Question

a crude of oil 0.97 poise and relative density 0.9 is flowing through a horizantal pipe of diameter 100 mm and of the length 10 m. Calculate the difference of pressure at the two ends of the pipe. if 100kg of oil is collected in a tank in 30 secondd​

Answer 1

Answer:

Explanation:

0.97poise = =0.097Ns

10

/m2

Relative density = 0.9 (of 0:1)

. Densityp, = 0.9 x 1000 (of oil)

= 900kg/m'

Dia. of pipe D = 100 mm = 0.1 m

Length L=10 m

Mass of Oil = m = 100 kg.

and timet=30 seconds.

P - Pa =?

The difference of pressure for viscous

flow is

32mUL

Avg. velocity

P1-P2 D?

Q

Area

100

Now, mass of oil per second 30 kg/s

=p x Q= 900 x Q

100 x Q

30

.Q=0.0037m/s

Now,Ü = 0.0037 0.0037

TxD2

Tx0.12

= 0.471m/s

For Laminar/viscous flow, Re< 2000

Now, Re = pVD

P=Po 900, v = U = 0.471m/s

D=0.1m, M = 0.097

Re = 900 x4XO = 436.91

0.097

Value of Re< 2000, hence flow is

laminar.

324UL

32x0.097x0.471x 10

.'. P1- P2=

(0.1)

D2

P1 P2 = 1462.28N/m2

= 1462.28x 10 4N/cm2

P1 P2 = 0.1462N/cm2