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ddrtrinity

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**1. Assume a cubical pot of water 1 m on each side. Heat is applied to the bottom of the pot. Assume the water at the bottom is held at 99 degrees C while the water at the top of the pot is 4 degrees C. Estimate the convective heat flow (using algebraic physics).**

Step 1: Compute the net buoyant force on a cube with 1.00 grams of water at 99 degrees C which is immersed in a bath of water at 98 degrees C.

Step 2: Compute the end velocity of the fluid (that is, how fast it is going after rising 99 cm in the pot) assuming it accelerates continuously due to the net buoyant force. Next, equate the net buoyancy force to the viscous drag force and solve for the upward velocity at which the two forces are equal. [This is where I'm having the most difficulty.]

**2. The list of equations is broken down into steps:**

Step 1: ΔV=βVΔT, as well as density = m/v, and F(net) = ρ(fluid)V(object)g – mg

Step 2: F = ma, v^2 = 2a(Δx), and F(net) = ηA(v/

*l*)

**3. Attempts so far in steps:**

New density of pot water (β = 210 x 10^-6) is 980.6 kg/m^3 and new density of cube is 980.4 kg/m^3. Since the volume of he cube is 1.02 x 10^-6 m^3 at that new density, then I found the net force to be 2.08 x 10^-6 N upwards.

[2.08 x 10^-6 N = (980.6 kg/m^3)(1.02 x 10^-6 m^3)(9.8 m/s^2) - (0.001 kg)(9.8 m/s^2)]

However, I contacted the professor and it was implied this wasn't correct.

Step 2: Assuming my result from above is correct, I had no difficulties finding the acceleration given mass 0.001 kg and then the resulting velocity at 99 cm (2.08 x 10^-3 m/s^2 = a and v at 99 cm = 0.0652 m/s).

However, I had no idea how to go about doing the viscosity equation; it was never covered in class, and I do not understand what A and

*l*in particular would be, nor what to use for the coefficient η especially since this doesn't seem like a thermically stable environment. I only know that I need to equate the result from Step 1 to the viscosity equation.

**If you're interested, the remaining steps in this problem:**

3. Compute the heat delivered by that 1 gram of water at 99 degrees C into the layer of water at the top of the pot, assuming the latter is 1 m square, 1 cm deep and 4 degrees C. Start by solving the caliormetry problem to determine the final temperature of the top layer plus hot 1 gram from below and then determine how much heat was delivered in the hot water.

4. Using the heat from 3 and the time to transport from 2, estimate the heat transported in watts by 1 gram of water in convective motion (simply divide the result from 3 by the result from 2).

5. Estimate how much water at 98 degrees C can be heated by conduction from the pot's bottom (which is steel and at 100 degrees C) to a temperature of 99 degrees C in one second. To do this, compute the conductive energy flow across 1 square meeter of steel that is 5 mm thick where the temperature on opposite sides are 98 and 100 degrees respectively. Then take that conductive energy flow and dividie it by the specific heat of water which should give you a number with units (kg degrees C/s). Dividing this by 1 degree C will give you a number of kilograms of water that can be heated per second from 98 to 99 degrees C.

6. Finally, compare the mass per second of water that can be heated to 99 (result from 5) to the total mass in the bottom 1 cm of the pot. If it's more than half the total mass, then use only half that total mass instead (you can't have all the water rising at once). Then take that mass of rising water and multiply it by the amount of heat delivered per second in each gram of warm water (your result from 4). At this point you should have a number in watts for the total convective heat transport in the pot.