Gasoline of density 1.32 slugs/ft3 and viscosity 3.26×10-3 lbf. s/ft2 flows through a 1-in diameter 10 ft long rubber hose of absolute roughness 0.0004 ft. Find the volumetric flow rate for a pressure drop of 1 psi

Q = 0.042 ft^3/s

Explanation:

Given data

pressure drop = 1 psi ( p1 -p2) =

Gasoline density = 1.32 slugs/ft^3

viscosity = 3.26 8 10^-3 Ibf.s/ft^2

correct viscosity of gasoline = 6.5 *10^-6 Ibf.s/ft^2

e = 0.0004 ft

assumptions

Since The base is of constant diameter then V1 = V2

also z1 = z2

therefore the steady  flow  energy equation will be

hf = pressure drop / ( gasoline density * g ) ------ equation 1

hf = head loss due to friction

g = 32.174 ft/s^2

equation 1 becomes

note : 1 ibf = slug ft / s^2

hf =  = 3.4 ft

Volume flow rate can be obtained through iteration process using (hf)

therefore hf can be expressed as

hf = f * l/d * v^2/2g

fv^2 = hf * d/l * 2g ------------- equation 2

d = 1/12 (ft)

l = 10 (ft )  

g = 32 .174 ft/s^2

insert the following values into equation 2

fv^2 = 1.823 (ft^2/s^2) -------------- equation 3

assuming full turbulent flow

e/d = 0.0004 / (1/12) = 0.0048 ft  and

assuming Reynolds number ( re ) = 10^8

therefore from Moody's chart ; f = 0.03

insert ; f = 0.03 into equation 3

v =  = 7.8 ft/s

Now we have V = 7.8 ft/s  then we have to calculate for Reynolds number

Re = Pvd / u

u = viscosity of  gasoline

Re  = 1.32 * 7.8 * (1/12) / (6.5 *10^-6)

Re = 1.32 * 10^5

hence at this Reynolds number the value of ; f = 0.031 as gotten when we assumed the Reynolds number

v  =  = 7.67 ft/s

therefore the Volumetric flow rate

= Q =

Q =  = 0.042 ft^3/s

answer:

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explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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answer:

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explanation: