Wow, a lot of sleuthing to do !
To begin with input sheet line 1:
STP
Standard temperature and pressure (STP) is usually defined as being at 0°C (273.15 K) and 1013.25 hPa (1 atm). However, standard temperature may also be specified as 20 °C or 25 °C. Sometimes these reference conditions may also be referred to as normal temperature and pressure (NTP). Special industrial branches may even have their own definitions, e.g. the gas industry may reference flow volume to a temperature of 70 °F.
So you want to check which is which for your case.
MMSCFD and kSm
3/d is easy: 1 SCFD = (12*0.0254)
3 m
3/d
(This factor should be in one of the two cells, probably in the kSm
3/d one)
(Provided the standard states are the same -- they are in your example)
A converter comes close: when I use
this one , 17.72 in 5 decimals yields 500.63 kSm
3/d
so 0.2% off for some reason -- different but close standard ?
Next : MW, molweight is 16.88 so mainly CH
4 (as opposed to
19.5 for natural gas)
No idea what the dG stands for (
@Chestermiller ?)You'll have to find out how the viscosity, Z and C
p/C
v are calculated (with a formula in the cell or with a VBA function call).
In case of VBA (which I would expect): press alt-F11 and search for the function name. Welcome to the powerful world of VBA.
With 61 bar and 55 ##^\circ##C you are
past the critical T and p of methane
Then come the pipe parameters: I can't find a wall thickness of (219.08 - 197.3)/2 = 10.89 mm
here, but I suppose the 197.3 mm is what you need and can trust. (I would expect 1/2 inch wall thickness, actually).
And roughness, an optimistic 0.05 mm.
-------------------------------------------------
We continue with output sheet line 1:
STP density is a lookup or again VBA
(
here it says for methane
Density, gas at STP; 32°F/0°C 1 atm | 44.7 | mol/m3 | 0.7168 | kg/m3 |
)
T&P density may come from the gas law with a formula using Z, or again through VBA.
Mass flow is straightforward:
STP volume flow 501.77 kSm
3/d * 1000 /(24 h/d)/(3600 s/h) * 0.71 kg/m3 = 4.12 kg/s
(why he gets 4.14 I don't know. 0.713 kg/m3 ?)
Then Q in m3/s from: mass flow / T&P density= 4.14/39.67 = 0.104 m3/s
And I would get a velocity of 0.104 / (¼π 0.1973
2) = 3.41 m/s, but he gets 3.54 !?
We skipped over the friction factor that follows from this calculation:
Churchill ,
Darcy
The resulting pressure drop of 1.92 kPa (0.02 Bar) per 100 m isn't all that much compared to the 61 Bar (I suppose the contraction from 18" to 8" isn't at hundreds of km from the delivery points ?)
----I did a lot of googling for you which you should be able to do yourself...
(I like reverse engineering, it apppears -- but to do a good job the .xls file is needed for that)
And I certainly second Chet's advice -- the 0.95 Z doesn't dgrade a preliminary answer all that much.