please help answer 10.8 and show work 10.8 A settling tank in a water treatment

Question

please help answer 10.8 and show work

10.8 A settling tank in a water treatment plant has an inflow of 2 m3/min and a solids concen- tration of 2100 mg/L. The effluent from this settling tank goes to sand filters. The concen- tration of sludge coming out of the bottom (the underflow) is 18,000 mg/L, and the flow to the filters is 1.8 m3/min. a. What is the underflow flow rate? b. What is the solids concentration in the effluent? C. How large must the sand filters be (in m2)? 45Sum e ! , ct of25, 109 / n, 10.9 A settling tank is 20 m long, 10 m deep, and 10 m wide. The flow rate to the tank is PL2

Explanation / Answer

Inflow rate (Q1) = 2 m3/min = (2/60) m3/sec = (1/30) m3/sec

Inflow solids concentration (C1) = 2100 mg/L = 2100 g/m3

Effluent flow rate (Q2) = 1.8 m3/min = (1.8/60) m3/sec = 0.03 m3/sec

Effluent solids concentration (C2) = ?

Underflow flow rate (Q3) = ?

Underflow solids concentration (C3) = 18000 mg/L = 18000 g/m3

a) Considering steady-state, incoming discharge is equal to outgoing discharge.

Q1 = Q2 + Q3

=> 1/30 = 0.03 + Q3

=> Q3 = 0.0033 m3/sec

So, Underflow flow rate = 0.0033 m3/sec = 0.2 m3/min

b) Since steady state is considered, mass flow rate of incoming fluid is equal to mass flow rate of outgoing fluid.

C1 Q1 = C2 Q2 + C3 Q3

=> 2100 x 2 = C2 x 1.8 + 18000 x 0.2

=> C2 x 1.8 = 600

=> C2 = (600/1.8) = 333.33 mg/L

So, solids concentration in the effluent = 333.33 mg/L

c) Flow to the filters = 1.8 m3/min

Filter loading = 4 gpm/ft2 = (4 / 264.173) m3/min/ft2

Area of sand filters required = Flow to the filters / Filter loading = 1.8 x 264.173 / 4 = 118.878 ft2

= 11.044 m2

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.