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krok68 [10]
3 years ago
14

A 280 km long pipeline connects two pumping stations. It is desired to pump 0.56 m3/s of oil through a 0.62 m diameter line, the

discharge station is 250 m lower in elevation than the upstream station, and the discharge pressure is to be maintained at 300 kPa. The oil has a dynamic viscosity of 3.645 x 10-3 Pa·s and a density of 810 kg/m3. The pipe is constructed of commercial steel and the inlet pressure may be taken as 1 atmosphere. Considering the mechanical-energy friction loss Ff, determine the power required to pump the oil if the power efficiency is 70%. Develop a sketch of the piping system considered.
Ignore energy losses due to contraction, expansion & 90° bends!!!
Engineering
1 answer:
Arlecino [84]3 years ago
8 0
Can u answer myn and if u do I’ll do urs
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Answer:

(a)

<em>d</em>Q = m<em>d</em>q

<em>d</em>q = C_p<em>d</em>T

q = \int\limits^{T_2}_{T_1} {C_p} \, dT   = C_p (T₂ - T₁)

From the above equations, the underlying assumption is that  C_p remains constant with change in temperature.

(b)

Given;

V = 2L

T₁ = 300 K

Q₁ = 16.73 KJ    ,   Q₂ = 6.14 KJ

ΔT = 3.10 K       ,   ΔT₂ = 3.10 K  for calorimeter

Let C_{cal} be heat constant of calorimeter

Q₂ = C_{cal} ΔT

Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂

Q₁ - Q₂ = m C_p ΔT

number of moles of n-C₆H₁₄, n = m/M

ρ = 650 kg/m³  at 300 K

M = 86.178 g/mol

m = ρv = 650 (2x10⁻³) = 1.3 kg

n = m/M => 1.3 / 0.086178 = 15.085 moles

Q₁ - Q₂ = m C_p' ΔT

C_p = (16.73 - 6.14) / (15.085 x 3.10)

C_p = 0.22646 KJ mol⁻¹ k⁻¹

6 0
3 years ago
Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori
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Answer:

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One of the arc welding process is most often used today for welding metal thicker than blank gauge
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How to measure the quality of the output signal in ADC?
hodyreva [135]

Answer:

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Explanation:

8 0
2 years ago
A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov
babunello [35]

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_pQ / 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

6 0
3 years ago
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