Practice questions
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.. |-| replace:: :math:`-` .. |+| replace:: :math:`+` .. |micron| replace:: :math:`\mu \text{m}`
.. 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 coming in leave 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}}` #. :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} = 468,000` kg per hour, or 468 tonnes per hour (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}}`.
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