# 1.9: Hydrostatic Equilibrium

As discussed before, pressure decreases with height. Any thin horizontal slice from a column of air would thus have greater pressure pushing up against the bottom than pushing down from the top (Figure 1.12). This is called a vertical pressure gradient, where the term gradient means change with distance. The net upward force acting on this slice of air, caused by the pressure gradient, is F = ΔP·A, where A is the horizontal cross section area of the column, and ΔP = Pbottom – Ptop. Figure 1.12. Hydrostatic balance of forces on a thin slice of air.

\ \begin{align}F=m\cdot g\tag{1.24}\end{align}

where g = – 9.8 m·s–2 is the gravitational acceleration. (See Appendix B for variation of g with latitude and altitude.) Negative g implies a negative (downward) force. (Remember that the unit of force is 1 N = 1 kg·m·s–2 , see Appendix A). The mass m of air in the slice equals the air density times the slice volume; namely, m = ρ · (A·Δz), where Δz is the slice thickness.

For situations where pressure gradient force approximately balances gravity force, the air is said to be in a state of hydrostatic equilibrium. The corresponding hydrostatic equation is:

\ \begin{align}\Delta P=\rho\cdot g\cdot \Delta z \tag{1.25a}\end{align}

or

\ \begin{align}\frac{\Delta P}{\Delta z}=- \rho\cdot |g| \tag{1.25b}\end{align}

The term hydrostatic is used because it describes a stationary (static) balance in a fluid (hydro) between pressure pushing up and gravity pulling down. The negative sign indicates that pressure decreases as height increases. This equilibrium is valid for most weather situations, except for vigorous storms with large vertical velocities.

Sample Application

What is the weight (force) of a person of mass 75 kg at the surface of the Earth? Find the Answer

Given: m = 75 kg

Find: F = ? N

Use eq. (1.24)

F = m·g = (75 kg)·(– 9.8 m·s–2) = – 735 kg·m·s–2 = – 735 N

Check: Units OK. Sketch OK. Physics OK.

Exposition: The negative sign means the person is pulled toward the Earth, not repelled away from it.

Sample Application

Near sea level, a height increase of 100 m corresponds to what pressure decrease? Find the Answer

Given: ρ= 1.225 kg·m–3 at sea level

Δz = 100 m

Find: ΔP = ? kPa

Use eq. (1.25a):

ΔP = ρ·g·Δz = ( 1.225 kg·m–3)·(–9.8 m·s–2)·(100 m) = – 1200.5 kg·m–1·s–2 = – 1.20 kPa

Check: Units OK. Sketch OK. Physics OK.

Exposition: This answer should not be extrapolated to greater heights.

A SCIENTIFIC PERSPECTIVE • Check for Errors

As a scientist or engineer you should always be very careful when you do your calculations and designs. Be precise. Check and double check your calculations and your units. Don’t take shortcuts, or make unjustifiable simplifications. Mistakes you make as a scientist or engineer can kill people and cause great financial loss.

Be careful whenever you encounter any equation that gives the change in one variable as a function of change of another. For example, in equations (1.25) P is changing with z. The “change of” operator (Δ) MUST be taken in the same direction for both variables. In this example ΔP/Δz means [ P(at z2) – P(at z1) ] / [ z2 – z1 ] . We often abbreviate this as [ P2 – P1 ] / [ z2 – z1 ].

If you change the denominator to be [ z1 – z2 ], then you must also change the numerator to be in the same direction [ P1 – P2 ] . It doesn’t matter which direction you use, so long as both the numerator and denominator (or both Δ variables as in eq. 1.25a) are in the same direction.

To help avoid errors in direction, you should always think of the subscripts by their relative positions in space or time. For example, subscripts 2 and 1 often mean top and bottom, or right and left, or later and earlier, etc. If you are not careful, then when you solve numerical problems using equations, your answer will have the wrong sign, which is sometimes difficult to catch.

HIGHER MATH • Physical Interpretation of Equations

Equations such as (1.25b) are finite-difference approximations to the original equations that are in differential form:

\ \begin{align}\frac{d P}{d z}=- \rho\cdot |g| \tag{1.25c}\end{align}

The calculus form (eq. 1.25c) is useful for derivations, and is the best description of the physics. The algebraic approximation eq. (1.25b) is often used in real life, where one can measure pressure at two different heights [i.e., ΔP/Δz = (P2 – P1)/ (z2 – z1)].

The left side of eq. (1.25c) describes the infinitesimal change of pressure P that is associated with an infinitesimal local change of height z. It is the vertical gradient of pressure. On a graph of P vs. z, it would be the slope of the line. The derivative symbol “d” has no units or dimensions, so the dimensions of the left side are kPa m–1.

Eq. (1.25b) has a similar physical interpretation. Namely, the left side is the change in pressure associated with a finite change in height. Again, it represents the slope of a line, but in this case, it is a straight line segment of finite length, as an approximation to a smooth curve.

Both eqs. (1.25b & c) state that rate of pressure decrease (because of the negative sign) with height is greater if the density ρ is greater, or if the magnitude of the gravitational acceleration |g| is greater. Namely, if factors ρ or |g| increase, then the whole right hand side (RHS) increases because ρ and |g| are in the numerator. Also, if the RHS increases, then the left hand side (LHS) must increase as well, to preserve the equality of LHS = RHS.