Question

A level pipe contains a nonviscous, incompressible fluid with a density 1200 kg/m3 that is flowing steadily. At one position within the pipe, the pressure is 300 kPa and the speed of the flow is 20.0 m/s. At another position, the pressure is 200 kPa. What is the speed of the flow at this second position?

Answer 1

Answer:

V2 = 23.8m/s

Explanation:

as we know That,  

p1/(r*g) + (v1^2)/2*g = p2/(r*g) + (v2^2)/2*g  

p1=300kPa=300000Pa  

p2=200kPa=200000Pa  

r=1200kG/m^3  

v1 = 20 m/s  

250+200 = 166.66+(v2^2)/2  

v2=(450-166.66)*2  

v2^2 =566.68  

v2=23.8m/s

Your answer is : V2 = 23.8m/s

Answer 2

The speed of the flow at this second position is 23.8 m/s.

The given parameters:

• Density of the fluid, ρ = 1200 kg/m³
• Pressure of the pipe, P₁ = 300 kPa
• Speed of the flow, v₁ = 20.0 m/s
• Pressure at second position, P₂ = 200 kPa

The speed of the flow at this second position is calculated by applying Bernoulli's equation as follows;

$P_1 + \frac{1}{2} \rho v_1^2= P_2 + \frac{1}{2} \rho v_2 ^2\\\\(P_1 - P_2) + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2 ^2\\\\2(P_1-P_2) + \rho v_1^2 = \rho v_2^2\\\\\frac{2(P_1-P_2) }{\rho} + v_1^2 = v_2^2\\\\\sqrt{\frac{2(P_1-P_2) }{\rho} + v_1^2} = v_2\\\\\sqrt{\frac{2(300,000-200,000) }{1200} + 20^2}= v_2 \\\\\sqrt{566.67} = v_2\\\\23.8\ m/s = \ v_2$

Thus, the speed of the flow at this second position is 23.8 m/s.

Learn more about Bernoulli's equation here: https://brainly.com/question/7463690