Question

A centrifuge processes 1000 L/h of water containing 2 g/L suspended solids. The thickened slurry leaving the centrifuge has 20 g per 100 mL of solids. If the centrifuge has 99% solids-separation efficiency, find the flow rate (in L/h, rounded to one decimal place) of the supernatant stream.

Hint:

Always convert slurry concentration to g/L and divide captured solids by this value to get slurry flow. Supernatant flow is simply feed minus slurry flow.

Answer 1

Correct Answer: 989

Step 1: Solids entering the centrifuge.
\[ \text{Feed flow} = 1000\ \text{L/h} \] \[ \text{Feed solids concentration} = 2\ \text{g/L} \] \[ \text{Solids in feed} = 1000 \times 2 = 2000\ \text{g/h} \] Step 2: Solids captured in thickened slurry (99% efficiency).
\[ \text{Captured solids} = 0.99 \times 2000 = 1980\ \text{g/h} \] Step 3: Slurry solids concentration.
Given: \[ 20\ \text{g per 100 mL} = 200\ \text{g per L} \] Thus slurry concentration: \[ C_s = 200\ \text{g/L} \] Slurry flow rate: \[ Q_s = \frac{1980\ \text{g/h}}{200\ \text{g/L}} = 9.9\ \text{L/h} \] Step 4: Supernatant flow rate.
\[ Q_{\text{sup}} = Q_{\text{feed}} - Q_s \] \[ Q_{\text{sup}} = 1000 - 9.9 = 990.1\ \text{L/h} \] Rounded to one decimal: \[ \boxed{990.1\ \text{L/h}} \]