Chapter 10, Problem 10.19P

To determine

The estimate of the power required to maintain a brass plate at $\Delta T_c=15°\text{C}$ , also determine the power requirement if the water is pressurized at $10\text{atm}$ and determine at what fraction of the critical heat flux is the plate operating.

The power required to maintain a brass plate at a temperature of $15°\text{C}$ is $4.7\times {10}^3\text{kW}/{\text{m}}^{\text{2}}$ , also the power required to maintain the temperature of the brass at $15°\text{C}$ and $10\text{atm}$ is $23.8\times {10}^3\text{kW}/{\text{m}}^{\text{2}}$ . The critical heat flux of the plate operating at $15°\text{C}$ is $1.26\times {10}^3\text{kW}/{\text{m}}^2$ and the critical heat flux for the plate operating at $10\text{atm}$ is $2.97\times {10}^3\text{kW}/{\text{m}}^2$.

Given:

The temperature of the brass plate is $\Delta T_c=15°\text{C}$ .

The temperature of the saturated water is $1\text{atm}$ .

Formula Used:

From the thermo-physical properties of the saturated water, the temperature of the saturated water is given by,

  $T_{\text{sat}}=453.4\text{K}$

From the thermo-physical properties of the saturated water, the value of enthalpy of vaporization is given by,

  $h_{fc}=2012\text{kJ}/\text{kg}$

From the thermo-physical properties of the saturated water, the density of the water liquid is given by,

  ${\rho }_l=886.7\text{kg}/{\text{m}}^3$

From the thermo-physical properties of the saturated water, the density of the saturated vapor is given by,

  ${\rho }_v=5.155\text{kg}/{\text{m}}^3$

From the thermo-physical properties of the saturated water, the expression for the dynamic viscosity of the liquid is given by,

  ${\mu }_l=149\times {10}^{-6}\text{N}\cdot \text{s}/{\text{m}}^2$

From the thermo-physical properties of the saturated water, the Prandtl number of the saturated liquid is given by,

  ${\text{Pr}}_l=0.98$

From the thermo-physical properties of the saturated water, the specific heat of the liquid is given by,

  $c_{p,l}=4410\text{J}/\text{kg}\cdot \text{K}$

From the thermo-physical properties of the saturated water, the surface tension of the liquid is given by,

  $\sigma =42.2\times {10}^{-3}\text{N}/\text{m}$

The expression for the nucleate boiling heat flux(power) at $1\text{atm}$ is given by,

  ${q^{″}}_s={\mu }_l{h}_{fg}{\left[\frac{g\left({\rho }_l-{\rho }_v\right)}{\sigma }\right]}^{\frac{1}{2}}{\left[\frac{c_{p,l}\Delta T_c}{C_{s,f}{h}_{fg}p{r}_l^n}\right]}^3$

The expression to determine the heat flux is given by,

  ${q^{″}}_{\text{max}}=C{h}_{fg}{\rho }_v{\left[\frac{\sigma g\left({\rho }_l-p_v\right)}{{\rho }_v^2}\right]}^{\frac{1}{4}}$

Calculation:

The power required to maintain a brass plate at $15°\text{C}$ excess temperature and $1\text{atm}$ is calculated as,

  $\begin{array}{c}{q^{″}}_s={\mu }_l{h}_{fg}{\left[\frac{g\left({\rho }_l-{\rho }_v\right)}{\sigma }\right]}^{\frac{1}{2}}{\left[\frac{c_{p,l}\Delta T_c}{C_{s,f}{h}_{fg}p{r}_l^n}\right]}^3\\ =\left[\begin{array}{l}\left(279\times {10}^{-6}\text{N}\cdot \text{s}/{\text{m}}^2\right)\left(2257\text{kJ}/\text{kg}\right){\left[\frac{9.8\text{m}/{\text{s}}^2\left(957.9\text{kg}/{\text{m}}^3-0.596\text{kg}/{\text{m}}^3\right)}{58.9\times {10}^{-3}\text{N}/\text{m}}\right]}^{\frac{1}{2}}\\ {\left[\frac{4217\text{kJ}/\text{kg}\cdot \text{K}\times 15°\text{C}}{0.006\times 2257\text{kJ}/\text{kg}\times {10}^3\times {1.76}^1}\right]}^3\end{array}\right]\\ =4.7\times {10}^3\text{kW}/{\text{m}}^{\text{2}}\end{array}$

The power required to maintain the boiling plate at $15°\text{C}$ and at a pressure of $10\text{atm}$ is calculated as,

  $\begin{array}{c}{q^{″}}_s={\mu }_l{h}_{fg}{\left[\frac{g\left({\rho }_l-{\rho }_v\right)}{\sigma }\right]}^{\frac{1}{2}}{\left[\frac{c_{p,l}\Delta T_c}{C_{s,f}{h}_{fg}p{r}_l^n}\right]}^3\\ =\left[\begin{array}{l}\left(149\times {10}^{-6}\text{N}\cdot \text{s}/{\text{m}}^2\right)\left(2012\text{kJ}/\text{kg}\right){\left[\frac{9.8\text{m}/{\text{s}}^2\left(886.7\text{kg}/{\text{m}}^3-5.155\text{kg}/{\text{m}}^3\right)}{42.2\times {10}^{-3}\text{N}/\text{m}}\right]}^{\frac{1}{2}}\\ {\left[\frac{4410\text{kJ}/\text{kg}\cdot \text{K}\times 15°\text{C}}{0.006\times 2012\text{kJ}/\text{kg}\times {10}^3\times {0.98}^1}\right]}^3\end{array}\right]\\ =23.8\times {10}^3\text{kW}/{\text{m}}^{\text{2}}\end{array}$

The value of the constant $C$ for long and horizontal plates is $0.149$ , the value of liquid copper surface combination $C_{s,f}$ is $0.0128$ and value of $n$ is $1$ .

The critical heat flux of the plate operating at $15°\text{C}$ is calculated as,

  $\begin{array}{c}{q^{″}}_{\text{max}}=C{h}_{fg}{\rho }_v{\left[\frac{\sigma g\left({\rho }_l-p_v\right)}{p_v^2}\right]}^{\frac{1}{4}}\\ =\left[\begin{array}{l}{\left[\frac{58.9\times {10}^{-3}\text{N}/\text{m}\left(9.8\text{m}/{\text{s}}^2\right)\left(957.9\text{kg}/{\text{m}}^3-0.596\text{kg}/{\text{m}}^3\right)}{{\left(0.596\text{kg}/{\text{m}}^3\right)}^2}\right]}^{\frac{1}{4}}\left(0.149\right)\\ \left(2257\text{kJ}/\text{kg}\right)\left(0.596\text{kg}/{\text{m}}^3\right)\end{array}\right]\\ =1.26\times {10}^3\text{kW}/{\text{m}}^2\end{array}$

The critical heat flux of the plat operating at $10\text{atm}$ is calculated as,

  $\begin{array}{c}{q^{″}}_{\text{max}}=C{h}_{fg}{\rho }_v{\left[\frac{\sigma g\left({\rho }_l-p_v\right)}{p_v^2}\right]}^{\frac{1}{4}}\\ =\left[\begin{array}{l}{\left[\frac{42.2\times {10}^{-3}\text{N}/\text{m}\left(9.8\text{m}/{\text{s}}^2\right)\left(886.7\text{kg}/{\text{m}}^3-5.155\text{kg}/{\text{m}}^3\right)}{{\left(5.155\text{kg}/{\text{m}}^3\right)}^2}\right]}^{\frac{1}{4}}\left(0.149\right)\\ \left(2012\text{kJ}/\text{kg}\right)\left(5.155\text{kg}/{\text{m}}^3\right)\end{array}\right]\\ =2.97\times {10}^3\text{kW}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the power required to maintain a brass plate at a temperature of $15°\text{C}$ is $4.7\times {10}^3\text{kW}/{\text{m}}^{\text{2}}$ , also the power required to maintain the temperature of the brass at $15°\text{C}$ and $10\text{atm}$ is $23.8\times {10}^3\text{kW}/{\text{m}}^{\text{2}}$ . The critical heat flux of the plate operating at $15°\text{C}$ is $1.26\times {10}^3\text{kW}/{\text{m}}^2$ and the critical heat flux for the plate operating at $10\text{atm}$ is $2.97\times {10}^3\text{kW}/{\text{m}}^2$.