Stratification of the Ocean: Profiles, Stability, Double Diffusion, Equation of State and Conservation Equations

This chapter opens the paper’s ocean physics. It answers Section XE7.2 of XE7 Atmospheric and Oceanic Sciences, one of the nine optional sections of the Engineering Sciences (XE) paper, any two of which a candidate chooses in the exam. The first part of that heading reads: “Vertical Profiles of Temperature and Salinity; Stability and Double Diffusion; Equation of State, Equations for Conservation of Mass, Momentum, Heat and Salt”. The chapter teaches the layered structure of the ocean (mixed layer, thermocline, halocline, pycnocline), the equation of state of seawater and its linearised form, static stability and the buoyancy frequency N, salt fingers and diffusive convection, and the conservation equations of mass, momentum, heat and salt. The numericals are density from the linearised equation of state, N² and its oscillation period from a profile, the density ratio that decides the double-diffusive regime, and the warming of a mixed layer from a heat budget. Every constant is given in the stem.

1. Vertical profiles: mixed layer, thermocline, halocline and pycnocline

Away from coasts and poles the ocean is layered. At the top is the mixed layer, where wind stirring, waves and cooling-driven overturning keep temperature, salinity and density nearly uniform with depth. It is tens of metres thick in the tropics and in summer, and hundreds of metres in the mid-latitude winter. Below it the thermocline is the layer in which temperature falls rapidly with depth, typically by 10 to 20 K over a few hundred metres in the tropics and subtropics, down to the cold deep water (about 1 to 4 °C) that fills most of the ocean’s volume. The halocline is the layer of rapid salinity change and the pycnocline the layer of rapid density change.

Which of these sets the stratification depends on the region. In the tropics and subtropics density follows temperature, so the thermocline is the pycnocline. In subpolar and polar seas, and wherever river runoff or heavy rain lays fresh water over salty water, salinity dominates: the halocline is the pycnocline and temperature may even rise with depth. Surface salinity follows evaporation minus precipitation (E − P) and runoff: maxima near 36 to 37 in the subtropical gyres and the Arabian Sea where E exceeds P, minima in the rainy equatorial belt, at high latitudes and in the Bay of Bengal, where river water freshens the surface. A fresh cap can sit above the top of the thermocline, giving a salinity-stratified barrier layer that is isothermal but not isohaline.

Typical stratification by region
RegionMixed layerWhat sets the pycnoclineRemark
TropicsShallow, warmTemperature (strong thermocline)Warm layer over cold deep water
Mid-latitudesThin in summer, deep in winterTemperature; seasonal thermocline over the permanent oneWinter cooling erases the seasonal one
Polar and subpolarCold, freshSalinity (halocline)Weak thermal stratification
Bay of BengalFresh capSalinity near the surfaceBarrier layer possible
ℹ️ In situ versus potential temperature
Water sinking to depth is compressed and warms slightly by adiabatic compression, so the measured (in situ) temperature in the deep ocean exceeds the potential temperature, the temperature the parcel would have if brought to the surface without exchanging heat or salt. The gap is a few tenths of a degree at 4000 m. Compare water masses on potential temperature and potential density, never on in situ values from different depths.

2. Density and the equation of state

Seawater density is a function of three variables, ρ = ρ(S, T, p), given by an empirical equation of state fitted to laboratory data (not derivable from theory). Surface values lie between about 1020 and 1029 kg m⁻³. Density increases with salinity, decreases with temperature, and increases with pressure, because seawater is slightly compressible (about 0.4 % per kilometre of depth). It is convenient to quote σ_t = ρ(S, T, 0) − 1000 kg m⁻³. Because compression changes density with depth, parcels at different depths are compared with the potential density, the density they would have at a common reference pressure.

Over small ranges the equation of state is linearised about a reference state (ρ₀, T₀, S₀): ρ = ρ₀ [ 1 − α (T − T₀) + β (S − S₀) ], with α = −(1/ρ) ∂ρ/∂T ≈ 2 × 10⁻⁴ K⁻¹ (thermal expansion) and β = (1/ρ) ∂ρ/∂S ≈ 7.6 × 10⁻⁴ per unit salinity (haline contraction).

Worked example. Take ρ₀ = 1027.0 kg m⁻³ at T₀ = 10 °C, S₀ = 35, α = 2.0 × 10⁻⁴ K⁻¹, β = 7.6 × 10⁻⁴. For T = 20 °C and S = 34.0: ρ = 1027.0 [1 − 2.0 × 10⁻⁴ (10) + 7.6 × 10⁻⁴ (−1)] = 1027.0 × 0.99724 = 1024.2 kg m⁻³. Note the equivalence α/β ≈ 0.26: a 1 K warming lowers density as much as a salinity fall of about 0.26.

⚠️ The coefficients are not constants
α is about 0.5 × 10⁻⁴ K⁻¹ near 0 °C but about 2.5 × 10⁻⁴ K⁻¹ at 25 °C, and it also rises with pressure, while β changes little. So temperature controls density in warm water, and salinity in cold water. This nonlinearity gives cabbeling: mixing two parcels of equal density but different (T, S) produces a mixture denser than either. Freshwater has a density maximum near 4 °C, but seawater with S above about 24.7 has none above its freezing point (near −1.9 °C at S = 35), so cooling keeps making it denser and convection continues to freezing.

3. Static stability and the buoyancy frequency

Displace a parcel upward by δz without mixing, so that it keeps its own (potential) density. The surroundings there have density ρ(z) + (dρ/dz) δz, so the net upward acceleration is (g/ρ₀)(dρ/dz) δz. This is simple harmonic motion, d²(δz)/dt² = −N² δz, with the buoyancy (Brunt–Väisälä) frequency N² = −(g/ρ₀) ∂ρ/∂z (z upward, ρ the potential density). With the linear equation of state this becomes N² = g (α ∂T/∂z − β ∂S/∂z). N² > 0 is stable, N² = 0 neutral, and N² < 0 statically unstable: the column overturns and mixes. The oscillation period is 2π/N, about 10 minutes in a strong thermocline (N ≈ 10⁻² s⁻¹) and an hour or two in the deep ocean (N ≈ 10⁻³ s⁻¹). Internal gravity waves can exist at frequencies between f and N, so a strong pycnocline supports and guides them.

Worked example. Between 50 m and 150 m the temperature falls from 24 °C to 12 °C and salinity from 35.2 to 35.0 (α = 2.0 × 10⁻⁴ K⁻¹, β = 7.6 × 10⁻⁴, g = 9.81 m s⁻²). With z upward, ∂T/∂z = +12/100 = 0.12 K m⁻¹ and ∂S/∂z = +0.2/100 = 0.002 m⁻¹. N² = 9.81 (2.0 × 10⁻⁴ × 0.12 − 7.6 × 10⁻⁴ × 0.002) = 9.81 × 2.248 × 10⁻⁵ = 2.2 × 10⁻⁴ s⁻². N = 1.49 × 10⁻² s⁻¹ and the period is 2π/N ≈ 423 s ≈ 7 min. The thermal term is positive (cooler below) and the haline term is negative (saltier above), so salinity here weakens the stability slightly.

⚠️ Sign and axis convention
N² = −(g/ρ₀) ∂ρ/∂z holds with z upward. If depth is measured downward, the sign flips: N² = +(g/ρ₀) ∂ρ/∂(depth). A decrease of temperature with depth is stabilising, but a salinity that decreases with depth is destabilising. “Temperature falls with depth, so the column is stable” is not safe without checking the salinity term.

4. Double diffusion: salt fingers and diffusive convection

Heat diffuses about a hundred times faster than salt (κ_T ≈ 1.4 × 10⁻⁷ m² s⁻¹, κ_S ≈ 1.1 × 10⁻⁹ m² s⁻¹). When temperature and salinity gradients push density in opposite senses, a column that is statically stable overall (N² > 0) can still be unstable to double diffusion, because the faster-diffusing property equilibrates with the surroundings while the slower one does not. Define the density ratio R_ρ as the stabilising density contribution divided by the destabilising one, for example αΔT/(βΔS) when temperature is the stabilising one.

The two double-diffusive regimes
FeatureSalt fingersDiffusive convection
ArrangementWarm, salty water over cooler, fresher water; T and S both fall with depthCold, fresh water over warmer, saltier water; T and S both rise with depth
Stabilising gradientTemperatureSalinity
MechanismA downward-displaced salty parcel loses heat quickly but keeps its salt, so it is denser than its surroundings and keeps sinking: narrow, centimetre-scale fingersHeat leaks upward across the interface faster than salt, so the lower layer warms and convects: well-mixed layers separated by sharp interfaces (staircases)
Condition1 < R_ρ < κ_T/κ_S ≈ 100Overall stable (R_ρ > 1), salinity stabilising
Where seenSubtropical thermocline, beneath outflows of warm saline waterBeneath the cold fresh surface layer of the Arctic, where warm saline Atlantic water lies below

Worked example. Over one step the temperature falls by 3.0 K and the salinity by 0.5 with depth (α = 2.0 × 10⁻⁴ K⁻¹, β = 7.6 × 10⁻⁴). Temperature stabilises by αΔT = 6.0 × 10⁻⁴ and salinity destabilises by βΔS = 3.8 × 10⁻⁴. The column is stable (N² > 0) and R_ρ = 6.0/3.8 = 1.58, which lies between 1 and 100: it is prone to salt fingers.

🧠 One memory rule
Both properties decrease with depth: fingers (the warm salty layer is on top). Both increase with depth: diffusive (the cold fresh layer is on top). In either case the net density must still increase downward; double diffusion adds mixing and layering, it is not ordinary convective overturning.

5. Conservation of mass, momentum, heat and salt

The ocean is governed by conservation laws plus the equation of state. Let D/Dt = ∂/∂t + u ∂/∂x + v ∂/∂y + w ∂/∂z. Seawater density varies by only a few per cent, so the Boussinesq approximation keeps the density variation only where it multiplies g.

  • Mass: ∂ρ/∂t + ∇·(ρu) = 0, which under Boussinesq becomes ∂u/∂x + ∂v/∂y + ∂w/∂z = 0. A horizontal divergence must be balanced by vertical motion.
  • Momentum on a rotating Earth (f = 2Ω sin φ, Ω = 7.292 × 10⁻⁵ rad s⁻¹): Du/Dt − f v = −(1/ρ₀) ∂p/∂x + F_x and Dv/Dt + f u = −(1/ρ₀) ∂p/∂y + F_y. Vertically, for large-scale flow, the hydrostatic balance ∂p/∂z = −ρ g holds, so p = p_atm + g ∫ρ dz and pressure in decibars is about the depth in metres.
  • Heat: DT/Dt = κ_T ∇²T + (1/ρ₀ c_p) ∂Q/∂z, where Q is the radiative flux absorbed in the water and c_p ≈ 4.0 × 10³ J kg⁻¹ K⁻¹. Surface heat flux enters as a boundary condition. Salt: DS/Dt = κ_S ∇²S. Salt is not created, so interior salinity changes only by mixing, and at the surface by evaporation, precipitation, runoff and ice formation.

Counting: u, v, w, p, ρ, T, S are seven unknowns, and three momentum equations, continuity, heat, salt and the equation of state are seven equations. In practice turbulence makes the effective (eddy) diffusivities and viscosities far larger than molecular ones, about 10⁻⁵ m² s⁻¹ vertically in the thermocline, and mixing along surfaces of constant density is much easier than across them. Worked example (heat budget). A net heat flux of 200 W m⁻² into a mixed layer 40 m thick for 10 days, with ρ = 1025 kg m⁻³ and c_p = 3990 J kg⁻¹ K⁻¹: ΔT = Q t/(ρ c_p h) = 200 × 8.64 × 10⁵/(1025 × 3990 × 40) = 1.06 K.

🎯 Why hydrostatic balance works
Ocean flows are thousands of kilometres wide but only a few kilometres deep, so vertical accelerations are tiny compared with g and the vertical momentum equation reduces to a balance between the pressure gradient and weight. This is what lets a map of density determine a map of pressure, the basis of geostrophic calculations in the next chapter.

Key takeaways

  • The ocean has a mixed layer, a thermocline, a halocline and a pycnocline; in the tropics temperature sets density and in polar and river-fed regions salinity does. Compare water masses on potential temperature and potential density.
  • Linearised equation of state: ρ = ρ₀[1 − α(T − T₀) + β(S − S₀)], with α ≈ 2 × 10⁻⁴ K⁻¹ and β ≈ 7.6 × 10⁻⁴; the true α grows with temperature and pressure, which causes cabbeling.
  • N² = −(g/ρ₀) ∂ρ/∂z = g(α ∂T/∂z − β ∂S/∂z) with z upward; N² > 0 is stable, N² < 0 overturns, and the buoyancy period is 2π/N.
  • Double diffusion needs a statically stable column with the faster-diffusing property (heat) stabilising or destabilising against salt: both falling with depth gives salt fingers (1 < R_ρ < about 100), both rising gives diffusive convection.
  • The governing set is continuity (∇·u = 0 under Boussinesq), momentum with Coriolis force and hydrostatic balance, heat, salt and the equation of state: seven equations for u, v, w, p, ρ, T, S.
  • Surface heat flux warms a mixed layer by ΔT = Q t/(ρ c_p h); salinity changes only through mixing and surface freshwater exchange.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In a subpolar sea a layer of cold, fresh surface water lies over warmer, saltier water. Which feature mainly controls the density stratification there?

    1. The thermocline, because temperature always dominates the change of seawater density with depth in every ocean region
    2. The halocline, because the salinity rise with depth outweighs the density effect of the warmer water
    3. The mixed layer, because its uniform density and well-stirred properties give the whole column its stability
    4. The abyssal layer, because pressure compression of the deep water sets the stratification in polar seas
    Show answer

    Answer: B — The halocline, because the salinity rise with depth outweighs the density effect of the warmer water

    Cold fresh water over warm salty water is stable only because salinity, which raises density with depth, outweighs the destabilising effect of the warmer water below. The salinity gradient (halocline) therefore sets the pycnocline, and temperature may even increase with depth. The tempting answer fails because the thermocline dominates only where density follows temperature, as in the tropics. A mixed layer is, by definition, nearly uniform and gives no stratification.

  2. Which pair of processes deepens the oceanic mixed layer?

    1. Wind stirring and surface cooling that drives convective overturning
    2. Strong solar heating of the surface layer together with heavy rainfall over the sea
    3. Weak winds together with the spreading of a layer of river water over the sea surface
    4. Surface warming together with the formation of a salinity-stratified barrier layer
    Show answer

    Answer: A — Wind stirring and surface cooling that drives convective overturning

    Mixing energy comes from wind stress and waves, and from cooling or evaporation that makes surface water denser than the water beneath and so drives convection; both push the base of the layer down. Heating and rain make the surface lighter and more stable, shoaling the layer, and a barrier layer is itself a stabilising fresh cap, so each of the other pairs works against deepening.

  3. Seawater of salinity 35 is cooled steadily from 5 °C to its freezing point at constant pressure. How does its density behave?

    1. It rises to a maximum near 4 °C and then falls on further cooling, as it does for freshwater
    2. It stays constant throughout, because the dissolved salt exactly cancels the thermal expansion
    3. It rises all the way to the freezing point, so cooling keeps driving convection
    4. It decreases steadily, because the water contracts less on cooling than the dissolved salt expands
    Show answer

    Answer: C — It rises all the way to the freezing point, so cooling keeps driving convection

    For salinity above about 24.7 the temperature of maximum density lies below the freezing point (about −1.9 °C at S = 35), so within the liquid range the density rises monotonically on cooling and cold surface water keeps sinking. The 4 °C maximum belongs to freshwater and is the tempting answer. The other two options contradict the positive thermal expansion of seawater in this range.

  4. Why are water parcels from different depths compared using potential density rather than in situ density when judging stability in the deep ocean?

    1. Potential density removes the effect of salinity from the comparison so that temperature alone is compared between depths
    2. Potential density is measured directly by a probe, while in situ density needs an additional salinity sensor
    3. In situ density is always lower than potential density at depth, so it cannot be used for stability
    4. It removes the density rise caused by compression alone, which would make deep water look heavier even if neutrally stable
    Show answer

    Answer: D — It removes the density rise caused by compression alone, which would make deep water look heavier even if neutrally stable

    Pressure compresses water, so in situ density increases with depth even in a perfectly neutral column. Referencing each parcel to a common pressure removes this trivial increase and leaves only the part that reflects real stratification in T and S. Potential density keeps the salinity effect, it is computed rather than measured, and at depth in situ density is higher, not lower, than the potential value.

  5. In the linearised equation of state take α to be the thermal expansion coefficient, which rises with temperature, and β the haline contraction coefficient, which changes little. In which water is density most sensitive to salinity relative to temperature?

    1. Warm tropical surface water near 28 °C, where α is largest
    2. Cold polar water near 0 °C, where α is small so salinity controls density
    3. Deep water at 4000 m, where pressure removes the dependence on temperature
    4. Water of salinity above 36, where β vanishes
    Show answer

    Answer: B — Cold polar water near 0 °C, where α is small so salinity controls density

    The relative importance of salinity is measured by β ΔS compared with α ΔT. Since α is small in cold water (about 5 × 10⁻⁵ K⁻¹ near 0 °C) and β is almost constant, the haline term dominates in polar waters. In warm water the large α makes temperature the controlling factor. Pressure does not remove the temperature dependence (α actually grows with pressure), and β does not vanish at high salinity.

  6. Across a layer the temperature falls by 2.0 K and the salinity falls by 0.90 with increasing depth. Take α = 2.0 × 10⁻⁴ K⁻¹ and β = 7.6 × 10⁻⁴ per unit salinity. Which statement is correct?

    1. The layer is statically unstable (N² < 0) and will overturn by ordinary convection
    2. The layer is stable and prone to salt fingers, because both properties fall with depth
    3. The layer is stable and prone to diffusive convection
    4. The layer is neutrally stable, with N² = 0
    Show answer

    Answer: A — The layer is statically unstable (N² < 0) and will overturn by ordinary convection

    The stabilising temperature term is αΔT = 4.0 × 10⁻⁴ and the destabilising salinity term is βΔS = 6.84 × 10⁻⁴. Since the destabilising term is larger, density decreases with depth (R_ρ = 0.58 < 1) and N² < 0, so the layer overturns. Salt fingers need the column to be stable overall (R_ρ > 1), which is the tempting but wrong answer here; the diffusive regime needs both properties to increase with depth.

  7. Two water parcels have the same density but different temperature and salinity. When they mix, the mixture is denser than either parent. Which property of the equation of state explains this (cabbeling)?

    1. The haline contraction coefficient β grows strongly with pressure, so deeper parcels contract more when mixed
    2. Mixing releases a large amount of heat to the surroundings, which then cools and so densifies the mixture
    3. α increases with temperature, so density is a curved function of T and S and the mixture is denser than the straight-line average
    4. Salinity is not conserved when two waters of different temperature are mixed, so the mixture becomes saltier
    Show answer

    Answer: C — α increases with temperature, so density is a curved function of T and S and the mixture is denser than the straight-line average

    Temperature and salinity mix linearly (they are conserved), so the mixture lies on the straight line between the parents in the T–S plane. Because the isopycnals are curved (α is larger in warm water), the straight line crosses to the denser side and the mixture contracts. Pressure dependence of β is not the cause, mixing of seawater is not accompanied by significant heating that would explain the effect, and salinity is conserved in mixing.

  8. Which statement correctly describes the Boussinesq approximation used in the ocean equations?

    1. Density is treated as exactly constant everywhere in the equations, including in the buoyancy force
    2. Viscous and diffusive terms are dropped so that the flow is treated as inviscid and non-diffusive
    3. The pressure is assumed to be uniform in the horizontal, so that no pressure-driven currents arise
    4. Density differences are kept only where they multiply g, so continuity becomes ∇·u = 0
    Show answer

    Answer: D — Density differences are kept only where they multiply g, so continuity becomes ∇·u = 0

    Seawater density varies by only a few per cent, so it is replaced by ρ₀ in the inertial terms and in continuity, giving ∇·u = 0, while the small variations are kept in the buoyancy term ρg because g makes them dynamically important. Setting density constant even in the buoyancy force would remove stratification altogether, and neither viscous terms nor horizontal pressure gradients are dropped.

  9. Why can the large-scale ocean circulation be treated as being in hydrostatic balance?

    1. Seawater is exactly incompressible, so the pressure cannot change with depth except through the weight of water above
    2. Horizontal scales of thousands of kilometres far exceed the depth, so vertical accelerations are negligible against g
    3. The Coriolis force vanishes in the vertical direction, which removes all other vertical forces
    4. Ocean currents are everywhere slower than 1 mm s⁻¹, so no acceleration can arise at all
    Show answer

    Answer: B — Horizontal scales of thousands of kilometres far exceed the depth, so vertical accelerations are negligible against g

    Because horizontal scales far exceed the depth, vertical velocities and their accelerations are tiny, and the vertical momentum equation reduces to ∂p/∂z = −ρg. Seawater is slightly compressible, so exact incompressibility is not the reason, the Coriolis effect on vertical motion is a separate small term, and ocean currents are much faster than 1 mm s⁻¹ in many places.

  10. Which of the following statements about static stability and the buoyancy frequency N (z upward, potential density used) are correct?

    1. P: N² > 0 means a vertically displaced parcel oscillates about its level with period 2π/N
    2. Q: N is typically larger in the main thermocline than in the deep ocean
    3. R: A temperature that falls with depth and a salinity that rises with depth both make positive contributions to N²
    4. S: N² < 0 is the condition for salt fingers in an otherwise stable column
    Show answer

    Answer: A — P: N² > 0 means a vertically displaced parcel oscillates about its level with period 2π/N; B — Q: N is typically larger in the main thermocline than in the deep ocean; C — R: A temperature that falls with depth and a salinity that rises with depth both make positive contributions to N²

    P is the definition of the buoyancy oscillation. Q holds because the density gradient is steepest in the pycnocline (N ≈ 10⁻² s⁻¹) and weak in the deep ocean (about 10⁻³ s⁻¹). R follows from N² = g(α ∂T/∂z − β ∂S/∂z): colder water below (∂T/∂z > 0) and saltier water below (∂S/∂z < 0) both stabilise. S is wrong: salt fingers occur when N² > 0, so the column is statically stable overall.

  11. Which of the following statements about double diffusion are correct?

    1. P: It depends on heat diffusing much faster than salt
    2. Q: Salt fingers form where warm, salty water overlies cooler, fresher water
    3. R: Diffusive convection can produce staircases of well-mixed layers separated by sharp interfaces
    4. S: Salt fingers require a density ratio R_ρ below 1
    Show answer

    Answer: A — P: It depends on heat diffusing much faster than salt; B — Q: Salt fingers form where warm, salty water overlies cooler, fresher water; C — R: Diffusive convection can produce staircases of well-mixed layers separated by sharp interfaces

    The contrast κ_T ≈ 100 κ_S is what makes the two regimes possible, so P is right. Q describes the finger configuration, where temperature stabilises and salinity destabilises. R describes the diffusive regime, with cold fresh water over warm salty water. S is wrong: fingers need 1 < R_ρ < about 100; a ratio below 1 means static instability and ordinary overturning.

  12. Which of the following statements about the conservation equations for the ocean are correct?

    1. P: Under the Boussinesq approximation the continuity equation becomes ∂u/∂x + ∂v/∂y + ∂w/∂z = 0
    2. Q: Interior salinity changes only through advection and mixing, because salt is neither created nor destroyed
    3. R: With the hydrostatic balance the pressure at depth equals the surface pressure plus g times the integral of density over depth
    4. S: For large-scale flow the vertical momentum equation is dominated by the vertical acceleration Dw/Dt
    Show answer

    Answer: A — P: Under the Boussinesq approximation the continuity equation becomes ∂u/∂x + ∂v/∂y + ∂w/∂z = 0; B — Q: Interior salinity changes only through advection and mixing, because salt is neither created nor destroyed; C — R: With the hydrostatic balance the pressure at depth equals the surface pressure plus g times the integral of density over depth

    P, Q and R are standard: continuity under Boussinesq, salt as a conserved tracer whose surface exchange (evaporation, rain, runoff) enters only as a boundary condition, and p = p_atm + g∫ρ dz. S is wrong, since for large-scale flow Dw/Dt is negligible and the vertical balance is between the pressure gradient and the weight of the water.

  13. Use the linearised equation of state ρ = ρ₀[1 − α(T − T₀) + β(S − S₀)] with ρ₀ = 1026.0 kg m⁻³ at T₀ = 15 °C and S₀ = 35.0, α = 2.0 × 10⁻⁴ K⁻¹ and β = 7.6 × 10⁻⁴ per unit salinity. Find the density, in kg m⁻³ to one decimal place, of surface water at T = 27 °C and S = 33.5.

    Numerical answer — type the value.

    Show answer

    Answer: 1022.4

    T − T₀ = 12 K and S − S₀ = −1.5. The bracket is 1 − 2.0 × 10⁻⁴ × 12 + 7.6 × 10⁻⁴ × (−1.5) = 1 − 0.0024 − 0.00114 = 0.99646. ρ = 1026.0 × 0.99646 = 1022.37, which is 1022.4 kg m⁻³ to one decimal place. Ignoring the salinity term gives 1023.5, and taking it with the wrong sign gives 1024.7.

  14. In a thermocline the temperature is 20.0 °C at 100 m and 8.0 °C at 200 m, and the salinity is 35.3 at 100 m and 35.0 at 200 m. Use g = 9.81 m s⁻², α = 2.0 × 10⁻⁴ K⁻¹ and β = 7.6 × 10⁻⁴ per unit salinity, with gradients taken as uniform between the two depths. The buoyancy frequency squared N² is ____ × 10⁻⁴ s⁻² (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.13

    With z upward, the temperature gradient is ∂T/∂z = (20.0 − 8.0)/100 = +0.12 K m⁻¹ and the salinity gradient is ∂S/∂z = (35.3 − 35.0)/100 = +0.003 m⁻¹. N² = g(α ∂T/∂z − β ∂S/∂z) = 9.81 × (2.0 × 10⁻⁴ × 0.12 − 7.6 × 10⁻⁴ × 0.003) = 9.81 × (2.4 × 10⁻⁵ − 2.28 × 10⁻⁶) = 9.81 × 2.172 × 10⁻⁵ = 2.13 × 10⁻⁴ s⁻². Dropping the salinity term gives 2.35, and using a downward axis without flipping the signs gives a negative value.

  15. Going down through a layer, the temperature falls by 4.0 K and the salinity falls by 0.40. With α = 2.0 × 10⁻⁴ K⁻¹ and β = 7.6 × 10⁻⁴ per unit salinity, define the density ratio R_ρ = αΔT/(βΔS), the stabilising thermal contribution divided by the destabilising haline one. Find R_ρ to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 2.63

    The thermal contribution is αΔT = 2.0 × 10⁻⁴ × 4.0 = 8.0 × 10⁻⁴ and the haline contribution is βΔS = 7.6 × 10⁻⁴ × 0.40 = 3.04 × 10⁻⁴. R_ρ = 8.0/3.04 = 2.63. Because R_ρ lies between 1 and about 100 and both properties fall with depth, the layer is statically stable yet prone to salt fingers.

  16. A mixed layer 25 m thick (ρ = 1025 kg m⁻³, c_p = 3990 J kg⁻¹ K⁻¹) receives 230 W m⁻² of solar radiation, of which 12 % passes through the base of the layer and warms the water below. The layer loses a further 70 W m⁻² to the atmosphere by longwave radiation and by sensible and latent heat flux. Assuming the fluxes are steady for 12 days (1 day = 86 400 s) and ignoring mixing across the base, the warming of the layer is ____ K (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.34

    Heat retained by the layer = 0.88 × 230 − 70 = 202.4 − 70 = 132.4 W m⁻². Energy per unit area over 12 days = 132.4 × 12 × 86 400 = 1.373 × 10⁸ J m⁻². ΔT = Q t/(ρ c_p h) = 1.373 × 10⁸/(1025 × 3990 × 25) = 1.373 × 10⁸/1.0224 × 10⁸ = 1.34 K. Using the full 230 W m⁻² with no penetration, or forgetting the 70 W m⁻² loss, gives a much larger answer.