The dynamic viscosity of carbon dioxide at 50°C and 200°Care 1.612 × 10 − 5 Pa ⋅ s and 2.276 × 10 − 5 Pa ⋅ s , respectively. Determine the constants a and b of Sutherland correlation for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100 ° C and compare your result against the value given Table A-10.

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Answer 1

Textbook Question

Chapter 2, Problem 85P

The dynamic viscosity of carbon dioxide at 50°C and 200°Care 1.612 × 10 − 5 Pa ⋅ s and 2.276 × 10 − 5 Pa ⋅ s , respectively. Determine the constants a and b of Sutherland correlation for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100 ° C and compare your result against the value given Table A-10.

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

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Answer 2

Textbook Question

Chapter 2, Problem 85P

The dynamic viscosity of carbon dioxide at 50°C and 200°Care 1.612 × 10 − 5 Pa ⋅ s and 2.276 × 10 − 5 Pa ⋅ s , respectively. Determine the constants a and b of Sutherland correlation for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100 ° C and compare your result against the value given Table A-10.

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

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Answer 3

Textbook Question

Chapter 2, Problem 85P

The dynamic viscosity of carbon dioxide at 50°C and 200°Care 1.612 × 10 − 5 Pa ⋅ s and 2.276 × 10 − 5 Pa ⋅ s , respectively. Determine the constants a and b of Sutherland correlation for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100 ° C and compare your result against the value given Table A-10.

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

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Answer 4

Textbook Question

Chapter 2, Problem 85P

The dynamic viscosity of carbon dioxide at 50°C and 200°Care 1.612 × 10 − 5 Pa ⋅ s and 2.276 × 10 − 5 Pa ⋅ s , respectively. Determine the constants a and b of Sutherland correlation for carbon dioxide at atmospheric pressure. Then predict the viscosity of carbon dioxide at 100 ° C and compare your result against the value given Table A-10.

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Answer 8

Expert Solution & Answer

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The viscosity of carbon dioxide at 100°C.

The change in viscosity.

Answer 12

Answer to Problem 85P

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Answer 13

Explanation of Solution

Given information:

The dynamic viscosity of carbon dioxide at 50°C is 1.612×10−5 Pa⋅s, the dynamic viscosity of carbon dioxide at 200°C is 22.276×10−5 Pa⋅s.

Write the expression for the Sutherland correlation

μ=aT121+bT........ (I)

Here, the dynamic viscosity is μ, the temperature is T, and the constants are a,b.

Write the expression for the change in viscosity.

Δμ=μ−μ′........ (II)

Here, the change in viscosity is Δμ, the viscosity of carbon dioxide is μ, and the viscosity of carbon dioxide by the table "A-10 The properties of gases" is μ′.

Calculation:

Substitute 1.612×10−5 Pa⋅s for μ and 50°C for T in Equation (I).

(1.612× 10 −5 Pa⋅s)=a ( 50°C ) 1 2 1+b ( 50°C )(1.612× 10 −5 Pa⋅s)=a ( ( 50°C+273 ) K ) 1 2 1+b ( 50°C+273 ) K323+b=360112955.587a ..... (III)

Substitute 2.276×10−5 Pa⋅s for μ and 200°C for T in Equation (I).

(2.276× 10 −5 Pa⋅s)=a ( 200°C ) 1 2 1+b ( 200°C )(2.276× 10 −5 Pa⋅s)=a ( ( 200°C+273 ) K ) 1 2 1+b ( 200°C+273 ) K473+b=451980228.47a ..... (IV)

Subtract Equation (III) from Equation (IV).

473+b−323−b=451980228.47a−360112955.587a150=98167272.88aa=1.63279×10−6

Substitute 1.63279×10−6 for a in Equation (III).

323+b=(360112955.587×( 1.63279× 10 −6 ))b=(360112955.587×( 1.63279× 10 −6 ))−323b=264.9889

Substitute 1.63279×10−6 for a and 100°C for T and 264.9889 for b in Equation (I).

μ=( 1.63279× 10 −6 ) ( 100°C ) 1 2 1+ 264.9889 ( 100°C )=( 1.63279× 10 −6 ) ( ( 100°C+273 ) K ) 1 2 1+ 264.9889 ( 100°C+273 ) K=1.8436×10−5 Pa

Refer to the table "A-10 Properties of gases at 1 atm "to find the value of viscosity of carbon dioxide as μ′=1.841×10−5 Pa⋅s.

Substitute 1.841×10−5 Pa⋅s for μ′, 1.8436×10 Pa⋅s for μ in Equation (5).

Δμ=(1.8436× 10 −5 Pa⋅s)−(1.841× 10 −5 Pa⋅s)=2.6×10−8 Pa⋅s

Conclusion:

The viscosity of carbon dioxide at 100°C is 1.8436×10−5 Pa.

The change in viscosity is 2.6×10−8 Pa⋅s.

Answer 14

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