Air is supplied through a ring sparger which has 6 holes of 1mm each. and suppose the air flow rate is 1 L/min. How can I calculate the superficial gas velocity.
If the reactor is tall, Ug is not calculated simply by dividing flowrate by area, since the volumetric gas flow rate changes (increases) as the bubbles rise to the surface. For an open top tank, Boyle's law states that P1V1=P2V2 and if time element is included, would become P1Q1=P2Q2 where subscripts 1 and 2 refer to the top and and bottom of reactor.
Combining Eq. A-1b in Section A.5.1 of the ASCE standard(2-06) Annex A and Eq. A-2b where they were written as:
Q1 = QP (T1PP / TPP1) (Eq. A-1b)
QS = Q1TSP1 / T1PS (Eq. A-2b)
where
Q1 = gas flow at the gas supply system
QP = gas flow at the point of flow measurement
P1 = ambient (gas supply inlet) atmospheric pressure
PP = gas pressure at the point of flow measurement
T1 = ambient (gas supply inlet) temperature, 0K (=0C + 273)
TP = gas temperature at the point of flow measurement
Substituting Eq. A-1b into Eq. A-2b, we have
QP = QS (PS / PP)(TP / TS)
Assuming the mass amount of gas is conserved, as the bubbles rise to the surface, Boyle’s Law states that the volume is increased as the liquid pressure decreases, giving the following:
Qtop = (PP /Pb) * QP
Therefore,
Qtop = (PP /Pb) * [(Ps /PP) (TP / TS)] * QS
The average flow rate is therefore given by
Qa = QS PS TP /2/Ts [1/PP + 1/Pb]
Since Ps = 1.01325 * 105 N/m2 and Ts = 293.15 0K
Therefore,
Qa = QS* 172.82*TP * [1/PP + 1/Pb]
Of course, if the tank is not open top, the pressure in the headspace varies, and so the average gas flow rate would become more complicated.
Well, probably Sohail's tank is not high enough to bother with pressure differences, or temperature differences when properly mixed.
For high tanks this is different, indeed, but mine approach is different and easier, provided that temperature differences are small, what would be the case when not ideally sparged.
The pressures I use are top pressure and bottom pressure. With fermentations the top pressure is in general not barometric, the 'bottom' pressure is the pressure at the sparger level. The ideal gas law at constant T, is pV constant or p FV or p×vGs constant. p (z) = ptop + rho g z. Combining and averaging by integration gives that the 'characteristic' superficial gas velocity corresponds with the ln-mean pressure. Using standaard pV denoted as po Vo or po×vGso, vGsc = po/pln×vGso. Because of the ln function it kan easily and conveniently be derived that the specific power input (W/kg) equals g×vGsc, quite useful.
Verification with a 10fold pressure difference demonstrated that the arithmetic mean is a very good and more easy approximation even at that pressure difference.
My previous answer holds for aerated (bubble column type) reactors. With stirred vessels it becomes more complex. Most of the mass transfer happens at the stirrer. So for correlations with mass transfer the pressure at that level would be the most relevant, holding for standard configurations (clearance = impeller diameter, etc). For non-standard configurations (multiple impellers; aspect ratio > 1) it is even more complex to find the relevant value for correlations. E.g. when using H/T = 2, and only one impeller (C = D), the lower part might be considered as a standard stirred vessel, but the upper half as a bubble column, and then using the local average superficial gas velocity.
For multiple impellers, e.g. two in case of H/T = 2, the superficial gas velocity at the upper stirrer is of course equal to the lower one (but for pressure). However, not all gas is entrained by the upper stirrer, so the relevant gas flow through the stirrer is significantly less than the lower stirrer.
The upper stirrer is often retrofitted to improve mixing, but on the contrary it will worsen at least in case of radial impellers.
Lets say your tank top is held at 1.5 atm (151.99 kN/m^2), hence your bottom pressure is 151.99 +9.789*3.05=181.8 kN/m2, assuming ur tank is 10 ft (3.05m) tall. At Qs=1nL/min, the arithmetic mean of the volumetric gas flow rate is:
Qa=172.82*303.15*(1L/min)*(1/151.99+1/181.8)*10^-3=0.63 L/min, assuming a test temperature of 30 degC and the media fluid is water. The superficial gas velocity is then Qa/S where S is the cross-sectional area of the bioreactor.
Correct result. But looks somewhat complex. Ok, my approach: given 1.5 bar top pressure, 3 m broth height, density 1050 kg/m3, the pressure difference is 1050×9.817×3 =3100 Pa = 0.31 bar. Bottom 1.81 bar. Arithmetic mean 1.655 bar, ln-mean 1.650 bar (well this reactor is quite small, so no significant difference). Now the flow rate would be 1/1.650 = 0.61 L/min. (1 NL/min × 1 bar / 1.65 bar).
If you want 3.05 cm/s, (0.30 W/kg) you need 5 cm/s at 1 bar. With H/T = 2 you need 5.3 Nm3/min (N: at 1 bar).
Up to now temperature is irrelevant, you only need gas density when converting to mass flow (meter).
If the test temp. Is 30C and the gas flow rate given at standard 20C then u do need to convert that as well unless u want to express the superficial vel. At standard conditions also.
When using volume flow control, yes. Correction for density, resulting from both deviating temperature ànd pressure from calibrated values inside the meter. And temperature correction for the meter itself.
Volume flow control is a must. When bubbles released from the diffuser form a cloud and coalesce to form larger bubbles, it is hard to measure the transfer rate. Analysis of bubble aeration therefore depends on average values. Oxygen transfer rate depends on the average surface area of the bubbles and thus on the mean bubble diameter Db. Eckenfelder used this to relate to the average gas flow rate. Since Db depends on temperature and pressure, Qa would require adjustment to temperature as well, otherwise the basic transfer model cannot be used correctly.
@Sohail: do you mind this discussion? Otherwise we will do it elsewhere.
Thank you everyone for the reply. Now it is fully clear that superficial gas velocity is dictated by the air flow rate entering the reactor and its crossectional area. Sparger hole dia. is irrelevant.
Regards,
Sohail
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