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Problem 8: Three states in a thermodynamic process are defined by P l


=1 bar and V 1

=2 L,P 2

=20 bar and V 2

=0.2 L, and P 3

=4 bar and V 3

=1 L. The quantity β is defined for the path of integration 1−2−3 by: β=−∫ 1
2

pdV−∫ 2
3

pdV Determine β (expressed in units of joules) connecting states 1 and 2 with a linear P−V process and connecting states 2 and 3 with a polytropic process. Remember that a polytropic process is one that can be described by the following relationship: PV n
=C

User Ernestasju
by
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1 Answer

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To determine β for the given thermodynamic process, we need to calculate the work done along the specified paths. Let's calculate the work done for each path and then sum them up to find β.

Path 1-2 (Linear P-V Process):

In a linear P-V process, the pressure and volume are related by a linear equation: PV = constant. We can use the given states 1 and 2 to determine the equation for this path.

State 1: P1 = 1 bar, V1 = 2 L

State 2: P2 = 20 bar, V2 = 0.2 L

Using the equation PV = constant, we can write:

P1V1 = P2V2

(1 bar)(2 L) = (20 bar)(0.2 L)

2 bar·L = 4 bar·L

Since the constant values on both sides are equal, we can say that the linear equation for this path is:

PV = 2 bar·L

Now, let's calculate the work done along this path:

β₁ = -∫(PdV) from V1 to V2

Since PV = 2 bar·L, we can express pressure P in terms of volume V:

P = 2/V

Substituting this into the integral, we have:

β₁ = -∫(2/V)dV from V1 to V2

Integrating, we get:

β₁ = -2ln(V)|V1 to V2

β₁ = -2[ln(V2) - ln(V1)]

β₁ = -2ln(V2/V1)

Substituting the given values, we have:

β₁ = -2ln(0.2 L/2 L)

β₁ = -2ln(0.1)

β₁ ≈ 4.605 J

Path 2-3 (Polytropic Process):

A polytropic process is described by the equation PV^n = constant, where n is a constant value. We can use the given states 2 and 3 to determine the equation for this path.

State 2: P2 = 20 bar, V2 = 0.2 L

State 3: P3 = 4 bar, V3 = 1 L

Using the equation PV^n = constant, we can write:

P2V2^n = P3V3^n

(20 bar)(0.2 L)^n = (4 bar)(1 L)^n

Simplifying, we have:

4(0.2^n) = 20

Dividing both sides by 4, we get:

0.2^n = 5

Taking the logarithm of both sides, we have:

n ln(0.2) = ln(5)

Solving for n, we find:

n = ln(5) / ln(0.2)

n ≈ -2.8616

The polytropic equation for this path is:

PV^(-2.8616) = constant

Now, let's calculate the work done along this path:

β₂ = -∫(PdV) from V2 to V3

Since PV^(-2.8616) = constant, we can express pressure P in terms of volume V:

P = constant / V^(-2.8616)

Substituting this into the integral, we have:

β₂ = -∫(constant / V^(-2.8616))dV from V2 to V3

β₂ = -constant ∫(V^2.8616)dV from V2 to V3

β₂ = -constant[(V^(2.8616 + 1))/(2.8616 + 1)]|V2 to V3

Substituting the given values, we have:

β₂ = -constant[(V3^(2.8616 + 1))/(2.8616 + 1)] - (-constant[(V2^(2.8616 + 1))/(2.8616 + 1)])

Since we don't have the constant value, we can't determine β₂ without additional information.

Therefore, the value of β connecting states 1 and 2 with a linear P-V process is approximately 4.605 joules. However, we can't determine the value of β connecting states 2 and 3 with a polytropic process without the constant value.

User Lhassan Baazzi
by
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