Exercise 1
multiply and divide
Two tubs in the same yard
Anahit and her grandmother set up two water tubs on the balcony in Malatia, on the same morning, with the same water from the same tap. One is a long shallow trough. The other is a tall plastic barrel, cut down. They put exactly the same amount of water in each. By August the fish in the barrel are gasping at the surface in the early morning and the fish in the trough are not, and nobody has fed them differently or put more fish in one than the other.
- Water in each container
- 180 L
- Trough, inside floor
- 0.90 m × 0.50 m
- Barrel, inside floor
- 0.60 m × 0.30 m
- Gas transfer velocity, both
- k = 0.5 m/day
Find:
- the mean depth of the water in each container, in meters;
- the time constant of re-oxygenation for each, in hours;
- how many times faster the trough refills its oxygen than the barrel does.
The two relations you need, mean depth from volume and surface, and the time constant from mean depth and the gas transfer velocity, are on the physics of an aquarium.
Exercise 2
multiply and subtract
The same tank in two towns
Two cousins keep the same 54 liter tank, bought from the same shop on the same day. One lives in Gyumri and one lives in Meghri. Both run their tanks at 25 degrees, both keep the same fish, and both insist their water is identical. One of them is right about the fish and wrong about the water.
- Tank volume, each
- 54 L
- Water temperature, each
- 25 °C
- Towns
- Gyumri and Meghri
- Both tanks
- fully saturated with oxygen
Find:
- the dissolved oxygen concentration at saturation in each town, in mg/L;
- the total mass of dissolved oxygen in each full tank, in milligrams;
- how much more oxygen, as a percentage, the Meghri tank holds.
The saturation table corrected for each town's station pressure is on the physics of an aquarium. You need no formula for this one, only the right row and the right two columns.
Exercise 3
multiply and divide
Correcting a sea level book for Yerevan
A borrowed aquarium book from the 1980s prints a dissolved oxygen table and says nothing about altitude, because it was written at sea level for readers at sea level. You want the 20 degree figure for a tank in Yerevan. A friend says just scale it by the ratio of the pressures. That is nearly right, and the reason it is only nearly right is that water vapour in the air above the surface takes up part of the pressure and does not thin out with altitude the way the dry gases do.
- Water temperature
- 20 °C
- Saturation vapour pressure of water at 20 °C
- 23.4 hPa2
- Standard sea level pressure
- 1013.25 hPa
- Yerevan station pressure
- look it up
Find:
- the saturation concentration of dissolved oxygen in Yerevan at 20 °C, in mg/L, using the altitude correction properly;
- the answer your friend's simple pressure ratio would have given;
- the difference between the two, and say which way the shortcut is wrong.
Then check your first answer against the printed table on the same page. They should agree.
The correction formula, the Yerevan station pressure and the sea level value at 20 °C are all on the physics of an aquarium.
Exercise 4
needs ex and ln
The morning after a hot night
A shallow tub stands in the shade of a walnut tree in a courtyard in Yerevan. Through a still July night the fish and the bacteria have been using oxygen faster than the surface could replace it, and at sunrise the water is badly down. There is no pump and no air stone. The question the owner actually has is whether to run out and buy one, or whether the morning will fix it by itself.
- Water volume
- 120 L
- Water surface area
- 0.75 m2
- Water temperature
- 25 °C, steady
- Where
- Yerevan
- Gas transfer velocity
- k = 0.3 m/day, still and sheltered
- Dissolved oxygen at sunrise
- 3.6 mg/L
Find:
- the mean depth and the time constant, in hours;
- how long after sunrise the water reaches 6.0 mg/L, in hours;
- the dissolved oxygen concentration twelve hours after sunrise, in mg/L.
The recovery curve, the time constant and the saturation value for Yerevan at 25 °C are on the physics of an aquarium. Rearranging the curve to give a time rather than a concentration is the part you do yourself.
What the fish can see of the room
A child lies on the floor beside a tank and notices that the underside of the water surface is a mirror almost everywhere, except for one bright round patch directly above the fish, through which the ceiling lamp and her own face are visible. She wants to know how big that patch is, and whether the fish can see the whole room through it or only part of it.
- Fish depth below the surface
- 0.30 m
- Tank inside width, front to back
- 0.60 m
- Water
- clean and still
Find:
- the radius of the bright circle on the surface, in centimeters;
- its diameter;
- whether the circle fits inside the tank, and say what the fish sees in the direction where it does not.
The angle that decides where the mirror starts is on optics. The circle's radius is that angle and the depth in a right-angled triangle, which is yours to draw.
Exercise 6
multiply and divide
The slope that looks flatter than it is
Tigran builds a substrate slope, low at the front and high at the back, and works entirely by looking down into the open tank from above, which is how almost everyone does it. When the tank is filled and he looks at it from the front, through the glass, the slope he was so pleased with looks disappointingly flat. He has not made a mistake with the sand. He has made one with the physics.
- Real water depth at the front
- 36 cm
- Real water depth at the back
- 48 cm
- Viewing
- straight down from above, through the water surface
Find:
- how deep the water appears at the front and at the back, seen from above;
- how large the height difference of the slope appears, in centimeters;
- how tall he would have to build the real difference so that it looks like 12 cm from above.
The factor by which water makes a depth look smaller is printed on optics, with the worked example of a 40 cm tank.
Exercise 7
needs ex and ln
Why the red plant at the bottom is not the red plant at the top
A tall tank has the same red stem plant growing at the surface and at the substrate. Under the same lamp, the one at the bottom photographs brown. The water is clear, there is no tint and no algae bloom, so the only thing between the lamp and the lower leaves is 45 centimeters of clean water. That turns out to be enough.
- Depth from surface to substrate
- 0.45 m
- Water
- pure, no color and no particles
- Colors to compare
- 700 nm deep red and 420 nm blue
Find:
- the half depth of each of the two colors, in meters;
- the percentage of each color that still reaches the substrate;
- how many times more deep red is lost over that 45 cm than blue is.
The attenuation law, the half depth relation and the absorption coefficient of pure water at each wavelength are all on optics.
Exercise 8
needs ex and ln
The stoneworts that used to grow at nineteen meters
Before the lake was drawn down, meadows of stoneworts grew on the floor of Lake Sevan at nineteen meters, which means enough light was reaching them there to live on. They do not grow there now. The water above them has changed: dissolved organic matter and suspended algae absorb light that used to get through, and the published thermal model of the lake uses a single extinction coefficient to describe it.
- Lake
- Sevan, present day
- Extinction coefficient
- the one the published lake model uses — look it up
- Depth the stoneworts once grew at
- 19 m
Find:
- the depth at which one percent of the surface light is left, in meters;
- the percentage of surface light reaching 19 m today;
- the depth at which half the surface light is left.
The attenuation law and the extinction coefficient for Sevan are both on optics. What the plants did with that light is on the Lake Sevan page.
Exercise 9
needs ex and ln
Will a leaf on the bottom pay for itself?
A leaf uses energy all day and all night just to stay alive, and it makes energy only while there is light on it. Below a certain brightness it spends more than it earns and slowly starves, however healthy it looks on the day you plant it. This tank has slightly tea-coloured water from a piece of driftwood, which absorbs more light than clean water does, and the owner wants to know whether a carpet plant will hold at the bottom.
- Light just under the surface
- 120 µmol m−2 s−1 of PAR
- Measured total attenuation coefficient of this water
- 1.4 per meter
- Depth from surface to substrate
- 0.45 m
Find:
- the PAR reaching the substrate;
- whether that is above or below the measured range in which a submerged leaf just breaks even;
- the depth at which the light falls to the bottom of that measured range, and say whether any part of this tank is below it.
Then answer in one sentence: would you put a demanding carpet plant on this substrate, and why?
The attenuation law is on optics. The measured break-even light range for a submerged leaf is on find a plant, and again on materials and water.
Exercise 10
multiply and divide
How far the tub swings in a July day
A wide open tub stands in a yard outside Yerevan through July. The owner has a thermometer in it and is startled by how far the reading travels between dawn and late afternoon. His neighbor has a deeper tub in the same yard, in the same sun, and hers barely moves. Neither of them has done anything clever. The difference is entirely the depth of the water, because the same sunlight has to warm a taller column.
- Sunlight taken in by the water
- 29.9 MJ per m2 over the day
- First tub, mean depth
- 0.20 m
- Second tub, mean depth
- 0.60 m
- Both
- unshaded, same yard, same day
Find:
- the daily temperature swing of each tub, in kelvin;
- the mean depth that would hold the swing to 4 K or less;
- and answer in a sentence: does making the tub deeper make the water cooler, or only steadier?
The swing equation, the fraction of the day's energy it uses and the volumetric heat capacity of water are all on the physics of an aquarium. Your first answer should land on a value that page has already printed for a 0.20 m tub.
Exercise 11
multiply and divide
The neighbor downstairs
A family in a Soviet-era block in Yerevan wants a 240 liter tank in the living room. A relative who once worked in construction tells them the floor is rated for far less than the tank will weigh and that the whole idea is dangerous. The relative has quoted a real number from a real code. The question is whether it is the right number to compare against.
- Tank
- 240 L
- Grand total mass, water, glass, substrate, hardscape and cabinet
- 339 kg
- Cabinet footprint
- 0.60 m2
- Room floor area
- 12 m2
- Gravity
- 9.81 m/s2
Find:
- the pressure under the cabinet, in kilopascals;
- how many times that is the residential live load figure in the code Armenia inherited;
- the pressure if the same mass were spread evenly over the whole room, and what percentage of the code figure that is.
Then say, in two sentences, which of those two comparisons is the fair one and why.
The pressure relation, the code figure and the reason the two comparisons differ are on floor loading and glass thickness. Your first and third answers should both match numbers printed there.
Exercise 12
multiply and divide
Three milliseconds, or twenty four seconds
Every underwater leaf wears a thin film of water that will not move, however hard the filter is blowing at the far side of the tank. Carbon dioxide has to cross that film by diffusion alone before the leaf can do anything with it. The same molecule crossing the same distance of still air would arrive almost instantly. That comparison is the whole reason underwater plants are carbon-starved and land plants are not.
- Thickness of the still film on the leaf
- 0.3 mm
- Rough time to diffuse a distance L
- t = L2 / (2D)3
- Diffusing substance
- carbon dioxide
Find:
- how many times larger the diffusion coefficient of carbon dioxide is in air than in water;
- the time to cross 0.3 mm of still water, in seconds;
- the time to cross 0.3 mm of still air, in seconds.
Then answer: if the filter doubles the water speed at the leaf and halves the thickness of that film, by what factor does the crossing time fall?
Both diffusion coefficients are printed on chemistry. What the film does to a leaf is on materials and water.