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. Take p= 1050 kg/m'. 15 mm Peace be upon you, and God's mercy and
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. Take p= 1050 kg/m'. 15 mm Peace be upon you, and God's mercy and
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
![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
Peace be upon you, and
God's mercy and
blessings. Please, at my
request, I want the same
solution and writę the
data in detail so that I
understand how the
result of the solution
came out in detail, I
mean, mathematics is a
solution in boring detail.
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,
+ gzn
Since V,<<<<V, the term
to the atmosphere, p, = 0. Here, P,
is negligible. Since the tip of the needle is exposed
= 5.659(10) F and the
(0.0075 m)
datum will coincide with the control streamline. Then
A,
5.659(10') F
1050 kg/m
V2
+ 0 +0 =0 + + 0
Va = (3.2831 VF) m/s
Va = (3.283 VF) m/s where F is in N
Ans.
The plot of V vs Fis shown in Fig. a.
Using the continuity equation,
-V,A, + V, A, = 0
-v,[#(0.0075 m)] + (3.2831 VF){=[0.3(10-3)m]²} = 0
V, = 0.00525 VF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74714912-2b0b-42ab-bb08-e607553d9d86%2F59def22d-dbf4-486e-9fe8-052065d243e3%2F4qbqfjk_processed.jpeg&w=3840&q=75)
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
Peace be upon you, and
God's mercy and
blessings. Please, at my
request, I want the same
solution and writę the
data in detail so that I
understand how the
result of the solution
came out in detail, I
mean, mathematics is a
solution in boring detail.
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,
+ gzn
Since V,<<<<V, the term
to the atmosphere, p, = 0. Here, P,
is negligible. Since the tip of the needle is exposed
= 5.659(10) F and the
(0.0075 m)
datum will coincide with the control streamline. Then
A,
5.659(10') F
1050 kg/m
V2
+ 0 +0 =0 + + 0
Va = (3.2831 VF) m/s
Va = (3.283 VF) m/s where F is in N
Ans.
The plot of V vs Fis shown in Fig. a.
Using the continuity equation,
-V,A, + V, A, = 0
-v,[#(0.0075 m)] + (3.2831 VF){=[0.3(10-3)m]²} = 0
V, = 0.00525 VF
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