The pressures at two sections of a horizontal pipe are 0.3 kgf/cm^{2} and 0.6 kgf/cm^{2} and the diameters are 7.5 cm, and 15 cm respectively. Determine the direction of flow if water flows at a rate of 8.5 kg/sec. State your assumptions.

Data:

P_{1 }= 0.3 kgf/cm^{2}

P_{2} = 0.6 kgf/cm^{2}

D_{1 }= 7.5 cm

D_{2} = 15 cm

Mass flow rate = 8.5 kg/sec

Formulae:

- Equation of continuity \(\displaystyle \rho_1A_1v_1 = \rho_2A_2v_2 \)
- Bernoulli's equation
- Mass flow rate = volumetric flow rate x density

For the flow direction from 1 to 2,

\(\displaystyle \frac{p_1}{\rho_1g} + \frac{v_1^2}{2g}+z_1 = \frac{p_2}{\rho_2g} + \frac{v_2^2}{2g} + z_2 + h + w - q \)Calculations:

Volumetric flow rate = 8.5/1000 = 8.5 x 10^{-3} m^{3}/sec

V_{1} = 8.5 x 10^{-3}/(πD_{1}^{2}/4) = 8.5 x 10^{-3}/(π x 0.075^{2}/4) = 8.5 x 10^{-3}/0.00441= 1.924 m/sec

V_{2} = 8.5 x 10^{-3}/(πD_{2}^{2}/4) = 8.5 x 10^{-3}/(π x 0.15^{2}/4) = 8.5 x 10^{-3}/0.01767 = 0.481 m/sec

P_{1} = 0.3 kgf/cm^{2} = 0.3 x 9.812 N/cm^{2} = 2.9436 x 10^{4} N/m^{2}

P_{2} = 0.6 kgf/cm^{2} = 0.6 x 9.812 N/cm^{2} = 5.8872 x 10^{4} N/m^{2}

Assuming the flow direction is from 1 to 2:

2.9436 x 10^{4}/1000 + 1.924^{2}/2 = 5.8872 x 10^{4}/1000 + 0.481^{2}/2 + h + w -q

29.436 + 1.851 = 58.872 + 0.116 + h + w -q

In the given problem work done by fluid (w) and pump work on fluid (q) are zero.

So to balance the above equation, the quantity h has to have negative values. This is not possible.

The above equation will be a correct one, if the flow is from 2 to 1.

i.e. 58.872 + 0.116 = 29.436 + 1.851 + h

Therefore the **flow direction is
from the end at which pressure is 0.6 kgf/cm ^{2} and diameter is
15 mm to the end at which pressure is 0.3 kgf/cm^{2} and diameter
is 7.5 mm.**

Last Modified on: 04-Feb-2022

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