To find the pressure difference between the inlet and throat in the venturimeter, we can use Bernoulli’s equation and continuity equation.

Given dataTo find the pressure difference between the inlet and throat in the venturimeter, we can use Bernoulli’s equation and continuity equation.

Given

• Inlet diameter (D₁) = 150 mm = 0.15 m
• Throat diameter (D₂) = 75 mm = 0.075 m
• Height difference ($h$) = 225 mm = 0.225 m
• Venturimeter constant (Cₓ) = 0.96
• Specific gravity of petrol = 0.78 → Density of petrol $\rho$ = 0.78 $\times$ 1000 = 780 kg/m³
• Flow rate ($Q$) = 0.029 m³/s

We need to find the pressure difference (ΔP) between the inlet and the throat.

Step 1: Calculate the velocities at the inlet and throat using the continuity equation.

The continuity equation states: $Q = A_1 V_1 = A_2 V_2$ where $A_1$ and $A_2$ are the cross-sectional areas at the inlet and throat, respectively, and $V_1$ and $V_2$ are the velocities at the inlet and throat.

The area of a circular section is given by: $A = \frac{\pi D^2}{4}$

1. Calculate $A_1$ and $A_2$:

   $A_1 = \frac{\pi \times (0.15)^2}{4} = 0.01767 \, \text{m}^2$ $A_2 = \frac{\pi \times (0.075)^2}{4} = 0.00442 \, \text{m}^2$

2. Calculate $V_1$ and $V_2$:

   $V_1 = \frac{Q}{A_1} = \frac{0.029}{0.01767} = 1.641 \, \text{m/s}$ $V_2 = \frac{Q}{A_2} = \frac{0.029}{0.00442} = 6.56 \, \text{m/s}$

Step 2: Apply Bernoulli's equation between the inlet and throat

Using Bernoulli's equation, including the height difference, we have: $P_1 + \frac{1}{2} \rho V_1^2 + \rho g z_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g z_2$ where:

• $P_1$ and $P_2$ are the pressures at the inlet and throat,
• $V_1$ and $V_2$ are the velocities at the inlet and throat,
• $z_1$ and $z_2$ are the heights of the inlet and throat, with $z_2 - z_1 = h$,
• $g$ is the gravitational acceleration (9.81 m/s²).

Rearrange to find $\Delta P = P_1 - P_2$: $\Delta P = \frac{1}{2} \rho (V_2^2 - V_1^2) + \rho g h$

Substituting the values:

1. Calculate the dynamic pressure term: $\frac{1}{2} \rho (V_2^2 - V_1^2) = \frac{1}{2} \times 780 \times (6.56^2 - 1.641^2) = 14809.4 \, \text{Pa}$

2. Calculate the hydrostatic pressure term: $\rho g h = 780 \times 9.81 \times 0.225 = 1715.4 \, \text{Pa}$

3. Total pressure difference $\Delta P$ (considering the Venturimeter constant $C_x = 0.96$): $\Delta P = C_x \left(14809.4 + 1715.4\right) = 0.96 \times 16524.8 = 15844.6 \, \text{Pa}$

Answer:

The pressure difference between the inlet and throat is approximately 15,844.6 Pa.

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Would you like further explanation, or have any questions?

Related Questions

1. How does the height difference affect the pressure drop in a Venturimeter?
2. What are the limitations of using the Venturimeter constant (Cₓ) in real-life applications?
3. How would the pressure difference change if a denser fluid were used?
4. Can a Venturimeter be used horizontally, and how would this affect the calculations?
5. Why is the throat diameter smaller than the inlet in a Venturimeter?

Tip

Remember, specific gravity relates the density of a fluid to water, which is useful in converting it directly into density (kg/m³) by multiplying by 1000.