Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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A hypodermic needle has an inside diameter of 0.3 mm and is 60 mm in length, as shown in Fig. 9-17. If the
piston moves to the right at a speed of 18 mm/s and there is no leakage, what force Fis needed on the piston?
The medicine in the hypodermic has a viscosity u of 0.980 x 10 Pa ·s and its density p is 800 kg/m'.
Consider flows in both needle and cylinder. Neglect exit losses from the needle as well as losses at the juncture
of the needle and cylinder.
I For cylinder:
Q = Av = [(x)(0.005)/4](0.018) = 3.534 x 10-"m/s
%3D
NR = pdu/u = (800)(0.005)(0.018)/(0.980 x 10) = 73
(laminar)
%3D
128µLQ
nd
(128)(0.980 x 10-(0.050)(3.534 x 10-).
(*)(0.005)*
P1=
=1.129 Pa
For needle:
v = Q/A = 3.534 x 10-7/[(7)(0.3/1000)*/4] = 5.000 m/s
Na = (800)(0.3/1000)(5.000)/(0.980 x 10-) = 1224 (laminar)
(128)(0.980 x 10 )(0.060)(3.534 x 10 ).
(x)(0.3/1000)*
%3D
%3D
P2
= 104 525 Pa
F = (Ap)(Agytinder) = (104 525 –
- 1.129)[(x)(0.005)/4] = 2.05 N
Cylinder
Needle
5 mm
V18 mm/s
d=0.3 mm
%3D
50 mm-
-60 mm
expand button
Transcribed Image Text:A hypodermic needle has an inside diameter of 0.3 mm and is 60 mm in length, as shown in Fig. 9-17. If the piston moves to the right at a speed of 18 mm/s and there is no leakage, what force Fis needed on the piston? The medicine in the hypodermic has a viscosity u of 0.980 x 10 Pa ·s and its density p is 800 kg/m'. Consider flows in both needle and cylinder. Neglect exit losses from the needle as well as losses at the juncture of the needle and cylinder. I For cylinder: Q = Av = [(x)(0.005)/4](0.018) = 3.534 x 10-"m/s %3D NR = pdu/u = (800)(0.005)(0.018)/(0.980 x 10) = 73 (laminar) %3D 128µLQ nd (128)(0.980 x 10-(0.050)(3.534 x 10-). (*)(0.005)* P1= =1.129 Pa For needle: v = Q/A = 3.534 x 10-7/[(7)(0.3/1000)*/4] = 5.000 m/s Na = (800)(0.3/1000)(5.000)/(0.980 x 10-) = 1224 (laminar) (128)(0.980 x 10 )(0.060)(3.534 x 10 ). (x)(0.3/1000)* %3D %3D P2 = 104 525 Pa F = (Ap)(Agytinder) = (104 525 – - 1.129)[(x)(0.005)/4] = 2.05 N Cylinder Needle 5 mm V18 mm/s d=0.3 mm %3D 50 mm- -60 mm
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