Thermal conduction, help?

Question

A circular pond of radius 12 m has a layer of ice 0.25 m thick. On a spring day, the temperature of the water below the ice is 4.0°C and the temperature of the air above is 11°C. What is the rate of heat transfer through the ice, in kW?


- The thermal conductivity of air k (W/mK) is 0.025 , and its thermal resistance RSI-value (m2 x K/W) (for 1 inch) is 1.00 and R-value (ft2 x F° x h/Btu) is 5.70.
- The thermal conductivity of water k (W/mK) is 0.598, and its thermal resistance RSI-value (m2 x K/W) (for 1 inch) is 0.04 and R-value (ft2 x F° x h/Btu) 0.24.
- The thermal conductivity of 0° C ice k (W/mK) is 2.14 , and its thermal resistance RSI-value (m2 x K/W) (for 1 inch) is 0.01 and R-value (ft2 x F° x h/Btu) 0.07.
-These values are at 10^5 Pa, 20°C.


Thus far, my answer of 25 kW was incorrect, help please?

Answer 1

Heat flows down through the ice from air at 11°c
to water at 4°c
Q1/t =λ A [11 - 4] / d.

Heat flows up through the ice from water at 4°c to ice at 0°c
Q2/t =λ A [4 -0] / d.

The net flow of heat is
Q1/t - Q2/t = λ A [11 - 4 - 4 +0] / d
[Q1-Q2] /t = λ A 3 / d=2.14 π 12^2 * 3 / 0.25
[Q1-Q2] /t = 11617. W= 11.617 kW.

Answer 2

For this problem, the air above the ice and the water below the ice can be considered infinite reservoirs.
Q = UA∆T/t
Q = (2.14 W/m°C)(π*12^2 m^2)(7°C) / (0.25 m)
Q ≈ 27,107.17 W ≈ 27 KW

Answer 3

Construct a 1-d resistance network. The low-temperature reservoir is at 4C and the high-temperature reservoir is at 11C. There are three resistances in between: (1) convection from the 4C water to the bottom surface of the ice, (2) conduction through the ice, and (3) convection from the top surface of the ice to the high-temperature reservoir. Note that there needs to be a temperature gradient to the ice for there to be any heat transfer, so the entire slab of ice - bottom surface and top - cannot possibly be uniform temperature. Also, the ice cannot possibly be 0C since this temperature does not lie between 4C and 11C and the ice is not generating any energy.

R1 = thermal resistance from 4C to bottom surface of ice
R2 = conduction thermal resistance through ice
R3 = thermal resistance from top surface of ice to 11C

Rtot = R1 + R2 + R3

Rtot = (11C - 4C)/Q

Q = 7C/Rtot

R1 = RSI1/A = 0.04/(144*pi) = 8.8419e-5 C/W
R2 = L/kA = 0.25/(2.14*144*pi) = 2.5823e-4 C/W
R3 = RSI2/A = 1.00/(144*pi) = 2.2105e-3 C/W

Rtot = 2.5571e-3 C/W

** Q = 2.737 kW **

Note:
T0 = 4 C
T1 = 4.24 C (bottom of ice)
T2 = 4.99 C (top of ice)
T3 = 11 C