Answer:
A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?
The flow rate is 3.1175×10⁻³ m³/s
Explanation:
To solve the question we rely on Bernoulli's principle as follows [tex]P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}[/tex]
thus where the pipe is horizontal we have
z₁ = z₂ hence the above equattion becomes
[tex]P_{1} +\frac{1}{2}\rho v^{2} _{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2}[/tex]
since the flow rate is constant then
Q = v₁A₁ = v₂A₂
Where is the area of the two sections given by A₁ = π·D₁²÷4 and
A₂ = π·D₂²÷4
Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²
and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²
v₁ = v₂A₂/A₁ =0.391×v₂
The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and
ρ = 1000 kg/m³
Plugging the values into the above equation we get
31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²
= 31000+76.3·v₂² =24000+500·v₂²
or 423.706·v₂² = 7000
v₂² = 7000/423.706 = 16.52 or v₂ = 4.065 m/s and v₁ 0.391×4.065 = 1.59 m/s
The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s
