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goldfiish [28.3K]

3 years ago

12

A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD

S). The drainage waters are the only source of water into the pond and evaporation is the only loss of water from the pond. Fluxes of water due to both evaporation and drainage water are 60,000 m^3 / year (the volume of the water in the lake will not change with time). If the lake is initially filled with water with 100 mg / L TDS, after how many years will the lake no longer be considered to contain freshwater?

2 answers:

lys-0071 [83]3 years ago

8 0

Answer:

<em>33.77 Years</em>

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable up to 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initial 100 mg / l

Code of tpr from drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr = 240,000 / 9,000

<em>= 33.77 Years</em>

Stels [109]3 years ago

5 0

Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

V = 160 , 000 x 103 l

The crainage gives 150 mg / l and lake has initialy 100 mg / l

Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

Cn = ( 2000 x 160, 00 x 1000 )

= 320, 000 kg

so, each year the rate increases, by 9000 kg / yr

Read level = ( 320, 000 - 16,000 )

Li = 304, 000 kg

Tr=<u>304,000</u>

900

=33.77

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Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

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$C= 80 m/min$

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$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

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$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

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$C_c= C_mT_c+\frac{C_e}{n_p}$

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$T_t-1min$

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Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

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