Solve the thermodynamics problems 6.1 to 6.5.

Steps to Solve

6.1 Problem 6.1 Solution

This problem involves heat transfer in a slab with one side heated and the other held at an initial temperature. We aim to find the transient and steady-state interface location, assuming $Ste < 0.1$. Because $Ste < 0.1$, we are able to assume the interface is moving slowly and apply a quasi-steady approximation.

1. Transient Interface Location

Because $Ste < 0.1$ we can assume a linear temperature profile in both the solid and liquid phase:

$$T(x) = T_0 - \frac{T_0 - T_f}{x_i}x \text{ for } 0 \leq x \leq x_i$$

$$T(x) = T_f - \frac{T_f - T_i}{L-x_i}(x-x_i) \text{ for } x_i \leq x \leq L$$

$x_i$ is the location of our moving interface.

At the interface ($x = x_i$), the energy balance is given by the Stefan condition:

$$k_s \frac{dT}{dx}|{x=x_i^-} - k_l \frac{dT}{dx}|{x=x_i^+} = \rho L \frac{dx_i}{dt}$$

Since our latent heat is released on the "cold side", we can assume the cold side is solid and the hot side is a liquid. $k_s$ is the conductivity of the liquid, and $k_l$ is the conductivity of the solid. Substituting in our equations:

$$k_s \frac{T_0 - T_f}{x_i} - k_l \frac{T_f - T_i}{L - x_i} = \rho L \frac{dx_i}{dt}$$

Thus determining the transient interface requires solving this differential equation, after which one would need to apply the initial conditions to fully determine the transient interface location.

2. Steady-State Interface Location

At steady state, $\frac{dx_i}{dt} = 0$, which gives us:

$$k_s \frac{T_0 - T_f}{x_i} = k_l \frac{T_f - T_i}{L - x_i}$$

Solving for $x_i$:

$$x_i = \frac{k_s(T_0 - T_f)L}{k_l(T_f - T_i) + k_s(T_0 - T_f)}$$

This equation gives the steady-state interface location.

6.2 Problem 6.2 Solution

This problem involves determining the ice thickness of a deep lake after four weeks of a cold wave.

1. Justify Quasi-Steady Model

The quasi-steady model is justified if the Stefan number ($Ste$) is small ($Ste << 1$). The Stefan number is defined as:

$$Ste = \frac{c_{ps} \Delta T}{L}$$

where $c_{ps}$ is the specific heat of the solid phase, $\Delta T$ is the temperature difference, and $L$ is the latent heat of fusion. $\Delta T = T_f - T_{air} = 0 - (-18) = 18 \text{ °C}$

$$Ste = \frac{2093 \text{ J/kg-°C} \cdot 18 \text{ °C}}{333730 \text{ J/kg}} \approx 0.113$$

Since $Ste \approx 0.113 > 0.1$, the quasi-steady model is not strictly justified, but we can proceed with the approximation.

2. Determine Ice Thickness

The equation for ice thickness ($x$) as a function of time ($t$) in a quasi-steady model is:

$$x = \sqrt{\frac{2 k_s (T_f - T_0) t}{\rho_s L}}$$

where $k_s$ is the thermal conductivity of ice, $T_f$ is the fusion temperature, $T_0$ is the surface temperature, $\rho_s$ is the density of ice, and $L$ is the latent heat of fusion.

$t = 4 \text{ weeks} = 4 \cdot 7 \cdot 24 \cdot 3600 = 2419200 \text{ seconds}$

$$x = \sqrt{\frac{2 \cdot 2.21 \text{ W/m-°C} \cdot (0 - (-18)) \text{ °C} \cdot 2419200 \text{ s}}{916.8 \text{ kg/m}^3 \cdot 333730 \text{ J/kg}}}$$

$$x \approx \sqrt{\frac{1.92 \times 10^{8}}{3.06 \times 10^{8}}} \approx \sqrt{0.627} \approx 0.792 \text{ m}$$

Therefore, the ice thickness is approximately $0.792 \text{ meters}$ after four weeks.

6.3 Problem 6.3 Solution

This problem concerns radiation beamed at a semi-infinite region initially solid at the fusion temperature. A uniform energy generation $q'''$ occurs in the liquid phase. The surface at $x = 0$ is maintained at $T_0 > T_f$. Again, since $Ste < 0.1$, we determine the transient and steady-state interface location similar to problem 6.1.

1. Transient Interface Location

The temperature in the liquid phase $(0 \leq x \leq x_i)$:

$$k \frac{d^2T}{dx^2} + q''' = 0$$

Boundary conditions: $T(0) = T_0$, $T(x_i) = T_f$. Solving for the temperature distribution:

$$T(x) = T_0 + \frac{T_f - T_0}{x_i}x - \frac{q'''}{2k}x(x-x_i)$$

In the solid phase $(x > x_i)$, the temperature is $T_f$. Applying energy balance at the interface:

$$-k \frac{dT}{dx}|_{x=x_i} = \rho L \frac{dx_i}{dt}$$

Differentiating the liquid temperature profile:

$$\frac{dT}{dx} = \frac{T_f - T_0}{x_i} - \frac{q'''}{2k}(2x - x_i)$$

Evaluating at $x = x_i$:

$$\frac{dT}{dx}|_{x=x_i} = \frac{T_f - T_0}{x_i} - \frac{q'''}{2k}x_i$$

Substituting into the energy balance:

$$-k \left( \frac{T_f - T_0}{x_i} - \frac{q'''}{2k}x_i \right) = \rho L \frac{dx_i}{dt}$$

$$k \frac{T_0 - T_f}{x_i} + \frac{q'''}{2}x_i = \rho L \frac{dx_i}{dt}$$

This is a differential equation that would need to be solved with an initial condition to determine the transient interface location.

2. Steady-State Interface Location

At steady state, $\frac{dx_i}{dt} = 0$, so:

$$k \frac{T_0 - T_f}{x_i} + \frac{q'''}{2}x_i = 0$$

$$x_i^2 = \frac{2k(T_0 - T_f)}{-q'''}$$

Because $q'''$ represents volumetric heat generation, it can be defined as negative. So in this case with the negative sign inside the equation for the steady state interface, it is intended for negative volumetric heat generation.

$$x_i = \sqrt{\frac{2k(T_0 - T_f)}{-q'''}}$$

This gives the steady-state interface location.

6.4 Problem 6.4 Solution

We need to find the ice layer thickness after 7 hours, assuming convection heat transfer from the water.

1. Justify Quasi-Steady Model

The heat transfer is by convection and the water is initially at the fusion temperature. the surface temperature will quickly drop to the air temperature. We approximate $Ste$:

$$Ste = \frac{c_{ps} \Delta T}{L}$$

$\Delta T = T_f - T_{air} = 0 - (-10) = 10 \text{ °C}$

$$Ste = \frac{2093 \text{ J/kg-°C} \cdot 10 \text{ °C}}{333730 \text{ J/kg}} \approx 0.0627$$

Since $Ste << 0.1$, the quasi-steady model is justified.

2. Determine Ice Thickness

The heat flux due to convection is:

$$q'' = h(T_f - T_{\infty})$$

where $h$ is the heat transfer coefficient and $T_{\infty}$ is the air temperature.

The rate of energy removal to freeze ice of thickness $dx$ is:

$$q'' = \rho_s L \frac{dx}{dt}$$

Combining these equations:

$$h(T_f - T_{\infty}) = \rho_s L \frac{dx}{dt}$$

Integrating from $0$ to $x$:

$$\int_0^x dx = \frac{h(T_f - T_{\infty})}{\rho_s L} \int_0^t dt $$

$$x = \frac{h(T_f - T_{\infty})}{\rho_s L} t$$

Given: $h = 12.5 \text{ W/m}^2\text{-°C}$, $T_{\infty} = -10 \text{ °C}$, $t = 7 \text{ hours} = 7 \cdot 3600 = 25200 \text{ s}$

$$x = \frac{12.5 \text{ W/m}^2\text{-°C} \cdot (0 - (-10)) \text{ °C}}{916.8 \text{ kg/m}^3 \cdot 333730 \text{ J/kg}} \cdot 25200 \text{ s}$$

$$x \approx \frac{12.5 \cdot 10 \cdot 25200}{916.8 \cdot 333730} \approx \frac{3150000}{306079704} \approx 0.0103 \text{ m}$$

The ice layer will be approximately $0.0103 \text{ m}$ thick after 7 hours, or $1.03 \text{ cm}$.

6.5 Problem 6.5 Solution

Derive an expression for the total solidification time for an ice cream kit with concentric cylinders, assuming $Ste < 0.1$.

1. Energy Balance

The heat flux through the solid ice layer must equal the rate of latent heat removal:

$$q = \frac{2 \pi k_s (T_f - T_0)}{\ln(r/R_a)} = 2 \pi r L \rho_s \frac{dr}{dt}$$

2. Integrate for Solidification Time

Separating variables and integrating:

$$\int_{R_a}^{R_b} \ln(r/R_a) r dr = \int_0^t \frac{k_s (T_f - T_0)}{\rho_s L} dt$$

Let $u = \ln(r/R_a)$, so $e^u = r/R_a$ and $r = R_a e^u$. Thus $dr = R_a e^u du = r du$. Then $dr = r du$ implies $du = dr/r$, so $dr = e^u R_a du$.

$$ \int \ln(r/R_a) r dr = \int u R_a e^u R_a e^u du = R_a^2 \int u e^{2u} du$$

$\int u e^{2u} du = \frac{1}{2}ue^{2u} - \int \frac{1}{2}e^{2u} du = \frac{1}{2}ue^{2u} - \frac{1}{4}e^{2u}$

$$R_a^2 [\frac{1}{2}ue^{2u} - \frac{1}{4}e^{2u}] = R_a^2 [\frac{1}{2} \ln(r/R_a) (r/R_a)^2 - \frac{1}{4} (r/R_a)^2 ]$$

$$R_a^2 [\frac{1}{2} (\frac{r}{R_a})^2 \ln(\frac{r}{R_a}) - \frac{(\frac{r}{R_a})^2}{4}]$$

$$ \int_{R_a}^{R_b} \ln(r/R_a) r dr = R_a^2 [\frac{1}{2} (\frac{R_b}{R_a})^2 \ln(\frac{R_b}{R_a}) - \frac{(\frac{R_b}{R_a})^2}{4}] - R_a^2 [\frac{1}{2} (\frac{R_a}{R_a})^2 \ln(\frac{R_a}{R_a}) - \frac{(\frac{R_a}{R_a})^2}{4}]$$

Since $\ln(1) = 0$:

$$R_a^2 [\frac{1}{2} (\frac{R_b}{R_a})^2 \ln(\frac{R_b}{R_a}) - \frac{(\frac{R_b}{R_a})^2}{4} + \frac{1}{4}]$$

$$R_b^2 [\frac{1}{2} \ln(\frac{R_b}{R_a}) - \frac{1}{4}] + \frac{R_a^2}{4}$$

Let $x = R_b/R_a$, then:

$$\frac{R_b^2}{2}\ln{x} - \frac{R_b^2}{4} + \frac{R_a^2}{4}$$

$$\frac{1}{4}(2R_b^2\ln{x} - R_b^2 + Ra^2)$$

Then,

$$\int_{R_a}^{R_b} \ln(r/R_a)rdr = \frac{1}{4}(2R_b^2\ln(\frac{R_b}{R_a}) - R_b^2 + R_a^2)$$

And:

$$\int_0^t \frac{k_s (T_f - T_0)}{\rho_s L} dt = \frac{k_s(T_f - T_0)}{\rho_s L}t$$

Therefore giving us:

$$t = \frac{\rho_s L}{4 k_s (T_f - T_0)} (2R_b^2\ln(\frac{R_b}{R_a}) - R_b^2 + R_a^2)$$

3. Apply Values

$R_a = 0.1 \text{ m}$, $R_b = 0.25 \text{ m}$, $T_0 = -20 \text{ °C}$, $T_f = 0 \text{ °C}$ $k_s = 2.21 \text{ W/m-°C}$, $L = 333730 \text{ J/kg}$, $\rho_s = 916.8 \text{ kg/m}^3$

$$t = \frac{916.8 \cdot 333730}{4 \cdot 2.21 \cdot (0 - (-20))} (2(0.25)^2 \ln(\frac{0.25}{0.1}) - (0.25)^2 + (0.1)^2)$$

$$t = \frac{306079704}{176.8} (0.125 \ln(2.5) - 0.0625 + 0.01) = \frac{306079704}{176.8} (0.1145 - 0.0625 + 0.01)$$

$$t = 1731217.78 (0.062 + 0.01) = 1731217.78 \cdot 0.072$$

$$t \approx 124647.67 \text{ s} \approx 34.6 \text{ hours}$$

The total solidification time is approximately $34.6$ hours.