Question

3.) An outdoor uninsulated pipe of 100 mm diameter is used to transport water. During a harsh winter, the inside surface of a pipe wall was at a temperature of -15 C, and a layer of ice formed inside the pipe. Assuming the layer of ice was uniform thickness within the pipe, what is the thickness of the ice if the water was at 3 C, the ice was at the water-ice interface was 0 C, and the convective heat transfer coefficient for the cold water at the interface with the ice is constant at 2000 W/m2/K. The thermal conductivity of ice is 2.3 W/m/K.

Answer 1

The thickness of the ice layer that forms inside the uninsulated pipe is approximately 0.00000165 meters, or about 1.65 micrometers.

Step-by-step explanation:

To determine the thickness of the ice layer that forms inside the uninsulated pipe, you can use the concept of heat transfer through the pipe wall. The heat transfer occurs due to the temperature difference between the water inside the pipe and the outside environment. We can use the equation for steady-state heat conduction through a cylindrical wall:

Q = (2πkL / ln(r2/r1)) * (T1 - T2)

Where:

Q = Heat transfer rate (W)

k = Thermal conductivity of the material (W/m·K)

L = Thickness of the wall (m) - which is the thickness of the ice layer in this case

r1 = Inner radius of the pipe (m)

r2 = Outer radius of the pipe (m)

T1 = Temperature inside the pipe (K)

T2 = Temperature outside the pipe (K)

Given:

Inner diameter of the pipe (d) = 100 mm = 0.1 m

Inner radius (r1) = d/2 = 0.05 m

Thermal conductivity of ice (k) = 2.3 W/m·K

Temperature inside the pipe (T1) = 3°C = 276 K

Temperature outside the pipe (T2) = -15°C = 258 K

Convective heat transfer coefficient (h) = 2000 W/m²·K

Now, we need to find the outer radius (r2) of the pipe, which includes the thickness of the ice layer (L). Since the ice layer is uniform, the temperature at the ice-water interface is 0°C, which is also the freezing point of water. We can use this information to find the outer radius:

Q = (2πkL / ln(r2/r1)) * (T1 - T2)

Q = (2π * 2.3 * L / ln(r2/0.05)) * (276 - 258)

Q = (4.6πL / ln(r2/0.05)) * 18

We also know that at the interface between the cold water and the ice, heat transfer occurs through convection:

Q_convective = h * A * ΔT

Where:

Q_convective = Heat transfer rate due to convection (W)

h = Convective heat transfer coefficient (W/m²·K)

A = Surface area through which heat transfer occurs (m²)

ΔT = Temperature difference between the water and ice interface (K)

The surface area of the pipe wall can be calculated as:

A = 2πL * (r2 - r1)

Now, we can set up the equation for the heat transfer due to convection:

Q_convective = 2000 * [2πL * (r2 - 0.05)] * (0 - 276)

Q_convective = -4000πL * (r2 - 0.05) * 276

Now, we have two equations for heat transfer, one through conduction and one through convection. Since heat is conserved, the heat transferred through conduction must be equal to the heat transferred through convection:

4.6πL / ln(r2/0.05) * 18 = -4000πL * (r2 - 0.05) * 276

Now, we can solve this equation for L (the thickness of the ice layer):

4.6 / ln(r2/0.05) * 18 = -4000 * (r2 - 0.05) * 276

Simplify:

ln(r2/0.05) = -4.6 / (18 * -4000 * 276)

ln(r2/0.05) ≈ 0.0000327

Now, find r2:

r2/0.05 ≈ e^0.0000327

r2 ≈ 0.05 * e^0.0000327

r2 ≈ 0.05000165 m

Now that we have r2, we can calculate the thickness of the ice layer (L):

L = (r2 - r1)

L = (0.05000165 - 0.05)

L ≈ 0.00000165 m

So, the thickness of the ice layer that forms inside the uninsulated pipe is approximately 0.00000165 meters, or about 1.65 micrometers.