Basic Mechanical Engineering (3110006)

Question-3c

BE | Semester-1 Summer-2019 | 04-06-2023

Q3) (c)

Given Data:

D = 150 mm = 0.15 m

L = 225 mm = 0.225 m

h_{s} = 4 m

h_{d} = 12 m

N = 80 rpm

Q_{act} = 0.61 \frac{m^{3}}{\min} = 0.0101 \frac{m^{3}}{s}

η = 80 %

Find:

% Slip

C_{d}

P

Solution:

% Slip

Q_{th} = \frac{2 \times A \times L \times N}{60} = \frac{\frac{π}{4} \times 0 . 15^{2} \times 0.225 \times 80}{60} = 0.0106 \frac{m^{3}}{s}

% Slip = \frac{Q_{th} - Q_{act}}{Q_{th}} \times 100 % = \frac{0.0106 - 0.0101}{0.0106} \times 100 % = 0.049

C_{d}

C_{d} = \frac{Q_{act}}{Q_{th}} = \frac{0.0101}{0.0106} = 0.95

P

P = \frac{2 \times ρg \times A \times L \times N \times (h_{s} + h_{d})}{Q_{th}}

P = \frac{2 \times 100 \times 9.81 \times \frac{π}{4} \times 0 . 15^{2} \times 0.225 \times 80 \times (4 + 12)}{Q_{th}}

P = 1.66 kW

Actual power required is,

P_{act} = \frac{P}{C_{d}} = \frac{1.66}{0.95} = 2.075 kW