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This section examines the Construction of Velocity Triagles and their use in the establishment of Manometric Head and Efficiency

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Introduction

The types of pump can be broadly classified under three headings. Centrifugal; Reciprocating; and axial. This submission only deals with the first category. Centrifugal Pumps are used in a huge number of applications and are found in all

industries where it is necessary to move liquids around. They vary from the simple impellers employed in domestic appliances to the huge machines used by electric pumped storage schemes. What ever the size or design they all have one thing in common. They will only work with a flooded input and a full casing. This fact determines the positioning of pumps within a machine or demands means whereby the pump casing is primed before use (i.e.Filled with liquid). Where the suction hose is long, it is usually necessary to fit a non return foot valve and to ensure that the hose is also primed. When pumps are required to be portable and this is particularly a requirement of the construction industry, it is possible to use self priming pumps

These contain enough liquid in the casing to prime the pump. This obviously comes at a price. Self priming pumps are both bulky and heavy and they will not work with large suction heads since there is not enough liquid in the pump body for a long suction hose.

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In recent years submersible pumps have become popular for a whole range of applications,many of which are domestic.

In their simplest form they are light and can be used for pumping out flooded cellars; Irrigating the garden and driving a fountain in a small fish pond. Larger and more powerful submersibles are used to pump water from wells and bore holes.

All centrifugal and reciprocating pumps rely on producing an area of low pressure into which liquid is sucked. This can be a major problem because if the pressure is too low the liquid in effect "boils" and the pump ceases to work. For this reason it is not possible to have a suction head of more than 28 ft. Associated with this is Cavitation. When a centrifugal pump can not get enough liquid small bubbles form in the areas of least pressure. As these move into an area of slightly higher pressure they implode violently. A cavitating pump sounds as if it is pumping gravel and considerable damage may be done to parts of the impeller. A submission on Cavitation can be found in our Fluid Mechanics section.

Centrifugal Pumps can be regarded as the inverse of a reaction turbine.The flow is always outwards and they convert mechanical energy into a head of liquid. Sadly they are less efficient than a turbine because of smaller losses when converting pressure energy to kinetic energy than visa versa. The rotating impeller creates a forced vortex in the passages of the pump and the efficiency depends largely upon the extent to which the high velocity head is converted to pressure head.

Recuperators

Whilst all centrifugal pumps share many features, the internal design allows them to be classified as follows.

Vortex Chamber

The gradual increase of area decreases the velocity of flow and increases the pressure.

Losses are high due to eddy currents caused by the radial component of flow from the impeller. There is also an out of balance radial thrust.

Parallel Vortex Chamber

A free spiral vortex is formed in the parallel vortex chamber.

This helps to steady the flow before discharging into the spiral vortex.

Diffuser Also known as a turbine pump.

In this type of pump there is a diffuser which consists of a fixed ring of guide vanes surrounding the impeller.

In this design velocity energy is converted into pressure energy in the diffuser which discharges into the volute.

The inlet angle of the guide vanes equals the outlet absolute velocity angle of the impeller.

Velocity triangles.

These are similar to those produced for turbines but the flow is now outwards.

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a) It is usual to assume that water enters the impeller radially. i.e. without whirl.

$\alpha = 90^0;\;\;\;\;V_w = 0;\;\;\;V = V_f.$

b) $\beta$ = The diffuser vane angle - if any

c) $\frac{V}{r} = \frac{V_1}{r_1} = \omega = \frac{2\pi N}{60}$

d) $V_f=\frac{Q}{\text{Circumferential Area}}=\frac{Q}{K\pi Db}$
e) And $V_{f1}=\frac{Q}{K\pi D_1b_1}$

The Euler Equation

The work done by the vanes on the water $= \frac{V_{w1}v_1 - V_wv}{g}\;ft.$

$= \frac{V_{w1}v_1}{g}\;ft.\;\;\;\;if\;\;V_w = 0$

(1)

Variation of pressure across the impeller.

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There are two methods of calculating the variation of Pressure across the Impeller. These are :-

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1. The Work done theory

The total energy at outlet = Total energy at input + Work done - Losses

$\frac{p_1}{w} + \frac{V_1^2}{2g} = \frac{p}{w} + \frac{V^2}{2g} + \frac{V_{w1}v_1}{g} - h_f$

(2)

$or\;\;\;\;\;\frac{p_1 - p}{w} = \frac{v^2 - V_1^2}{2g} + \frac{V_{w1}v_1}{g} - h_f$

(3)

2. The Vortex theory

Difference in pressure head = centrifugal head + difference in velocity head - losses

$\frac{p_1 - p}{w} = \frac{v_1^2 - v^2}{2g} + \frac{V_r^2 - V_{r1}^2}{2g} - h_f$

(4)

The Efficiency of Centrifugal Pumps.

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Apply Bernoulli's equation at (0) and (1)

$0 = \frac{p_1}{w} + \frac{V_1^2}{2g} + h_s + h_{fs}$

(5)

Therefore the reading of the pressure gauge at the suction flange is :-

$\frac{p_1}{w}= - \left( \frac{V_1^2}{2g} + h_s + h_{fs} \right)$

(6)

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Applying Bernoulli's equation at (2) and (3)

$\frac{p_2}{w} + \frac{V_2^2}{2g} = \frac{V_d^2}{2g}\;+h_d + h_{fd}$

(7)

Therefore the pressure gauge reading at the delivery flange is:-

$\frac{p_2}{w} = \frac{V_d^2}{2g}\;+h_d + h_{fd} - \frac{V_2^2}{2g}$

(8)

Thus the reading of the pressure gauge across the pump flanges equals:-

$\frac{p_2 - p_1}{w} = \left(h_d + h_s \right)+\left(h_{fd} + H_{fs} \right) + \frac{V_1^2 - V_2^2}{2g}+\frac{V_d^2}{2g}$

(9)

= (Actual lift) +(Friction head lost in pipe) + (Difference of velocity head across pump) + (the velocity head at discharge)

The Manometric Head

This is defined by British Standards as the sum of the actual lift (H) + the friction losses in the pipes + the discharge velocity head. However for special pumps, allowance must also be made for the velocity of flow towards the suction intake and any pressure differences at the water surfaces in the supply and receiving tanks.

$Thus\;\;\;\;\;H_m = H + h_f + \frac{V_d^2}{2g}$

(10)

$= \frac{p_2 - p_1}{w} + \frac{v_2^2 - V_1^2}{2g}$

(11)

Commonly the suction and delivery pipes are of equal diameter. In which case:-

$H_m = \frac{p_2 - p_1}{w}$

(12)

Note If the two pressures are registered on different gauges. A correction must be made for any difference in the datum heights of the gauges.

The Manometric efficiency

This is defined as the Manometric head divided by the work done by the impeller on the water.

$\therefore\;\;\;\;\;\eta = H_m\div \frac{V_{w1}v_1 - V_wv}{g}$

(13)

and $V_w\;is\;usually\;zero$

The Total Work Done by the Pump.

This differs by from the Euler head input $\left(\frac{V_{w1}v_1}{g} \right)$ by

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The mechanical losses in the bearings.

Disc friction on the outside of the impeller shroud.

Leakage losses in pumping water that leaks through the neck rings to waste or

back into the suction side.

The Hydraulic efficiency

This equals the Manometric head divided by the total hydraulic work done. ( This is seldom used)

The Actual Efficiency

This is obtained by dividing the water power output by the driving motor power input.

It can be shown that the Gross Lift (Manometric Head) $= AN^2 + BNQ + CQ^2$ Where N = Speed; Q = Delivery and A;B;C;are Constants.

Worked Examples

The solutions to the following worked examples have been hidden. They can be seen by clicking on the red button.

Example 1

If the static lift of a centrifugal pump is $h_s$ ft. The speed of rotation N r.p.m. and the external diameter of the Impeller id D ft. Deduce that $N=153.4\;\sqrt{\frac{h_s}{D}}$ for the speed at which the pumping begins assuming only rotation of the water in the impeller at the "no flow" condition.

Such a pump delivers 280 galls of water per min. at 1200 r.p.m.. The impeller diameter is 14 in and the breadth of the outlet $\frac{1}{2}\;in.$ . The pressure difference between the inlet and the outlet flange is $40\;lb./in^2$ . Taking the manometric efficiency as 63% , Calculate the impeller exit angle. (B.Sc. Part 1)

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To see the solution please click on the red button

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Example 2

A centrifugal Pump impeller has an external diameter of 12 inches and a discharge area of 1.2ft. The blades are bent backwards so that the direction of the the relative velocity at the discharge surface makes an angle of 145 degrees with the tangent to this surface drawn in the direction of the the impeller rotation. The diameters of the suction and delivery pipes are 12in. and 9in. respectively.

Gauges at points on the suction and delivery pipes close to the pump and each 5ft. above the level of the supply sump show heads of 12 ft. below and 62ft. above atmospheric pressure when the pump is delivering 7.2 cusec. of water at 1200r.p.m. It requires 96 h.p. to drive the pump. (B.Sc. Part 2)

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Find a) the overall efficiency.

b) The manometric or hydraulic efficiency assuming that the water enters the impeller without shock or whirl.

c) The loss of head in the suction pipe.

a) Overall efficiency

Suction Pipe Velocity $V_s = 7.2\div \frac{\pi \times 1^2}{4} = 9.17 ft/sec$

Delivery Pipe Velocity $V_d = 7.2\div \frac{\pi \times 9}{4\times 16} = 16.3 ft/sec.$

$\therefore\;\;\;\;\;Manometric\;Head = \frac{p_d - p_s}{w} + \frac{V_d^2 - V_s^2}{2g}$

(28)

$= 74 + \frac{16.3^2 - 9.17^2}{2g} = 76.82\;ft.$

(29)

$\therefore\;\;\;\;\;Pump\;out\;put = \frac{WH_m}{550} = \frac{62.4\times 7.2\times 76.82}{2g} = 62.8\;h.p.$

(30)

$Thus\;\;\;\;\;the\;overall\;\eta = \frac{62.8}{96}\equiv 65.4\%$

(31)

b) The Manometric efficiency equals the Manometric head divided by the Euler Head.

$v_1 = \frac{\pi d_1N}{60} = \frac{\pi \times 1\times 1200}{60} = 62.84\;ft/sec.$

(32)

The velocity of flow = $\frac{Q}{Area} = \frac{7.2}{1.2} = 6\;ft/sec.$

From Velocity triangle $v_1 - V_{w1} = \frac{V_{f1}}{tan35} = 8.57\;ft/sec.$

$\therefore\;\;\;\;\;V_{W1} = 62.84 - 8.57 = 54.27\;ft/sec.$

(33)

Work done per lb. of water $= \frac{54.27\times 62.84}{32.2} = 105.9\;ft.$

$\therefore\;\;\;\;\;The Manometric\;\eta = \frac{76.8}{105.9}\;\equiv \;72.5\%$

(34)

c) The Head Lost in the Suction Pipe.

Apply Bernoulli's equation to the supply sump and at the suction flange.

$0 = \frac{p_s}{w} + \frac{V_s^2}{2g} + 2s + h_{fs}$

(35)

$= - 12 + \frac{9.17^2}{64.4} + 5 + h_{fs}$

(36)

$\therefore\;\;\;\;\;h_{fs} = 5.69\;ft.$

(37)

Example 3

A Centrifugal Pump has vanes curved backwards so that the tangent to the vanes makes an angle of 30 degrees with the tangent to the periphery. The impeller diameter is 10 in. and the width 3/4 in. Vane thickness takes up 15% of the peripheral area.

Assuming that 50% of the kinetic energy at exit from the impeller is lost in eddies, calculate, on the usual assumptions, what will be the lift and manometric efficiency when discharging 400 gallons/min. at a speed of 1100r.p.m. (B.Sc. Part 2)

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$Q = 400\;galls/min. = 4000\;lb/min. = \frac{4000}{60\times 62.4} = 1.068\;ft^3/sec.$

(38)

$Thus\;\;\;\;\;V_{f1} = \frac{Q}{K\times area} = \frac{1.068}{0.85\times \pi \times \frac{10}{12}\times \frac{3}{4\times 12}} = 7.68\;ft/sec.$

(39)

$v_1 = \frac{\pi d_1N}{60} = \frac{\pi \times \frac{10}{12}\times 1100}{60} = 48\;ft/sec.$

(40)

$and\;\;\;\;\;(v_1 - V_{w1}) = V_{f1}\div tan\,30 = 13.3\;ft/sec.$

(41)

$\therefore\;\;\;\;\;V_{w1} = 48.0 - 13.3 = 34.7\;ft/sec.$

(42)

$\therefore\;\;\;\;\;Work\;done\;per\;second = \frac{V_{w1}v_1}{g}\;\;\;(Assuming\;V_w = 0)$

(43)

$= \frac{34.7\times 48}{32.2} = 51.7\;ft.$

(44)

The Casing losses $= \frac{1}{2}\frac{V_1^2}{2g} = \frac{1}{2}\frac{V_{f1}^2 + V_{w1}^2}{2g} = 9.8\;ft.$

Therefore, neglecting the impeller loss.

$H_m = 51.7 - 9.8 = 41.9\;ft.$

(45)

$And\;\;\;\;\;The\; manometric\;\eta = \frac{41.9}{51.7}\;\equiv 81\%$

(46)

Example 4

The axis of a centrifugal pump is 8 ft. above the water level in a sump and the static lift from the pump centre is 110 ft. The friction losses in the suction and delivery pipes are $3\;ft.lb./lb.$ and $25\;ft.lb./lb.$ respectively. The suction and delivery pipes are each 5 in. in diameter. The impeller is 12 in. in diameter and 0.7 in. wide at outlet and its speed is 1700 r.p.m. The water at inlet has radial flow and the blade angle at outlet is $32^0$ to the tangent to the periphery.

Calculate the horse-power to be supplied and the discharge in gals/min. assuming a manometric efficiency of 72%. (B.Sc. Part 2)

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$U_2=\frac{\pi d_2N}{60}=\frac{\pi \times 1\times 1700}{60}=89\;ft./sec.$

(47)

From the velocity diagram it can be seen that:-

$\frac{v_{f2}}{U_2-v_{W2}}=\tan 32$

(48)

Inserting known values

$\frac{v_{f2}}{89-v_{W2}}=0.625$

(49)

$\text{Thus}\;\;\;\;\;v_{W2}=89-1.6v_{f2}$

(50)

$\text{The work done per lb.}\;\;\;=\frac{U_2v_{W2}}{g}=\frac{89}{g}\left ( 89-1.6v_{f2} \right )$

(51)

$\text{But}\;\;\;\;\;Q=A_2v_{f2}$

(52)

$\therefore \;\;\;\;\;Q=\frac{\pi \times 1\times 0.7}{4\times 12}\times v_{f2}$

(53)

$\text{Thus}\;\;\;\;\;v_{f2}=21.8\times Q$

(54)

From Equations (51) and (4)

$\text{The work done per lb.}\;\;=96.5(2.55-Q)$

(55)

$\text{The manometric efficiency }\;\;=\frac{\text{The manometric head}}{\text{The work done per lb.}}=0.77$

(56)

From Equations (55) and (56)

$H_m=0.77\times 96.5(2.55-Q)$

(57)

$\text{But}\;\;\;\;H_m=H_0+h_f+\frac{{v_{d}}^{2}}{2g}=118+28+\frac{{v_{d}}^{2}}{2g}$

(58)

$\text{And}\;\;\;\;\;Q=v_d\times a=v_d\times \left ( \frac{\pi \times 25}{4\times 12^2} \right )$

(59)

$\text{Thus}\;\;\;v_d=7.33Q\;\;\;\;\text{And}\;\;\;\;\frac{{v_{d}}^{2}}{2g}=0.835Q^2$

(60)

From Equation (58)

$H_m=146+0.835Q^2$

(61)

From Equations (57) and (61)

$146+0.835Q^2=74.5(2.55-Q)$

(62)

The solution to this quadratic can be estimated by assuming that $Q^2$ is negligible compared to 146 and hence:-

$Q=0.59\;ft^3/sec.$

(63)

$\text{Hence}\;\;\;\;\;H_m\approx 146$

(64)

$\text{The mechanical efficiency }\;(0.72)=\frac{W\times H_m}{550\times \text{required horse-power}}=\frac{w\times Q\times H_m}{550\times \text{required horse-power}}$

(65)

$\therefore \;\;\;\;\text{The required horse-power}=\frac{62.4\times 0.59\times 146}{550\times 0.72}=13.6\;h.p.$

(66)

Example 5

The impeller of a centrifugal pump has an outer diameter of 10 inches and an effective outer area of $26.4\;in.^2$ . The blades are bent back so that the angle at the outlet is $148^0$ to the tangent drawn in the direction of impeller rotation.

The diameters of suction and delivery openings are 6 in. and 5 in. respectively At 1450 r.p.m. and with a delivery of $1.1\;ft^3/sec.$ the pressure heads at suction and delivery were found to be respectively 15 ft. below and 45 ft. above atmospheric pressure the points at which the pressure heads were measured being at the same level. The motor driving the pump supplied 11.1.b.h.p.