Question 9.2: Nucleate pool boiling takes place in a liquid consisting of ...
Nucleate pool boiling takes place in a liquid consisting of 54.5 mole percent SF_{6} (1) and 45.5 mole percent CCl _{2} F _{2}(2) at a pressure of 23 bar with a heat flux of 10^{4} W/m². The vapor in equilibrium with the liquid contains 71.1 mole percent SF_{6}. The liquid density of the mixture is 10 kmol/m³. At 23 bar, the boiling point of SF _{6} is 22°C and that of CCl _{2} F _{2} is 80ºC. Experimental values for the pure component heat-transfer coefficients at this pressure are 11,787 W/m²· K for SF_{6} and 6,105 W/m²· K for CCl _{2} F _{2}. Additional data are given in the table below. Estimate the heat-transfer coefficient for the mixture using:
(a) Schlünder’s method.
(b) The method of Thome and Shakir.
(c) Palen’s method.
| Property | SF _{6} | CCl _{2} F _{2} |
| Critical temperature (ºC) | 45.5 | 111.5 |
| Critical pressure (bar) | 37.6 | 41.2 |
| Normal boiling point (ºC) | – 63.5 | – 29.8 |
| Latent heat of vaporization at normal | 16,790 | 20,207 |
| boiling point (J/mol) |
(a) Since CCl _{2} F _{2} has the higher boiling point, it is designated as component number 2 and SF _{6} as component number 1. Values of all the parameters appearing in Equation (9.14) are given except for the latent heat of vaporization of the mixture. In order to estimate this value, the pure component latent heats must first be determined at the system pressure of 23 bar. The Watson correlation, which relates the latent heats at two different temperatures, is used for this purpose
\lambda\left(T_{2}\right)=\lambda\left(T_{1}\right)\left[\frac{T_{c} – T_{2}}{T_{c} – T_{1}}\right]^{0.38}
In the present application, T_{1} is taken as the normal boiling point and T_{2} is taken as the boiling point at system pressure. Thus, for SF _{6} we have:
\lambda_{1}\left(22^{\circ} C \right)=16,790\left[\frac{45.5 – 22}{45.5 – ( – 63.5)}\right]^{0.38}
\lambda_{1}=9372 J / mol
Similarly, for CCl _{2} F _{2} we have:
\lambda_{2}\left(80^{\circ} C \right)=20,207\left[\frac{111.5 – 80}{111.5 – ( – 29.8)}\right]^{0.38}
\lambda_{2}=11,424 J / mol
For the mixture, the mole fraction weighted average of the pure component latent heats is used:
\lambda=x_{1} \lambda_{1}+x_{2} \lambda_{2}=0.545 \times 9372+0.455 \times 11,424
\lambda \cong 10,300 J / mol
Next, h_{\text {ideal }} is computed using Equation (9.15):
h_{ideal} \left[\sum\limits_{i = 1}^{n} x_{i}/h_{nb,i}\right]^{- 1} (9.15)
h_{\text {ideal }}=\left(x_{1} / h_{n b, 1}+x_{2} / h_{n b, 2}\right)^{-1}
=(0.545 / 11,787+0.455 / 6105)^{-1}
h_{\text {ideal }}=8280 W / m ^{2} \cdot K
The heat-transfer coefficient for the mixture can now be obtained by substituting the data into Equation (9.14):
h_{n b}=h_{\text {ideal }}\left\{1+\left(h_{\text {ideal }} / \hat{q}\right)\left[1-\exp \left(\frac{-\hat{q}}{\rho_{L} \lambda \beta}\right)\right] \sum\limits_{i=1}^{n-1}\left(T_{\text {sat }, n}-T_{\text {sat }, i}\right)\left(y_{i}-x_{i}\right)\right\}^{-1}
=8280\left\{1+(8280 / 10,000)\left[1-\exp \left(\frac{ – 10,000}{10,000 \times 10,300 \times 0.0002}\right)\right] \times (80 – 22)(0.711 – 0.545)\right\}^{-1}
h_{n b}=2036 W / m ^{2} \cdot K
The measured heat-transfer coefficient for this mixture is 2096 W/m²· K [12], which differs from the calculated value by about 3%.
(b) Since the dew-point and bubble-point temperatures for the mixture were not given in Ref. [12], the PRO/II (SimSci-Esscor) process simulator was used to obtain the following values at a pressure of 23 bars: T_{D} = 326.5 K and T_{B} = 316.9 K. The boiling range for the mixture is, therefore
B R=T_{D}-T_{B}=326.5-316.9=9.6 K
The heat-transfer coefficient for the mixture is obtained from Equation (9.16) using the values of h_{\text {ideal }} and λ from part (a):
h_{n b}=h_{\text {ideal }}\left\{1+\left(B R \cdot h_{\text {ideal }} / \hat{q}\right)\left[1-\exp \left(\frac{-\hat{q}}{\rho_{L} \lambda \beta}\right)\right]\right\}^{-1}
=8280\left\{1+\left(\frac{9.6 \times 8280}{10,000}\right)\left[1-\exp \left(\frac{ – 10,000}{10,000 \times 10,300 \times 0.0003}\right)\right]\right\}^{-1}
h_{n b}=2589 W / m ^{2} \cdot K
This value is about 24% higher than the experimental value of 2096 W/m²·K.
(c) For a mixture, the pseudo-critical pressure, P_{p c}, can be used in place of the critical pressure in the Mostinski correlation. The pseudo-critical pressure is the mole fraction weighted average of the pure component critical pressures. Thus
P_{p c}=x_{1} P_{c 1}+x_{2} P_{c 2}=0.545 \times 37.6 + 0.435 \times 41.2
P_{p c} \cong 39.24 \text { bar }=3924 kPa
The pseudo-reduced pressure is then:
P_{p r}=P / P_{p c}=23 / 39.24=0.586
Since this value is greater than 0.2, Equation (9.18) is used to calculate the pressure correction factor with P_{p r} in place of P_{r}.
F_{p}=1.8 P_{ r}^{0.17} (9.18)
F_{p}=1.8 P_{p r}^{0.17}=1.8(0.586)^{0.17}=1.6437
The mixture correction factor is obtained from Equation (9.17b).
F_{m}=\left(1+0.023 \hat{q}^{0.15} B R^{0.75}\right)^{-1}
=\left[1+0.023(10,000)^{0.15}(9.6)^{0.75}\right]^{-1}
F_{m}=0.6669
Including this factor in Equation (9.2b) gives:
h_{n b}=0.00417 P_{c}^{0.69} \hat{q}^{0.7} F_{P} (9.2b)
h_{n b}=0.00417 P_{c}^{0.69} \hat{q}^{0.7} F_{P} F_{m}
=0.00417(3924)^{0.69}(10,000)^{0.7} \times 1.6437 \times 0.6669
h_{n b}=870 W / m ^{2} \cdot K
This value is very conservative compared with the measured value of 2096 W/m²·K for this system. Note, however, that the mixture correction factor itself is not conservative, since F_{m} · h_{ideal} = 5522 W/m²·K which is much higher than the measured heat-transfer coefficient. This result emphasizes the point that Palen’s mixture correction factor is intended for use only with the Mostinski correlation for calculating h_{n b}.