Question 9.7: The fluid of Example 9.1 boils while flowing vertically upwa...

(a) The first step is to calculate the heat-transfer coefficient for forced convection using Equation (9.75). The Reynolds number is based on the flow rate of the liquid phase alone, and fluid properties are obtained from Example 9.1.

R e_{L}=\frac{D_{i} G_{L}}{\mu_{L}}=\frac{0.0212  \times  0.8  \times  300}{156  \times  10^{-6}}=32,615

P r_{L}=\frac{C_{P, L} \mu_{L}}{k_{L}}=\frac{2730  \times  156  \times  10^{-6}}{0.086}=4.9521

h_{L}=0.023\left(k_{L} / D_{i}\right) \operatorname{Re}_{L}^{0.8} \operatorname{Pr}_{L}^{0 . 4}

=0.023(0.086 / 0.0212)(32,615)^{0.8}(4.9521)^{0.4}

h_{L}=722  W / m ^{2} \cdot K

Next, the Lockhart–Martinelli parameter is calculated using Equation (9.37).

X_{ tt }=\left(\frac{1  –  x}{x}\right)^{0.9}\left(\rho_{V} / \rho_{L}\right)^{0.5}\left(\mu_{L} / \mu_{V}\right)^{0.1}

X_{t t}=0.847

Equation (9.71) is used to calculate F(X_{t t}). SinceX_{t t} < 10, we have:

F\left(X_{t t}\right)=2.35\left(X_{t t}^{-1}+0.213\right)^{0.736}=2.35\left[(0.847)^{-1}+0.213\right]^{0.736}

F\left(X_{t t}\right)=3.00

Equations (9.72) and (9.70) are used to calculate the suppression factor.

R e=R e_{L}\left[F\left(X_{ tt }\right)\right]^{1.25}=32,615(3.00)^{1.25}=128,771

S_{C H}=\left(1+2.53 \times 10^{-6} R e^{1.17}\right)^{-1}=\left[1+2.53 \times 10^{-6}(128,771)^{1.17}\right]^{-1}

S_{C H}=0.2935

In Example 9.1, the Forster–Zuber correlation was used to obtain h_{nb} = 5,512 W/m²· K, and this value is still valid here. Therefore, h_{b} can be found by substituting in Equation (9.69):

h_{b}=S_{C H} h_{n b}+F\left(X_{t t}\right) h_{L}=0.2935 \times 5512+3.00 \times 722

h_{b}=3784  W / m ^{2} \cdot K

(b) The values of h_{L} ,F\left(X_{t t}\right), and S_{C H} are the same as in part (a). The Mostinski correlation in the form of Equation (9.3) was used in Example 9.1 to calculate the nucleate boiling heat-transfer coefficient. However, Equation (9.3) was derived from Equation (9.2) by substituting \hat{q}=h_{b} \Delta T_{e}, which for convective boiling is replaced by \hat{q}=h_{b} \Delta T_{e}. Therefore, the calculation must be repeated using Equation (9.2b).

h_{n b}=0.00417 P_{c}^{0.69} \hat{q}^{0.7} F_{P}

h_{nb} = 4.33  \times  10^{-8}  P_{c}^{2.3} \Delta T_{e}^{2.333} F_{P}^{3.333}                 (9.3a)

h_{nb} = 1.167  \times  10^{-8}  P_{c}^{2.3} \Delta T_{e}^{2.333} F_{P}^{3.333}                 (9.3b)

Using Palen’s recommendation for the pressure correction factor, we obtain F_{P} = 1.3375 from Example 9.1, part (c). Also, P_{c} = 2500 kPa as given in Example 9.1. Thus,

h_{n b}=0.00417(2500)^{0.69} \hat{q}^{0.7} \times 1.3375

h_{n b}=1.2331 \hat{q}^{0.7}

Substituting in Equation (9.69) gives:

h_{b}=S_{C H} h_{n b}+F\left(X_{t t}\right) h_{L}

=0.2935 \times 1.2331 \hat{q}^{0.7}+3.00 \times 722

h_{b}=0.3619 \hat{q}^{0.7}+2166

Since \Delta T_{e} = 16.2 K from Example 9.1,

\hat{q}=h_{b} \Delta T_{e}=\left(0.3619 \hat{q}^{0.7}+2166\right) \times 16.2

\hat{q}=5.8628 \hat{q}^{0.7}+35,089.2

\hat{q}-5.8628 \hat{q}^{0.7}-35,089.2=0

Using the nonlinear equation solver on a TI calculator, the solution to this equation is found to be \hat{q} = 45, 826  W/m². The heat-transfer coefficient is obtained from this value as follows:

h_{b}=\frac{\hat{q}}{\Delta T_{e}}=\frac{45,826}{16.2}=2829  W / m ^{2} \cdot K

(c) The following results from part (a) apply here as well:

R e_{L}=32,615

h_{L}=722  W / m ^{2} \cdot k

X_{ tt }=0.847

Also, from Example 9.1, the latent heat of vaporization is 272,000 J/kg.
The boiling number is given by Equation (9.79):

B o=\frac{\hat{q}}{\lambda G}=\frac{\hat{q}}{272,000  \times  300}=1.2255  \times  10^{-8} \hat{q}

The convective enhancement factor is computed by substituting in Equation (9.78).

E_{ GW }=1+24,000 B o^{1.16}+1.37 X_{ tt }^{-0.86}

=1+24,000\left(1.2255 \times 10^{-8} \hat{q}\right)^{1.16}+1.37(0.847)^{-0.86}

=1+1.5946 \times 10^{-5} \hat{q}^{1.16}+1.5803

E_{G W}=2.5803+1.5946 \times 10^{-5} \hat{q}^{1.16}           (i)

The suppression factor is computed using Equation (9.77).

S_{G W}=\left(1+1.15 \times 10^{-6} E_{G W}^{2} R e_{L}^{1.17}\right)^{-1}

=\left(1+1.15 \times 10^{-6} E_{G W}^{2}(32,615)^{1.17}\right)^{-1}

S_{ GW }=\left(1+0.2195 E_{ GW }^{2}\right)^{-1}            (ii)

The Cooper nucleate boiling correlation in SI units is given by Equation (9.6b):

h_{n b}=55 \hat{q}^{0.67} P_{r}^{0.12}\left(-\log _{10} P_{r}\right)^{-0.55} M^{-0.5}

From Example 9.1, P_{r} = 0.1217 and the molecular weight is 110.37. Substitution gives:

Substituting in Equation (9.76) gives the following result for h_{b}:

h_{b}=S_{G W} h_{n b}+E_{G W} h_{L}

h_{b}=4.2704 S_{G W} \hat{q}^{0.67}+722 E_{G W}

The equation for the heat flux is obtained as follows:

\hat{q}=h_{b} \Delta T_{e}=h_{b} \times 16.2

\hat{q}=69.1805 S_{G W} \hat{q}^{0.67}+11,696.4 E_{G W}

\hat{q}-69.1805 S_{ GW } \hat{q}^{0.67}-11,696.4 E_{ GW }=0                 (iii)

Equation (iii) is to be solved in conjunction with Equations (i) and (ii) for the heat flux. However, these equations have no real solution, as the reader can verify using the nonlinear equation solver on a TI calculator or by graphing the left side of Equation (iii). The crux of the problem is the term involving the boiling number in the enhancement factor, which, as previously noted, is not physically realistic. Therefore, the Gungor–Winterton correlation cannot be used to solve this problem.
(d) The following results from part (a) are applicable:

R e_{ L }=32,615 \quad h_{L}=722  W / m ^{2} \cdot K \quad \operatorname{Pr_{L}}=4.9521

The enhancement and suppression factors are calculated from Equations (9.82) and (9.81).

E_{L W}=\left[1+x Pr_{L}\left(\rho_{L}-\rho_{V}\right) / \rho_{V}\right]^{0.35}

=[1+0.2 \times 4.9521(567-18.09) / 18.09]^{0.35}

E_{L W}=3.3284

S_{L W}=\left[1+0.055 E_{L W}^{0.1} R e_{L}^{0.16}\right]^{-1}

=\left[1+0.055(3.3284)^{0.1}(32,615)^{0.16}\right]^{-1}

S_{L W}=0.7535

From part (c), the Cooper correlation gives:

h_{n b}=4.2704 \hat{q}^{0.67}

Substituting in Equation (9.80) for the convective boiling coefficient gives:

h_{b}=\left\{\left(S_{L w} h_{n b}\right)^{2}+\left(E_{L w} h_{L}\right)^{2}\right\}^{1 / 2}

=\left\{\left(0.7535 \times 4.2704 \hat{q}^{0.67}\right)^{2}+(3.3284 \times 722)^{2}\right\}^{1 / 2}

h_{b}=\left\{10.3539 \hat{q}^{1.34}+5,774,913\right\}^{1 / 2}

Setting h_{b}= \hat{q}/ \Delta T_{e} gives the following nonlinear equation for the heat flux:

\frac{\hat{q}}{16.2}-\left(10.3539 \hat{q}^{1.34}+5,774,913\right)^{0.5}=0

Using the nonlinear equation solver on a TI calculator, the solution is found to be \hat{q} = 172, 788 W/m². Thus, the heat-transfer coefficient is:

h_{b}=\hat{q} / \Delta T_{e}=172,788 / 16.2=10,666  W / m ^{2} \cdot K

This value is much higher than the one obtained with the Chen correlation, primarily due to the difference in the nucleate boiling terms. Recall that in Example 9.1 the Cooper correlation predicted a much higher value for the nucleate boiling heat-transfer coefficient than either the Forster–Zuber or Mostinski correlation.