Work saved in percentage For double acting pump

𝑄 = 2𝐴𝐿𝑁

60 = 2 𝐴 𝑎

_𝑠

𝜔𝑟 𝑃

₁

− 𝑃

₂

𝜋

𝑃

₁

× 100%

= 3

2 × 2

3 − 4

𝜋

²

× 100% = 𝟑𝟗. 𝟐%

Effect of Air Vessels

Problem–3: A single acting reciprocating pump has a plunger diameter of 125 mm and stroke length of 300 mm. The length of the suction pipe is 10 m and diameter 75 mm. (i) Find acceleration head at the beginning, middle and end of suction stroke. (ii) If the suction head is 3 m, determine the pressure head in the cylinder at the beginning of stroke when the pump runs at 30 rpm. (iii) Under the circumstance, what can be the maximum running speed of the pump without separation (cavitation). Take atmospheric pressure as 10.3 m of water, and the water vapor pressure as 2.6 m of water abs.

Solution: (i) 𝐻_𝑎𝑠 = ^𝑙_𝑔^𝑠_𝑎^𝐴

𝑠𝜔²𝑟𝑐𝑜𝑠𝜃 = _9.81¹⁰ × ^0.125_0.075²₂ × ^2𝜋×30₆₀ ² × 0.15 × 𝑐𝑜𝑠𝜃 = 4.192𝑐𝑜𝑠𝜃

(ii)

𝐻_{𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟} 𝑎𝑏𝑠 = 𝐻_𝑎𝑡𝑚 − 𝐻_𝑠 − 𝐻_𝑎𝑠 − 𝐻_𝑓𝑠 = 10.3 − 3 − 4.192 − 0 = 3.108 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 (𝑎𝑏𝑠)

(iii)

𝐻_{𝑠𝑢𝑐𝑡𝑖𝑜𝑛} 𝑎𝑏𝑠 = 𝐻_𝑎𝑡𝑚 − 𝐻_𝑠 − 𝐻_𝑎𝑠 − 𝐻_𝑓𝑠 = 2.6 (𝑎𝑏𝑠) 𝐻_𝑎𝑡𝑚−𝐻_𝑠 − ^𝑙_𝑔^𝑠_𝑎^𝐴

𝑠

2𝜋𝑁_𝑚𝑎𝑥 60

2 𝑟 = 2.6 (𝑎𝑏𝑠)

⇒ 10.3 − 3 − _9.81¹⁰ × ^0.125_0.075²₂ × ^2𝜋𝑁₆₀^{𝑚𝑎𝑥 2} × 0.15 = 2.6 ∴ 𝑁_𝑚𝑎𝑥 = 31 𝑟𝑝𝑚

Example Problems

Condition 𝜽 𝑯_𝒂𝒔

Beginning of stroke 0° 4.192 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

Middle of stroke 90° 0

End of stroke 180° −4.192 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

Problem–4: A single acting reciprocating pump has a piston of diameter 75 mm and a stroke of 150 mm. It draws water from a sump 3.5 m below the pump through a pipe of 5 m long. If separation occurs at 78.46 kPa below atmospheric pressure when the pump runs at 45 rpm, find the diameter of suction pipe for no separation. Assume simple harmonic motion of the piston.

Solution:

−𝐻

_𝑠

−

^𝑙𝑠

𝑔 𝐴 𝑎_𝑠

2𝜋𝑁_𝑚𝑎𝑥 60

2

𝑟 ≥

^{−78.46×103}

9810

−3.5 − 5

9.81 × 0.075

²

𝑑

_𝑠²

× 2𝜋 × 45 60

2

× 0.075 ≥ −8 𝑑

_𝑠

≥ 32.6 𝑚𝑚

Example Problems

Problem–5: A single acting reciprocating pump has the following characteristics:

Cylinder diameter = 225 mm; Stroke length = 450 mm.

Suction head = 4.5 m; Diameter of suction pipe = 225 mm;

Suction pipe length = 20 m;

Atmospheric pressure = 10 m water (abs) Cavitation pressure = 2 m water (abs)

Determine the maximum speed at which the pump can be run without cavitation.

Solution:

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

− 𝑙

_𝑠

𝑔

𝐴 𝑎

_𝑠

2𝜋𝑁

_𝑚𝑎𝑥

60

2

𝑟 = 2 (abs) 10 − 4.5 − 20

9.81 × 0.225

²

0.225

²

× 2𝜋 × 𝑁

_𝑚𝑎𝑥

60

2

× 0.225 = 2 𝑁

_𝑚𝑎𝑥

= 26 𝑟𝑝𝑚

Example Problems

Problem–6: A single acting reciprocating pump has the following characteristics:

Piston diameter = 100 mm; Stroke length = 300 mm.

Suction head = 4 m; Diameter of suction pipe = 75 mm;

Suction pipe length = 4 m;

Atmospheric pressure = 10 m water (abs) Cavitation pressure = 2.5 m water (abs)

Determine the maximum speed at which the pump can be run without cavitation.

Assume Frictional losses = 1 m Solution:

𝐻

_𝑐𝑦𝑙

𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │

_{𝑠𝑢𝑐𝑡𝑖𝑜𝑛}

= 𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

− 𝐻

_𝑎𝑠

− 𝐻

_𝑓𝑠

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

− 𝑙

_𝑠

𝑔 𝐴 𝑎

_𝑠

2𝜋𝑁

_𝑚𝑎𝑥

60

2

𝑟 − 𝐻

_𝑓𝑠

= 2.5 (abs) 10 − 4 − 4

9.81 × 0.1

²

0.075

²

× 2𝜋 × 𝑁

_𝑚𝑎𝑥

60

2

× 0.15 − 1 = 2.5 𝑁

_𝑚𝑎𝑥

= 45 𝑟𝑝𝑚

Example Problems

Problem–7: A double acting reciprocating pump has the following characteristics:

Cylinder diameter = 200 mm; Stroke length = 200 mm.

Diameter of suction pipe = 200 mm;

Suction pipe length = 15 m;

Atmospheric pressure = 10 m water (abs) Cavitation pressure = 2.5 m water (abs) Speed of the pump = 60 rpm

Determine the maximum suction lift without separation.

Neglect frictional losses.

Solution:

𝐻

_𝑐𝑦𝑙

𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │

_{𝑠𝑢𝑐𝑡𝑖𝑜𝑛}

= 𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

− 𝐻

_𝑎𝑠

− 𝐻

_𝑓𝑠

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

− 𝑙

_𝑠

𝑔 𝐴 𝑎

_𝑠

2𝜋𝑁 60

2

𝑟 − 𝐻

_𝑓𝑠

= 2.5 (abs) 10 − 𝐻

_𝑠

− 15

9.81 × 0.2

²

0.2

²

× 2𝜋 × 60 60

2

× 0.1 − 0 = 2.5 𝐻

_𝑠

= 1.46 𝑚

Example Problems

Example Problems

𝐻

_𝑐𝑦𝑙

𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 │

_{𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑦}

= 𝐻

_𝑎𝑡𝑚

+ 𝐻

_𝑑

+ 𝐻

_𝑎𝑑

+ 𝐻

_𝑓𝑑

≥ 2.75 (𝑎𝑏𝑠) 𝐻

_𝑎𝑡𝑚

+ 𝐻

_𝑑

− 𝑙

_𝑑

𝑔 𝐴

𝑎

_𝑑

𝜔

²

𝑟 + 0 ≥ 2.75 9.75 + 40 − 45

9.81 × 0.2

²

0.1

²

× 2𝜋𝑁

_𝑚𝑎𝑥

60

2

× 0.4 = 2.75 𝑁

_𝑚𝑎𝑥

= 24 𝑟𝑝𝑚

Problem–8: A single acting reciprocating pump has a piston of diameter 200 mm with a crank of radius 400 mm. It delivers water to a tank with 100 mm diameter and 45 m long delivery pipe. Water is lifted to a height of 40 m above the axis of the cylinder. Find the maximum speed at which the pump can be run without cavitation. Assume atmospheric pressure 9.75 m water abs, and cavitation occurs at 2.75 m water abs.

Solution:

Example Problems

Solution:

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

−

^𝑙𝑠

𝑔 𝐴 𝑎_𝑠

2𝜋𝑁_𝑚𝑎𝑥 60

2

𝑟 = 2.5

10 − 3.5 − 7.5

9.81 × 0.3

²

𝑑

_𝑠²

× 2𝜋 × 40 60

2

× 0.1875 = 2.5 𝑑

_𝑠

= 237.9 𝑚𝑚

Problem–9: A double acting single cylinder reciprocating pump has a piston of diameter 300 mm and a stroke of 375 mm. The pump axis is 3.5 m above the water level in the sump. The suction pipe is 7.5 m long. Find the diameter of the suction pipe so that minimum stipulated pressure head from cavitation considerations is not violated. Assume atmospheric pressure 10 m water abs, and cavitation occurs at 2.5 m water abs.

Problem–10: A single acting reciprocating pump has a piston of diameter 200 mm and a stroke of 300 mm. It draws water from a sump 3.5 m below the pump through a pipe of 5.5 m long. The separation pressure head is 2.5 m water abs, and atmospheric pressure is 10.3 m abs. When the pump runs at 60 rpm, find the minimum diameter of suction pipe for no separation. Assume simple harmonic motion of the piston.

Solution:

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

−

^𝑙𝑠

𝑔 𝐴 𝑎_𝑠

2𝜋𝑁 60

2

𝑟 ≥ 2.5

10.3 − 3.5 − 5.5

9.81 × 0.2

²

𝑑

_{𝑠│𝑚𝑖𝑛}²

× 2𝜋 × 60 60

2

× 0.15 = 2.5 𝑑

_{𝑠│𝑚𝑖𝑛}

= 0.1757 𝑚 = 176 𝑚𝑚

Example Problems

Problem–11: A double acting reciprocating pump has a 250 mm cylinder with a stroke of 400 mm. The suction pipe is 5 m long and the suction lift is 3 m. If the speed of the crank is 25 rpm, determine the minimum diameter of the suction pipe to prevent occurrence of cavitation. The minimum pressure from cavitation consideration is limited to 2.5 m of water abs. Assume atmospheric pressure as 10 m of water abs.

Solution:

𝐻

_𝑎𝑡𝑚

− 𝐻

_𝑠

−

^𝑙𝑠

𝑔 𝐴 𝑎_𝑠

2𝜋𝑁 60

2

𝑟 ≥ 2.5

10 − 3 − 5

9.81 × 0.25

²

𝑑

_{𝑠│𝑚𝑖𝑛}²

× 2𝜋 × 25 60

2

× 0.2 = 2.5 𝑑

_{𝑠│𝑚𝑖𝑛}

= 0.0985 𝑚 = 98.5 𝑚𝑚

Example Problems

Problem–12: A single acting reciprocating pump has 125 mm diameter cylinder with a stroke of 500 mm. The length and diameter of the suction pipe are 5.2 m and 100 mm respectively. The suction lift is 3.25 m and the delivery lift is 12 m.

The pump speed is 45 rpm. If an air vessel is fitted very close to the cylinder in the delivery side, calculate power required to pump water. Assume the frictional head in the delivery pipe to be 0.15 m and the velocity heads in the pipes can be neglected. Take, 𝜂_{𝑝𝑢𝑚𝑝} = 0.9 𝑎𝑛𝑑 𝑓 = 0.02.

Example Problems

𝑃 = 50.16 × [(3.25 + 4.6 +2

3 × 0.18 + 0) + (12 + 0.15 + 0)]

= 1009.22 𝑊𝑎𝑡𝑡 = 𝟏. 𝟎𝟏 𝒌𝑾 𝑨𝒏𝒔.

Solution: Power required,

2𝛾𝐴𝑟𝑁

60𝜂_{𝑝𝑢𝑚𝑝} = 2 × 9810 ×𝜋

4 × 0.125² × 0.25 × 45

60 × 0.9 = 50.16 𝑁/𝑠

Here,

𝑃 = 2𝛾𝐴𝑟𝑁

60𝜂_{𝑝𝑢𝑚𝑝} 𝐻_𝑠 + 𝐻_𝑎𝑠 +2

3𝐻_𝑓𝑠+𝑉_𝑠²

2𝑔 + 𝐻_𝑑 +2

3𝐻_𝑓𝑑2 + 𝐻_𝑓𝑑1+𝑉_𝑑² 2𝑔

𝐻_𝑓𝑠 = 𝑓 𝑙_𝑠 𝑑_𝑠

1 2𝑔

𝐴 𝑎_𝑠𝜔𝑟

2

= 0.02 × 5.2

0.1 × 1

2 × 9.81 × 0.125²

0.1² ×2𝜋 × 45

60 × 0.25

2

= 0.18 𝑚 𝐻_𝑎𝑠 = 𝑙_𝑠

𝑔 𝐴

𝑎_𝑠𝜔² 𝑟 = 5.2

9.81 × 0.125²

0.1² × 2𝜋 × 45 60

2

× 0.25 = 4.6 𝑚

2

3𝐻_𝑓𝑑2+ 𝐻_𝑓𝑑1 = 0.15 𝑚

Therefore,

Problem–13: A single acting reciprocating pump has 100 mm diameter piston with a stroke of 200 mm. The static suction head is 4 m, the diameter of the suction pipe is 75 mm and the length of the suction pipe is 8 m. The diameter and length of the delivery pipe are 100 mm and 30 m respectively. The static delivery head is 25 m.

An air vessel is fitted very near to the cylinder on the suction side and another at a distance of 3 m from the cylinder on the delivery side. If the pump speed is 60 rpm, estimate the power required to drive the pump. Take, 𝜂_{𝑝𝑢𝑚𝑝} = 0.9 𝑎𝑛𝑑 𝑓 = 0.025

.

Example Problems

𝑃 = 17.12 × 4 + 0.017 + 25 + 1.207 + 2

3× 0.015 + 0.014 = 517.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟏𝟖 𝒌𝑾 𝑨𝒏𝒔

Solution: Power required,

2𝛾𝐴𝑟𝑁

60𝜂_{𝑝𝑢𝑚𝑝} = 2 × 9810 ×𝜋

4 × 0.1² × 0.1 × 60

60 × 0.9 = 17.12 𝑁/𝑠

Here

𝑃 = 2𝛾𝐴𝑟𝑁

60𝜂_{𝑝𝑢𝑚𝑝} 𝐻_𝑠 + 𝐻_𝑓𝑠1 + 𝐻_𝑑 + 𝐻_𝑎𝑑2 +2

3𝐻_𝑓𝑑2 + 𝐻_𝑓𝑑1

𝐻_𝑓𝑠1 = 𝑓 𝑙_𝑠 𝑑_𝑠

1 2𝑔

𝐴 𝑎_𝑠

𝜔𝑟 𝜋

2

= 0.025 × 8

0.075 × 1

2 × 9.81 × 0.1²

0.075² ×2𝜋 × 60

60 × 𝜋 × 0.1

2

= 0.017 𝑚 𝐻_𝑎𝑑2 = 𝑙_𝑑2

𝑔 𝐴

𝑎_𝑑𝜔² 𝑟 = 3

9.81 × 0.1²

0.1² × 2𝜋 × 60 60

2

× 0.1 = 1.207 𝑚 𝐻_𝑓𝑑2 = 𝑓𝑙_𝑑2

𝑑_𝑑 1 2𝑔

𝐴 𝑎_𝑑𝜔𝑟

2

= 0.025 × 3

0.1 × 1

2 × 9.81 × 0.1²

0.1² ×2𝜋 × 60

60 × 0.1

2

= 0.015 𝑚 𝐻_𝑓𝑑1 = 𝑓𝑙_𝑑1

𝑑_𝑑 1 2𝑔

𝐴 𝑎_𝑑

𝜔𝑟 𝜋

2

= 0.025 × 27

0.1 × 1

2 × 9.81 × 0.1²

0.1² ×2𝜋 × 60

60 × 𝜋 × 0.1

2

= 0.014 𝑚 (Neglecting velocity heads in the pipes)

Example Problems

= 2 × 9810 ×𝜋

4 × 0.3² × 0.225 0.03 × 40

0.2× 1

2 × 9.81× 0.3²

0.2² ×2𝜋 × 60

60 × 0.225

2

× 2 3 − 1

𝜋²

= 545.8 𝑊𝑎𝑡𝑡 = 𝟎. 𝟓𝟒𝟓 𝒌𝑾

𝑃

₁

− 𝑃

₂

= 2𝛾𝐴𝑟 𝑓

^𝑙𝑑

𝑑_𝑑 1 2𝑔

𝐴

𝑎_𝑑

𝜔𝑟

^{2 2}

3

−

¹

𝜋²

𝑃

₂

= 2𝛾𝐴𝑟 𝐻

_𝑓𝑑

= 2𝛾𝐴𝑟 𝑓 𝑙

_𝑑

𝑑

_𝑑

1 2𝑔

𝐴 𝑎

_𝑑

𝜔𝑟 𝜋

2

Work (Power) saved in overcoming friction

, 𝑃

₁

= 2𝛾𝐴𝑟 2

3 𝐻

_𝑓𝑑𝑚

= 2𝛾𝐴𝑟 3

2

3 𝑓 𝑙

_𝑑

𝑑

_𝑑

1 2𝑔

𝐴

𝑎

_𝑑

𝜔𝑟

2

Problem–14: A single acting reciprocating pump has an air vessel in the delivery side fitted very close to the cylinder. The cylinder has a diameter of 300 mm and a stroke length of 450 mm. The delivery pipe is 40 m long and has a diameter of 200 mm. The speed of the pump is 60 rpm. Determine the power saved by the air vessel in overcoming friction in the delivery pipe. Take friction factor, f = 0.03.

Solution: Friction Work without air vessel,

Friction Work with air vessel,

Exercises

Problem-1: A single acting reciprocating pump has the piston diameter of 300 mm, stroke length of 500 mm, rotational speed of 40 rpm and the total lift (of water) of 25 m. If the actual discharge delivered at the pump outlet is 1380 liters/minute, calculate the slip, coefficient of discharge and theoretical power in kW required to drive the pump. [Ans. Slip = 2.34%, Coefficient of discharge = 0.9766, Power = 5.78 kW]

Problem-2: A double acting reciprocating pump running at 40 rpm is discharging 1 m³/min. The pump has a stroke of 400 mm and the diameter of the piston is 200 mm. The delivery and suction heads are 20 m and 5 m respectively. Find the slip of the pump and the power required to drive the pump.

[Ans. Slip = 0.53%, Power = 4.08 kW]

Problem-3: For a single acting reciprocating pump, piston diameter is 150 mm, stroke length is 300 mm, rotational speed is 50 rpm and the water is to be raised through 18 m. Determine (i) theoretical discharge, (ii) if the actual discharge is 4 liters/s, determine volumetric efficiency, slip and actual power required. Take the mechanical efficiency as 80%. [Ans. 4.4185 l/s, 90.53%, 94.72%, 882.9 watt]

Problem-4: A double acting reciprocating pump has a cylinder of diameter 200 mm and stroke of 300 mm.

The piston makes 30 strokes/minute. Estimate the maximum velocity and acceleration in the suction pipe of diameter 200 mm and delivery pipe of diameter 250 mm.

[Ans. Suction = 0.471 m/s and 1.480 m/s² ; Delivery = 0.301 m/s and 0.947 m/s²]

Problem-5: A Plunger is fitted to a vertical pipe filled with water. The lower end of the pipe is submerged in a sump. If the plunger is drawn up with an acceleration of 5 m/s², find the maximum height above the sump level at which the plunger will work without separation. Assume atmospheric pressure = 10 m water abs, and separation occurs at 2 m water abs. Take acceleration due to gravity as 10 m/s². [Ans. H_s = 5.33 m]

Problem-6: A single acting reciprocating pump has plunger diameter 200 mm and stroke 300 mm. The suction pipe is 100 mm in diameter and 8 m long. The water surface from which the pump draws water is 5 m below the pump cylinder axis. If the pump is working at 30 rpm, find the pressure in the cylinder at the beginning, middle and end of suction stroke. Take the friction factor, f = 0.04.

[Ans. 0.4708 (beginning), 4.7205 (middle) and 10.129 (end) m water abs]