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
Answer:
speed by mass attain is 55.86 m/s
Explanation:
given data
glucose = 10 g
mass = 100 kg
to find out
speed by mass attain
solution
we know glucose have 180 g molecular weight and
that 1 g glucose produce energy = 2816/180 × 10³ J
so here 10 g of glucose produce energy = 1.56 × J
so here energy release = 1/2 × mv²
1.56 × = 1/2 × (100)v²
v² = 3.12 × 10³
and v = 55.86 m/s
so speed by mass attain is 55.86 m/s