# Calculate the time it takes for river gravel to take on the same temperature as the a

#### evan

Heres the scenario:

Water level drops in a river. The previously saturated gravel slowly takes on the same temperature as the air. Assuming the gravel was originally the same temperature as the water, and the surface is flat, How long does it take?

I believe these are the relevant formulas :

Q=mC∆T

Q/t = kA∆T/d

Where Q = energy [J], m = mass (kg), t = time(secs), ∆T = temperature difference, d = thickness (meters) of material conducting energy, A = surface area of interface between bodies exchanging heat (m^2).

Given:
drop in water surface elevation = 1m (for simplicity's sake),
water temp = 10°C,
air temp = 15°C,
specific heat capacity of gravel (C) = 1175 J/kg\cdot K,
thermal conductivity of gravel (k) = 3.5 W/m\cdot K,
density of gravel (relevant to m) = 1950 kg/m^3,

This is how I've gone about it so far:

Use Q = mC∆T to calculate the amount of energy required to heat 1 cubic meter of gravel (cube dimensions are based off drop in elevation, 1m)

Then rearrange Q/t = (kA∆T)/d as
t = (d/(kA∆T))(Q)

Solve for t, inputing values consistent with the cube dimensions for surface area (A) = 1, and thickness (d) = 1

Is this the correct approach?

EDIT *** Bonus question

Would the approach be the same if the temperature was lower in the air than the water, ie; ∆T was negative?

#### oz93666

When the water level drops the moist gravel is exposed to the air ... and since the moist exposed area is now greater than the level river surface , evaporation will increase ...

the gravel temperature will initially drop ...

What are the size of the gravel particles ?? How porous is it ? what is the relative humidity of the air ??? is there a breeze affecting evaporation ???how quickly is moisture moving up to the surface by capillary action through the gravel bed ???

#### studiot

Is this the correct approach?
Not really.

I see this as a radiative heat transfer issue.

Further complicated by the fact that during daylight the exposed gravel will be absorbing radiant energy and warming up - conversely it will be loosing radiant energy during the hours of darkness. This introduces the temperature achieved as a function of time.

Oz has introduced another complication. Gravel is granular so its internal heat transfer coefficients (by conduction) will be from the solid rock material, as will any evaporative effects due to seepage.

You may be able to adapt standard building calcualtions to suit your base materials to achieve a sensible model.

#### ChipB

PHF Helper
You seem to be modelling the gravel as a slab, but calculating heat flow out of a slab is more complicated than this. Your formula assumes that the gravel always has a uniform temperature throughout as it warms. But in reality the surface approaches air temperature before the interior does, so heat flow through the surface of the gravel pile slows over time, and your method will underestimate the time required for the entire mass to reach air temperature. The temp at any point in the mass of gravel will be a logarithmic function of time, asymptotically approaching air temp. Rather than ask how long until the gravel reaches air temperature, you need to ask something along the lines of "how long before the interior is within 1 degree C of air temperature."

#### mscfd

It's clearly not a simple problem to be solved. Before thinking about how to solve the problem, you'll need an analysis of Biot.