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Question : Ethyl formate is to be produced from ethanol and formic acid...

Ethyl formate is to be produced from ethanol and formic acid in a continuous flow tubular reactor operated at a constant temperature of 303 K (30^{\circ}C). The reactants will be fed to the reactor in the proportions 1 mole HCOOH: 5 moles C_{2}H_{5}OH at a combined flowrate of 0.0002 m^{3}/s (0.72 m^{3}/h). The reaction will be catalysed by a small amount of sulphuric acid. At the temperature, mole ratio, and catalyst concentration to be used, the rate equation determined from small-scale batch experiments has been found to be:
R=KC_{F}^{2}
where: R is formic acid reacting/(kmol/m^{3}s)
C_{F} is concentration of formic acid kmol/m^{3} , and
K = 2.8 × 10-4 m^{3}/kmol s.
The density of the mixture is 820 kg/m^{3} and this may be assumed constant throughout.
Estimate the volume of the reactor required to convert 70 per cent of the formic acid to the ester.
If the reactor consists of a pipe of 50 mm i.d. what will be the total length required?
Determine also whether the flow will be laminar or turbulent and comment on the significance of this in relation to the estimate of reactor volume. The viscosity of the solution is 1.4 × 10^{-3} N s/m^{2}.

 

 

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Although the ethanol fed to the reactor is partly consumed in the reaction, the rate equation indicates that the rate of reaction depends only on the concentration of the formic acid.

Thus: R = kC_{F}^{2}

In this liquid phase reaction, it may be assumed that the mass density of the liquid is unaffected by the reaction, allowing the material balance for the tubular reactor to be applied on a volume basis with plug flow.

Thus: \frac{V_{t}}{\upsilon }=\int_{0}^{X_{f}}\frac{dX}{R} (equation 1.37)

In this case: R = kC_{F}^{2}=K(C_{0}-X)^{2},

where C_{0} is the concentration of formic acid in the feed.

Thus: \mathbf{\frac{V_{t}}{\upsilon }=\int_{0}^{X_{f}}\frac{dX}{K(C_{0}-X)^{2}}=\frac{1}{K}\left [ \frac{1}{(C_{0}-X_{f})}-\frac{1}{C_{0}} \right ]}

In terms of fractional conversion X_{f}=\alpha _{f}C_{0}, then:

\mathbf{\frac{V_{t}}{\upsilon }=\frac{1}{KC_{0}}\left [ \frac{1}{(1-\alpha _{f})} -1\right ]=\frac{1}{KC_{0}}\left ( \frac{\alpha _{f}}{1-\alpha _{f}} \right )}

For HCOOH, the molecular mass = 46 kg/kmol.

For C_{2}H_{5}OH, the molecular mass = 46 kg/kmol.

Thus: 1 kmol HCOOH is present in (1 × 46) + (5 × 46)

= 276 kg feed mixture.

or: (276/820) = 0.337 m^{3} feed mixture.

Thus: C_{0} = (1/0.337) = 2.97 kmol/m^{3}

The volume of the reactor required to convert a fraction of the feed of 0.7 is given by:

V_{t} = (0.72/3600)(1/(2.8 × 10^{-4} × 2.97))(0.7/(1 – 0.7))

= 0.561 m^{3}

The equivalent length of a 50 mm ID pipe is then:

0.561/[(\pi /4)0.050^{2}] =286 m

say 29 lengths, each of 10 m length, connected by U-bends.

The mean velocity in the tube, u = (0.72/3600)/(\pi /4)0.050^{2} = 0.102 m/s and, the Reynolds Number is then:

Re=u\rho d/\mu =(0.102\times 820\times 0.050)/(1.4\times 10^{-3})

= 2980

confirming turbulent flow and the validity of the assumed plug flow.