Solved problem on pressure measurement

SOLVED PROBLEM ON PRESSURE MEASUREMENT

Example - 47

Express the pressure intensity of 0.7356 N/mm2 gauge absolute pressure in (i) KN/m2 of and (ii) in m of water.

Given data:

Gauge pressure = 0.7356 N/mm2

= 0.7356 × 106 N/m2.

Solution:

Case (i) in KN/m2

Absolute pressure = Atmosphere pressure + Gauge pressure

W.K.T Atmospheric pressure = 1.014 × 105 N/mm2

Absolute pressure = 1.014 × 105 + 0.7356 × 106 

= 8.37 × 105 N/m2

= 837 KN/m2

Absolute pressure = 837 KN/m2.

Case (ii) In m of water

Gauge pressure = 0.7356 × 106 N/m2

Gauge pressure = 0.7356 × 106 / 9810

= 74.985 m of water

W.K.T

Atmospheric pressure = 10.33 m of water

Absolute pressure = atmospheric pressure + gauge pressure

= 10.33 + 74.985

= 85.315 m of water

Absolute pressure = 85.315 m of water.

Result:

Case (i) Absolute pressure = 837 KN/m2

Case (ii) Absolute pressure = 85.315 m of water.

Example - 48

A gauge records a pressure of 24.52 KN/m2 in vacuum. Compute the corresponding absolute pressure in (a) KN/m2. (b) m of water. The local atmospheric pressure is 0.75 m of mercury; specific gravity of mercury is 13.6. 

Given data:

Vacuum pressure = 24.52 KN/m2

Atmospheric pressure = 0.75 m of mercury

Solution:

case (i) in KN/m2

Atmospheric pressure = 0.75 m of mercury

= (13.6 × 9810) × 0.75      (p = WH)

= 100062 N/m2

= 100.062 KN/m2.

Absolute pressure = atmospheric pressure - vacuum pressure

= 100.062 - 24.52

= 75.542 KN/m2

Absolute pressure = 75.542 KN/m2

case (ii) In m of water

Atmospheric pressure = 0.75 m of mercury

= 13.6 × 0.75

'= 10.2 m of water

Vacuum pressure = 24.52 KN/m2

= 24.52 × 103 N/m2

= 2.4995 m of water.

Absolute pressure = Atmospheric pressure - Vaccum pressure

= 10.2 - 2.4995 = 7.7 m of water.

Absolute pressure = 7.7 m of water

Result:

Case (i) Absolute pressure = 75.542 KN/m2

Case (ii) Absolute pressure = 7.7 m of water

Example - 49

What is the gauge pressure in mm of mercury when the pressure at a point is (a) 85 absolute (b) 18 m of water absolute?

Given data:

Absolute pressure = 85 KN/m2 = 85 × 103 N/m2 

Specific weight of mercury = 13.6 × 9810 N/m2

Solution:

= 0.637 m of Hg

= 637.10 mm of Hg

Case (i)

Atmospheric pressure = 760 mm Hg

But absolute pressure = atmospheric pressure + gauge pressure 

Gauge pressure = absolute pressure - atmospheric pressure

= 637.10 - 760

= -122.9 mm of Hg

Case (ii)

Absolute pressure = 18 m of water

= 1323.53 mm of Hg

Gauge pressure = Absolute pressure - atmospheric pressure

= 1323.53 - 760

Result:

Case (i) Gauge pressure = 122.9 mm of Hg vacuum

Case (ii) Gauge pressure = 563.5 mm of Hg

Example - 50

Pressure Indicated by a column of water is 8.75 m what is the absolute pressure KN/m2. Take atmospheric pressure as 101.325 KPa.

Given data:

Water column H = 8.75 m

Absolute pressure Pabs = ?

Atmospheric pressure Patm = 101.325 KPa = 101.325 KN/m2

Solution:

Pabs = Patm + Pgauge

P = hw

Gauge pressure = 8.75 × 9810 N/m2 = 85837.5 N/m2

= 85.837 KN/m2

Result:

Gauge pressure = 85.837 KN/m2.

Example - 51

A gauge is fitted to a cylinder records a pressure of 24.52 KN/m2 vacuum compute the corresponding absolute pressure in (i) KN/m2 (ii) m of water. The local atmospheric pressure is 755 mm of Hg.

Given data:

Solution:

Case (i) Absolute pressure in KN/m2

Pabs = Patm + Pgauge

= 101.658 - 24.52

Case (ii) Absolute pressure in m of water

Result:

Case (i) Absolute pressure = 77.138 KN/m2

Case (ii) Absolute pressure = 7.863 m of water