Exam (elaborations)
Main Assignment for semester 1
- Course
- Institution
These are solutions for main assignment 1 in 2020
[Show more]
Seller
Follow
victormckate
Content preview
Mark: %MAIN ASSIGNMENT
DATE: 20 JULY 2020
SUBJECT: MECHANICAL ENGINEERING DESIGN I CODE: MED21A
MAXIMUM MARKS: 50
INSTRUCTIONAL PROGRAMME: DIP.ENG.TECH: MECHANICAL ENGINEERING
N.DIP ENGINEERING: MECHANICAL
ASSESSOR: DR. L. MUGWAGWA
Question: 1 2 3 Total
Allocated marks:
/25 /15 /10 /50
STATIONARY:
Programmable alphanumerical calculators are NOT allowed
Textbook: Mechanical Engineering Design – Peter RN Childs
INSTRUCTIONS:
Answer all questions in this question paper
Show all calculations with the necessary steps
Work written in pencil will not be marked.
No half marks will be allocated
Round off to 3 decimal places
Student to complete this block:
STUDENT NUMBER:
SURNAME AND INITIALS:
(Write in Block Letters)
SIGNATURE OF STUDENT:
(You have read the instructions on this page)
THIS PAPER CONSISTS OF 10 PAGES INCLUDING THIS COVER PAGE.
, Page |2
QUESTION 1: BEARING DESIGN [25]
A full journal bearing has a nominal diameter of 50 mm and bearing length 25 mm. The bearing
supports a load of 1450 N and the journal design speed is 2500 rpm. The radial clearance is 0.025mm.
An SAE 10 oil has been chosen and the lubricant supply temperature is 50°C. Using the standard
Raimondi and Boyd design charts, find the temperature rise of the lubricant, taking ΔT = 20°C as the
initial assumption. (NB: Carry out the necessary number of iterations) [25]
Given data:
D=50.0 mm, L=25 mm, W=1450 N, N=2500 rpm, c=0.025 mm, SAE 10, and T1=50 °C
Assuming lubricant temperature rise ∆𝑇 across the bearing to be ∆𝑇 = 20 °C,
∆𝑇 20
𝑇𝑎𝑣 = 𝑇1 + = 50 + = 60 °𝐶 (1 𝑚𝑎𝑟𝑘)
2 2
From Figure 4.7 for SAE 10 at 60 oC, µ=0.014 Pa s. (1 mark)
2500
𝑁𝑠 = = 41.67 𝑟𝑝𝑠 (1 𝑚𝑎𝑟𝑘)
60
𝐿 25
= = 0.5 (1 𝑚𝑎𝑟𝑘)
𝐷 50
𝑊 1450
𝑃= = = 1.16 × 106 𝑁/𝑚2 (1 𝑚𝑎𝑟𝑘)
𝐿𝐷 0.025 × 0.05
The next step is to determine the Somerfield number (bearing characteristic number)
𝑟 2 µ𝑁𝑠
𝑆=( )
𝑐 𝑃
2
25 × 10−3 0.014 × 41.67
=( ) = 0.5029 (1 𝑚𝑎𝑟𝑘)
0.025 × 10−3 1.16 × 106
From Figure 4.12 with S = 0.5029 and L/D = 0.5; (𝑟/𝑐)𝑓 = 12 (1 mark)
From Figure 4.13, with S = 0.5029 and L/D = 0.5; 𝑄/𝑟𝑐𝑁𝑠 𝐿 = 4.6 (1 mark)
From Figure 4.14, with S = 0.5029 and L/D = 0.5; 𝑄𝑠 /𝑄 = 0.65 (1 mark)
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller victormckate. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.57. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
72042 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now