I Percentage change: A percentage change is a way of scaling quantities that are in different units so that they can be compared with one another in a meaningful way. A percentage change (P) of quantity Q is:
P = 100*(Q2 - Q1)/Qbar
Where Q1 and Q2 are values of Q at different times or places, and Qbar is the mean value of Q. As you can see, the units of Q cancel out so that P is unitless. II Unit conversions: Convert the following quantities to the units specified. Show your work:
  1. g = 9.81 m/s^2 in m/hr^2
  2. 101325 Pa in dbar (see below for help)
  3. 1.0 g/cm^3 in kg/m^3
  4. 1 m/s in cm/s
  5. 1 knot in cm/s (a knot is a nautical mile per hour)
  6. 4 n.m. in km
  7. 1 mm/day in cm/year
Remember that F = ma,  so a force has units of kg m/s^2.  Pressure is a force per unit area, which would give it units such as  (kg m/s^2)/m^2.  Obviously, some of these terms cancel out giving us units of kg/(m s^2) for pressure in the mks system.  1 Pascal = 1 Pa = 1 kg/(m s^2), making the Pascal the standard unit of pressure in the mks system.  Unfortunately, a column of water one-tenth of 1 millimeter deep exerts a pressure of 1 Pa, so Pascals are not a very useful unit of measure for oceanographers.  Instead, we start with a unit called the bar, which is equal to 1x10^5 kg/(m s^2), and make use of it as the decibar (dbar), where 1 dbar is one-tenth of a bar.  You might have heard of the meteorological unit the millibar which, you guessed it is 1/1000 of a bar.  In the problem above, you are asked to convert from Pascals (Pa) to decibars (dbar), and you have right here all the information you need to do it.

Bonus Question: It just so happens that standard atmosphere pressure at sea-level is 101325 Pa.  That means that the pressure created by weight of all the air in the atmosphere, from outer space down to sea level is 101325 Pa. Given that the density of seawater is approximately 1000 kg/m^3, and gravitational acceleration, g, is just about 10 m/s^2, approximately how deep do you have to go in the ocean to increase the pressure by that much again?  Show your work.

III Potential temperature: Seawater is slightly compressible, meaning that it becomes more dense under increasing pressure. This increase in density packs the molecules in the seawater more closely together, and the temperature goes up accordingly as there are collisions between molecules occur more frequently. However, this temperature change is reversible. So, a parcel of water that sinks will warm up, and if it rises back to its original level it will return to its original temperature. This kind of temperature change is called adiabatic. Potential temperature, then is the temperature that water would have if it were to be raised to sea level (pressure = 0).

Although the amount of adiabatic temperature change in the water column is slight, typically less than 1C, temperature strongly influences seawater density and the actual stability of the water column is much more accurately reflected by the potential density (which is simply the density calculated using potential temperature) than by the in situ density.

Calculate the potential temperature, the in situ density, and the potential density of the following samples.  You can use software that you have, or  use this calculator .   But be careful if you do!  The calculator wants your input in units of 10 kPa (ten kiloPascals, or 1x10^4  Pa).  Based on your work in Part I, how will you convert from dbar to 10 kPa units? (If 10 kPa is a confusing unit, think of it as a using ratio, such as 4:1 or 1:100.  How many dbar to make 10 kPa?).

  1. S = 34.5, T = 0C, P = 3500 dbar
  2. S = 32.0, T = 15C, P = 20 dbar
  3. S = 42.0, T = 20C, P = 1000 dbar
  4. S = 34.4, T = -1.8C, P = 1800 dbar
  5. S = 35.6, T = 16C, P = 400 dbar
  6. S = 31.2, T = 12C, P = 60 dbar