Practice questions

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.. |-| replace:: :math:`-` .. |+| replace:: :math:`+` .. |micron| replace:: :math:`\mu \text{m}`

Several practice questions have been covered in class time also, usually on Fridays. Please review your notes for extra problems.

.. question::

A sample of material was settled in a graduated lab cylinder 300mm tall. The interface dropped from 500mL to 215mL on the graduations during a 4 minute period.

#. Give a preliminary estimate of the clarifier diameter required to treat a waste stream of 2100 L per minute. Over-design by a factor of 2, based on the settling rate, and account for about 7 $\text{m}^2$ of entry area used to eliminate turbulence in the entering stream. #. If the feed concentration is 3.5 kg per :math:`\text{m}^3` feed, what is the loading rate? Is it within the typical thickener range of 50 to 120 kg per day per square meter (Perry, 8ed, p22-79)?

.. answer::

#. Settling rate = 171 mm per 4 minutes = 42.8 mm/min.

Area = :math:`\displaystyle \frac{2.1~\text{m}^3.\text{min}^{-1}}{\left(\frac{1}{2}\right)\left(42.8 \times 10^{-3} \text{m}.\text{min}^{-1}\right)} = 98 + 7 \text{m}^2`

#. :math:`\psi = C_0 v = 3.5 \displaystyle\frac{\text{kg}}{\text{m}^3} \cdot 0.022 \displaystyle \frac{\text{m}}{\text{min}} \cdot \displaystyle\frac{60 \times 24 \text{min}}{\text{day}} = 106 \displaystyle \frac{\text{kg}}{\text{day}.\text{m}^2}`

.. question::

#. Calculate the minimum area and diameter of a circular thickener to treat 720 :math:`\text{m}^3` per hour of slurry containing 20 |micron| particles of silica, whose density is about 2600 :math:`\text{kg.m}^{-3}`. The particles are suspended in water at a concentration of 650 :math:`\text{kg.m}^{-3}`. The slurry cannot be tested in a lab. Use an over-design factor of 1.5 on the settling velocity. #. If it is desired to have an underflow concentration of 1560 kg solids per :math:`\text{m}^{3}` underflow; what is the underflow volumetric flow rate if total separation of solids occurs? #. Calculate the separation factor.

.. answer::

#. The settling velocity = :math:`\displaystyle \frac{(20 \times 10^{-6})^2(2600-1000)(9.81)}{(18)(0.001} = 350 \times 10^{-6} \text{m.s}^{-1}` , assuming ambient conditions for physical property estimates.

Over-designing by a factor of 1.5, implies a settling velocity of :math:`233 \times 10^{-6} \text{m.s}^{-1}` should be used. This leads to an area of :math:`A = \displaystyle \frac{Q}{v_\text{TSV}} = \frac{720}{(233 \times 10^{-6})(60)(60)}` = 858 :math:`\text{m}^2`, or a diameter of about 33m.

#. If all the solids entering leave again in the underflow, a mass balance implies there are :math:`QC_0 =\displaystyle \frac{720 \text{kg solids}}{\text{hour}}\cdot \frac{650~\text{kg solids}}{\text{m}^3~\text{of feed}} = 468,000` kg per hour, or 468 tonnes per hour of solids leaving (a high amount, which explains the large vessel diameter). If the desired density of the underflow is 1560 kg solids per :math:`\text{m}^{3}` underflow, the volumetric flow of this stream must be :math:`\displaystyle \frac{468000~\text{kg solids}}{\text{hour}} \cdot \displaystyle \frac{\text{m}^3~\text{underflow}}{1560~\text{kg solids}} = 300 \frac{\text{m}^3}{\text{hour}}`, or about 0.083 :math:`\text{m}^3.\text{s}^{-1}`, or only about 5 :math:`\text{m}^3.\text{min}^{-1}`, a relatively small and much more compact amount.

#. The separation factor is infinite, because all the solids are removed from the overflow.

.. question::

There are many practice questions on adsorption in the Geankoplis textbook.

.. question::

A pigment is being extracted from an aqueous feed stream to an organic solvent stream using a continuous counter-current liquid-liquid extractor. This device gives the equivalent of 3 theoretical equilibrium stages. The feed flow rate is 120 kg/min and has a pigment concentration of 0.18 kg/kg feed. The molecular weight of the pigment is 90 kg/kmole. The molecular weight of the organic solvent is 150 kg/kmol and enters the extractor at a flow rate of 55 kg/min. The equilibrium relationship is given by :math:`y = 45 x` where :math:`x` and :math:`y` are the concentrations in the raffinate and the extract and respectively and are expressed in molar ratio. Calculate the concentration of the pigment in the extract and the raffinate. (Molar ratio = moles of solute/moles of solvent).

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