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3 years ago

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The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint

ain a constant solids residence time in the reactor for proper waste treatment. The maximum water content acceptable for disposal of the sludge is 78% and the specific gravity of the solids is 2.5. The density of the water is 1000 kg/m3. What volume of biological sludge from the activated sludge process will require disposal?

1 answer:

snow_lady [41]3 years ago

6 0

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = $\frac{100}{100 - P} M$ kg per day

put her value

mass of sludge produced = $\frac{100}{100 - 78} 7240$ kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge = $\frac{\rho sludge}{\rho water }$

and as we know that

$\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}$

$\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}$

$S sludge$ = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = $\frac{mass sludge produce}{\rho sludge}$

volume of biological sludge = $\frac{32909.09}{1152}$

volume of biological sludge = 28.566 m³ per day

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Answer:

The specific volume is reduced in 80 per cent due to isothermal compression.

Specific enthalpy remains constant.

Explanation:

Let suppose that neon behaves ideally, the equation of state for ideal gases is:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in cubic meters.

$n$ - Molar quantity, measured in kilomoles,

$T$ - Temperature, measured in kelvins.

$R_{u}$ - Ideal gas constant, measured in $\frac{kPa\cdot m^{3}}{kmol\cdot K}$ .

On the other hand, the molar quantity ( $n$ ) and specific volume ( $\nu$ ), measured in cubic meter per kilogram, are defined as:

$n = \frac{m}{M}$ and $\nu = \frac{V}{m}$

Where:

$m$ - Mass of neon, measured in kilograms.

$M$ - Molar mass of neon, measured in kilograms per kilomoles.

After replacing in the equation of state, the resulting expression is therefore simplified in term of specific volume:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$P\cdot \nu = \frac{R_{u}\cdot T}{M}$

Since the neon is compressed isothermally, the following relation is constructed herein:

$P_{1}\cdot \nu_{1} = P_{2}\cdot \nu_{2}$

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$\nu_{1}$ , $\nu_{2}$ - Initial and final specific volume, measured in cubic meters per kilogram.

The change in specific volume is given by the following expression:

$\frac{\nu_{2}}{\nu_{1}} = \frac{P_{1}}{P_{2}}$

Given that $P_{1} = 100\,kPa$ and $P_{2} = 500\,kPa$ , the change in specific volume is:

$\frac{\nu_{2}}{\nu_{1}} = \frac{100\,kPa}{500\,kPa}$

$\frac{\nu_{2}}{\nu_{1}} = \frac{1}{5}$

The specific volume is reduced in 80 per cent due to isothermal compression.

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Specific enthalpy remains constant.

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Answer:

total time = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - $e^{(-kt^n)}$ ...........1

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}

solve it we get

k = 0.003786 = 37.86 × $10^{4}$

so here

y = 1 - $e^{(-kt^n)}$

1 - y = $e^{(-kt^n)}$

take ln both side

ln(1-y) = -k × $t^n$

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Answer:

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Explanation:

given data:

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Due to incompressibility , the density remain constant.

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Answer:

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$-q_{out} - w_{out} + h_{in}-h_{out} = 0$

$w_{out} = h_{in} - h_{out}-q_{out}$

$w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}$

$w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}$

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$-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0$

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3 years ago

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Ludmilka [50]

Known :

Q = 300 L/s = 0.3 m³/s

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