Answer: 3002.86kg
Explanation:
Hydraulic cylinder diameter =125mm
Ambient pressure =1bar
Pressure =2500kpa
Piston Mass (MP) =?
F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg
Po=1 bar=100kpa
A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2
Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665
Mp=3002.86kg.
Answer:
a) Friction factor for this duct = 0.0239
b) ε = 0.006 ft
Explanation:
Given data :
Flow rate = 11000 ft^3 /min
Pressure drop = 1.2 in per 1500 ft of duct
<u>a) Determine the value of the friction factor for this duct</u>
Friction factor for this duct = 0.0239
<u>b) Determine the approximate size of the equivalent roughness of the surface of the duct</u>
ε = 0.006 ft
attached below is the detailed solution to the given problem
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