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aleksklad [387]
3 years ago
11

A sludge pump pumps at a rate of 40 GPM. The raw sludge density is 4.5 percent solids. How many minutes per hour should the pump

operate if the inflow to the clarifier is 1.8 MGD, the influent suspended solids are 210 mg/L, and the effluent suspended solids are 60 mg/L
Engineering
1 answer:
Sergio039 [100]3 years ago
3 0

Answer:

The pump should operate at 6.25 minutes per hour

Explanation:

We have to find the minute per day pumping rate first.

Therefore,

Sludge removal lbs/ day = Sludge pumped lbs/day

Formula is given as:

(Suspended solids removed (mg/L) × Clarifier inflow(MGD) × 8.34lbs/gal) ÷ ( solids contents in % ÷ 100) = GPM Sludge pumping rate × minute per day pumping rate × 8.34 lbs/gal

From the question,

Suspended solids removed is calculated as : Influent suspended solids concentration - Effluent suspended solids concentration

= 210 mg/L - 60mg/L

Suspended solids removed = 150mg/L

Clarifier Inflow(MGD) = 1.8 MGD

Sludge density = 4.5% solids

GPM = 40 GPM

Minute per day pumping rate = is unknown so we represent it as y

Hence,

(150 mg/L × 1.8 MGD Flow × 8.34lbs/gallon) ÷ ( 4.5% ÷ 100) = 40 GPM × y minute per day pumping rate × 8.34 lbs/gallon

(150 mg/L × 1.8 MGD Flow × 8.34lbs/gallon) ÷ 0.045 = 40 GPM × y minute per day pumping rate × 8.34 lbs/gallon

50040 = 333.6y minute per day

y = 50040 ÷ 333.6

y = 150 minute per day pumping rate

In other to get the minute per hour pumping rate, we would convert the minute per day to minute per hour,

Therefore,

24 hours = 1 day

So we have,

150 minute per day ÷ 24 hours

= 6.25minute per hour

The pump should operate at 6.25 minutes per hour.

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A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
3 years ago
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Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

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K= rate constant;

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E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

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