At one end of the Venturi Tube the pressure and speed of flowing water is 21 kPa and 0.05m/s respectively. At another point the pressure is 5 kPa. Find the speed of the water at the second point. Answer in m/s
p1+1/2 dV1^2=p2+cV2^2 d=densitywater
23k+1/2 *1E3*.0025=5k+1/2*1E3*V2^2
V2^2=(23E3 +500*.0025-5000)/500
= 36
V2=6m/s
check my math.
The math was right, but 23k should be 2k Pa
23 should be 21
So, the answer is 0.0351240737 m/s
To find the speed of the water at the second point, we can use Bernoulli's equation, which relates the pressure and speed of a fluid along a streamline. Bernoulli's equation is given as:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
Where:
P₁ and P₂ are the pressures at the two points,
ρ is the density of the fluid, and
v₁ and v₂ are the speeds at the two points.
In this case, P₁ = 21 kPa, v₁ = 0.05 m/s, and P₂ = 5 kPa. We need to convert the pressures from kilopascals (kPa) to pascals (Pa) since the SI unit of pressure is the pascal.
1 kPa = 1000 Pa.
So, P₁ = 21 kPa = 21,000 Pa
And, P₂ = 5 kPa = 5,000 Pa.
Substituting these values into Bernoulli's equation, we get:
21,000 + ½ρ(0.05)² = 5,000 + ½ρv₂²
Simplifying the equation by substituting the known values, we have:
21,000 + 0.5ρ(0.0025) = 5,000 + 0.5ρv₂²
Simplifying further, we can cancel out the ρ factor:
21,000 + 0.00125 = 5,000 + 0.5v₂²
Rearranging the equation, we isolate v₂:
0.5v₂² = 21,000 - 5,000 + 0.00125
0.5v₂² = 16,000 + 0.00125
0.5v₂² = 16,000.00125
v₂² = (16,000.00125) / 0.5
v₂² = 32,000.0025
Taking the square root of both sides to solve for v₂:
v₂ = √(32,000.0025)
v₂ ≈ 178.89 m/s
Therefore, the speed of the water at the second point is approximately 178.89 m/s.