Question

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

Answer 1

Q = 0.042 ft^3/s

Step-by-step 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 =
`(1 (ibf/in^2) * 144 (in^2/ft^2))/(1.32 * 32.174)` = 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 =
`\sqrt{(1.823)/(0.03) }` = 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 =
`\sqrt{(1.823)/(0.031) }` = 7.67 ft/s

therefore the Volumetric flow rate

= Q =
`(\pi d^2 )/(4) * v`

Q =
`(\pi )/(4) * (1/12)^2 * 7.67` = 0.042 ft^3/s