Two pressure gauges measure a pressure drop of 16.3 psi (lb/in.2) at the entrance and exit of an old buried pipeline. The original drawings have been lost. If the 6-in. galvanized iron pipe carries water at 68°F with a flow rate of 1.64 cfs (ft3/s), determine the length of the horizontal underground line ignoring minor losses.

866.25ft

Explanation:

Pressure drop= 16.3 psi

1 feet of water column= 0.4335psi

x= 16.3psi

Therefore, head loss of water= 16.3/0.4335=37.6ft

In inches, head loss of water= 37.6 × 12=451.2 inches

Q= 1.64ft^2/s

A= pi/4 D^2

Where D= 6 inches, 0.5ft

A= 3.142/4 × (0.5)^2

A= 0.1962ft^2

V= Q/A= 1.64/ 0.1962

V= 8.35ft/s, 100.28 in/s

REGNOLD'S NO

Re= SvD/ μ=

Where, D= 6 inches, V= 100.28 inches/s, S=62.4lb/ft^3= 0.03611lb/inches^3

At 68F, Dynamic viscosity μ= 1.0016× 10^-3NS/m^2

1psi= 6895N/m^3

μ= 1.0016× 10^-3NS/m^2/6895N/m^3

μ= 1.452× 10^-7lbs/ins^2

Re= 0.03611× 100. 28×6/1.42×10^-7

Re= 1.49×10^8

e/D Relative roughness

e= 0.0005ft

e/D= 0.001

Therefore, for Re= 1.49×10^8 and e/D= 0.001

F=0.02

Head loss is given by Dancy-Weisbach formula

hL= fLV^2/2gD

Wherr g= 9.81m/s^2= 386.12inches/s^2

451.2= 0.02×L×(100.28^2)/2 × 386.12inches/s^2× 6

L= 10395inches

L= 10395/12= 866.25ft

Therefore, the length of pipe is 866.25ft.

answer:

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