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A horizontally placed venturi meter has a throat of diameter 200 mm. It is installed in a pipe of diameter 300 mm. The reading from the throat tappings is 135 mm mercury. Calculate the flow of water in the pipe.

Physics

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Answer

0.199 m3/s\text{0.199 m}^3/\text{s}

Step-by-step solution

4 steps

Here's how to calculate the flow of water in the pipe using the Venturi meter:

Given:

  • Throat diameter, d2=200 mm=0.200 md_2 = 200 \text{ mm} = 0.200 \text{ m}
  • Pipe diameter, d1=300 mm=0.300 md_1 = 300 \text{ mm} = 0.300 \text{ m}
  • Manometer reading (mercury), hm=135 mm=0.135 mh_m = 135 \text{ mm} = 0.135 \text{ m}
  • Fluid in pipe: Water
  • Manometer fluid: Mercury

Assumptions:

  • Density of water, ρw=1000 kg/m3\rho_w = 1000 \text{ kg/m}^3
  • Density of mercury, ρm=13600 kg/m3\rho_m = 13600 \text{ kg/m}^3
  • Acceleration due to gravity, g=9.81 m/s2g = 9.81 \text{ m/s}^2
  • Coefficient of discharge for the Venturi meter, Cd=0.98C_d = 0.98 (a typical value when not specified).

  1. Calculate the cross-sectional areas of the pipe and the throat.

    The area of the pipe (A1A_1) and the throat (A2A_2) are calculated using the formula for the area of a circle, A=π4d2A = \frac{\pi}{4}d^2.

    A1=π4(0.300 m)2=π4×0.09 m2≈0.070686 m2A_1 = \frac{\pi}{4} (0.300 \text{ m})^2 = \frac{\pi}{4} \times 0.09 \text{ m}^2 \approx 0.070686 \text{ m}^2 A2=π4(0.200 m)2=π4×0.04 m2≈0.031416 m2A_2 = \frac{\pi}{4} (0.200 \text{ m})^2 = \frac{\pi}{4} \times 0.04 \text{ m}^2 \approx 0.031416 \text{ m}^2

  2. Convert the manometer reading from mm of mercury to an equivalent head of water (hh).

    The differential head hh in terms of the flowing fluid (water) is given by:

    h=hm(ρmρw−1)h = h_m \left( \frac{\rho_m}{\rho_w} - 1 \right) h=0.135 m(13600 kg/m31000 kg/m3−1)h = 0.135 \text{ m} \left( \frac{13600 \text{ kg/m}^3}{1000 \text{ kg/m}^3} - 1 \right) h=0.135 m(13.6−1)h = 0.135 \text{ m} (13.6 - 1) h=0.135 m×12.6h = 0.135 \text{ m} \times 12.6 h=1.701 m of waterh = 1.701 \text{ m of water}

  3. Apply the Venturi meter flow rate formula.

    The formula for the volumetric flow rate (QQ) through a Venturi meter is:

    Q=CdA1A22ghA12−A22Q = C_d \frac{A_1 A_2 \sqrt{2gh}}{\sqrt{A_1^2 - A_2^2}}

  4. Substitute the calculated values and solve for QQ.

    First, calculate the term 2gh\sqrt{2gh}:

    2gh=2×9.81 m/s2×1.701 m\sqrt{2gh} = \sqrt{2 \times 9.81 \text{ m/s}^2 \times 1.701 \text{ m}} 2gh=33.37062 m2/s2≈5.7767 m/s\sqrt{2gh} = \sqrt{33.37062 \text{ m}^2/\text{s}^2} \approx 5.7767 \text{ m/s}

    Next, calculate the denominator term A12−A22\sqrt{A_1^2 - A_2^2}:

    A12=(0.070686 m2)2≈0.0049964 m4A_1^2 = (0.070686 \text{ m}^2)^2 \approx 0.0049964 \text{ m}^4 A22=(0.031416 m2)2≈0.00098696 m4A_2^2 = (0.031416 \text{ m}^2)^2 \approx 0.00098696 \text{ m}^4 A12−A22=0.0049964 m4−0.00098696 m4\sqrt{A_1^2 - A_2^2} = \sqrt{0.0049964 \text{ m}^4 - 0.00098696 \text{ m}^4} A12−A22=0.00400944 m4≈0.063320 m2\sqrt{A_1^2 - A_2^2} = \sqrt{0.00400944 \text{ m}^4} \approx 0.063320 \text{ m}^2

    Now, substitute all values into the flow rate formula:

    Q=0.98×(0.070686 m2)×(0.031416 m2)×(5.7767 m/s)0.063320 m2Q = 0.98 \times \frac{(0.070686 \text{ m}^2) \times (0.031416 \text{ m}^2) \times (5.7767 \text{ m/s})}{0.063320 \text{ m}^2} Q=0.98×0.012829 m3/s0.063320Q = 0.98 \times \frac{0.012829 \text{ m}^3/\text{s}}{0.063320} Q=0.98×0.20260 m3/sQ = 0.98 \times 0.20260 \text{ m}^3/\text{s} Q≈0.19855 m3/sQ \approx 0.19855 \text{ m}^3/\text{s}

    Rounding to three significant figures:

    Q≈0.199 m3/sQ \approx \boxed{\text{0.199 m}^3/\text{s}}
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