Engineering Fundamentals: An Introduction to Engineering
Engineering Fundamentals: An Introduction to Engineering
6th Edition
ISBN: 9780357112311
Author: Saeed Moaveni
Publisher: Cengage Learning US
Question
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Chapter 9, Problem 17P

(a)

To determine

Measure the density of a cooking oil and express the findings in kg/m3,slugs/ft3,andlbm/ft3.

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The density of cooking oil in kg/m3 is 918kg/m3_.

The density of cooking oil in slugs/ft3 is 1.781slugs/ft3_.

The density of cooking oil in lbm/ft3 is 57.30lbm/ft3_.

The specific gravity of the cooking oil is 0.918_.

Explanation of Solution

Find the density of the cooking oil using the procedure as follows:

  • Measure the weight of the empty test container.
  • Take the sample of cooking oil in a container. Measure the weight of container with oil.
  • Find the mass of the oil by subtracting the weight of oil container with weight of empty container.
  • Measure the volume of the liquid using a graduated cylinder.
  • Find the density of the oil using the value of mass and volume.
  • Density=massVolume
  • The density of cooking oil is ρoil=918kg/m3.

Therefore, the density of cooking oil in kg/m3 is 918kg/m3_.

Convert the unit of density from kg/m3 to slugs/ft3.

ρoil=918kg/m3×1slug14.594kg×(1m3.281ft)3=1.781slugs/ft3

Therefore, the density of cooking oil in slugs/ft3 is 1.781slugs/ft3_.

Convert the unit of density from kg/m3 to lbm/ft3.

ρoil=918kg/m3×1lbm0.45359kg×(1m3.281ft)3=57.30lbm/ft3

Therefore, the density of cooking oil in lbm/ft3 is 57.30lbm/ft3_.

Find the specific gravity (Goil) of cooking oil using the relation.

Goil=ρoilρw

Substitute 918kg/m3 for ρoil and 1,000kg/m3 for ρw.

Goil=9181,000=0.918

Therefore, the specific gravity of the cooking oil is 0.918_.

(b)

To determine

Measure the density of SAE 10W-40 engine oil and express the findings in kg/m3,slugs/ft3,andlbm/ft3.

(b)

Expert Solution
Check Mark

Answer to Problem 17P

The density of SAE 10W-40 engine oil in kg/m3 is 865kg/m3_.

The density of SAE 10W-40 engine oil in slugs/ft3 is 1.678slugs/ft3_.

The density of SAE 10W-40 engine oil in lbm/ft3 is 53.99lbm/ft3_.

The specific gravity of the SAE 10W-40 engine oil is 0.865_.

Explanation of Solution

Find the density of the SAE 10W-40 engine oil using the procedure as follows:

  • Measure the weight of the empty test container.
  • Take the sample of SAE 10W-40 engine oil in a container. Measure the weight of container with oil.
  • Find the mass of the oil by subtracting the weight of oil container with weight of empty container.
  • Measure the volume of the liquid using a graduated cylinder.
  • Find the density of the oil using the value of mass and volume.
  • Density=massVolume
  • The density of SAE 10W-40 engine oil is ρSAE10W-40=865kg/m3

Therefore, the density of SAE 10W-40 engine oil in kg/m3 is 865kg/m3_.

Convert the unit of density from kg/m3 to slugs/ft3.

ρSAE 10W-40=865kg/m3×1slug14.594kg×(1m3.281ft)3=1.678slugs/ft3

Therefore, the density of SAE 10W-40 engine oil in slugs/ft3 is 1.678slugs/ft3_.

Convert the unit of density from kg/m3 to lbm/ft3.

ρSAE 10W-40=865kg/m3×1lbm0.45359kg×(1m3.281ft)3=53.99lbm/ft3

Therefore, the density of SAE 10W-40 engine oil in lbm/ft3 is 53.99lbm/ft3_.

Find the specific gravity (GSAE 10W-40) of SAE 10W-40 engine oil using the relation.

GSAE 10W-40=ρSAE 10W-40ρw

Substitute 865kg/m3 for ρSAE 10W-40 and 1,000kg/m3 for ρw.

GSAE 10W-40=8651,000=0.865

Therefore, the specific gravity of the SAE 10W-40 engine oil is 0.865_.

(c)

To determine

Measure the density of water and express the findings in kg/m3,slugs/ft3,andlbm/ft3.

(c)

Expert Solution
Check Mark

Answer to Problem 17P

The density of water in kg/m3 is 999.94kg/m3_.

The density of water in slugs/ft3 is 1.94slugs/ft3_.

The density of water in lbm/ft3 is 62.41lbm/ft3_.

The specific gravity of the water is 1_.

Explanation of Solution

Find the density of the water using the procedure as follows:

  • Measure the weight of the empty test container.
  • Take the sample of water in a container. Measure the weight of container with water.
  • Find the mass of the water by subtracting the weight of water container with weight of empty container.
  • Measure the volume of the liquid using a graduated cylinder.
  • Find the density of the water using the value of mass and volume.
  • Density=massVolume
  • The density of water is ρwater=999.94kg/m3.

Therefore, the density of water in kg/m3 is 999.94kg/m3_.

Convert the unit of density from kg/m3 to slugs/ft3.

ρwater=999.94kg/m3×1slug14.594kg×(1m3.281ft)3=1.94slugs/ft3

Therefore, the density of water in slugs/ft3 is 1.94slugs/ft3_.

Convert the unit of density from kg/m3 to lbm/ft3.

ρwater=999.94kg/m3×1lbm0.45359kg×(1m3.281ft)3=62.41lbm/ft3

Therefore, the density of water in lbm/ft3 is 62.41lbm/ft3_.

Find the specific gravity (Gwater) of water using the relation.

Gwater=ρwaterρw

Substitute 999.94kg/m3 for ρwater and 1,000kg/m3 for ρw.

Gwater=999.941,000=1

Therefore, the specific gravity of the water is 1_.

(d)

To determine

Measure the density of milk and express the findings in kg/m3,slugs/ft3,andlbm/ft3.

(d)

Expert Solution
Check Mark

Answer to Problem 17P

The density of milk in kg/m3 is 1,030kg/m3_.

The density of milk in slugs/ft3 is 1.998slugs/ft3_.

The density of milk in lbm/ft3 is 64.29lbm/ft3_.

The specific gravity of the milk is 1.03_.

Explanation of Solution

Find the density of the milk using the procedure as follows:

  • Measure the weight of the empty test container.
  • Take the sample of milk in a container. Measure the weight of container with water.
  • Find the mass of the milk by subtracting the weight of milk container with weight of empty container.
  • Measure the volume of the liquid using a graduated cylinder.
  • Find the density of the milk using the value of mass and volume.
  • Density=massVolume
  • The density of milk is ρmilk=1,030kg/m3.

Therefore, the density of milk in kg/m3 is 1,030kg/m3_.

Convert the unit of density from kg/m3 to slugs/ft3.

ρmilk=1,030kg/m3×1slug14.594kg×(1m3.281ft)3=1.998slugs/ft3

Therefore, the density of milk in slugs/ft3 is 1.998slugs/ft3_.

Convert the unit of density from kg/m3 to lbm/ft3.

ρmilk=1,030kg/m3×1lbm0.45359kg×(1m3.281ft)3=64.29lbm/ft3

Therefore, the density of milk in lbm/ft3 is 64.29lbm/ft3_.

Find the specific gravity (Gmilk) of milk using the relation.

Gmilk=ρmilkρw

Substitute 1,030kg/m3 for ρmilk and 1,000kg/m3 for ρw.

Gmilk=1,0301,000=1.03

Therefore, the specific gravity of the milk is 1.03_.

(e)

To determine

Measure the density of ethylene glycol (antifreeze) and express the findings in kg/m3,slugs/ft3,andlbm/ft3.

(e)

Expert Solution
Check Mark

Answer to Problem 17P

The density of ethylene glycol (antifreeze) in kg/m3 is 1,125kg/m3_.

The density of ethylene glycol (antifreeze) in slugs/ft3 is 2.18slugs/ft3_.

The density of ethylene glycol (antifreeze) in lbm/ft3 is 70.22lbm/ft3_.

The specific gravity of the ethylene glycol (antifreeze) is 1.125_.

Explanation of Solution

Find the density of the ethylene glycol using the procedure as follows:

  • Measure the weight of the empty test container.
  • Take the sample of ethylene glycol in a container. Measure the weight of container with ethylene glycol.
  • Find the mass of the ethylene glycol by subtracting the weight of ethylene glycol container with weight of empty container.
  • Measure the volume of the liquid using a graduated cylinder.
  • Find the density of the ethylene glycol using the value of mass and volume.
  • Density=massVolume
  • The density of ethylene glycol (antifreeze) is ρethylene=1,125kg/m3.

Therefore, the density of ethylene glycol (antifreeze) in kg/m3 is 1,125kg/m3_.

Convert the unit of density from kg/m3 to slugs/ft3.

ρethylene=1,125kg/m3×1slug14.594kg×(1m3.281ft)3=2.18slugs/ft3

Therefore, the density of ethylene glycol (antifreeze) in slugs/ft3 is 2.18slugs/ft3_.

Convert the unit of density from kg/m3 to lbm/ft3.

ρethylene=1,125kg/m3×1lbm0.45359kg×(1m3.281ft)3=70.22lbm/ft3

Therefore, the density of ethylene glycol (antifreeze) in lbm/ft3 is 70.22lbm/ft3_.

Find the specific gravity (Gethylene) of ethylene glycol (antifreeze) using the relation.

Gethylene=ρethyleneρw

Substitute 1,125kg/m3 for ρethylene and 1,000kg/m3 for ρw.

Gethylene=1,1251,000=1.125

Therefore, the specific gravity of the ethylene glycol (antifreeze) is 1.125_.

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