Water requirements miscellaneous
- With reference to a standard Cartesian (x, y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are separated by distance, h, is given by the expression
u = - h² dp 
1 - 4 
y 
² 
8μ dx h
In this equation, the y = 0 axis lies equidistant between the plates at a distance h/2 from the two plates, p is the pressure variable and m is the dynamic vicosity term. The maximum and average velocities are, respectively
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dQ = Area × velocity
Q = A × V= - h² dp = (h × 1) × Umax 12μ dx ∴ Uavg = - h² dp 12μ dx
At y = 0, U = Umax∴ Umax = h² dp 8μ dx ∴ = - h² dp Uavg 12μ dx = 2 Umax - h² dp 3 8μ dx Correct Option: A
dQ = Area × velocity
Q = A × V= - h² dp = (h × 1) × Umax 12μ dx ∴ Uavg = - h² dp 12μ dx
At y = 0, U = Umax∴ Umax = h² dp 8μ dx ∴ = - h² dp Uavg 12μ dx = 2 Umax - h² dp 3 8μ dx
- For a sample of water with the ionic composition shown in the figure below, the carbonate and non-carbonate hardness concentration (in mg/I as CaCO3), respectively are:
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Carbonate hardness = 3.5 × 10–3 g-eq.
= 3.5 × 10–3 × 50g as CaCO3 l
= 175 mg/l as CaCO3
Non carbonate Hardness = Total hardness – carbonate hardness
Total hardness = 5 × 50 mg/l as CaCO3
= 250 mg/l as CaCO3
∴ Non carbonate hardness = 250 – 175 = 75 mg/l as CaCO3Correct Option: C
Carbonate hardness = 3.5 × 10–3 g-eq.
= 3.5 × 10–3 × 50g as CaCO3 l
= 175 mg/l as CaCO3
Non carbonate Hardness = Total hardness – carbonate hardness
Total hardness = 5 × 50 mg/l as CaCO3
= 250 mg/l as CaCO3
∴ Non carbonate hardness = 250 – 175 = 75 mg/l as CaCO3
- A student began experiment for determination of 5-day, 20° C BOD on Monday. Since the 5th day fell on Saturday, the final DO readings were taken on next Monday. On calculation, BOD (i.e. 7 day, 20° C) was found to be 150 mg/L. What would be the 5-day, 20°C BOD (in mg/L)? Assume value of BOD rate constant (k) at standard temperature of 20° C as 0.23/day (base e).
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128.0979 mg/L
kD = 0.434 × 0.23 = 0.0998
BOD7 = L[1 - 10-kDt]
∴ 150 = L[1 – 10(–0.0998 × 7)]
⇒ L = 187.54 mg/L
∴ BOD5 = L[1 - 10-kDt]
= 187.539 × [1 – 10–0.0998 ×5]
= 128.0979 mg/LCorrect Option: C
128.0979 mg/L
kD = 0.434 × 0.23 = 0.0998
BOD7 = L[1 - 10-kDt]
∴ 150 = L[1 – 10(–0.0998 × 7)]
⇒ L = 187.54 mg/L
∴ BOD5 = L[1 - 10-kDt]
= 187.539 × [1 – 10–0.0998 ×5]
= 128.0979 mg/L
- A settling tank in a water treatment plant is designed for a surface overflow rate of 30(m³/day.m²) . Assume specific gravity of sediment particles = 2.65, density of water (ρ) = 1000 kg/m³, dynamic viscosity of water (m) = 0.001 N.s/m², and Stokes law is valid. The approximate minimum size of particles that would be completely removed is
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To calculate minimum size of particles equating settling velocity to overflow rate, we get
Vo = VsVo = (Gsl)&gaamma;wd² 18μ Now Vo = 30 m³ = 30 m day.² 24 × 60 × 60 s ⇒ 30 = (2.65 - 1) × 9.81 × 10³ × d² 24 × 60 × 60 18 × 0.001
Solving we get d = 1.965 × 10–5m ≈ 0.02 mmCorrect Option: B
To calculate minimum size of particles equating settling velocity to overflow rate, we get
Vo = VsVo = (Gsl)&gaamma;wd² 18μ Now Vo = 30 m³ = 30 m day.² 24 × 60 × 60 s ⇒ 30 = (2.65 - 1) × 9.81 × 10³ × d² 24 × 60 × 60 18 × 0.001
Solving we get d = 1.965 × 10–5m ≈ 0.02 mm
- A town is required to treat 4.2 m³/min of raw water for daily domestic supply. Flocculating particles are to be produced by chemical coagulation. A column analysis indicated that an overflow rate of 0.2 mm/s will produce satisfactory particle removal in a settling basin at a depth of 3.5 m. The required surface area (in m²) for settling is
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Q = 4.2 m³/min = 0.07 m³/s
Vo = 0.2 mm/s = 0.2 × 10–3 m/sVo = Q A ∴ A = Q Vo = 0.07 = 350 m² .2 × 10-3 Correct Option: B
Q = 4.2 m³/min = 0.07 m³/s
Vo = 0.2 mm/s = 0.2 × 10–3 m/sVo = Q A ∴ A = Q Vo = 0.07 = 350 m² .2 × 10-3
