Answer:
(a) 3.43 m/s
(b) 3.43 m/s
(c) 95.8 kPa
Step-by-step explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.
(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube. I'll say 0 elevation is at point Z, and I'll use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0
ρ (9.8 m/s²) (0.60 m) = ½ ρ v²
5.88 m²/s² = ½ v²
v = 3.43 m/s
(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.
(c) Use Bernoulli equation again, choosing point 2 to be at Y. I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)
0 = P + 4116 Pa + 1372 Pa
P = -5488 Pa
The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.