Question

2. Liquid ethylene oxide at 0 °C is evacuated through a circular pipe with a diameter of 2.5 cm and an average velocity of 10 m/s to the open air at a pressure of 105 kPa. The measured pressure (absolute) 20 m before the exit using a pressure gage is 192 kPa. Calculate the volumetric flow rate of ethylene oxide if (a) the pipe is horizontal, (b) the pipe is bent 10° up, and (c) 15° down from the horizontal (peo = 882 kg/mº and peo=6.35x102 kg/m*s) Assumptions: The flow is steady and incompressible. 2. The flow section considered is away from the entrance, and thus the flow is fully developed. 3. The roughness effects and reducing energy components are negligible, and thus the inner surfaces are considered to be smooth. 4. No turbine or pump in the system.

Answer 1

A) The volumetric flow rate when pipe is horizontal ≈  6.57 * 10⁻⁸ m³/sec

B) The volumetric flow rate when pipe is bent 10° upwards ≈ 6.35 * 10⁻⁸ m³/s

C) The Volumetric flow rate when pipe is 15° downwards ≈ 6.07 * 10⁻⁸ m³/s

Given data :

P1 = 105 kPa

P2 = 192 kPa

Radius ( r ) = 2.5 cm / 2 = 0.0125 m

Δ x ( change in length ) = 20 m

μeo = 6.35 * 10² kg / m.s

Determine the Volumetric flow rate

The general equation to determine volumetric flow rate

Q = $\frac{-\pi }{8u} (\frac{dP}{dx}) r^{4}$  ------ ( 1 )

a) When the pipe is horizontal

ΔP = ( 192 - 105 ) = 87 kPa

Q =  π / 8  * 6.35 * 10² * ( $\frac{87 * 10^{3} }{20}$ ) * 0.0125⁴

   ≈ 6.57 * 10 ⁻⁸ m³/sec

b) When the pipe is bent 10° up

ΔP = - P1 + P2 cos 10°

      = 84.08 kPa

insert value into equation ( 1 )

Q ≈  6.35 * 10⁻⁸ m³/s

c) When pipe is 15° downwards

ΔP =  P2 cos 15° - P1

    = 192 * cos 15° - 105

    = 192 * 0.9659 - 105  =  80.45 kPa

Insert value into equation ( 1 )

Q =  π / ( 8  * 6.35 * 10²)  *  4022.5  * 0.0125⁴

   = π / ( 5080 ) * 0.00009820556

   = 6.07 * 10⁻⁸ m³/s

Hence we can conclude that  The volumetric flow rate when pipe is horizontal ≈  6.57 * 10⁻⁸ m³/sec,  The volumetric flow rate when pipe is bent 10° upwards ≈ 6.35 * 10⁻⁸ m³/s,  The Volumetric flow rate when pipe is 15° downwards ≈ 6.07 * 10⁻⁸ m³/s

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