Solved Examples Based on Inclined Venturimeter

SOLVED EXAMPLES BASED ON INCLINED VENTURIMETER 

Example - 78

Find the discharge of water through a pipe of 40cm diameter placed in an inclined position where a venturimeter is inserted, having a throat diameter of 20cm. The difference of pressure between the main and throat is measured by a liquid of specific gravity 0.7, in an inverted U-tube which gives a reading of 20cm. The loss of head between the main and throat is 0.4 times the kinetic head of the pipe.

Given data:

Pipe diameter (d1) = 40cm = 0.4m

Throat diameter (d2) = 20cm = 0.2m 

Specific gravity of oil (S) = 0.7

Manometer reading (x) = 20cm = 0.2m

Loss of head (hL) = 0.4 X Kinetic head of the pipe( V1)2/2g

To find:

Rate of flow of water (Q)

Solution:

Result :

Discharge(Q) = 0.0346 m3/s

Example - 79

In vertical pipe conveying oil of specific gravity 0.8 to pressure gauge have been installed at A and B where the diameter are 18 cm and 6 cm respectively A is 2m above B. The pressure gauge readings have shown that the pressure of B is greatest than A by -1 N/cm2. Neglecting all losses. Calculate the flow rate if the gauge at A and B are replaced by tubes filled with the same liquid and connected to a U tube containing mercury. Calculate the difference of level of mercury in the two limbs of the tube.

Given data:

Specific gravity of oil (So) = 0.8

Diameter of the pipe at point A (dA) = 18cm = 0.18m

Diameter of the pipe at point B (dB) = 6cm = 0.06m

Datum head at point A (ZA) = 2m

Datum head at point B (ZB) = 0

Difference of pressure (PB - PA) = -1 × 104 N/m2

To find:

(1) Rate of flow while using pressure gauge

(2) Difference of level of mercury while using U tube manometer

Solution:

7. Difference level of mercury in the U-tube

Result :

(i) Rate of flow while using pressure gauge (Q) = 0.0112 m3/s.

(ii) Difference of mercury level (x) = 0.04536 m

Example - 80

A 30 × 15cm Venturimeter is provided in a vertical pipe line carrying oil of specific gravity 0.6, the flow being upwards. The difference in elevation of the throat section and entrance section of the Venturimeter is 40cm. The differential U-tube mercury manometer shows a gauge deflection of 25cm. Calculate 

(i) The discharge of oil.

(ii) The pressure difference between the entrance section and throat section. Take co-efficient of discharge as 0.98.

Given data:

Pipe diameter (d1) = 30 cm = 0.3 m

Throat diameter (d2) = 15 cm = 0.15 m

Specific gravity of oil (So) = 0.6

Difference of datum head (Z2 - Z1) = 40 cm = 0.4 m

Differential manometer reading (x)= 25 cm = 0.25 m

Co-efficient of discharge (Cd) = 0.98

To find:

1. Discharge of oil (Q) 

2. Pressure difference between two sections (P1 – P2)

Solution:

5. Venturimeter head (h) = (P1/w + Z1) − (P2/w + Z2)

h = (P1/w - P2/w) + (Z1 − Z2)

h = (P1 - P2/w) + (−0.40)

(P1 - P2/w) = h + 0.40

= 5.416 + 0.40

(P1 – P2/2/w ) = 5.816

(P1 - P2) = 5.816 × w

Specific weight of oil = specific gravity of oil × specific weight of water

= 0.6 × 9810

w = 5886 N/m3

P1 – P2 = 5.816 × 5886 = 34232.976 N/m2

P1 - P2 = 34.232 KN/m2

Result :

(i) Discharge Q = 0.1844 m3/s 

(ii) Pressure difference P1 - P2 = 34.232 KN/m2

Example - 81

The oil of specific gravity 0.85 flows upwards of a volume rate of flow of 50 litre per second through a vertical venturimeter with an inlet diameter of 300mm and a throat diameter of 200mm. The co-efficient of discharge of venturimeter is 0.90. The vertical distance between pressure tapping is 200mm. Determine the difference of readings of the two pressure gauges. And also determine the manometer reading when replacing the U-tube manometer instead of pressure gauge in the pipe.

Given data:

Specific gravity of oil (So) = 0.85

Rate of flow (Q) = 50 litres = 0.05m3/s 

Inlet diameter (d1) = 300 mm = 0.3m 

Throat diameter (d2) = 200 mm = 0.2m

Co-efficient of discharge (Cd) = 0.90

Difference in datum head (Z2 – Z1) = 200mm = 0.2m

To find:

(i) Pressure difference (P1 - P2)

(ii) Manometer reading (x)

Solution:

4. Venturimeter head (h) = (P1/w + Z1) - (P2/w + Z2)

Specific weight of oil (w) = specific gravity of oil × specific weight of water

= 0.85 × 9810

w = 8338.5 N/m3

h = (P1/w - P2/w) + (Z1 − Z2)

h = (P1 - P2/w) + (-0.2) 

(P1 − P2) = (h × w) − 0.2 = 0.1280 × 8338.5 − 0.2

(P1 - P2) = 1067.19 N/m2

Result:

(i) Pressure difference (P1 - P2) = 1067.19 N/m2

(ii) Monometer Reading (x) = 8.53 × 10-3 m