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2 years ago

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A complete mix of an activated sludge system without primary clarification is used for treatment of municipal wastewater with a

flowrate of 12 MGD. The plant components include four clarifiers with a diameter of 80ft and a side-water depth of 14ft. The aeration basin volume is 3 million gallons, the mixed liquor suspended solid concentration of 3,000 mg/L, and the solid's residence time is 5 days. The return activated sludge flow rate is 10 MGD. The effluent suspended solids concentration is negligible. Sludge is wasted from the aeration basin. What is the return of suspended solids concentration (mg/L)

1 answer:

Hitman42 [59]2 years ago

4 0

Answer:

sorry di ko alam

Explanation:

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3 years ago

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness

Bingel [31]

Answer:

a) heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

Explanation:

Assumptions:

Constant properties
Steady state conditions
Negligible effect of radiation
Negligible constant resistance between tube and insulation
one dimensional radial conduction

a) What is the heat gain per unit tube length

$R_{conv,i}'=\frac{1}{2\pi r_1h_i}$

$d_1=36mm$ Therefore $r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}$

$r_2=2mm=2*10^{-3}m$

$k_{st}=14.2W/m.k$

$h_o=6W/m^2$

$h_i=400W/m^2$

$R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W$

$R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W$

$R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W$

$R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W$

heat gain per unit tube length = $\frac{23-6}{1.35} = 12.6W/m$

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

$r_3=r_1+r_2+10mm=30mm=0.03m$

$R_{conv,i}'$ and $R_{cond,st}'$ are the same, but $R_{conv,o}'$ changes.

Therefore:

$R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W$

$R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W$

The total resistance $R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W$

heat gain per unit tube length = $\frac{23-6}{2.20} = 7.7W/m$

8 0

3 years ago

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