Question 9.8: The fluid of Example 9.1 boils while flowing through a verti...
The fluid of Example 9.1 boils while flowing through a vertical 1-in. 14 BWG tube (ID = 2.12 cm) with a mass flux of 300 kg/s · m².The tube length is 10 ft (3.048 m) and the inlet subcooling of the fluid is 23,260 J/kg. The tube is uniformly heated over its entire length, and the average internal pressure is 310 kPa. Estimate the critical heat flux using:
(a) The Palen correlation.
(b) The Katto-Ohno correlation
(a) From Example 9.1, the critical pressure is 2550 kPa. Hence, the reduced pressure is:
P_{r}=P / P_{c}=310 / 2550 \cong 0.12157
Equation (9.84b) is used to calculate the critical heat flux.
\hat{q}_{c}=23,660\left(D^{2} / L\right)^{0.35} P_{c}^{0.61} P_{ r }^{0.25}\left(1-P_{ r }\right)
=23,660\left[(0.0212)^{2} / 3.048\right]^{0.35}(2550)^{0.61}(0.12157)^{0.25}(1-0.12157)
\hat{q}_{c}=66,980 W / m ^{2}
Notice that this method does not take into account either the inlet subcooling or the flow rate of the fluid.
(b) From Example 9.1, the liquid and vapor densities are 567 kg/m³ and 18.09 kg/m³, respectively. Hence
\rho_{V} / \rho_{L}=18.09 / 567=0.031905
Since this value is less than 0.15, Equations (9.95) and (9.96) are used to evaluate \hat{q}_{0} and Γ. We first calculate \hat{q}_{oA} and \hat{q}_{oB} using Equations (9.86) and (9.87). Since L/D = 3.048/0.0212 = 143.8, Equation (9.91) gives the value of C_{2} as:
\begin{matrix} \hat q_{o} = \hat q_{oA} & \quad & (\hat q_{oA} \leq \hat q_{oB})\\ = Min(\hat q_{oB} , \hat q_{oC}) & \quad & (\hat q_{oA} \gt \hat q_{oB})\end{matrix} (9.95)
Γ = Max(Γ_{A} , Γ_{B} ) (9.96)
\begin{matrix}G_{2} = 0.25 & & (L/D < 50)\\ = 0.25 + 0.0009 (L/D – 50) && (50\leqslant L/D \leqslant 150) \\ = 0.34 && (L/D) > 150 )\end{matrix} (9.91)
C_{2}=0.25+0.0009(143.8-50)=0.334
From Example 9.1, σ = 8.2 \times 10^{-3} N/m. Hence,
\frac{g_{c} \sigma \rho_{L}}{G^{2} L}=\frac{1.0 \times 8.2 \times 10^{-3} \times 567}{(300)^{2} \times 3.048}=1.6949 \times 10^{-5}
\hat{q}_{o A} / G \lambda=C_{2}\left(\frac{g_{C} \sigma \rho_{L}}{G^{2} L}\right)^{0.043}(L / D)^{-1}
=0.334\left(1.6949 \times 10^{-5}\right)^{0.043}(143.8)^{-1}
\hat{q}_{o A} / G \lambda=1.4482 \times 10^{-3}
\hat{q}_{o A}=1.4482 \times 10^{-3} \times 300 \times 272,000=118,176 W / m ^{2}
\hat{q}_{o B} / G \lambda=0.10\left(\rho_{V} / \rho_{L}\right)^{0.133}\left(\frac{g_{C} \sigma \rho_{L}}{G^{2} L}\right)^{1 / 3}(1+0.0031 L / D)^{-1}
=0.10(0.031905)^{0.133}\left(1.6949 \times 10^{-5}\right)^{1 / 3}(1+0.0031 \times 143.8)^{-1}
\hat{q}_{o B} / G \lambda=1.1236 \times 10^{-3}
\hat{q}_{o B}=1.1236 \times 10^{-3} \times 300 \times 272,000=91,688 W / m ^{2}
Since \hat{q}_{oA} > \hat{q}_{oB} we need to calculate \hat{q}_{oC} from Equation (9.88).
\hat{q}_{o C} / G \lambda=0.098\left(\rho_{V} / \rho_{L}\right)^{0.133}\left(\frac{g_{c} \sigma \rho_{L}}{G^{2} L}\right)^{0.433}(L / D)^{0.27}(1+0.0031 L / D)^{-1}
=0.098(0.031905)^{0.133}\left(1.6949 \times 10^{-5}\right)^{0.433}(143.8)^{0.27}(1+0.0031 \times 143.8)^{-1}
\hat{q}_{o C} / G \lambda=1.4091 \times 10^{-3}
\hat{q}_{o C}=1.4091 \times 10^{-3} \times 300 \times 272,000=114,985 W / m ^{2}
From Equation (9.95), since \hat{q}_{oA} > \hat{q}_{oB} , we have:
\hat{q}_{o}=\operatorname{Min}\left(\hat{q}_{o B}, \hat{q}_{o C}\right)=\operatorname{Min}(91,688 ; 114,985)=91,688 W / m ^{2}
The next step is to calculate \Gamma_{A} and \Gamma_{B} from Equations (9.92) and (9.93).
\Gamma_{A}=\frac{1.043}{4 C_{2}\left(g_{c} \sigma \rho_{L} / G^{2} L\right)^{0.043}}=\frac{1.043}{4 \times 0.334\left(1.6949 \times 10^{-5}\right)^{0.043}}
\Gamma_{A}=1.2521
\Gamma_{B}=\frac{(5 / 6)(0.0124 + D / L)}{\left(\rho_{V} / \rho_{L}\right)^{0.133}\left(g_{c} \sigma \rho_{L} / G^{2} L\right)^{1 / 3}}
=\frac{(5 / 6)(0.0124 + 0.0212 / 3.048)}{(0.031905)^{0.133}\left(1.6949 \times 10^{-5}\right)^{1 / 3}}
\Gamma_{B}=0.9929
From Equation (9.96) we have:
\Gamma=\operatorname{Max}\left(\Gamma_{A}, \Gamma_{B}\right)=\operatorname{Max}(1.2521,0.9929)=1.2521
The critical heat flux is given by Equation (9.85):
\hat{q}_{c}=\hat{q}_{o}\left(1+\Gamma \Delta H_{\text {in }} / \lambda\right)=91,688(1+1.2521 \times 23,260 / 272,000)
\hat{q}_{c} \cong 101,500 W / m ^{2}
This value is about 50% higher than the result obtained using Palen’s correlation.