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Bingel [31]
3 years ago
12

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced press

ure PR approaches zero. How do you explain this behavior?
Engineering
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

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g A 12,000 m3/day treatment plant has a rectangular sedimentation basin with dimensions 12 meters wide, 3 meters deep, and 25 me
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Answer:

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

Explanation:

Given;

volumetric flow rate of the treatment, Q₀ = 12,000 m³/day

length of the rectangular tank, L = 25 m

width of the tank, W = 12 m

height of the tank, H = 3 m

settling velocity of the particles, V_s = 6 x 10⁻³ m/s

The overflow rate of the sediments are calculated as follows;

V_o = \frac{Q_o}{A_s}

where;

As is the surface area of the tank, m²

Q₀ is the flow rate, m³/s

As = 2LW + 2LH + 2WH

As = (2 x 25 x 12) + (2 x 25 x 3) + (2 x 12 x 3)

As = 822 m²

Q_0 (m^3/s)= \frac{12,000 \ m^3}{day} \times \frac{1 \ day}{24 \ hr} \times \frac{1 \ hour}{60 \ \min} \times \frac{1 \ \min}{60 \ s} = \frac{12,000}{24 \times 60 \times 60}  (m^3/s)= \frac{12,000}{86,400} \ m^3/s\\\\Q_o = 0.139 \ m^3/s

The overflow rate;

V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

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