User:Malosse/Equation of the updraught speed in a cloud

The equation of the updraught speed in a convective cloud describes the updraught vertical speed as a function of the airmass instability and the pressure difference between a parcel of air and the surrounding environment. The derivation of the equation is based on William Cotton's book.

It is often said that an air parcel will rise if it is warmer than the surrounding air. However, this model is insufficient because it ignores the effect of a pressure deficit that can be large under a supercell cumulonimbus. So, the movement equation of the air parcel is separated into 2 terms. The first term corresponds to the buoyancy, the second term corresponds to the pressure deficit (see below).

The second term in the Equation below is often ignored in the equation of the vertical motion of an air parcel and can be large in a cloud such as a cumulonimbus where even a colder air parcel can be sucked up due a pressure deficit aloft. The pressure deficit can reach 1 hPa (or more) and this deficit can be sufficient to counteract the negative buoyancy also known as convective inhibition energy. In an extreme case in Oklahoma, the pressure deficit reached 4 hPa, The thunderstorm having generated very large hail, a tornado and tubas.

When the first term of the Equation below is dominant (thermal updraught), very strong turbulence can occur, in particular at altitude. When the second term is dominant (dynamic updraught), the updraughts are laminar in particular close to the ground. These smooth updraughts can deceive an aircraft pilot, in particular a glider, the pilot not realising that he is caught in a dangerous updraught.

"Anomalies" associated to cumulonimbus
Few glider pilots know that the updraught under a cumulonimbus are often dynamically generated and are not aware of the temperature anomalies discussed in what follows.

Mild and smooth updraughts
It is often stated that the updraughts associated to cumulonimbus are almost always turbulent and can break up an aircraft. However, updraughts under a cumulonimbus are often smooth and laminar which seems to be totally at odds with which what has just been said. However, the contradiction is only apparent since in fact the air under the cumulonimbus is colder that the surrounding airmass. Thus, the updraught must be dynamic; dynamic updraughts usually being laminar. However, the turbulence usually becomes from very severe to extreme at higher altitude (around 6 km) since at this level air parcels are warmer than the surrounding airmass (negative lifted index); moreover a phase change occurs from the liquid phase to the solid phase for water droplets releasing latent heat.

Observation of temperature anomalies
It has been noted that under a cumulonimbus, the rising airmass can be colder than the surrounding air by 1 to 3 Kelvin and that consequently the updraught close to the ground is not thermal but dynamic. This observation is counter-intuitive and is little known by glider pilots who still remember the concept of Thermal index''. This concept is based on the incorrect assumption that the air in the updraught column is warmer than the outside air in the full length of that column. The equation only based on buoyancy is therefore insufficient to model updraughts and must be extended by taking into account the pressure differences generating uplifting convection.

Vertical acceleration formula
In the following, we give two formulations of the acceleration of an air parcel under and inside a convective cloud. These formulae are almost equivalent, List's formula is more general.

List's formula
The general formula giving the acceleration is the following:
 * $${d \vec{v} \over d t} = \vec{a} = g \left({T' \over T_0} - {p' \over p} + \gamma'\right) \vec{k} - {1 \over \rho_0} \vec{\nabla} p'$$


 * $$\vec{v}$$ is th velocity vector of the air parcel;
 * g is the gravity acceleration;
 * p' is the pressure difference between the air parcel and the surrounding air;
 * p is the atmospheric pressure;
 * ρ0 is the specific mass of dry air;
 * T' is the temperature difference between the parcel and the surrounding;
 * T is the temperature of the surrounding air.

The horizontal acceleration is the following:
 * $${d u \over d t} = a_x = - {1 \over \rho_0} {\partial p' \over \partial x}$$


 * u is the horizontal speed of the parcel.

The vertical acceleration is the following:
 * $${d w \over d t} = a = g \left({T' \over T} - {p' \over p} \right) - {1 \over \rho_0} {\partial p' \over \partial z} $$


 * w is the vertical speed;
 * g is the gravity acceleration;
 * cv,a is the heat capacity at constant volume;
 * cp,a is the heat capacity at constant pressure.

Cotton's formula
The vertical acceleration of an air parcel is the following: :
 * $${d w \over d t} = a = g \left({\theta'_v \over \theta_v} - {c_{v,a} \over c_{p,a}} {p' \over p} + \gamma'\right) - {1 \over \rho_0} {\partial p' \over \partial z} $$


 * w is the vertical speed;
 * g is the gravity acceleration;
 * cv,a is the heat capacity at constant volume;
 * cp,a is the heat capacity at constant pressure.
 * p' is the pressure difference between the air parcel and the surrounding air;
 * p is the atmospheric pressure;
 * ρ0 is the specific mass of dry air;
 * $$\theta_v$$ is the potential temperature of the surrouunding air;
 * $$\theta'_v$$ is the ptential temperature difference between the
 * parcel and the surrounding;
 * γ' is the correcting factor relative to the difference between the temperature and the virtual temperature.

{{hidden|align=left|title=Proof of List and Cotton Formulae|content=

Proof of the formula for dry air
We consider an air parcel at temperature T et at altitude z rising inside an airmass of temperature T0. The pressure of surrounding air is p0 and the pressure inside the air parcel is p. We define p' = p - p0 and  T' = T - T0

We assume that the air parcel has a horizontal area S (that is infinitly small) and a thicknesse e also infinitely small.

The forces balance
The parcel mass is:
 * $$m = S e \rho$$

where ρ is the specific mass of the parcel.

The weight of the parcel is: $$W = - \rho S e g$$

where g is the gravity acceleration.

At the altitude z the pressure is p(z) and at the altitude z + e, the pressure is p(z + e).the applied force applied to the air parcel is the following:
 * $$F = W + p(z) S - p(z+e) S$$

We then obtain (e is infintel small):
 * $$ p(x,y,z+e) = p(x,y,z) + {\partial p \over \partial z} e$$

In a cumulonimbus, the hydrostatic equation cannot be used. We develop.
 * $$p(x,y,z) = p_0(z) + p'(x,y,z) $$

Then,
 * $${\partial p(x,y,z) \over \partial z} = {d p_0(z) \over d z} + {\partial p'(x,y,z) \over \partial z} $$

We then obtain:
 * $$ p(x,y,z) - p(x,y,z+e) = p_0(z) - p_0(z+e) + p'(x,y,z) - \left(p'(x,y,z) + {\partial p' \over \partial z} e \right) $$

We use the hydrostatic equation and we obtain:
 * $$ p_0(z) - p_0(z+e) = + \rho_0 g e$$

Then after substitution, we obtain:
 * $$ p(x,y,z) - p(x,y,z+e) = + \rho_0 g e - {\partial p' \over \partial z} e $$

Then,
 * $$ p(x,y,z) - p(x,y,z+e) = \left(\rho_0 g - {\partial p' \over \partial z}\right) e $$

The acceleration a is
 * $$a = {F \over m}$$

We subtitute and then:
 * $$a = {W + p(x,y,z) S - p(x,y,z+e) S \over m }$$

Then,
 * $$a = {-\rho S e g + p(x,y,z) S - p(x,y,z+e) S \over \rho S e }$$

Then,
 * $$a = -g + {p(x,y,z) - p(x,y,z+e) \over \rho e }$$

We subtitute and then:
 * $$a = -g - \left(-\rho_0 g + {\partial p' \over \partial z}\right) e \times {1 \over \rho e} $$

Then,
 * $$a = - g - \left(-\rho_0 g + {\partial p' \over \partial z}\right) \times {1 \over \rho} $$

Then,
 * $$a = -g + g {\rho_0 \over \rho} - {\partial p' \over \partial z} \times {1 \over \rho} $$

At the first order we have $$\rho \approx \rho_0$$.

Then at the first order,
 * $$a = -g + g {\rho_0 \over \rho} - {\partial p' \over \partial z} \times {1 \over \rho_0} $$

We have:
 * $$\rho = \rho_0 + \rho'$$

Then,
 * $${\rho_0 \over \rho} = {\rho_0 \over \rho_0 + \rho'}$$

At the first order we then obtain:
 * $${\rho_0 \over \rho} = 1 - {\rho' \over \rho_0} $$

Then,
 * $$a = -g + g \left(1 - {\rho' \over \rho_0}\right) - {\partial p' \over \partial z} \times {1 \over \rho_0} $$

Then,
 * $$a = - {\rho' \over \rho_0} g - {\partial p' \over \partial z} \times {1 \over \rho_0} $$

Ideal gas law
From the ideal gas law, we have:
 * $$p = \rho R T$$

Then,
 * $$\rho = {p \over R T} $$


 * $$\rho = {p_0 + p' \over R (T_0+T') } $$

Then,
 * $$\rho = {p_0 \left(1 + {p' \over p_0}\right) \over R T_0\left(1+{T' \over T}\right)} $$

Then at the first order,
 * $$\rho = \rho_0 {1-{T' \over T_0} \over 1 + {p' \over p_0}} $$

At the first order, we then obtain:
 * $$\rho = \rho_0 \left(1 - {T' \over T_0}\right) \times \left(1 - {p' \over p_0}\right) $$

Then,
 * $$\rho = \rho_0 \left(1 - {T' \over T_0} + {p' \over p_0}\right) $$

Then,
 * $$\rho' = \rho - \rho_0 = \rho_0 \left(1 - {T' \over T_0} + {p' \over p_0} - 1\right) $$

Finally,
 * $${\rho' \over \rho_0} = - {T' \over T_0} + {p' \over p_0} $$

The acceleration is then:
 * $${d w \over d t} = a = g \left({T' \over T_0} - {p' \over p_0}\right) - {1 \over \rho_0} {\partial p' \over \partial z} $$

This proves List's formula.

Ideal gas law
Let R_a be the ideal gas constant for air and R_v the ideal gas constant for water vapour.

The pressure of the mix air + water vapour is the sum of partial pressures. Let ρa the specific mass of air and ρv the specific mass of water vapour. From the ideal gas law, the total pressure is the following:
 * $$p = p_a + p_v = \rho_a R_a T + \rho_v R_v T$$

Then,
 * $$p = \rho_a R_a T\left(1 + {\rho_v R_v \over \rho_a R_a} \right)T$$

We define $$ \epsilon = {R_a \over R_v} \approx 0.622$$ and $$r_v = {\rho_v \over \rho_a}$$.

We then obtain:
 * $$p = \rho_a R_a T\left(1 + {r_v \over \epsilon} \right)T$$

Let $$\rho_m = \rho_a + \rho_v$$.

We then have:
 * $$\rho_m = \rho_a + \rho_a {\rho_v \over \rho_a} = \rho_a (1 + r_v) $$

We subtitute and then,
 * $$p = {\rho_m \over 1 + r_v} R_a T\left(1 + {r_v \over \epsilon} \right)T$$

We define the virtual temperature as follows:
 * $$T_v = \left({1 + {r_v \over \epsilon} \over 1 + r_v} \right)T$$

We then obtain the following exact relation:
 * $$p = \rho_m p T_v$$

We note that Tv > T and that humid air is less dense than dry air.

Let $$\rho = \rho_a + \rho_v + \rho_l + \rho_g$$ the total specific mass of the air parcel. ρl et ρg are the respective masses by unit of volume for water and ice.

We define the potential temperature in density T_ρ so that: $$p = \rho R_a T_{\rho}$$. We then obtain:


 * $$T_\rho = \left({1 + {r_v \over \epsilon} \over 1 + r_t} \right)T$$

where rt is the mixing ratio of water.

Exner function
We define the Exner function as follows:

$$ \pi = \left({p \over p_{00}}\right)^{R_a \over c_{p,a}}$$

where Ra is the ideal gas constant for dry air and cp,a is the specific heat at constant pressure. p00 est the pressure at sea level.

The potential temperature is then defined as follows:
 * $$ \theta = {T \over \pi} $$

The virtual potential temperature is defined as follows:
 * $$ \Theta_v = {T_v \over \pi} $$

We then obtain:
 * $$ \theta_v = T_v \left({p \over p_{00}}\right)^{-{R_a \over c_{p,a}}}$$

Then,
 * $$ {\delta \theta_v \over \theta_v} = {\delta T_v \over T_v} - {\delta p \over p} {R_a \over c_{p,a}}$$

We note that $$R_a = c_{p,a} - c_{v,a}$$. Then,
 * $$ {\delta \theta_v \over \theta_v} = {\delta T_v \over T_v} - {c_{p,a} - c_{v,a} \over c_{p,a}} \ {\delta p \over p}$$

Acceleration inside a cloud
We remind that:
 * $$a = - {\rho' \over \rho_0} g - {\partial p' \over \partial z} \times {1 \over \rho_0} $$

We write:
 * $$\rho' = \rho_m - \rho_0$$

We have:
 * $$\rho_m = {p \over R_a T_v} = {p_0 + p' \over R_a (T_{0,v} + T'_v)} $$
 * $$\rho_0 = {p_0 \over R_a T_0}$$

As well, we have:
 * $$\rho' = {p_0 + p' \over R_a (T_{0,v} + T'_v)} - {p_0 \over R_a T_0}$$

We develop:
 * $$\rho' = {p_0 \over R_a (T_{0,v} + T'_v)} + {p' \over R_a (T_{0,v} + T'_v)} - {p_0 \over R_a T_0}$$

At the first order we have:
 * $$\rho' = {p_0 \over R_a T_{0,v} \left(1 + {T'_v \over T_{0,v}}\right)} + {p' \over R_a T_{0,v}} - {p_0 \over R_a T_0}$$

Then at the first order:
 * $$\rho' = {p_0 \over R_a T_{0,v}} \left(1 - {T'_v \over T_{0,v}}\right) + {p' \over R_a T_{0,v}} - {p_0 \over R_a T_0}$$

Then,
 * $$\rho' = {p_0 \over R_a T_{0,v}} - {T'_v \over T_{0,v}} {p_0 \over R_a T_{0,v}} + {p' \over R_a T_{0,v}} - {p_0 \over R_a T_0}$$

Then after permuting:
 * $$\rho' = {p_0 \over R_a T_{0,v}} - {p_0 \over R_a T_0} - {T'_v \over T_{0,v}} {p_0 \over R_a T_{0,v}} + {p' \over R_a T_{0,v}} $$

Then,
 * $$\rho' = {p_0 \over R_a T_{0,v}} - {p_0 \over R_a T_0} - {T'_v \over T_{0,v}} \rho_m + {p' \over p_0} \rho_m $$

At the first order, we can write:
 * $$\rho' = {p_0 \over R_a T_{0,v}} - {p_0 \over R_a T_0} - {T'_v \over T_{0,v}} \rho_0 + {p' \over p_0} \rho_0 $$

We remind that:
 * $$T_{0,v} = T_0 (1 + \gamma')$$

We substitue and then:
 * $$\rho' = {p_0 \over R_a T_0 (1 + \gamma')} - {p_0 \over R_a T_0} -\left( {T'_v \over T_{0,v}} - {p' \over p_0} \right)\rho_0 $$

Then at the first order,
 * $$\rho' = {p_0 \over R_a T_0 }(1 - \gamma') - {p_0 \over R_a T_0} -\left( {T'_v \over T_{0,v}} - {p' \over p_0} \right)\rho_0 $$

Then,
 * $$\rho' = \left(-\gamma' - {p_0 \over R_a T_0} -\left( {T'_v \over T_{0,v}} \right) - {p' \over p_0} \right)\rho_0 $$

Then,
 * $$-{\rho' \over \rho_0} = {T'_v \over T_{0,v}} - {p' \over p_0} + \gamma'$$

We recall that using the preceding notation that:


 * $$ {\delta \theta_v \over \theta_v} = {\delta T_v \over T_v} - {c_{p,a} - c_{v,a} \over c_{p,a}} \ {\delta p \over p}$$

Then,
 * $$ {\theta'_v \over \theta_v} = {T'_v \over T_v} - \left(1 - {c_{v,a} \over c_{p,a}} \right) \ {p' \over p}$$

Then,
 * $$ {T'_v \over T_v} - {p' \over p} = {\theta'_v \over \theta_v} - {c_{v,a} \over c_{p,a}} {p' \over p}$$

We then finally obtain:
 * $$-{\rho' \over \rho_0} = {\theta'_v \over \theta_v} - {c_{v,a} \over c_{p,a}} {p' \over p} + \gamma'$$

And then the acceleration becomes:
 * $${d w \over d t} = a = - {\partial p' \over \partial z} \times {1 \over \rho_0} + g \left({\theta'_v \over \theta_v} - {c_{v,a} \over c_{p,a}} {p' \over p} + \gamma' \right)$$

}}

Analysis of a flight perform through a cumulonimbus
Marwitz analysed a flight performed across a cumulonimbus and has been able to determine the pressure deficit p'.

Estimate of horizontal speed change
We assume that the vertical speed is small compared to the horizontal speed since the vertical acceleration is counteracted by the negative buoyancy.

The horizontal acceleration of a parcel is the following:
 * $$a_x = -{1 \over \rho_0} {\partial p \over \partial x}$$

Therefore, the change in horizontal speed will be:
 * $$ \Delta u = \sqrt{2} \sqrt{p' \over \rho_0}$$

and then,
 * $$p' = {1 \over 2} \rho_0 (\Delta u)^2$$

Thus, one can estimate the pressure deficit by measuring the speed change at ground level between the outside and the updraught column.

During this flight, horizontal speed changes in the order of 15m/s have been observed. Assuming that $$\rho_0 \approx 1 {\rm kg/m^3}$$ we then obtain $$p' \approx 100 {\rm Pa}$$.

Estimate of the vertical acceleration
The height of the maximum pressure deficit has been estimated to 1600 m. We then can estimate that
 * $${\partial p' \over \partial z} \approx {100 \over 1600} \approx 6 \times 10^{-2}$$

and thus the vertical acceleration due to the pressure deficit has been estimated to be 0.06m/s2.

Discussion
In the case of the flight mentionned above, the effect of pressure deficit is in the order of 0.06m/s2. The air parcel being colder by 2K, the downward acceleration caused by negative buoyancy is
 * $$a_b = - g {T' \over T} = 10 \times {2 \over 300} = - 6.7 \times 10^{-2}$$

We note that the pressure deficit almost counteracts the negative buoyancy and thus the air parcel will easily ascend to great heights.

Conclusion
The above flight performed in a low severity thunderstorm (no large hail or tornado) demonstrates that the updraughts are mainly dynamic under a cumulonimbus. Consequently, an upper air sounding that shows that the air is stable at low level (the virtual temperature increasing with the altitude) does not guarantee in the impossibility of thunderstorms to form. Moreover if the thunderstorm is severe and/or the cloud base is low, then $${\partial p' \over \partial z}$$ becomes larger and the dynamic effect will increase.

List's formula is applicable to thermal updraughts below a fair weather cumulus and thus the change of airspeed will provide an estimate of p'. The airspeed change is usually in the order of 5 kn. The pressure change is then only 2Pa. The effect of the pressure gradient then becomes totally negligible and thus the updraught is purely thermal. To conclude, if at the beginning of the afternoon, updraughts are weak and later in the afternoon so-called "thermal" updraughts become smooth and stronger, then there is a phenomenon of dynamic sucking taking place due to a thunderstorm buildup.