Question 12.9: A fluid whose properties are essentially those of o-dichloro...

A fluid whose properties are essentially those of o-dichlorobenzene is vaporised in the tubes of a forced convection reboiler. Estimate the local heat-transfer coefficient at a point where 5 per cent of the liquid has been vaporised. The liquid velocity at the tube inlet is 2 m/s and the operating pressure is 0.3 bar. The tube inside diameter is 16 mm and the local wall temperature is estimated to be 120^{\circ} C.

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Physical properties:

 

\text { boiling point } 136^{\circ} C

 

\rho_{L}=1170 kg / m ^{3}

 

\mu_{L}=0.45 mNs / m ^{2}

 

\mu_{v}=0.01 mNs / m ^{2}

 

\rho_{v}=1.31 kg / m ^{3}

 

k_{L}=0.11 W / m ^{\circ} C

 

C_{p L}=1.25 kJ / kg ^{\circ} C

 

P_{c}=41 bar

 

The forced-convective boiling coefficient will be estimated using Chen’s method.

With 5 per cent vapour, liquid velocity (for liquid flow in tube alone)

 

=2 \times 0.95=1.90 m / s

 

R e_{L}=\frac{1170 \times 1.90 \times 16 \times 10^{-3}}{0.45 \times 10^{-3}}=79,040

 

\text { From Figure } 12.23, j_{h}=3.3 \times 10^{-3}

 

P r=\frac{1.25 \times 10^{3} \times 0.45 \times 10^{-3}}{0.11}=5.1

 

Neglect viscosity correction term.

 

\begin{aligned}h_{f c} &=\frac{0.11}{16 \times 10^{-3}} \times 3.3 \times 10^{-3}(79,040)(5.1)^{0.33} \\&=3070 W / m ^{2 \circ} C\end{aligned} (12.15)

 

\begin{aligned}\frac{1}{X_{t t}} &=\left[\frac{0.05}{1-0.05}\right]^{0.9}\left[\frac{1170}{1.31}\right]^{0.5}\left[\frac{0.01 \times 10^{-3}}{0.45 \times 10^{-3}}\right]^{0.1} \\&=1.44\end{aligned} (12.69)

 

\text { From Figure 12.56, } f_{c}=3.2

 

h_{f c}^{\prime}=3.2 \times 3070=9824 W / m ^{2 \circ} C

 

Using Mostinski’s correlation to estimate the nucleate boiling coefficient

 

h_{n b}=0.104 \times 41^{0.69}\left[h_{n b}(136-120)\right]^{0.7} (12.63)

 

\times\left[1.8\left(\frac{0.3}{41}\right)^{0.17}+4\left(\frac{0.3}{41}\right)^{1.2}+10\left(\frac{0.3}{41}\right)^{10}\right]

 

h_{n b}=7.43 h_{n b}^{0.7}

 

h_{n b}=800 W / m ^{2 \circ} C

 

R e_{L} f_{c}^{1.25}=79,040 \times 3.2^{1.25}=338,286

 

\text { From Figure } 12.57, f_{s}=0.13 \text {, }

 

h_{n b}^{\prime}=0.13 \times 800=104 W / m ^{2}{ }^{\circ} C

 

h_{c b}=9824+104=\underline{\underline{9928} W / m ^{2}{ }^{\circ} C }
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