Question

A fluid, density ρ=960 kg/m3 is flowing steadily through the below tube. The section diameters are d1​=10 cm and d2​=8 cm. The gauge pressure at 1 is P1​=200kN/m2 The ​ velocity at 1 is u1​=5 m/s. The tube is horizontal (z1​=z2​). What is the gauge pressure at section 2?​

Answer 1

211,520 N/m^2

Step-by-step explanation:

To calculate the gauge pressure at section 2, we can apply Bernoulli's equation, which states that the total energy of a fluid in a horizontal flow remains constant. Bernoulli's equation is expressed as:

P1 + 0.5 * ρ * u1^2 + ρ * g * z1 = P2 + 0.5 * ρ * u2^2 + ρ * g * z2

Given the information provided:

P1 = 200 kN/m^2

u1 = 5 m/s

d1 = 10 cm = 0.1 m (converted to meters)

d2 = 8 cm = 0.08 m (converted to meters)

z1 = z2 (since the tube is horizontal)

ρ = 960 kg/m^3 (density of the fluid)

We can calculate the velocity at section 2 (u2) using the continuity equation, which states that the mass flow rate is constant in an incompressible fluid:

A1 * u1 = A2 * u2

A1 = (π/4) * d1^2 (area at section 1)

A2 = (π/4) * d2^2 (area at section 2)

Substituting the values and solving for u2:

(π/4) * d1^2 * u1 = (π/4) * d2^2 * u2

(0.785) * (0.1)^2 * 5 = (0.785) * (0.08)^2 * u2

0.03925 = 0.03925 * u2

u2 = 1 m/s

Now we can substitute all the known values into Bernoulli's equation:

200 kN/m^2 + 0.5 * 960 kg/m^3 * (5 m/s)^2 = P2 + 0.5 * 960 kg/m^3 * (1 m/s)^2

Simplifying the equation:

200000 N/m^2 + 0.5 * 960 kg/m^3 * 25 m^2/s^2 = P2 + 0.5 * 960 kg/m^3 * 1 m^2/s^2

200000 N/m^2 + 12000 N/m^2 = P2 + 480 N/m^2

212000 N/m^2 = P2 + 480 N/m^2

P2 = 211520 N/m^2

Therefore, the gauge pressure at section 2 is 211,520 N/m^2.