solved problem on pressure measurement: Fluid Properties and Flow Characteristics - Fluid Mechanics and Machinery
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
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