Question

Question 23 Glycerine at 30°C is flowing between two large reservoirs. These reservoirs have differences in the level of 10 m and connected through a plastic pipe. This pipe is 75 m long and goes up to a height of 3 m above the upper reservoir for a distance of 25 m where its diameter is 50 cm from the sharp-edged' entrance. The average velocity of glycerine along this distance is 62.79 m/s. After 5 m declining from the highest point of the pipe, its diameter is suddenly expanded to 100 cm before enters the 'rounded' exit of the lower reservoir. a) Sketch a diagram for the problem above and list the location where losses occurred in that system, b) Calculate the 'major losses that occurred in the system, c) Calculate the 'minor losses' and total head loss that occurred in the system, d) Pressure (P) at the highest point of the pipe.

Answer 1

loss of pressure or “head” that occurs in pipe or duct flow due to the effect of the fluid's viscosity near the surface of the pipe or duct.[1] In mechanical systems such as internal combustion engines, the term refers to the power lost in overcoming the friction between two moving surfaces, a different phenomenon.

${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ ${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$ In the following discussion, we define volumetric flow rate V̇ (i.e. volume of fluid flowing) V̇ = πr2v

where

r = radius of the pipe (for a pipe of circular section, the internal radius of the pipe).

v = mean velocity of fluid flowing through the pipe.

A = cross sectional area of the pipe.

In long pipes, the loss in pressure (assuming the pipe is level) is proportional to the length of pipe involved. Friction loss is then the change in pressure Δp per unit length of pipe L

${\displaystyle {\frac {\Delta p}{L}}.}$

When the pressure is expressed in terms of the equivalent height of a column of that fluid, as is common with water, the friction loss is expressed as S, the "head loss" per length of pipe, a dimensionless quantity also known as the hydraulic slope.

${\displaystyle S={\frac {h_{f}}{L}}={\frac {1}{\rho \mathrm {g} }}{\frac {\Delta p}{L}}.}$

ρ = density of the fluid, (SI kg / m3)

g = the local acceleration due to gravity;