Engineering designEngineering Design
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This Article shows Past Questions and Answer workings in Engineering Design for Higher National Diploma (HND) Students of Mechanical Engineering Technology Auchi Polytechnic, Auchi.

This Past Questions and Answers is helpful to students in the department of mechanical Engineering offering this course; Engineering Design (MEC 312). For those currently writing their on going exams in the Department of Mechanical Engineering Technology, this is the past questions and answers for you to study.

N.B – These questions and answers are uploaded only as a guide and ease for your studies.

Below are the Questions:

QUESTION ONE

A compressor requires 20 kW to run about 400rpm. The drive is by v-belt from an electric motor at about 1000 rpm. The diameter of the pulley on the compressor shaft must be greater than 1m while the centre distance between pulleys is limited to 1.75m. The belt speed should not exceed 1800 m/min. Determine the numbers of v belts require to transmits power if each bed has a cross-sectional area of 350mm square, density 1000 kg/m³, and allowable tensile stress of 2.50MPa. The groove angle of the cross sectional area of pulley is 35°. The coefficient of friction between the belts is 0.25. Calculate the length required for each. 

QUESTION TWO

a), Briefly describe the following design concepts (i). Adaptive design (ii). Development design (iii). New design

b) Highlights the important of the following cost of materials in material selection process

(i) cost of materials (ii) Mechanical properties iii) Corrosion resistance iv) identify three materials that are widely used in manufacturing of design products

See also  DROWN AND BURN.

 QUESTION THREE

(a) What is ergonomics in engineering design?

(ii) What are the importance of ergonomics to the product design process?

(b) Briefly explain the following areas of application of ergonomics principles in design.

(i) instruments and control

(ii) Working environment

(iii) Work place

QUESTION FOUR

(a)(i) List four properties of materials used for shaft

(ii) Find the diameter of a solid shaft to transmit 300 kW at 250rpm. The ultimate shear stress of

the steel may be taken as 360 MPa and a factor of safety of 8. If a hollow shaft is to be used in

place of the solid shaft, find the outside and inside diameter when the ratio of the outside to inside diameter is 0.5.

QUESTION FIVE

a) (i) Enumerate three stresses acting on shafts

(ii). A line shaft at 300 rpm is to transmit 25kW. The shaft is made of mild steel with an acted upon by an ultimate stress of 252 MPa. Using a factor of safety of 6 determine the diameter of the shaft neglecting the bending moment on the shaft.

QUESTION SIX

a). i). Why are design solutions initiated from consideration of multiple alternatives? (3 marks)

(ii). Consider the following properties of different phone bands in the market. Do a ranking for the mobile phones using 10 marks for each property

FeaturesSamsung galaxy S 23 ultraTechno Canon 20 PROInfinix ZeroRedmi 12Huawei P40 PRO
RAM (GB)88848
ROM (GB)512256256128256
Version (Android)1313111310
Battery (mAH)48855000500050004200
Camera (Mega pixel)200641085050
Cost (Naira)1,400,000220,000600,000140,000500,000

i. Which of the phones has the highest ranking?

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ii. Which of the phone is most economically competitive in relation to performance?

iii. Give three (3) reasons for your choice

QUESTION SEVEN

Consider the design of safety boots for engineers working in swamping territories,

i. Identify three ergonomics factors that are essential to the product design process and briefly explain them

ii. identify three safety factors relevant to the success of the design

iii. identify three performance factors that will promote the sales of the product

Below are the answers:

ANSWER ONE

Solution
Data:
L = 1m, X = distance = 1.75m
V’ = 1600m/mm = 1600/60
V = 26.7m/s
F = 90kw = 90 x 10³ W
¶ = density= 1000kg/m³
d1 = ?, d2 = 1
N1 = 750, N2 = 250
t = 2.5mpa, μ = 0.25
20/2 = 35/2 = 17.5
Area = 3.75mm²

i. To calculate for d1
N1/N2 = d2/d1 ==> 750/250 = 1/d1
Crossmultiply…
750d1 = 250
d1= 250/750, ==> d1 =0.33m

ii. For an open belt
Sin ∞ = r2 – r1/ x, for radius.
Sin ∞ = d2 – d1/ 2x, for diameter.
Sin ∞ = 1 – 0.33 / 2(1.75)
Sin ∞ = 0.1914
∞ = Sin-¹ 0.1914
∞ = 11.03°

iii. For angle between the pulley of the belt
(180 – 2 ∞) π / 180
(180 – 2 (11.03) ) 3.142 / 180
= 2.76rad

iv. Mass of belt per meter length
M = Area x length x density
= 375 x 10-⁶ x 1 x 1000
M = 0.3475kg/m

v. Centrifugal tension
Tc = M.V²
= 0.375 x 26.7² ==> 267.3N

vi. For Max tension in belt
T = ₱ x A
T = 2.5 x 10⁶ N x 375 x 10-⁶
T = 2.5 x 375
T = 937.5N

vii. Tension on the tight side
T1 = T – Tc
T1 = 937.5 – 267.3
T1 = 670.2N


viii. Tension on the slack side T2
2.3log (T1/T2) = μ × ∅ cossec ¢
2.3log (T1/T2) = 0.25 × 2.76 cossec 17.5
2.3log (T1/T2) = 0.25 × 2.76 × 3.326
2.3log (T1/T2) = 2.295
log T1/T2 = 2.295/2.3
T1/T2 = log-¹ 0.9978
T1/T2 = 9.95

See also  WOUNDED HEALER

670.2/T2 = 9.95/1
T2 = 67.3N

ix. Power transmitted for the belt
P = (T1 – T2) V
P = 670.2 – 67.3 × 26.7
= 16097.4 W
P = 16.09kW

x. Calculate for numbers of V-belt
№ of V-belt = Total power transmitted/ power transmitted for belt
= 9000/16.09 × 10³ = 5.59 ≈ 6
xi. Length on each belt
T1 = d1/2 = 0.33/2
= 0.165m
r2 = d1 /2 = 0.5m

xii. L = π (r2 + r1) + 2x + (r2 – r1)² / x
L = 3.142 (0.5 + 0.165) + 2(1.75) + (0.5 – 0.165)² / 1.75
L = 3.142 (0.665) + 3.5 + (0.335)²/1.75
L = 2.089 + 3.5 + 0.064
L = 5.653

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