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4 millibars over 2 minutes - gas cost?
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The disastrous news is that my gas pipes are leaky, and I've been disconnected. *shiver*
The good news is that it's set me to thinking, my gas bills have been really high despite my frugality measures to reduce them.
So, does anyone know how much it was costing me if my gas pipes were leaking 4 millibars over two minutes(or whatever the time measure they use is)
Thank you for any help!
The good news is that it's set me to thinking, my gas bills have been really high despite my frugality measures to reduce them.
So, does anyone know how much it was costing me if my gas pipes were leaking 4 millibars over two minutes(or whatever the time measure they use is)
Thank you for any help!
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Comments
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A millibar is a measure of pressure, so presumably someone has done a pressure test and found it lost 2 millibars.
It is not possible, from just the information you have given, to work out what volume(quantity) of gas you have been losing. However it is likely to be only a very small amount.0 -
So, does anyone know how much it was costing me if my gas pipes were leaking 4 millibars over two minutes(or whatever the time measure they use is)
Thank you for any help!
Did you have a really strong gas smell in and around the house? If not, then the cost of the leak would be minimal.0 -
No, of course there was no smell, otherwise I would reported it.
The question is straightforward. Gas was leaking at a rate of 4 mbar per x units of time.
I was hoping to find someone here who knew enough about the industry to know:
- what the unit of time they use for this was
- how to calculate what the volume of gas lost in this manner would be over a day/week/month would be.0 -
Sorry cattkitt but as Cardew explained "Gas was leaking at a rate of 4 mbar per x units of time" doesn't actually make any sense. It isn't possible to work out a volume of gas from this information.
It's simply not mathematically possible. We're not just being difficult.0 -
No, of course there was no smell, otherwise I would reported it.
The question is straightforward. Gas was leaking at a rate of 4 mbar per x units of time.
I was hoping to find someone here who knew enough about the industry to know:
- what the unit of time they use for this was
- how to calculate what the volume of gas lost in this manner would be over a day/week/month would be.
Atmospheric pressure = 1000 mb = 10^5 Nm-2
Natural gas pressure normally 1020 mb = 1.02 * 10^5 Nm-2
after 2 mins 1016 mb = 1.016 * 10^5 Nm-2
P*V = k*N*T [combined ideal gas law]
where
P = Pressure Nm-2
V = Volume m3 [of your pipe work, of course]
k = Boltzmann constant, 1.381×10−23J·K−1
N = Number of gas molecules
T = Temperature K - absolute
Initially
P0 * V = k * N0 * T
After 2 mins
P1 * V = k * N1 * T
The difference is the gas law applied to the gas you have lost. So assuming no temperature change:
[P0 - P1] * V = k * [N0 - N1] * T
[N0 - N1] = [[P0 - P1] * V] / [ k * T]
Now
[N0 - N1] is the number of molecules of gas you have lost in the period of the test. Assume a temperature of 15 deg C, your absolute temperature is 273 + 15 = 288.
[N0 -N1] = [[1.02 * 10^5 Nm-2 - 1.016 * 10^5 Nm-2] * V] / [Boltzmann constant * 288]
So if you measure the volume of your pipework, you can calculate how many gas molecules you are losing in the period of the gas test.
I'll do you a deal. If you measure the volume of your pipework and calculate how many molecules you have lost, I will calculate the volume of gas you have lost in the period of the test.
Over to you.Hi, we’ve had to remove your signature. If you’re not sure why please read the forum rules or email the forum team if you’re still unsure - MSE ForumTeam0 -
LOL :rotfl:0
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DVardysShadow wrote: »k = Boltzmann constant, 1.381×10−23J·K−1
.
It would perhaps have made it easier for the OP if you had expressed the Boltzmann's constant in Planck units - as below.Planck's system of natural units is one system constructed such that the Boltzmann constant is 1. This gives
as the average kinetic energy of a gas molecule per degree of freedom; and makes the definition of thermodynamic entropy coincide with that of information entropy:
The value chosen for the Planck unit of temperature is that corresponding to the energy of the Planck mass—a staggering 1.416 785(71) × 1032 K.0 -
DVardysShadow wrote: »OK, here goes:
Atmospheric pressure = 1000 mb = 10^5 Nm-2
Natural gas pressure normally 1020 mb = 1.02 * 10^5 Nm-2
after 2 mins 1016 mb = 1.016 * 10^5 Nm-2
P*V = k*N*T [combined ideal gas law]
where
P = Pressure Nm-2
V = Volume m3 [of your pipe work, of course]
k = Boltzmann constant, 1.381×10−23J·K−1
N = Number of gas molecules
T = Temperature K - absolute
Initially
P0 * V = k * N0 * T
After 2 mins
P1 * V = k * N1 * T
The difference is the gas law applied to the gas you have lost. So assuming no temperature change:
[P0 - P1] * V = k * [N0 - N1] * T
[N0 - N1] = [[P0 - P1] * V] / [ k * T]
Now
[N0 - N1] is the number of molecules of gas you have lost in the period of the test. Assume a temperature of 15 deg C, your absolute temperature is 273 + 15 = 288.
[N0 -N1] = [[1.02 * 10^5 Nm-2 - 1.016 * 10^5 Nm-2] * V] / [Boltzmann constant * 288]
So if you measure the volume of your pipework, you can calculate how many gas molecules you are losing in the period of the gas test.
I'll do you a deal. If you measure the volume of your pipework and calculate how many molecules you have lost, I will calculate the volume of gas you have lost in the period of the test.
Over to you.
Any chance you can explain that in a bit more detail please?;)No free lunch, and no free laptop0 -
It would perhaps have made it easier for the OP if you had expressed the Boltzmann's constant in Planck units - as below.
I think you and I must be as thick as 2 Plancks with this stuff.Hi, we’ve had to remove your signature. If you’re not sure why please read the forum rules or email the forum team if you’re still unsure - MSE ForumTeam0
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