Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil's a. Specific weight b. Density Specific gravity

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
icon
Related questions
Topic Video
Question
### Problem Statement:

Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil’s:

a. Specific weight  
b. Density  
c. Specific gravity

#### Explanation:

- **Specific Weight (γ):** 
  The specific weight is defined as the weight per unit volume. It can be calculated using the following formula:
  \[
  \gamma = \frac{W}{V}
  \]
  where:
  - \( \gamma \) is the specific weight (N/m³)
  - \( W \) is the weight (N)
  - \( V \) is the volume (m³)

- **Density (ρ):** 
  The density of a substance is defined as its mass per unit volume. It can be calculated using the following formula:
  \[
  \rho = \frac{m}{V}
  \]
  where:
  - \( \rho \) is the density (kg/m³)
  - \( m \) is the mass (kg)
  - \( V \) is the volume (m³)
  
  The mass can also be derived from weight using the relationship:
  \[
  W = mg
  \]
  where:
  - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²)

- **Specific Gravity (SG):** 
  The specific gravity is the ratio of the density of the substance to the density of water. It is a dimensionless quantity:
  \[
  SG = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}}
  \]
  where:
  - \( \rho_{\text{oil}} \) is the density of the oil
  - \( \rho_{\text{water}} \) is the density of water (approximately 1000 kg/m³ at 4°C)

### Note:
To solve the calculations, you will need to convert the volume from liters to cubic meters (1 liter = 0.001 cubic meters). 

### Example Calculation:

1. Convert the volume of oil from liters to cubic meters:
   \[
   V = 11 \, \text{liters} \times 0.001 \, \frac{\text{m}^3}{\text{liter}} = 0.011 \, \text{m}^3
   \]

2. Calculate the specific weight:
Transcribed Image Text:### Problem Statement: Eleven liters of SAE 30 oil weighs 92 N. Calculate the oil’s: a. Specific weight b. Density c. Specific gravity #### Explanation: - **Specific Weight (γ):** The specific weight is defined as the weight per unit volume. It can be calculated using the following formula: \[ \gamma = \frac{W}{V} \] where: - \( \gamma \) is the specific weight (N/m³) - \( W \) is the weight (N) - \( V \) is the volume (m³) - **Density (ρ):** The density of a substance is defined as its mass per unit volume. It can be calculated using the following formula: \[ \rho = \frac{m}{V} \] where: - \( \rho \) is the density (kg/m³) - \( m \) is the mass (kg) - \( V \) is the volume (m³) The mass can also be derived from weight using the relationship: \[ W = mg \] where: - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²) - **Specific Gravity (SG):** The specific gravity is the ratio of the density of the substance to the density of water. It is a dimensionless quantity: \[ SG = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} \] where: - \( \rho_{\text{oil}} \) is the density of the oil - \( \rho_{\text{water}} \) is the density of water (approximately 1000 kg/m³ at 4°C) ### Note: To solve the calculations, you will need to convert the volume from liters to cubic meters (1 liter = 0.001 cubic meters). ### Example Calculation: 1. Convert the volume of oil from liters to cubic meters: \[ V = 11 \, \text{liters} \times 0.001 \, \frac{\text{m}^3}{\text{liter}} = 0.011 \, \text{m}^3 \] 2. Calculate the specific weight:
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Fluid Statics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Elements Of Electromagnetics
Elements Of Electromagnetics
Mechanical Engineering
ISBN:
9780190698614
Author:
Sadiku, Matthew N. O.
Publisher:
Oxford University Press
Mechanics of Materials (10th Edition)
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:
9780134319650
Author:
Russell C. Hibbeler
Publisher:
PEARSON
Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:
9781259822674
Author:
Yunus A. Cengel Dr., Michael A. Boles
Publisher:
McGraw-Hill Education
Control Systems Engineering
Control Systems Engineering
Mechanical Engineering
ISBN:
9781118170519
Author:
Norman S. Nise
Publisher:
WILEY
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning
Engineering Mechanics: Statics
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:
9781118807330
Author:
James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:
WILEY