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3 years ago

8

A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov

e that of the first. the total length of pipes required is 1000 ft. All pipes are 12 in. in diameter and are made of galvanized iron with relative roughness equal to 0.0005 (you may assume fully-rough flow). the pump performance curves suggest that the H-Q relationship is of the form: H_pump=665-0.051Q^2 (Q in ft) the expected flow rate the brake horsepower required to drive the pump (assume an efficiency of 78%). the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

1 answer:

babunello [35]3 years ago

6 0

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H $_{pump$ = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H $_p$ = (p/ρq)₂ + (v²/2g)₂ + z₂ + h $_L$

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH $_p$ Q / η

where; H $_p$ = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ( $P_{atom$ / ρg) - h $_s$ - ( P $_v$ / ρg )

we substitute

25 = ( 2116 / 62.42 ) - h $_s$ - ( 30 / 62.42 )

h $_s$ = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

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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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Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

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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,

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

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

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$\Rightarrow n_p=13$

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$T_t-1min$

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(f) The proportion of time spent actually cutting metal

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To know more about Rankine cycle, visit: brainly.com/question/13040242

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