Question

Final Semi 20182019 03 04.pdt - Adobe Acrobat Reader DC File Edit View Window Help Home Tools ilovepdf merged (4). Final Sem 1 2018201... x Sign In (1 of 2) 125% Jo Share Search 'Crop Page (b) D. Export PDF Edit PDF A pump operates at 1200 rpm with blade thickness occupied 12 percent of the circumference. The specifications of the pump are as below: Parameter Inlet Outlet Radius, r (mm) 150 Blade width, b (mm) 10 7 Blade angle, (deg) 25 45 90 Create PDF Comment Combine Files 20 Organize Pages 1. Redact Draw the velocity diagram and determine the; (i) discharge Q. (ii) outlet absolute flow angle a, (iii) theoretical pressure head rise across the pump H, (iv) brake power. Protect Es Compress PDF 2. Fill & Sign Send for comme Note that the blade is designed so that the liquid enters the pump impeller radially. Take pwater = 1000 kg/m3 (8 marks) Create, edit and sign PDF forms & agreements Start Free Trial El ENG 2:35 PM 1/7/2020

Answer 1

Solutions

Given Data :-

T1 = 90 mm

r2 = 150 mm

bi 10 mm

b2 = 7 mm

0 B1 = 25

s = 45°

N = 1200 rpm

C - 90*

t1 0

.

Velocity Diagram at inlet :

Vrel,1 V1 = Vr1 B1 αι U

.

Velocity Diagram at outlet :

Vrel.2 V2 Vr2 \B2 d2 Viz U2

.

The actual inlet flow area

:. A1 = (21 – 0.12(21))rıbı = 1.76arib

:. A1 = 1.767(0.09) (0.01)

1. A1 = 4.9763 x 10-3 m

.

The actual outlet flow area

A2 = 1.767122

:. A1 = 1.767(0.15)(0.007

:. 42 = 5.805664 x 10-3 m? 2

.

The speed of the blade at inlet and exit:

.:. Uι 2πη,Ν 60 2π(0.09) (1200) 60 Ξ 11.3098 m/s

::U2 = 21r2N 60 27 (0.15) (1200) 60 = 18.85 m/s

.

The velocity in the radial direction at inlet is written as:

V1 = V, = Uitan 31

.: V1 = (11.3098)tan 25

:: V1 = 5.27385 m/s

.

Part i)

The discharge of the pump is written as:

::Q = A1V1

::Q = (4.9763 x 10-) (5.27385)

3 ::Q = 0.026244 m S

.

From discharge we get, the radial component of velocity at outlet;

:: V-2 2

.:: Vr2 = 0.026244 5.805664 x 10-3

:: Vr2 = 4.52042 m/s

.

Part ii)

From the velocity triangle at outlet we get;

.: V2 = U2 - Vr2 cot B2

.: V2 = 18.85 18.85 – 4.52042 cot 45

:. Vt2 = 14.3296 m/s

.

Vr2 As we know, tan 02 = V12

.. tan 02 = 4.52042 14.3296

..0 = 17.51°

.

Part iii)

The theoretical pump head is given as:

U2V12 U1V11 :: Hth = U2Vt2 9 9

:: Hth = (18.85) (14.3296) 9.81

Нth 27.5345 m

.

$\textbf{Part iv)}$

$\textup{The brake power of pump is given as:}$

$\therefore \dot{W}_{pump}=\rho g Q H_{th}$

. W pump = 7088.845 W = 7.09 kW = 9.5063 hp