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Vitek1552 [10]
3 years ago
6

A sedimentation basin in a water treatment plant has a length = 48 m, width = 12 m, and depth = 3 m. The flow rate = 4 m 3 /s; p

article specific gravity = 1.1; water density = 10 3 kg/m 3 ; and dynamic viscosity = 1.30710 -3 N.sec/m 2 . What is the minimum particle diameter that is removed at 85%?
Engineering
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

The minimum particle diameter that is removed at 85% is 1.474 * 10 ^⁻4 meters.

Solution

Given:

Length = 48 m

Width = 12 m

Depth = 3m

Flow rate = 4 m 3 /s

Water density = 10 3 kg/m 3

Dynamic viscosity = 1.30710 -3 N.sec/m

Now,

At the minimum particular diameter it is stated as follows:

The Reynolds number= 0.1

Thus,

0.1 =ρVTD/μ

VT = Dp² ( ρp- ρ) g/ 10μ²

Where

gn = The case/issue of sedimentation

VT = Terminal velocity

So,

0.1 = Dp³ ( ρp- ρ) g/ 10μ²

This becomes,

0.1 = 1000 * dp³ (1100-1000) g 0.1/ 10 *(1.307 * 10 ^⁻3)²

= 3.074 * 10 ^⁻6 = dp³ (.g01 * 10^6)

dp³=3.1343 * 10 ^⁻12

Dp minimum= 1.474 * 10 ^⁻4 meters.

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Answer:

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Explanation:

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Step 1: Determine the rate of heat transfer in the heat exchanger

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Q=1.2*4.18*(80-20)

Q=1.2*4.18*(80-20)

Q=300.96kW

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Q=m_{h}C_{h}(T_{h,in}-T_{h,out})

300.96=2*4.18*(160-T_{h,out})

T_{h,out}=124\°C

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

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dT_{1}=160-80

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Explanation:

Given

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R=\frac{12100}{100}

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Answer:

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Explanation:

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From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

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Thus we get that the strength coefficient is K = 591.87 MPa

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