Two-Phase Heat Transfer

Nucleation superheat

Phase change: liquid -> gas / gas -> liquid

• Homogeneous nucleation 均匀成核 (A-B')
• Heterogeneous nucleation 异相成核 (A-B-C-D)

Specific volume = density${}^{-1}$

Clausius-Clapeyron relation [final-term]

Mechanical equilibrium:

$$(p_b-p_l)\pi r^{\ast 2}=\sigma 2\pi r^{\ast }\text{ or }\mathrm{\Delta }p=p_b-p_l=\frac{2\sigma }{r^{\ast }}$$

• $p_b$: Internal pressure in bubble
• $p_l$: Pressure outside the bubble
• $r^{\ast }$: radius of bubble
• $\sigma$: coefficient of surface tension 表面张力

Clausius-Clapeyron relation b/w $p_{\text{sat}}$ & $T_{\text{sat}}$ (where $h_{\mathrm{fg}}$ is Latent heat of phase change 相变潜热 / enthalpy of vaporization):

The Clausius-Clapeyron equation specifies the temperature dependencies of pressure, most importantly, vapor pressure, at a discontinuous phase transition b/w two phase of matter of a single contituent

$$\begin{array}{rl}& {\left(\frac{\mathrm{d}p}{\mathrm{d}T}\right)}_{\text{sat }}=\frac{h_{\mathrm{fg}}}{T_{\text{sat }}({\nu }_{\mathrm{g}}-{\nu }_{\mathrm{f}})}\\ \underset{{\nu }_{\mathrm{g}}\gg {\nu }_{\mathrm{f}}}{\overset{\text{ Saturated vapor }}{\to }}& \frac{\mathrm{d}{p}_{\mathrm{g}}}{\mathrm{d}{T}_{\mathrm{g}}}=\frac{h_{\mathrm{fg}}}{T_{\mathrm{g}}\left({\nu }_{\mathrm{g}}\right)}\stackrel{p_{\mathrm{g}}{\nu }_{\mathrm{g}}=R{T}_{\mathrm{g}}}{\to }\frac{\mathrm{d}{p}_{\mathrm{g}}}{p_{\mathrm{g}}}=\frac{h_{\mathrm{fg}}}{R{T}_{\mathrm{g}}^2}\mathrm{d}{T}_{\mathrm{g}}\\ & \ln \left(\frac{p_{\mathrm{b}}}{p_l}\right)=-\frac{h_{\mathrm{fg}}}{R}(\frac{1}{T_{\mathrm{b}}}-\frac{1}{T_{\text{sat }}})\\ & T_{\mathrm{b}}-T_{\text{sat }}=\frac{R{T}_{\mathrm{b}}{T}_{\text{sat }}}{h_{\mathrm{fg}}}\ln \left(\frac{p_{\mathrm{b}}}{p_l}\right)\\ \underset{T_{\text{sat }}:\text{ Saturation temperature at }{p}_{\mathrm{l}}}{\overset{T_{\mathrm{b}}:\text{ Saturation temperature at }{p}_{\mathrm{b}}}{\to }}& T_{\mathrm{b}}-T_{\text{sat }}=\frac{R{T}_{\mathrm{b}}{T}_{\text{sat }}}{h_{\mathrm{fg}}}\ln (1+\frac{2\sigma }{p_l{r}^{\ast }})\end{array}$$

Superheat ($=T_{\mathrm{b}}-T_{\text{sat }}$ ) required to maintain a bubble

when $2\sigma /p_l{r}^{\ast }\ll 1,p_{\mathrm{b}}\simeq p_l.R{T}_{\mathrm{b}}/p_{\mathrm{b}}={\nu }_{\mathrm{b}}\simeq {\nu }_{\mathrm{fg}}$ .

$$T_{\mathrm{b}}-T_{\text{sat }}\simeq \frac{R{T}_{\mathrm{b}}{T}_{\text{sat}}2\sigma }{h_{\mathrm{fg}}{p}_{\mathrm{b}}{r}^{\ast }}\simeq \frac{2\sigma T_{\text{sat }}{\nu }_{\mathrm{fg}}}{h_{\mathrm{fg}}}\left(\frac{1}{r^{\ast }}\right)p_l\ll p_{\mathrm{c}}$$

where $p_c$ is Critical pressure.

Superheat for homogeneous nucleation of water at $p=0.1\mathrm{MPa}:{220}^{\circ }\mathrm{C}$

Heterogeneous nucleation: dissolved gas & microcavities at surface

Pool boiling

• Boiling curve: Heat-transfer regime of pool boiling
• Critical heat flux: Linked with onset of pool fluidization (suspension of liquid by vapor stream)

Hydrodynamic instability:

• Kelvin- Helmholtz instability: 2 fluid flow counterpart: critical heat flux =...

$$q_{\mathrm{cr}}^{\prime \prime }={\rho }_g{h}_{\mathrm{fg}}{j}_g=C_1{\rho }_g{h}_{\mathrm{fg}}{\left[\frac{\sigma ({\rho }_f-{\rho }_g)g}{{\rho }_g^2}\right]}^{1/4}$$

where $C_1=0.13$.

Flow boiling

• Heat transfer regime: Mass flow rate, fluids, geometry, heat flux magnitude and distribution
• Dry-out type CHF: High quality
• DNB (departure from nucleate boiling) type CHF (similar to CHF in pool boiling): Low quality

Enthalpy 焓 [$\mathrm{J}/\mathrm{kg}$] at $z$ from inlet [core]:

$$h=h_{\text{l,in}}+\frac{4q_w^{″}z}{DG}$$

where:

• $q_w^{″}$: heat flux
• $z$: height w.r.t. the inlet
• $D$: diameter of pipe
• $G={\rho }_g⟨j_g⟩+{\rho }_f⟨j_f⟩={\rho }_m{v}_m$: mass flux

$\frac{4q_w^{″}z}{DG}$ comes from

1. The area of pipe wall: $z\pi D$
2. The whole heat: $\mathrm{\Delta }Q=q_w^{″}\cdot z\pi D$
3. The enthalpy increase: $\mathrm{\Delta }Q/W=\mathrm{\Delta }Q/(G\cdot \pi D^2/4)=\frac{4q_w^{″}z}{DG}$

definition: Thermal equilibrium

Two physical systems are in thermal equilibrium if there is no net flow of thermal energy between them when they are connected by a path permeable to heat (From wikipedia)

definition: Thermal equilibrium quality

Thermal equilibrium quality: ratio of the actual enthalpy of a fluid mixture to the enthalpy if it were completely vaporized, assuming the mixture is in thermal equilibrium at the local pressure.

Thermal equilibrium quality at $z$ from inlet [core]:

$$x_{eq}=\frac{h-h_{l,\text{sat}}}{h_{fg}}=\frac{h_{\text{l,in}}+\frac{4q_w^{″}z}{DG}-h_{l,\text{sat}}}{h_{fg}}$$

• $h_{l,\text{sat}}$: Saturated enthalpy
• $h_{fg}$: enthalpy of vaporization
• When $h=h_{fg}+h_{l,\text{sat}},x_{eq}=1$

For boiling curve:

• single-phase flow: $q^{″}=h(T_w-T_{\text{sat}})$
• The corner: $(T_w-T_{\text{sat}}{)}^{m-1}$
• 2-phase flow: $q^{″}=h(T_w-T_{\text{sat}}{)}^m,m\simeq 3$

Subcooled boiling

1. Region I: Subcooled nucleate boiling (Onset of nucleate boiling (核态沸腾起始点), $Z_{\mathrm{NB}}$: mean or bulk temperature below $T_{\text{sat}}$)
2. Region II: Subcooled nucleate boiling (Onset of significant void (显着空洞), $Z_{\mathrm{D}}$)
3. Region III: Saturated boiling ($x_{eq}=0$: since liquid does not reach saturation condition due to the bubble presence and non-equilibrium)
4. Region IV: Saturated boiling (thermal equilibrium condition at $Z_{\mathrm{E}}$)

• For $Z_{\mathrm{D}}$, that's Saha-Zuber correlation to determine
• For the curve: profile-fit method

Net vapor generation (Bubble departure)

Net vapor generation: The point at which bubbles can depart from the wall before they suffer condensation ($Z_{\mathrm{D}}$)

在流体中，蒸汽的“产生”速度开始大于“凝结”速度，从而出现净增的蒸汽

• Thermally controlled departure: The wall heat flux is balanced by heat removal due to liquid subcooling at $Z_{\mathrm{D}}$.
• Hydrodynamically controlled departure: The bubble detachment is primarily the result of drag (or shear) force overcoming the surface tension force

Saha-Zuber correlation: (empirical)

At first, introduce Peclet number [core]:

$$\begin{array}{rl}\mathrm{St}=\frac{\mathrm{Nu}}{\mathrm{Re}\mathrm{Pr}}\sim \mathrm{Pe}& =\frac{\mathrm{Nu}}{\mathrm{St}}\\ ⟹& =\frac{G{D}_n{C}_{pf}}{k_f}\end{array}$$

where:

• $C_{pf}$: Saturated liquid isobaric heat capacity
• $k_f$: Saturated liquid thermal conductivity

no use

$$x_{eq}(Z_D)=\{\begin{array}{ll}-0.0022\frac{q^{″}{D}_n{c}_{pf}}{k_f{h}_{fg}}& \mathrm{Pe}\le 70000\\ -153.85\frac{q^{″}}{G{h}_{fg}}& \mathrm{Pe}\ge 70000\end{array}$$

For temperature:

$$T_{\text{sat}}-T_D=\{\begin{array}{ll}0.0022\frac{q^{″}{D}_n}{k_f}& \mathrm{Pe}\le 70000\\ 153.85\frac{q^{″}}{G{c}_{pl}}& \mathrm{Pe}\ge 70000\end{array}$$

and $Z_D$ is calculated as:

$$Z_D=\frac{\frac{\pi D_h^2}{4}G{C}_{pl}(T_D-T_{in})}{\pi D_h{q}_w^{\prime \prime }}=\frac{D_{h}G{C}_{pl}(T_D-T_{in})}{4q_w^{\prime \prime }}$$

• 分子：将总质量流量的流体从 $T_{in}$ 显热加热到 $T_D$ 所需要的总功率
• 分母：单位长度加热功率

Subcooled flow quality and void fraction

Profile-fit approach: Currently accepted approach for determining the flow quality in subcooled region [core]

$$x(z)=x_{eq}(z)-x_{eq}(Z_{\mathrm{D}})\exp [\frac{x_{eq}(z)}{x_{eq}(Z_{\mathrm{D}})}-1]$$

The flow quality $x(z)$:

• $=0$ at $Z_{\mathrm{D}}$
• approaches $x_{eq}(z)$ asymptotically as $z$ increases
• When $x_{eq}(z)=0,x(z)=-\frac{1}{e}{x}_{eq}(Z_D)\approx -0.368x_{eq}(Z_D)$

For quality:

$$x\ne x_{eq}$$

• well-saturated boiling flow:

$$x\approx x_{eq}$$

• Subcooled boiling flow:

$$x\ge 0,x_{eq}<0$$

Using profile-fit approach to obtain $x_{eq}$ How to determine $Z_{\mathrm{D}}$: Profile-fit method

pump $\to$

• Bernoulli's Equation
  • Major loss
  • Minor loss

Mission: prediction of void fraction for heat transfer, etc.

1. To predict the void fraction $⟨\alpha ⟩$, the Drift-flux model is used (need $⟨x⟩$)

Drift-flux model:

$$⟨\alpha ⟩=\frac{⟨j_g⟩}{C_0(⟨j_g⟩+⟨j_f⟩)+⟨⟨v_{fg}⟩⟩}$$

where

$$⟨j_g⟩=\frac{G⟨x_f⟩}{{\rho }_g},⟨j_f⟩=\frac{G(1-⟨x_f⟩)}{{\rho }_f}$$

and $x_f$ is the flow quality

• For HEM, $⟨\alpha ⟩=⟨j_g⟩/(⟨j_g⟩+⟨j_f⟩)$ and 20% overprediction

2. Distribution parameter $C_0$ and drift velocity $⟨⟨v_{fg}⟩⟩$ can be obtained by formulas in the last chapter
3. To obtain the flow quality $⟨x_f⟩$, Profile-fit approach ($x_f\sim x_{eq}$) and OSV model (onset of significant void) are used $(\text{need }{x}_{eq}(z),x_{eq}(Z_D))$
   1. To obtain $x_{eq}$, Thermal equilibrium quality is used
   2. To obtain $x_{eq}(Z_D)$, the Saha-Zuber correlation is used

Saturated Cooling

Heat transfer:

• $1\varphi :q^{″}=h_{1\varphi }(T_w-T_{\mathrm{bulk}})$
  • $\mathrm{Nu}=\frac{h_{1\varphi }D}{k_f}$ where $k_f$ is the thermal conductivity
  • Dittus-Boelton equation for $1\varphi$ convective heat transfer (Turbulent flow):$$\mathrm{Nu}=0.023{\mathrm{Re}}^{0.8}{\mathrm{Pr}}^m,m=0.3(\text{cooling}),0.4(\text{heating})$$
  • $\mathrm{Re}={\rho }_f{j}_{f}D/{\mu }_f$ where $j_f=G(1-x)/{\rho }_f$
• $2\varphi :q^{″}=h_{2\varphi }(T_w-T_{\text{sat}})$ (Saturated boiling flow)
  • $h_{2\varphi }=\underset{\text{nucleate boiling}}{\underbrace{h_{\mathrm{NB}}}}+\underset{\text{forced convective flow}}{\underbrace{h_C}}(\text{Chen correlation})$

Boling incipience

Fourier's law:

$$q^{″}=-k_l\frac{\partial T}{\partial r}\sim k_l\frac{T_b-T_{\text{sat}}}{r^{\ast }}$$

Clausius-Clapeyron relation:

$$T_w=T_{\text{sat}}+\frac{2\sigma T_{\text{sat}}{v}_{fg}}{h_{fg}}\frac{1}{r^{\ast }}.$$

Thus

$$q^{″}=\frac{k_l{h}_{fg}}{8\sigma T_{\text{sat}}{v}_{fg}}(T_w-T_{\text{sat}}{)}^2.$$

Quiz3

Water is injected with a velocity of $1.5\mathrm{m}/\mathrm{s}$ into a uniformly heated vertical pipe with a diameter of $0.05\mathrm{m}$ and length of $10\mathrm{m}$ receiving a heat flux of $5\times {10}^6\mathrm{W}/{\mathrm{m}}^2$. Inlet pressure and temperature of water are $4.64\mathrm{MPa}$ and ${25}^{\circ }\mathrm{C}$. Saturation temperature, $T_{\text{sat}}$, at $4.64\mathrm{MPa}$ is ${259}^{\circ }\mathrm{C}$. Saturated liquid enthalpy, $h_{l,sat}$, and enthalpy of vaporization, $h_{fg}$, are $1132\mathrm{kJ}/\mathrm{kg}$ and $1665\mathrm{kJ}/\mathrm{kg}$. Saturated liquid density and saturated steam density are $785\mathrm{kg}/{\mathrm{m}}^3$ and $23.4\mathrm{kg}/{\mathrm{m}}^3$. Enthalpy of water at $4.64\mathrm{MPa}$ and ${25}^{\circ }\mathrm{C}$ is $123\mathrm{kJ}/\mathrm{kg}$. Density of the saturated liquid can be used as a good estimation of density of subcooled water. Other average fluid properties are $C_{pl}=4980\mathrm{J}/\mathrm{kg}\cdot \mathrm{K},k_l=0.570\mathrm{W}/\mathrm{m}\cdot \mathrm{K}$, ${\mu }_l=9.4\times {10}^{-5}\mathrm{kg}/(\mathrm{m}\cdot \mathrm{s})$. Surface tension at the saturation pressure can be used approximately for the calculation $(\sigma =0.0329\mathrm{N}/\mathrm{m})$.

1. Calculate the mixture mass flux (unit in $\mathrm{kg}/{\mathrm{m}}^2\mathrm{s}$). In the uniformly heated pipe:

$$G={\rho }_{\text{l,in}}{v}_{\text{l,in}}=785\times 1.5\mathrm{kg}/{\mathrm{m}}^2\mathrm{s}=1177.5\mathrm{kg}/{\mathrm{m}}^2\mathrm{s}.$$

$1119-1236$

2. Calculate the enthalpy at $4\mathrm{m}$ from the pipe inlet (unit in $\mathrm{kJ}/\mathrm{kg}$).

$$h=h_{\text{l,in}}+\frac{4q_w^{″}z}{DG}=(123+\frac{4\times 5\times {10}^3\times 4}{0.05\times 1177.5})\mathrm{kJ}/\mathrm{kg}\approx 1481.811\mathrm{kJ}/\mathrm{kg}.$$

$1407-1555$

3. Calculate the thermal equilibrium quality located at the $4\mathrm{m}$ from the pipe inlet. Thermal equilibrium quality is calculated as

$$x_e(z){|}_{z=4}=\frac{h(z){|}_{z=4}-h_{l,\text{sat}}}{h_{fg}}=\frac{1481.811-1132}{1665}\approx 0.210.$$

$0.2$

4. Calculate the Peclet number for the inlet liquid.

$$\mathrm{Pe}=\frac{G{C}_{pl}D}{k_l}=\frac{1177.5\times 4980\times 0.05}{0.570}\approx 5.143816\times {10}^5$$

$488680-540120$

5. Calculate the temperature for onset of significant void, $T_D$, (unit in $\mathrm{K}$). According to Saha-Zuber correlation, when $\mathrm{Pe}>70000$:

$$T_{\text{sat}}-T_D=153.85\frac{q^{″}}{G{C}_{pl}}=153.85\times \frac{5\times {10}^6}{1177.5\times 4980}\mathrm{K}\approx 131.183\mathrm{K}$$

Thus $T_D\approx T_{\text{sat}}-131.3\mathrm{K}\approx (259+273.15-131.183)\mathrm{K}\approx 400.967\mathrm{K}$

$381-421$

6. Calculate the length at which onset of significant void occurs, $Z_D$, (unit in $\mathrm{m}$).

$$Z_D=\frac{\frac{\pi D_h^2}{4}G{C}_{pl}(T_D-T_{in})}{\pi D_h{q}_w^{\prime \prime }}=\frac{D_{h}G{C}_{pl}(T_D-T_{in})}{4q_w^{\prime \prime }}\approx 1.507\mathrm{m}.$$

$1.43-1.59$

7. Calculate the equilibrium quality at $Z_D,x_e(Z_D)$.

Based on Saha-Zuber correlation, when $\mathrm{Pe}>70000$: (Not right while using this formula)

$$x_e(Z_D)=-153.85\frac{q^{″}}{G{h}_{fg}}=-153.85\times \frac{5\times {10}^6}{1177.5\times 1665\times {10}^3}\approx -0.392.$$

$$\begin{array}{c}h{|}_{z=Z_D}=h_{\text{l,in}}+\frac{4q_w^{″}{Z}_D}{DG}\approx (123+\frac{4\times 5\times {10}^3\times 1.507}{0.05\times 1177.5})\mathrm{kJ}/\mathrm{kg}\approx 634.392\mathrm{kJ}/\mathrm{kg}\\ x_e(Z_D)=\frac{h{|}_{z=Z_D}-h_{l,\text{sat}}}{h_{fg}}\approx -0.299.\end{array}$$

$-0.313--0.283$

8. Calculate the quality at $Z=2\mathrm{m},x(Z=2\mathrm{m})$. The quality is calculated based on the profile-fit approach:

$$x(z){|}_{z=2\mathrm{m}}=x_e(z){|}_{z=2\mathrm{m}}-x_e(Z_{\mathrm{D}})\exp [\frac{x_e(z){|}_{z=2\mathrm{m}}}{x_e(Z_{\mathrm{D}})}-1]$$

where

$$x_e(z){|}_{z=2\mathrm{m}}=\frac{h_{l,\mathrm{in}}+\frac{4q^{″}z{|}_{z=2\mathrm{m}}}{GD}-h_{l,\text{sat}}}{h_{fg}}\approx -0.198$$

Thus

(Wrong answer in 7th leads to wrong answer here)

$$x(z){|}_{z=2\mathrm{m}}\approx -0.198-(-0.392)\times \exp [\frac{-0.198}{-0.392}-1]\approx 0.0411$$

$$x(z){|}_{z=2\mathrm{m}}\approx -0.198-(-0.299)\times \exp [\frac{-0.198}{-0.299}-1]\approx 0.0153.$$

$0.0143-0.0158$

9. Calculate the distribution parameter, $C_0$, at $Z=2\mathrm{m}$. Given:

$$\begin{array}{rl}& C_0=\beta [1+(1/\beta -1{)}^b]\\ & b=({\rho }_g/{\rho }_f{)}^{0.1}\\ & \beta =(x/{\rho }_g)/(x/{\rho }_g+(1-x)/{\rho }_f)\end{array}$$

According to the given formula:

Wrong answer

$$\begin{array}{rl}& \beta \approx \frac{0.0411}{23.4}/(\frac{0.0411}{23.4}+\frac{1-0.0411}{785})\approx 0.590\\ & b=(23.4/785{)}^{0.1}\approx 0.704\\ & C_0\approx 0.590\times [1+(1/0.590-1{)}^{0.704}]\approx 1.047\end{array}$$

$$\begin{array}{rl}& \beta \approx \frac{0.0153}{23.4}/(\frac{0.0153}{23.4}+\frac{1-0.0153}{785})\approx 0.342\\ & b=(23.4/785{)}^{0.1}\approx 0.704\\ & C_0\approx 0.342\times [1+(1/0.342-1{)}^{0.704}]\approx 0.884\end{array}$$

$0.837-0.925$

10. Calculate the void fraction at $Z=2\mathrm{m},\alpha (Z=2\mathrm{m})$.

The void fraction is calculated by

$$⟨\alpha ⟩=\frac{⟨j_g⟩}{C_0⟨j⟩+⟨⟨v_{gj}⟩⟩}$$

where

Wrong answer

$$\begin{array}{rl}& j_g=\frac{Gx}{{\rho }_g}\approx \frac{1177.5\times 0.0411}{23.4}\mathrm{m}/\mathrm{s}\approx 2.068\mathrm{m}/\mathrm{s}\\ & j_f=\frac{G(1-x)}{{\rho }_f}\approx \frac{1177.5\times (1-0.0411)}{785}\mathrm{m}/\mathrm{s}\approx 1.438\mathrm{m}/\mathrm{s}\\ & j=j_g+j_f\approx (2.068+1.438)\mathrm{m}/\mathrm{s}\approx 3.507\mathrm{m}/\mathrm{s}.\end{array}$$

$$\begin{array}{rl}& j_g=\frac{Gx}{{\rho }_g}\approx \frac{1177.5\times 0.153}{23.4}\mathrm{m}/\mathrm{s}\approx 0.769\mathrm{m}/\mathrm{s}\\ & j_f=\frac{G(1-x)}{{\rho }_f}\approx \frac{1177.5\times (1-0.153)}{785}\mathrm{m}/\mathrm{s}\approx 1.477\mathrm{m}/\mathrm{s}\\ & j=j_g+j_f\approx (0.769+1.477)\mathrm{m}/\mathrm{s}\approx 2.246\mathrm{m}/\mathrm{s}\\ & ⟨⟨v_{gj}⟩⟩=2.9{\left(\frac{\mathrm{\Delta }\rho g\sigma }{{\rho }_f^2}\right)}^{0.25}=2.9\times {\left[\frac{(785-23.4)\times 9.8\times 0.0329}{{785}^2}\right]}^{0.25}\mathrm{m}/\mathrm{s}\approx 0.410\mathrm{m}/\mathrm{s}.\end{array}$$

Thus

Wrong answer

$$⟨\alpha ⟩\approx \frac{0.769}{0.884\times 2.246+0.410}\approx 0.507.$$

$$⟨\alpha ⟩\approx \frac{2.068}{1.047\times 3.507+0.410}\approx 0.321.$$

$0.302-0.334$

Quiz4

Liquid water with the mass flux $(G)$ of $2000\mathrm{kg}/{\mathrm{m}}^2/\mathrm{s}$ is injected into a uniformly heated vertical pipe at pressure ( $P$ ) condition of $15.5\mathrm{MPa}$ and inlet temperature ( $T$ ) of ${300}^{\circ }\mathrm{C}$. The pipe has the diameter $\left(D_h\right)$ of $0.05\mathrm{m}$ and height of $15\mathrm{m}$ . It is heated by the heat source through the outside wall with a heat flux $\left(q_w^{\prime \prime }\right)$ of $8\times {10}^5\mathrm{W}/{\mathrm{m}}^2$.

Key information are given below:

• Mass flux: $G=2000\mathrm{kg}/({\mathrm{m}}^2\cdot \mathrm{s})$;
• Pressure condition: $P=15.5\mathrm{MPa}$;
• Inlet temperature condition: $T_{\text{in }}={300}^{\circ }\mathrm{C}$;
• Saturation temperature: $T_{\text{sat }}={345}^{\circ }\mathrm{C}$;
• Liquid enthalpy at inlet: $h_{\mathrm{i}\mathrm{n}}=1338\mathrm{kJ}/\mathrm{kg}$;
• Saturated liquid enthalpy: $h_{\mathrm{s}\mathrm{a}\mathrm{t},f}=1623\mathrm{kJ}/\mathrm{kg}$;
• Saturated gas enthalpy: $h_{\mathrm{s}\mathrm{a}\mathrm{t},g}=2599\mathrm{kJ}/\mathrm{kg}$;
• Latent heat: $h_{fg}=976\mathrm{kJ}/\mathrm{kg}$;
• Saturated liquid density: ${\rho }_f=598\mathrm{kg}/{\mathrm{m}}^3$;
• Saturated liquid dynamic viscosity: ${\mu }_f=6.88\times {10}^{-5}\mathrm{Pa}\cdot \mathrm{s}$;
• Saturated liquid thermal conductivity: $k_f=0.452\mathrm{W}/\mathrm{m}/\mathrm{K}$;
• Saturated liquid isobaric heat capacity: $C_{pf}=8.74\mathrm{kJ}/\mathrm{kg}/\mathrm{K}$;
• Saturated gas density: ${\rho }_g=101\mathrm{kg}/{\mathrm{m}}^3$;
• Saturated gas dynamic viscosity: ${\mu }_g=2.31\times {10}^{-5}\mathrm{Pa}\cdot \mathrm{s}$;
• Steam-water surface tension: $\sigma =0.0047\mathrm{N}/\mathrm{m}$;
• Steel-water contact angle: ${\varphi }_s={38}^{\circ }$.

Assumptions: (1) Ignore the pressure loss in two-phase mixing section. (2) Air and water fluid compressibility is negligible. (3) Water properties change with temperature is negligible.

1. Assume that water properties do not vary significantly in single-phase flow section. Calculate the liquid convective heat transfer coefficient using the Dittus-Boelter equation.

$$\begin{array}{rl}& {\mathrm{N}\mathrm{u}}_f=0.023{\mathrm{R}\mathrm{e}}_f^{0.8}{\mathrm{P}\mathrm{r}}_f^{0.4}\\ & {\mathrm{R}\mathrm{e}}_f=\left(G{D}_h\right)/{\mu }_f\\ & {\mathrm{P}\mathrm{r}}_f=\left(C_{pf}{\mu }_f\right)/k_f\\ & {\mathrm{N}\mathrm{u}}_f=\left(h_f{D}_h\right)/k_f\end{array}$$

The answer of $h_f$ should be given in $\mathrm{W}/({\mathrm{m}}^2\mathrm{K})$.

Answer:

$$\begin{array}{rl}& {\mathrm{Nu}}_f=0.023{\mathrm{R}\mathrm{e}}_f^{0.8}{\mathrm{Pr}}_f^{0.4}=0.023{\left(\frac{G{D}_h}{{\mu }_f}\right)}^{0.8}{\left(\frac{C_{pf}{\mu }_f}{k_f}\right)}^{0.4}=\frac{h_f{D}_h}{k_f}\\ ⟹& h_f=0.023\frac{k_f}{D_h}{\left(\frac{G{D}_h}{{\mu }_f}\right)}^{0.8}{\left(\frac{C_{pf}{\mu }_f}{k_f}\right)}^{0.4}\approx 1.98340\times {10}^4\mathrm{W}/({\mathrm{m}}^2\cdot \mathrm{K}).\end{array}$$

$18810-20790$

2. What is the temperature difference between the wall temperature and mean bulk liquid temperature (wall super heat [${}^{\circ }\mathrm{C}$])?

$$q_x^{″}=h\mathrm{\Delta }T$$

The answer of $\mathrm{\Delta }T$ should be given in ${}^{\circ }\mathrm{C}$.

Answer:

$$q_w^{″}=h_f\mathrm{\Delta }T⟹\mathrm{\Delta }T=\frac{q_w^{″}}{h_f}\approx {40.335}^{\circ }\mathrm{C}.$$

$38.4-42.4$

3. Determine the temperature for the onset of nucleate boiling, $T_{\mathrm{ONB}}$, using Basu et al. (2004) correlation.

$$\mathrm{\Delta }{T}_{\mathrm{ONB}}=\frac{\sqrt{2}}{F({\varphi }_s)}{\left(\frac{\sigma T_{\text{sat}}{q}_w^{″}}{{\rho }_g{h}_{fg}{k}_f}\right)}^{0.5}$$

where

$$F({\varphi }_s)=1-\exp [-{\left(\frac{\pi {\varphi }_s}{180}\right)}^3-0.5\left(\frac{\pi {\varphi }_s}{180}\right)],{\varphi }_s={38}^{\circ }$$

$T_{\text{sat}}$ in the correlation should be used in the unit $\mathrm{K}$. The answer of $T_{\text{ONB}}$ should be given in ${}^{\circ }\mathrm{C}$.

Answer:

$$\begin{array}{rl}& \mathrm{\Delta }{T}_{\text{ONB}}=T_{\text{ONB}}-T_{\text{sat}}=\frac{\sqrt{2}}{F({\varphi }_s)}{\left(\frac{\sigma T_{\text{sat}}{q}_w^{″}}{{\rho }_g{h}_{fg}{k}_f}\right)}^{0.5}\\ ⟹& T_{\text{ONB}}=T_{\text{sat}}+\frac{\sqrt{2}}{F({\varphi }_s)}{\left(\frac{\sigma T_{\text{sat}}{q}_w^{″}}{{\rho }_g{h}_{fg}{k}_f}\right)}^{0.5}\end{array}$$

where

$$F({\varphi }_s){|}_{{\varphi }_s={38}^{\circ }}=1-\exp [-{\left(\frac{\pi {\varphi }_s}{180}\right)}^3-0.5\left(\frac{\pi {\varphi }_s}{180}\right)]\approx 0.464.$$

Thus

$$T_{\text{ONB}}\approx {345.696}^{\circ }\mathrm{C}.$$

$329-363$

4. Calculate the temperature of significant void, $T_D$.

$$\begin{array}{rl}& \mathrm{P}\mathrm{e}=\left(G{D}_h{C}_{pf}\right)/k_f\\ & T_{\text{sat}}-T_D=0.0022\left(\left(q_w^{\prime \prime }{D}_h\right)/k_f\right),\mathrm{P}\mathrm{e}<7\times {10}^4\\ & T_{\text{sat}}-T_D=154\left(\frac{q_w^{\prime \prime }}{G{C}_{pf}}\right),\mathrm{P}\mathrm{e}>7\times {10}^4\end{array}$$

The answer of $T_D$ should be given in ${}^{\circ }\mathrm{C}$.

Answer:

$$\mathrm{P}\mathrm{e}=\left(G{D}_h{C}_{pf}\right)/k_f\approx 1.9336283\times {10}^6.$$

Thus

$$T_{\mathrm{s}\mathrm{a}\mathrm{t}}-T_D=154\left(\frac{q_w^{\prime \prime }}{G{C}_{pf}}\right)⟹T_D=T_{\mathrm{s}\mathrm{a}\mathrm{t}}-154\left(\frac{q_w^{\prime \prime }}{G{C}_{pf}}\right)\approx {337.952}^{\circ }\mathrm{C}.$$

$321-355$

5. Calculate the length for the onset of significant void, $Z_D$.

$$Z_D=\frac{\frac{\pi D_h^2}{4}G{C}_{pf}(T_D-T_{in})}{\pi D_h{q}_w^{\prime \prime }}$$

The answer of $Z_D$ should be given in $\mathrm{m}$ .

Answer:

$$Z_D=\frac{\frac{\pi D_h^2}{4}G{C}_{pf}(T_D-T_{in})}{\pi D_h{q}_w^{\prime \prime }}=\frac{D_{h}G{C}_{pf}(T_D-T_{in})}{4q_w^{\prime \prime }}\approx 10.366\mathrm{m}.$$

$9.88-10.9$

6. In another scenario, assume that the heat flux is unknown and the wall temperature is fixed at ${370}^{\circ }\mathrm{C}$. Chen's correlation is used to calculate two-phase heat transfer coefficient and heat flux at $x=0.2$. Saturation pressure at ${370}^{\circ }\mathrm{C}$ is $21.0\mathrm{MPa}$ . Calculate the nucleate boiling heat transfer coefficient in Chen's correlation $h_{NB}$.

$$h_{NB}=S\times 0.00122\left[\frac{{\left(k^{0.79}{C}_p^{0.45}{\rho }^{0.49}\right)}_f}{{\sigma }^{0.5}{\mu }_f^{0.29}{h}_{fg}^{0.24}{\rho }_g^{0.24}}\right]\mathrm{\Delta }{T}_{\mathrm{s}\mathrm{a}\mathrm{t}}^{0.24}\mathrm{\Delta }{P}^{0.75}$$

where

$$\begin{array}{rl}& S=\frac{1}{1+2.53\times {10}^{-6}{\mathrm{R}\mathrm{e}}^{1.17}}\\ & \mathrm{R}\mathrm{e}={\mathrm{R}\mathrm{e}}_f{F}^{1.25}\\ & {\mathrm{R}\mathrm{e}}_f=\frac{G(1-x)D_h}{{\mu }_f}\\ & F=1\text{ for }\frac{1}{X_{tt}}<0.1\\ & F=2.35{(0.213+\frac{1}{X_{tt}})}^{0.736}\text{ for }\frac{1}{X_{tt}}>0.1\\ & X_{tt}={\left(\frac{1-x}{x}\right)}^{0.9}{\left(\frac{{\rho }_g}{{\rho }_f}\right)}^{0.5}{\left(\frac{{\mu }_f}{{\mu }_g}\right)}^{0.1}\\ & \mathrm{\Delta }{T}_{\mathrm{s}\mathrm{a}\mathrm{t}}=T_w-T_{\mathrm{s}\mathrm{a}\mathrm{t}}\\ & \mathrm{\Delta }P=P\left(T_w\right)-P\left(T_{\mathrm{s}\mathrm{a}\mathrm{t}}\right)\end{array}$$

The answer of $h_{NB}$ should be given in $\mathrm{W}/\left({\mathrm{m}}^2\mathrm{K}\right)$.

Answer:

$$\begin{array}{rl}& \mathrm{\Delta }{T}_{\mathrm{s}\mathrm{a}\mathrm{t}}=T_w-T_{\mathrm{s}\mathrm{a}\mathrm{t}}=(370-345{)}^{\circ }\mathrm{C}={25}^{\circ }\mathrm{C}\\ & \mathrm{\Delta }P=P\left(T_w\right)-P(T_{\mathrm{s}\mathrm{a}\mathrm{t}})=(21.0-15.5)\mathrm{MPa}=5.5\times {10}^6\mathrm{Pa}\\ & X_{tt}{|}_{x=0.2}={{\left(\frac{1-x}{x}\right)}^{0.9}{\left(\frac{{\rho }_g}{{\rho }_f}\right)}^{0.5}{\left(\frac{{\mu }_f}{{\mu }_g}\right)}^{0.1}|}_{x=0.2}\approx 1.596⟹\frac{1}{X_{tt}}\approx 0.617>0.1\\ ⟹& F=2.35{(0.213+\frac{1}{X_{tt}})}^{0.736}\approx 2.066\\ & {\mathrm{R}\mathrm{e}}_f=\frac{G(1-x)D_h}{{\mu }_f}\approx 1.163\times {10}^6\\ ⟹& \mathrm{R}\mathrm{e}={\mathrm{Re}}_f{F}^{1.25}\approx 2.880\times {10}^6\\ ⟹& S=\frac{1}{1+2.53\times {10}^{-6}{\mathrm{R}\mathrm{e}}^{1.17}}\approx 0.0108\\ ⟹& h_{NB}=S\times 0.00122\left[\frac{{\left(k^{0.79}{C}_p^{0.45}{\rho }^{0.49}\right)}_f}{{\sigma }^{0.5}{\mu }_f^{0.29}{h}_{fg}^{0.24}{\rho }_g^{0.24}}\right]\mathrm{\Delta }{T}_{\mathrm{s}\mathrm{a}\mathrm{t}}^{0.24}\mathrm{\Delta }{P}^{0.75}\approx 6700.441\mathrm{W}/({\mathrm{m}}^2\cdot \mathrm{K}).\end{array}$$

$6407-7081$

7. Calculate the convective boiling heat transfer coefficient in Chen's correlation $h_c$. Flow conditions are the same as in Question 6.

$$h_c=0.023{\mathrm{R}\mathrm{e}}_f^{0.8}{\mathrm{Pr}}_f^{0.4}\frac{k_f}{D_h}F$$

The answer of $h_c$ should be given in $\mathrm{W}/\left({\mathrm{m}}^2\mathrm{K}\right)$.

Answer:

$$h_c=0.023{\mathrm{R}\mathrm{e}}_f^{0.8}{\mathrm{Pr}}_f^{0.4}\frac{k_f}{D_h}F\approx 34279.827\mathrm{W}/({\mathrm{m}}^2\cdot \mathrm{K}).$$

$32395-35805$

8. Calculate the total boiling heat transfer coefficient in Chen's correlation $h_{2\varphi }$.

$$h_{2\varphi }=h_{NB}+h_c$$

The answer of $h_{2\varphi }$ should be given in $\mathrm{W}/\left({\mathrm{m}}^2\mathrm{K}\right)$.

Answer:

$$h_{2\varphi }=h_{NB}+h_c\approx 40980.268\mathrm{W}/({\mathrm{m}}^2\cdot \mathrm{K}).$$

$38760-42840$

9. Calculate the saturation boiling heat flux using Chen's correlation and temperature given in Question 6.

$$q_w^{\prime \prime }=h_{2\mathrm{\Phi }}(T_w-T_{\text{sat}})$$

The answer of $q_w^{\prime \prime }$ should be given in $\mathrm{W}/{\mathrm{m}}^2$.

Answer:

$$q_w^{\prime \prime }=h_{2\mathrm{\Phi }}(T_w-T_{\mathrm{s}\mathrm{a}\mathrm{t}})=h_{2\mathrm{\Phi }}(370-345)\mathrm{K}\approx 1024506.694\mathrm{W}/{\mathrm{m}}^2.$$

$969000-1071000$

10. Assume that the liquid velocity is zero (pool conditions), calculate the critical heat flux using Zuber's correlation:

$$q_{\text{cr}}^{\prime \prime }={\rho }_g{j}_g{h}_{fg}$$

where the vapor velocity leading to the onset of fluidization $j_g$ is given by:

$$j_g=0.13{\left[\frac{\sigma ({\rho }_f-{\rho }_g)g}{{\rho }_g^2}\right]}^{1/4}$$

The answer of $q_{\text{cr}}^{\prime \prime }$ should be given in $\mathrm{W}/{\mathrm{m}}^2$.

Answer:

$$\begin{array}{rl}& j_g=0.13{\left[\frac{\sigma ({\rho }_f-{\rho }_g)g}{{\rho }_g^2}\right]}^{1/4}\approx 0.0283\mathrm{m}/\mathrm{s}\\ ⟹& q_{\text{cr}}^{\prime \prime }={\rho }_g{j}_g{h}_{fg}\approx 2789163.270\mathrm{W}/{\mathrm{m}}^2.\end{array}$$

$2650500-2929500$