Answers to some common Air Separation Unit, Oxygen and Nitrogen Liquefaction Questions
Question 1: If I change the delta T of the main vaporizer from 2.5 °C to 1.5 °C, how much energy consumption can I save?
Answer 1: Your energy saving is:
Energy saved (kW) = 0.0478 * Q * (2.5 – 1.5) = 0.0478 * Q
Where Q is the main vaporizer duty in kW
Question 2: If I increase 10 mbar pressure drop in the waste nitrogen circuit, how much energy will increase on the total energy?
Answer 2: since the impact of 13.6 mbar increase on WN circuit can be compensated by reducing 0.1 °C vaporizer DT, 10 mbar will be equivalent as 0.0735 °C
Energy impact will be 0.0478 * Q * 0.0735 = 0.00351 * Q (kW) per 10 mbar increase in WN circuit
Where Q is the main vaporizer duty in kW
Question 3: If I increase 10 mbar pressure drop in the MPAir circuit (around 6 bara), how much energy will increase on the total energy?
Answer 3: since the impact of 42 mbar increase on MPAir circuit can be compensated by reducing 0.1 °C vaporizer DT, 10 mbar will be equivalent as 0.02381 °C
Energy impact will be 0.0478 * Q * 0.02381 = 0.001138 * Q (kW) per 10 mbar increase in MPAir circuit
Where Q is the main vaporizer duty in kW
Question 4: For a process scheme with only one MAC (around 6 bara), If I reduce 10 mbar on the WN circuit and 10 mbar on the MP Air circuit and 0.1 °C on the main vaporizer DT respectively, what’s the energy saving respectively?
Answer 4: We can easily calculate. 10 mbar on the WN circuit will save 0.29%; while 10 mbar on the MP Air circuit will only save 0.095%. since 0.1 °C DT is equivalent 13.6 mbar pressure drop on LP circuit, so by reducing 0.1 °C DT, we will save 0.29% * 13.6 / 10 = 0.394%. Please note, this is just the saving on the MAC, not include any boosters or nitrogen compressors. According to the conclusion mentioned here, jumping to a film vaporizer (0.6 °C DT) from a bath vaporizer (1.2 °C DT) will save us 0.394% * (1.2 – 0.6) / 0.1 = 2.36%. You may notice this 2.36% is bigger than the calculation by Benoit DAVIDIAN (see the end of the paper). The reason is we save 2.36% on the MAC (around 6 bara), while on the boosters or nitrogen compressors, we will consume more energy due to a lower suction pressure, then the global energy saving will be less than 2.36%.
Question 5: If the main vaporizer duty is 2650 kW, what’s the power on the MAC?
Answer 5: according to the conclusion above, 0.1 °C will save 0.394% energy on the MAC. At the same time, 0.1 °C saves energy 0.0478 * Q * 0.1 = 0.0478 * 2650 * 0.1 = 12.667 kW. So the MAC power is 12.667 / 0.394% = 3215 kW
Exergy loss of main vaporizer:
One can use the excel software “Balance” to calculate the exergy loss of the main vaporizer. But here a very easy formula is recommended to calculate the exergy loss.
Exergy loss vapo (kW) = Q * (T0 + 273.15) * (1 / (273.15 + Tv) – 1 / (273.15 + Tc))
Where:
Q: Main vaporizer duty in kW
T0: Cooling water temperature in °C
Tv: Lox temperature in the bath in °C
Tc: Lin condensation temperature in the vaporizer in °C
Note: The formula is applicable for pure lox vaporization and pure Lin condensation. Use “Balance” to calculate exergy loss of impure fluid vaporization and condensation.
Energy impact on the global plant:
Considering the global plant of ASU, except the main air compressor and suppressor, all the other equipments are consuming exergy, including valves, mixers, booster expander, booster aftercooler, liquid expander (It saves energy if it replaces a throttle valve. But it still consumes exergy because efficiency is not 100%), pumps, MHE, subcooler, columns, all the other vaporizers and all the other equipments inside the cold box.
Main air compressor and suppressor are the only equipments that gain exergy. The gained exergy needs to compensate all the exergy loss of the cold box.
The compressor and suppressor efficiency is not 100%, so here I introduce another conception of “exergy efficiency” (= gained exergy / input power). After checking many compressors, I found the exergy efficiency almost equals isothermal efficiency (around 70%).
All the exergy loss in cold box should equal the exergy gained from compressors. So
Exergy loss (kW) = Energy consumed (kW) * isoeff, then
Energy consumed (kW) = Exergy loss (kW) / isoeff
Where
Exergy loss (kW): Exergy loss of cold box
Energy consumed (kW): Consumed energy in main air compressor and suppressor.
isoeff: Compressor isothermal efficiency.
Let us assume 2 different delta T with subscript 1 and 2
Energy consumed1 (kW) = Exergy loss1 (kW) / isoeff
Energy consumed2 (kW) = Exergy loss2 (kW) / isoeff
After we finish the exergy analysis with “Balance”, we find except that exergy loss of main vaporizer is different, the loss of all the other equipments keep almost the same.
So here we make a big assumption: when you change the delta T of the main vaporizer, the exergy loss difference of the cold box equals the exergy loss difference of the main vaporizer. (Actually it’s not completely true when you delta T ramps too much)
ΔEnergy consumed
= Energy consumed2 - Energy consumed1
= Exergy loss2 / isoeff - Exergy loss1 / isoeff
= (Exergy loss vapo2 - Exergy loss vapo1) / isoeff
= {Q2 * (T0 + 273.15) * (1 / (273.15 + Tv) – 1 / (273.15 + Tc2)) - Q 1 * (T0 + 273.15) * (1 / (273.15 + Tv) – 1 / (273.15 + Tc1))} / isoeff
= {Q2 * (T0 + 273.15) * (Tc2 – Tv) / [(273.15 + Tv) * (273.15 + Tc2)] – Q1 * (T0 + 273.15) * (Tc1 – Tv) / [(273.15 + Tv) * (273.15 + Tc1)]} / isoeff
Considering Q2 = Q1 = Q when you delta T does not change too much.
To be simple, the formula above equals formula below if (Tc – Tv) / [(273.15 + Tv) * (273.15 + Tc)] is small and Tc2 almost equals Tc1
= {Q * (T0 + 273.15) * [(Tc2 – Tv) - (Tc1 – Tv)] / [(273.15 + Tv) * (273.15 + Tc)]} / isoeff
= {Q * (T0 + 273.15) * Δ(ΔT) / [(273.15 + Tv) * (273.15 + Tc)]} / isoeff
= Coef * Q * Δ(ΔT)
Where:
Coef = (T0 + 273.15) / [(273.15 + Tv) * (273.15 + Tc)] / isoeff
Δ(ΔT) means the difference of 2 vaporizer delta T.
Formula applies to common ASU:
For a normal double column pure Lox (99.5%) and pure Lin (100%) ASU, Lox bath is operated at 1.4 bara and -179.785 °C. Delta T 2 °C is assumed, then MP top is operated at 5.563 bara and -177.785 °C. Cooling water temperature 25 °C and isoeff 70% are considered. Then the formula above becomes below:
ΔEnergy consumed = Q * 298.15 * Δ(ΔT) / [(273.15 – 179.785) * (273.15 – 177.785)] / 0.7
= 0.0478 * Q * Δ(ΔT)
Note: formula above only applies to classical double columns, in which pure Lox (or GOX) and pure Lin (or MPGAN from MP column) are produced.
