Water flows at a rate of 0.011 m3/s in a horizontal pipe whose diameter increases from 6 to 11 cm by an enlargement section. if the head loss across the enlargement section is 0.66 m and the kinetic energy correction factor at both the inlet and the outlet is 1.05, determine the pressure change. take the density of water to be 1000 kg/m3.
the pressure change is 30.44 kpa.

Pressure change in pipe is 766.96 N/m²

Explanation:

The continuity equation is stated below,

AV = Q

A1V1 = A2V2

Where A is the cross-sectional area, V is the velocity and Q is the fluid flow rate.

To calculate the inlet velocity of the pipe,

V1 = Q/A1

V1 = Q/(π x d1²)

d1 is the inlet diameter of the pipe

Substituting values,

V1 = 0.011/(π x 1/4 x 0.06²)

V1 = 3.89 ms-¹

To determine the outlet velocity,

V2 = Q/A2

d2 is the outlet diameter of pipe

V2 = 0.011/(π x 1/4 x 0.11²)

V2 = 1.157 ms-²

Applying Bernoulli's equation for steady flow between the points,

P1/pg + a1(V1²/2g) + z1 + hp = P2/pg + a2(V2²/2g) + z2 + ht + hL

Collecting like terms,

The kinetic energy correction factor, a = a1 = a2

a((V1² - V2²)/2g) - hL = (P2 - P1)/pg

apg((V1² - V2²)/2g) - hLpg = P2 - P1

ap((V1² - V2²)/2) - hLpg = ∆P

p - density of water, g is the acceleration due to gravity and hL is the head loss due to friction in pipe.

Substituting values,

a = 1.05, p = 1000kg/m³, g = 9.81m/s², hL = 0.66m

∆P = (1000)(1.05)((3.89² - 1.157²)/2) - (0.66 x 1000 x 9.81)

∆P = 7241.56 - 6474.6

∆P = 766.96 N/m²

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.

answer:

explanation:

sbbdbdbdbdbdndjd