Page 26 - BASIC CONCEPTUAL OF THERMOFLUID

CHAPTER 2: FLUID APPLICATION

Example 2.6:

Oil flows through a pipe RS and split into two pipes, which are ST and SU as show in Figure 2.5. The following
information as given;
Diameter pipe,

RS = 250 mm Diameter pipe,
ST = 200 mm Specific gravity,

Soil = 0.95 Calculate:

i. Discharge and mass flow rate of oil at pipe RS if velocity is 2.5 m/s

ii. Diameter pipe SU if velocity at pipe ST is 1.5 m/s and at pipe, SU is 3 m/s.

Figure 2.5

Solution:

A1 = 0.04908 m² , A2 = 0.031416 m²

Q1 = 2.5 x 0.04908
Q1 = 0.1227 m³/s

Q1 = Q2 + Q3
Q1 = 1.5 (0.031416) + Q3

Q3 = 0.075576 m³/s

Q3 = A3 x v

0.075576
3 =
3

A3 = 0.025192 m²
= 0.179

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