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Question 2 (10 points) Retake question (1)/(f^((1)/(2)))~~-1.8log[(6.9)/(Re_(d))+((epsi lo(n)/(d))/(3.7))^(1.11)] mu _(water )=0.001(kg)/(m-s) ho _(water )=998(kg)/(m^(3)) Pipe (epsi )/(D)=0.008 For laminar flow: alpha =2

Question 2 (10 points) Retake question\

(1)/(f^((1)/(2)))~~-1.8log[(6.9)/(Re_(d))+((\\\\epsi lo(n)/(d))/(3.7))^(1.11)]\ \\\\mu _(water )=0.001(kg)/(m-s)\ \\\ ho _(water )=998(kg)/(m^(3))

\ Pipe\

(\\\\epsi )/(D)=0.008

\ For laminar flow:

\\\\alpha =2

\ For turbulent flow:

\\\\alpha =1.058

\ Water flows through the piping system shown above

3.2m

, and

L_(3)=1.5m

) at a volumetric flow rate of

0.202(m^(3))/(s)

. It enters the system inside the pipe at

A

and exits to the atmosphere at

C

. The gage pressure at

A

has

image text in transcribed
Question 2 (10 points) Retake question f1/211.8log[Red6.9+(3.7/d)1.11] water=0.001mskg water=998m3kg Pipe /D=0.008 For laminar flow: =2 For turbulent flow: =1.058 *Drawing Not to Scale* Pump Water flows through the piping system shown above (D=200mm,L1=9m,L2= 3.2m, and L3=1.5m ) at a volumetric flow rate of 0.202m3/s. It enters the system inside the pipe at A and exits to the atmosphere at C. The gage pressure at A has

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