# 4.3: Total Water

- Page ID
- 9549

# 4.3.1. Liquid and Solid Water

In clouds, fog, or air containing falling precipitation, one measure of the amount of liquid water in the air is the **liquid water content (LWC)**. It is defined as

\(\ \begin{align} \rho_{L W C}=\frac{m_{l i q . \text { water }}}{V o l}\tag{4.28}\end{align}\)

where m_{liq.water} is mass of liquid water suspended or falling through the air, and Vol is the air volume. Typical values in cumulus clouds are 0 ≤ ρ_{LWC} ≤ 5 g m^{–3}. It can also be expressed in units of kgliq.water m^{–3}. LWC is the liquid-water analogy to the absolute humidity for water vapor.

Another measure is the liquid-water mixing ratio:

\(\ \begin{align} r_{L}=\frac{m_{l i q . \text { water }}}{m_\text{dry air}} \tag{4.29}\end{align}\)

where m_{liq.water} is the mass of liquid water that is embedded as droplets within an air parcel that contains m_{dry}_{ air} mass of dry air. A similar ice mixing ratio can be defined:

\(\ \begin{align} r_{i}=\frac{m_{i c e}}{m_\text{dry air}}\tag{4.30}\end{align}\)

Both mixing ratios have units of kg_{water} kg_{air}^{–1}, or g_{water} kg_{air}^{–1}.

Liquid water content is related to liquid-water mixing ratio by

\(\ \begin{align} r_{L}=\frac{\rho_{L W C}}{\rho_{a i r}}\tag{4.31}\end{align}\)

where ρ_{air}_{ }is air density.

**Sample Application**

Find the liquid water mixing ratio in air at sea level, given a liquid water content of 3 g m^{–3}.

**Find the Answer**

Given: ρ_{LWC} = 3 gwater m^{–3}. Sea level.

Find: r_{L} = ? gwater kg_{dry air}^{–1}.

Assume standard atmosphere, and use Table 1-5 from Chapter 1 to get: ρ_{air} = 1.225 kg_{air} m^{–3} at sea level.

Use eq. (4.31):

r_{L} = (3 g_{water} m^{–3})/(1.225 kg_{air} m^{–3}) = 2.45 gwater kg_{dry air}^{–1}. = __ 2.45__ g kg

^{–1}

**Check:** Physics, units & magnitude are reasonable.

**Exposition:** Liquid, solid and water vapor might exist together in a cloud.

# 4.3.2. Mixing Ratio of Total Water

The **total-water mixing ratio** r_{T} is defined as the sum of masses of all phases of water (vapor, liquid, solid) per dry-air mass:

\(\ \begin{align}r_{T}=r_{S}+r_{L}+r_{i}\tag{4.32a}\end{align}\)

where r is **mixing ratio for water-vapor**, r_{L} is **mixing ratio for liquid-water**, and r_{i} is mixing ratio for ice. Be sure to use common units for all terms in this equation; namely kg_{water} kg_{air}^{–1}, or g_{water} kg_{air}^{–1}. Total-water absolute humidity and total-water specific humidity are similarly defined.

Liquid-water cloud droplets can exist unfrozen in air of temperature less than 0°C. Thus, it is possible for ice and liquid water to co-exist in the same air parcel at the same time, along with water vapor.

Eq. (4.32a) can be simplified if there is no precipitation:

\(\ \begin{align} r_{T}=r \quad\text{if the air is not cloudy}\tag{4.32b}\end{align}\)

\(\ \begin{align} r_{T}=r+r_{L}+r_{i} \quad \text{if the air is cloudy}\tag{4.32c}\end{align}\)

By “not cloudy” we mean air that is **unsaturated** (i.e., r < r_{s}). By “cloudy” we mean air that is **saturated** (i.e., r = r_{s}) and has liquid water drops and/or ice crystals suspended in it.

Suppose an air parcel has some total number of water molecules in it. Consider an idealized situation where an air parcel does not mix with its environment. For this case, we anticipate that all the water molecules in the parcel must move with the parcel. It makes no difference if some of these molecules are in the form of vapor, or liquid droplets, or solid ice crystals — all the water molecules must still be accounted for.

Hence, for this idealized parcel with no precipitation falling into or out of it, the amount of total water rT must be constant. Any changes in rT must be directly associated with precipitation falling into or out of the air parcel.

Suppose that the air parcel is initially unsaturated, for which case we can solve for the total water using eq. (4.32b). If this air parcel rises and cools and can hold less vapor at saturation (see eq. 4.5), it might reach an altitude where r_{s} < r_{T}. For this situation, eq. (4.32c) tells us that r_{L} + r_{i} = r_{T} – r_{s}. Namely, we can anticipate that liquid water droplets and/or ice crystals suspended in the air parcel must have formed to maintain the constant total number of water molecules.

**Sample Application**

At a pressure altitude of 50 kPa inside a thunderstorm, suppose that the air temperature is –5°C, and that each kg of air contains 4 g of water droplets and 2 g of ice crystals suspended in the air. What is the totalwater mixing ratio value?

**Find the Answer**

Given: P = 50 kPa, T = –5°C, r_{L} = 4 g kg^{–1}, r_{i} = 2 g kg^{–1}

Find: r_{T} = ? g kg^{–1}.

Assume cloudy (saturated) air inside the thunderstorm.

We need to solve eq. (4.32c), for which we have everything we need except rs. To use eq. (4.5) to get r_{s}, we can read e_{s} = 0.4222 kPa at T = –5°C from Table 4-1.

Now apply eq. (4.5): r_{s} = ε·e_{s} /(P–e_{s}) = (622 g·kg^{–1})·(0.4222 kPa)/(50– 0.4222 kPa) = 5.30 g kg^{–1}.

Finally, apply eq. (4.32c): r_{T} = r_{s} + r_{L} + r_{i} = 5.3 + 4 + 2 = 11.3 g·kg^{–1}

**Check:** Physics and units are reasonable.

**Exposition:** For altitudes where –40°C < T < 0°C, it is often the case in thunderstorms that both liquid water and ice crystals can be present in the air, along with water vapor.

# 4.3.3. Precipitable Water

Consider an air column between (top, bottom) altitudes as given by their respective air pressures (P_{T}, P_{B}). Suppose all the water molecules within that column were to fall to the bottom of the column and form a puddle. The depth of this puddle (namely, the precipitable water) is

\(\ \begin{align} d_{W}=\frac{r_{T}}{|g| \cdot \rho_{l i q}} \cdot\left(P_{B}-P_{T}\right)\tag{4.33}\end{align}\)

where the magnitude of gravitation acceleration is|g|= 9.8 m·s^{–2} , the liquid-water density is ρ_{liq} = 1000 kg·m^{–3}, and a column-average of the total-water mixing ratio is r_{T} . For a column where r_{T} varies with altitude, split it into column segments each having unique r_{T} average, and sum over all segments.

Precipitable water is sometimes used as a humidity variable. The bottom of the atmosphere is warmer than the mid and upper troposphere, and can hold the most water vapor. In a pre-storm cloudless environment, contributions to the total-column precipitable water thus come mostly from the boundary layer. Hence, precipitable water can serve as one measure of boundary-layer total water that could serve as the fuel for thunderstorms later in the day. See the Thunderstorm chapters for a sample map of precipitable water.

Note that the American Meteorological Society Glossary of Meteorology considers only the water __vapor__ in an air column for calculation of precipitable-water depth. However, some satellites can detect total water over a range of altitudes, for which eq. (4.33) would be applicable.

It is possible to have more precipitation reach the ground during a storm than the value of precipitable water. This occurs where moisture advection by the winds can replenish water vapor in a region.

**Sample Application**

mple Application Find the precipitable water in the troposphere if the bottom portion (below P = 50 kPa) has an average total water mixing ratio of 10 g kg^{–1} while the portion of troposphere between 50 and 20 kPa has an average total water mixing ratio of 2 g kg^{–1}.

**Find the Answer**

Assume: Bottom of troposphere is at 100 kPa.

Given: r_{T} = 0.010 kg kg^{–1}, PB = 100 kPa, P_{T} = 50 kPa, r_{T} = 0.002 kg kg^{–1}, PB = 50 kPa, P_{T} = 20 kPa

Find: d_{W} = ? m

Apply eq. (4.33) for the bottom and top segments of the column of tropospheric air, and sum the result.

d_{W} = [(9.8 m·s^{–2})·(1000 kg_{water}·m^{–3})]^{–1} · { (0.010 kg_{water} kg_{air}^{–1})·[100 – 50 kPa] + (0.002 kg_{water} kg_{air}^{–1})·[50 – 20 kPa] }

= [1.02x10^{–4} m^{2} s^{2} kg_{water} ^{–1}] · {(0.5 + 0.06) kPa·kg_{water}·kg_{air}^{–1}} = 5.71x10^{–5} m^{2} s^{2} kPa kg_{air}^{–1} .

To convert the units, use Appendix A info:

1 kPa = 1000 kg_{air}·m·s^{–2}

Thus, d_{w} = [(1000 kg_{air}·m·s^{–2})/(1 kPa)] · (5.71x10^{–5} m^{2} s^{2} kPa kg_{air}^{–1}) = 0.057 m

**Check:** Physics and units are reasonable.

**Exposition:** The puddle depth of 5.7 cm is large because the total water mixing ratios were large.