To determine minor losses due to a 90° elbow by using fluid friction apparatus.

MINOR LOSSES IN THE PIPE SYSTEM

Object

To determine minor losses due to a 90o elbow.

Apparatus

fluid friction apparatus; hydraulic bench; and stopwatch.

Word Image 2942 1

Fig. Fluid friction apparatus

Theory

The minor loss due to a 90o elbow is a function of velocity head and is written as

V 2

h K

l

e90 2g

(1)

where

hl = loss of head due to Elbow

V = velocity of flow = Q/A where, Area of flow =

A   d 2

4

 3.14

4

(1.7)2

 2.27

cm2

(since d = 17 mm = 1.7 cm)

Putting values in Eq.(1),

Procedure:

Ke90

 2g

V 2

.hl

 2  981

(Q / 2.27)2

.hl

 10108.2 hl

Q2

(2)

  1. Fill the tank of the hydraulic bench with clean water.
  2. Connect the fluid friction apparatus with the hydraulic bench.
  3. Open the valve(s) that allow flow to the section under consideration and close all the irrelevant valves of the pipe friction apparatus.
  4. Connect the two ends of the tubes to the pressure-tapping nipples at either side of the pair of elbows and the manometer.
  5. Slightly open the flow control valve at the hydraulic bench.
  6. Remove air bubbles from the tubes by opening the vent valve and drain valves of the differential manometer.
  7. Read the heads in the monometers corresponding to the pressure along the entry and exit of the elbows.
  8. Collect water in the volumetric measuring section of the hydraulic bench. Read the volume collected as well as the time taken to collect that volume of water.
  9. Open the flow control valve at the hydraulic bench slightly more to take new readings.
  10. Repeat steps 7 and 8 to observe new readings.
  11. Carry out computations as per the table below and compare the average value of Ke90 for 90o elbow with that given in literature i.e. Ke90 = 0.3.

Observations and Calculations:

Table

S.

No.

Head (cm)

Discharge (cm3/sec)

K  10108.2 hl

e90 Q2

h1

h2

hl

= h1 – h2

V

V

T

Q

= V/t

(cm)

(cm)

(cm)

(lit)

(cm3)

(sec)

(cm3/sec)

1

2

3

4

5

Average Ke90 =

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