Yokogawa 415 Steam and Gas User Manual

Page 23

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Example 3

If the mass flow is defined at non- standard base conditions and it
is required to display the corrected volume at standard conditions,
then it is first necessary to convert the mass to an equivalent mass
at standard conditions.

Using example 2 for a differential pressure device, the
corresponding mass at 15°C and 101.325 kPa can be determined
from equation 7 as:

S

M1

= S

MB

.

P

1

P

B

.

T

B

T

1

.

Z

B

Z

1

where S

M1

= the new span at 15°C and 101.325 kPa

and

with the input A = 1

Therefore, the new span S

M1

, with Z

B

= Z

1

= 1, is:

S

M1

= 1000 x

101.325

220

x

30 + 273.2
15 + 273.2

= 696.1 kg/hr

Hence, the span would be programmed as 696.1 kg/hr, the base
temperature as 15°C and the base pressure as 101.325 kPa. The
corrected volume will now be displayed at standard conditions.

Operation 21

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