Chapter 10, Problem 10.37P

To determine

The rate of heat transfer per unit surface area from the strip to the wall jet.

The value of heat transfer rate per unit surface area is $0.0412\times {10}^{-6}\text{W}/{\text{m}}^2$ .

Given:

The diameter of the cylinder is $D=1\text{m}$ .

The emissivity of the blanket is $\epsilon =0.35$ .

The temperature of the vapor blanket $T_s=907\text{K}$ .

The saturated temperature of the water jet is $T_{\text{sat}}=100°\text{C}$ .

The value of the specific heat of vaporization at the temperature $373\text{K}$ is $h_{fg}=2257\times {10}^3\text{J}/\text{kg}$ .

The value of the specific volume of the saturated temperature of water is $v_f=1.044\times {10}^{-3}{\text{m}}^3/\text{kg}$ .

The density of the film temperature at $640\text{K}$ is ${\rho }_v=0.3434\text{kg}/{\text{m}}^3$ .

The thermal conductivity of vapor film temperature $640\text{K}$ is $k_v=0.0456\text{W}/\text{m}\cdot \text{K}$ .

The kinematic viscosity vapor at film temperature $640\text{K}$ is $v_v=64.50\times {10}^{-6}{\text{m}}^2/\text{s}$ .

The given diagram is shown in Figure 1

  

Figure 1

Formula used:

The expression for the density of the water is given by,

  $\rho =\frac{1}{v}$

The expression for the film temperature for the liquid water is given by,

  $T_f=\frac{T_s+T_{\text{sat}}}{2}$

The expression to determine the value of the excess temperature is given by,

  $\Delta T_e=T_s-T_{\text{sat}}$

The expression to determine the latent heat of vaporization is given by,

  ${h^{\prime }}_{fg}=h_{fg}+0.8c_{p,v}\left(T_s-T_{\text{sat}}\right)$

The expression for the value of the Nusslet number is given by,

  $\overline{N}{u}_D=C{\left[\frac{g\left({\rho }_l-{\rho }_v\right){h^{\prime }}_{fg}{D}^3}{v_v{k}_v\left(T_s-T_{\text{sat}}\right)}\right]}^{\frac{1}{4}}$

The expression to determine the value of the average radiation heat transfer coefficient is given by,

  ${\overline{h}}_{\text{conv}}=\overline{N}{u}_D\frac{k_v}{D}$

The expression for the radiation heat transfer coefficient is given by,

  ${\overline{h}}_{\text{rad}}=\frac{\epsilon \sigma \left(T_s^4-T_{\text{sat}}^4\right)}{T_s-T_{\text{sat}}}$

The expression to determine the value of the heat transfer coefficient is given by,

  $\overline{h}={\overline{h}}_{\mathrm{co}\text{nv}}+\frac{3}{4}{\overline{h}}_{\text{rad}}$

The expression to determine the value of initial heat transfer rate from the bar is given by,

  $q_s=\overline{h}\pi DL\left(T_{\text{sat}}-T_s\right)$

The expression for the volumetric heat generation rate is given by,

  $\dot{q}=\frac{4q_s}{D}$

Calculation:

The density of the water is calculated as,

  $\begin{array}{c}\rho =\frac{1}{v}\\ =\frac{1}{1.044\times {10}^{-3}{\text{m}}^3/\text{kg}}\\ =957.9\text{kg}/{\text{m}}^3\end{array}$

The film temperature for the liquid water is calculated as,

  $\begin{array}{c}{T}_f=\frac{T_s+T_{\text{sat}}}{2}\\ =\frac{907\text{K}+373\text{K}}{2}\\ =640\text{K}\end{array}$

The value of the excess temperature is calculated as,

  $\begin{array}{c}\Delta T_e=T_s-T_{\text{sat}}\\ =907\text{K}-373\text{K}\\ =\text{534}\text{K}\end{array}$

The latent heat of vaporization is calculated as,

  $\begin{array}{c}{h^{\prime }}_{fg}=h_{fg}+0.8c_{p,v}\left(T_s-T_{\text{sat}}\right)\\ =2257\times {10}^3\text{J}/\text{kg}+0.8\left(\mathrm{2..050}\times {10}^3\text{J}/\text{kg}\cdot \text{K}\right)\left(\text{534}\text{K}\right)\\ =3.13\times {10}^3\text{J}/\text{kg}\end{array}$

The value of the Nusslet number is calculated as,

  $\begin{array}{c}\overline{N}{u}_D=C{\left[\frac{g\left({\rho }_l-{\rho }_v\right){h^{\prime }}_{fg}{D}^3}{v_v{k}_v\left(T_s-T_{\text{sat}}\right)}\right]}^{\frac{1}{4}}\\ =0.62{\left[\frac{9.8\text{m}/{\text{s}}^2\left(959.7\text{kg}/{\text{m}}^3-0.3434\text{kg}/{\text{m}}^3\right)\left(3.13\times {10}^3\text{J}/\text{kg}\right){\left(1\text{m}\right)}^3}{\left(64.50\times {10}^{-6}{\text{m}}^2/\text{s}\right)\left(0.456\text{W}/\text{m}\cdot \text{K}\right)\left(907\text{K}-\text{373}\text{K}\right)}\right]}^{\frac{1}{4}}\\ =1290\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The value of the average radiation heat transfer coefficient is given by,

  $\begin{array}{c}{\overline{h}}_{\text{conv}}=\overline{N}{u}_D\frac{k_v}{D}\\ =\left(1290\text{W}/{\text{m}}^2\cdot \text{K}\right)\frac{\left(0.0456\text{W}/\text{m}\cdot \text{K}\right)}{\left(1\text{m}\right)}\\ =58.8\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The effective radiation heat transfer coefficient is calculated as,

  $\begin{array}{c}{\overline{h}}_{\text{rad}}=\frac{\epsilon \sigma \left(T_s^4-T_{\text{sat}}^4\right)}{T_s-T_{\text{sat}}}\\ =\frac{0.35\left(5.67\times {10}^{-8}\right)\left({\left(907\text{K}\right)}^4-{\left(373\text{K}\right)}^4\right)}{\text{2045}\text{K}-\text{373}\text{K}}\\ =24.42\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The value of the heat transfer coefficient is calculated as,

  $\begin{array}{c}\overline{h}={\overline{h}}_{\mathrm{co}\text{nv}}+\frac{3}{4}{\overline{h}}_{\text{rad}}\\ =58.8\text{W}/{\text{m}}^2\cdot \text{K}+\frac{3}{4}\left(24.42\text{W}/{\text{m}}^2\cdot \text{K}\right)\\ =77.11\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

The rate of heat transfer per unit surface area is calculated as,

  $\begin{array}{c}{q}_s=\overline{h}\left(T_{\text{sat}}-T_s\right)\\ =\left(77.11\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(907\text{K}-373\text{K}\right)\\ =0.0412\times {10}^{-6}\text{W}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the value of heat transfer rate per unit surface area is $0.0412\times {10}^{-6}\text{W}/{\text{m}}^2$ .