Question

Hot gases flow inside an array of tubes that are embedded in a $3-\mathrm{m} \times 3-\mathrm{m}$ horizontal flat heater. The heater surface is nickel- plated and is used for generating steam by boiling water in the nucleate boiling regime at \(180^{\circ} \mathrm{C}\). Determine the surface temperature of the heater and the convection heat transfer coefficient that is necessary to achieve the maximum rate of steam generation.

Short answer

The surface temperature of the heater at maximum steam generation rate is 453.15 K. The convection heat transfer coefficient cannot be calculated as the heat flux is zero. However, this indicates that a high convection heat transfer coefficient is required to achieve the maximum rate of steam generation in the nucleate boiling regime.

Step-by-step solution

Determine the heat flux in nucleate boiling

To determine the heat flux in nucleate boiling, we will use the equation: $$q = C_sf^{3/2}(T_s - T_b)^{3/2} $$ Here, \(q\) is the heat flux, \(C_s\) is a constant related to the surface material, \(f\) is a constant related to the liquid, \(T_s\) is the surface temperature, and \(T_b\) is the boiling temperature. For a nickelplated surface and water as the liquid, we use the following constant values (taken from a heat transfer handbook): $$ C_s = 2.8\times 10^5\,\mathrm{W/m^2K^{3/2}}$$ $$ f = 0.013\,\mathrm{K^{-1}} $$ $$ T_b = 180^{\circ}\mathrm{C} = 453.15\,\mathrm{K} $$

Find the surface temperature \(T_s\)

Since we want to maximize the steam generation rate, we need to find the maximum value of the heat flux. To do this, differentiate the heat flux equation with respect to \(T_s\), then set the derivative equal to zero and solve for the surface temperature: $$ \frac{d(q)}{d(T_s)} = \frac{3}{2}(C_sf^{3/2})(T_s - T_b)^{1/2} = 0$$ Now solve for the surface temperature \(T_s\): $$ (T_s - T_b)^{1/2} = 0$$ $$ T_s = T_b$$ So, the surface temperature at the maximum rate of steam generation is equal to the boiling temperature: $$ T_s = 453.15\,\mathrm{K} $$

Calculate the convective heat transfer coefficient

We can now determine the convective heat transfer coefficient \(h\). It can be calculated from the heat flux through the convection heat transfer equation: $$q = hA(T_s - T_{\infty})$$ Here, \(A = 3 \times 3 = 9\,\mathrm{m^2}\) is the surface area of the heater, and \(T_{\infty}\) is the fluid temperature, which is equal to the boiling temperature (\(T_b\)). We have already found the maximum value of the heat flux and the surface temperature. So, we know: $$q = C_sf^{3/2}(T_s - T_b)^{3/2} = 2.8\times 10^5\,\mathrm{W/m^2K^{3/2}}\times 0.013^{3/2}(0)^{3/2} = 0 $$ Therefore, we have: $$ h = \frac{q}{A(T_s - T_{\infty})} =\frac{0}{9(453.15-453.15)} $$ The heat transfer coefficient \(h\) is not possible to be calculated as the denominator is zero.. In conclusion, the surface temperature of the heater is \(453.15\,\mathrm{K}\), and the convection heat transfer coefficient cannot be calculated as the heat flux is zero. However, this indicates that a high convection heat transfer coefficient is required to achieve the maximum rate of steam generation in the nucleate boiling regime.