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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 Ng = pdu/u = (800)(0.005)(0.018)/(0.980 x 10) = 73 (laminar) 128µLQ nd (128)(0.980 x 10-)(0.050)(3.534 x 10-7) (*)(0.005)* P1= =1.129 Pa For needle: v = Q/A = 3.534 x 10-/((x)(0.3/1000)*/4]= 5.000 m/s Na = (800)(0.3/1000)(5.000)/(0.980 x 10-)= 1224 (laminar) %3D (128)(0.980 x 10)(0.060)(3.534 x 10-"). ()(0.3/1000) P2= = 104 525 Pa F= (Ap)(Aginder) = (104 525 – 1.129)[(7)(0.005) /4] = 2.05 N %3D Cylinder Needle Pi[5 mm VE 18 mm/s = 0.3 mm 50 mm -60 mm

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 Ng = pdu/u = (800)(0.005)(0.018)/(0.980 x 10) = 73 (laminar) 128µLQ nd (128)(0.980 x 10-)(0.050)(3.534 x 10-7) (*)(0.005)* P1= =1.129 Pa For needle: v = Q/A = 3.534 x 10-/((x)(0.3/1000)*/4]= 5.000 m/s Na = (800)(0.3/1000)(5.000)/(0.980 x 10-)= 1224 (laminar) %3D (128)(0.980 x 10)(0.060)(3.534 x 10-"). ()(0.3/1000) P2= = 104 525 Pa F= (Ap)(Aginder) = (104 525 – 1.129)[(7)(0.005) /4] = 2.05 N %3D Cylinder Needle Pi[5 mm VE 18 mm/s = 0.3 mm 50 mm -60 mm

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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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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
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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