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Nataly_w [17]
3 years ago
11

Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter

under a potential difference of 100 V. The zeta potential at the capillary wall is 80 mV
Engineering
1 answer:
natita [175]3 years ago
8 0

Answer:

Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface \xi =80mV=0.08volt

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as \epsilon =10

Viscosity of glass is \eta =10^8

Electroosmotic velocity is given as v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}

v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec

So Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

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3 years ago
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an
ser-zykov [4K]

Answer:

volumetric flow rate = 0.0251 m^3/s

Velocity in pipe section 1 = 6.513m/s

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = 0.0251 m^3/s

volumetric flow rate = 0.0251 m^3/s

Average velocity calculations:

<em>Pipe section A:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2

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velocity = mass flow rate /(density X cross-sectional area)

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<em>Pipe section B:</em>

cross-sectional area =

\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2

mass flow rate = density X cross-sectional area X velocity

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7 0
2 years ago
A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3
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Answer:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

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Answer:

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Answer:

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