm-diameter needle. Determine the average velocity of the solution through the needle as a function of the force Fapplied to the plunger. Plot this velocity (vertical axis) as a function of the force for 0s Fs 20 N. Give values for increments of AF = 5 N. Take p = 1050 kg/m. 15 mm SOLUTION The saline solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli's equation is applicable. Applying this equation between a point in the syringe and the other at the tip of the needle of which both points are on the central streamline, P. V? Pa gz Hello, professor, solved in a detailed + gza Since V<<<<V, the term to the atmosphere, pa = 0. Here, P, = is negligible. Since the tip of the needle is exposed = 5.659(10) F and the way, the same solution, steps, but it is a detailed solution 7(0.0075 m) datum will coincide with the control streamline. Then A, 5.659(10') F 1050 kg/m V2 + 0 +0=0 + + 0 for mathematics. V = (3.2831 VF) m/s Va = (3.283 VF) m/s where Fis in N Ans. The plot of V vs Fis shown in Fig. a. Using the continuity equation, -V,A, + V A = 0 -v,[7(0.0075 m)] + (3.2831 VF){=[0.3(103)m]²} = 0 V, = 0.00525 VF
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