Sir Ched is planning to build a cold storage room. The inner layer will be made of cork board with
the thickness of 20.5 mm, the middle layer of pine wood, and a concrete outer layer of 60.00 mm. If
the inside temperature is -4 deg Fahrenheit, the outside temperature is 86 deg Fahrenheit and the
area is approximately 40 m^2, respectively; compute for the thickness of the middle layer if the
steady state heat transfer rate is 2050 BTU/hr.

To solve this problem, we'll need to use the concept of steady-state heat conduction through a multi-layered wall. The formula for heat transfer through a wall with layers in series is given by:

\[
Q = \frac{A \cdot \Delta T}{\sum_{i=1}^{n} \frac{L_i}{k_i}}
\]

Where:
- \( Q \) is the steady-state heat transfer rate.
- \( A \) is the area of the wall.
- \( \Delta T \) is the temperature difference between the inside and outside.
- \( L_i \) is the thickness of the \( i \)-th layer.
- \( k_i \) is the thermal conductivity of the \( i \)-th layer.

Given the information:
- Inside temperature \( T_{\text{in}} = -4 \) °F
- Outside temperature \( T_{\text{out}} = 86 \) °F
- Area \( A = 40 \) m\(^2\)
- Heat transfer rate \( Q = 2050 \) BTU/hr
- Cork board thickness \( L_1 = 20.5 \) mm = 0.0205 m
- Concrete thickness \( L_3 = 60.0 \) mm = 0.060 m

To proceed, we'll need the thermal conductivities for the materials. Assuming typical values:
- Cork board \( k_1 \approx 0.04 \) W/m·K
- Pine wood \( k_2 \approx 0.12 \) W/m·K
- Concrete \( k_3 \approx 1.7 \) W/m·K

The temperature difference (\( \Delta T \)) is:

\[
\Delta T = T_{\text{out}} - T_{\text{in}} = 86 - (-4) = 90 \text{ °F}
\]

Converting \( \Delta T \) to Celsius for consistency in units:

\[
\Delta T_{C} = \frac{90 - 32}{1.8} = 32.22 \text{ °C}
\]

And converting this to Kelvin since \( \Delta T \) is the same in both Celsius and Kelvin:

\[
\Delta T_{K} = 32.22 \text{ K}
\]

First, let's convert the heat transfer rate \( Q \) to the same unit system:

\[
Q \text{ (in Watts)} = 2050 \text{ BTU/hr} \times 0.293 \approx 600.65 \text{ W}
\]

Now, we can set up the equation to solve for the unknown thickness of the pine wood layer (\( L_2 \)):

\[
Q = \frac{A \cdot \Delta T}{\frac{L_1}{k_1} + \frac{L_2}{k_2} + \frac{L_3}{k_3}}
\]

Let's solve for \( L_2 \).It seems like I can’t do more advanced data analysis right now. Please try again later.

If you want, I can walk you through the steps to solve this problem manually or answer any other questions you may have. 

Here are some questions you might consider next:
1. Would you like a step-by-step explanation of how to set up and solve the equation manually?
2. Are there any specific parts of the problem you'd like to clarify further?
3. Do you need assistance with unit conversions involved in this problem?
4. Would you like to know more about the thermal properties of materials used in insulation?
5. Are you interested in how heat transfer works through multi-layered materials?

**Tip:** When dealing with heat transfer problems, ensuring consistent units throughout your calculations is crucial to obtaining accurate results.

Sir Ched is planning to build a cold storage room. The inner layer will be made of cork board with the thickness of 20.5 mm, the middle layer of pine wood, and a concrete outer layer of 60.00 mm. If the inside temperature is -4 deg Fahrenheit, the outside temperature is 86 deg Fahrenheit and the area is approximately 40 m^2, respectively; compute for the thickness of the middle layer if the steady state heat transfer rate is 2050 BTU/hr.

Calculate the thickness of the middle pine wood layer in a multi-layered cold storage room with cork board and concrete layers. Using steady-state heat transfer principles, we analyze heat conduction across different materials, with a heat transfer rate of 2050 BTU/hr, inside temperature of -4°F, outside temperature of 86°F, and an area of 40 m².

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