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Home / Company Information / Technical Information / Gas Examples

WORKED EXAMPLES FOR GASES & VAPOURS


Example 1

Given a safety valve of seat diameter do = 13mm, set at 7.5 bar.g with coefficient of discharge Kdr = 0.78.
Question:

  • What mass of air will it discharge, when flowing at 40°C?
  • What will be the discharge expressed in standard litres/sec.?

The flow area will be;

A=\frac{\pi}{4}\cdot 13^2=132.7\;\;\;\;\;\;\;\;\;\;(mm^2)

From expression (4)

q_{ma}=14.54\cdot 0.78\cdot 132.7\cdot \frac{9.25}{\sqrt{313}}
=786.86\;\;\;\;\;\;\;\;\;\;(kg/hr)

Also, from expression (13) and converting from 15°C to 40°C

Q_{air}=0.193\cdot 0.78\cdot 132.7\cdot 9.25\cdot \sqrt{\frac{288}{313}}
=177.25\;\;\;\;\;\;\;\;\;\;(std. litres/s)

For Seetru valves the accurate value of the flow area is given in the datasheets.

Example 2

Given carbon dioxide (CO2) stored at 24 bar gauge and -20°C. Given also a safety valve of seat bore do = 13mm and coefficient if discharge Kdr = 0.78 and set to 25 bar gauge pressure.

Question:

  • What mass of gas discharging when flowing in kg/hr.?
  • What is the volume of gas discharging expressed at standard conditions, i.e. 1.013 bar abs. and 15°C in std. litres/s?
  • What is the volume of gas discharging at the actual flowing conditions, 27.5 bar gauge at -20°C?

The flow area will be;

A=\frac{\pi}{4}\cdot 10^2=78.54\;\;\;\;\;\;\;\;\;\;(mm^2)

From expression (2) and gas data, C = 2.63, M = 44, and taking Z = 1

q_{mg}=0.78\cdot 78.54\cdot 28.5\cdot 2.63\cdot \sqrt{\frac{44}{253}}
=1915\;\;\;\;\;\;\;\;\;\;(kg/hr)

Furthermore, from expression (12)

Q_{gas}=6.566\cdot \frac{0.78\cdot 2.63\cdot 28.5\cdot 78.54}{\sqrt{44}\cdot \sqrt{253}}
=285.8\;\;\;\;\;\;\;\;\;\;(std. litres/s)

Finally, by using the gas law (9), the above value converts to litres/sec at 28.5 bar absolute and -20°C

=285.8\cdot \frac{1.013}{28.5}\cdot \frac{253}{288}=8.924\;\;\;\;\;\;\;\;\;\;(litres/s)

Example 3

Given a required discharge of R22 Freon gas of 7000 kg/hr at 28.5 bar gauge safety valve set pressure and 140°C

Question:

  • What is the minimum safety valve seat bore diameter for a coefficient of discharge Kdr of 0.73?

Turning to expression (8), and using the gas data F = 1.632

d_o\geq \sqrt{\frac{0.08756}{1.623\cdot 0.7\cdot 32.35}\cdot \sqrt{413}\cdot 7000}
\geq 18.36\;\;\;\;\;\;\;\;\;\;(mm)

Example 4

Required to discharge Helium gas at 1500 lb/in2 gauge pressure and 0°C. Minimum volume to flow 2.45 cu ft/min

Question:

  • What is the minimum safety valve seat bore diameter for a coefficient of discharge Kdr of 0.55?

Converting to metric units

2.45\;cuft/min=\frac{\sqrt{2.45\cdot 28.3168}}{60}
=1.156\;\;\;\;\;\;\;\;\;\;(litres/s)

Also

1500\;lb/in^2=1500\cdot 0.068948
=103.4\;\;\;\;\;\;\;\;\;\;(bar\;gauge)
=104.4\;\;\;\;\;\;\;\;\;\;(bar\;absolute)

The volume flow needs to be converted to standard conditions of 1.013 bar absolute and 15°C, by using the gas laws (9)

1.156\;litres/s=1.156\cdot \frac{(103.4+1)}{1.013}\cdot \frac{288}{273}
=125.7\;\;\;\;\;\;\;\;\;\;(std.litres/s)

The given discharge pressure (104.4 bar absolute) is taken as the actual flowing pressure P. Hence to obtain the set pressure p of the valve

P=104.4=(1.1p+1)
\therefore\;p=\frac{(104.4-1)}{1.1}=94\;\;\;\;\;\;\;\;\;\;(bar)

Using expression (16) and taking from the gas data f = 2.842

d_o\geq\sqrt{\frac{0.38665\cdot \sqrt{273}}{2.842\cdot 0.55\cdot 104.4}\cdot 125.7}
\geq2.2\;\;\;\;\;\;\;\;\;\;(mm)

Example 5

Given a safety valve of do = 8mm set at 105 bar gauge, discharging hydrogen gas at 20°C

Question:

  • What is the discharge in Nm3/hr, given according to DIN AD-Merkblatt A2 (TÜV)?

where, for expression (2) the following values should be used,

Kdr
= αw = 0.66
P
= 105 x 1.1 + 1 = 116.5 bar abs.
C
= 2.7
M
= 2
Z
= 1
T
= 293°K

and

A=\frac{\pi}{4}\cdot 6.4=50.27\;\;\;\;\;\;\;\;\;\;(mm^2)
\therefore\;q_{mg}=50.27\cdot 0.66\cdot 116.5\cdot 2.7\cdot \sqrt{\frac{2}{293}}
=862.23\;\;\;\;\;\;\;\;\;\;(kg/hr)

from gas equation (9) converted to m3

W
= 784.52
T
= 273°K (std. temperature)
M
= 2
P
= 1.013 bar abs. (std. pressure)
\therefore\;V=\frac{83.14}{1000}\cdot \frac{862.23\cdot 273}{2\cdot 1.013}=9659.5\;\;\;\;\;\;\;\;\;\;(Nm^3/hr)

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