# Steam in a piston–cylinder assembly undergoes a poly- tropic process, with n = 2, from an initial state where p1 = 3.45 MPa, v1 = 0.106 m3/kg, u1 = 3,171.1 kJ/kg, to a final state where u2 = 2,303.9 kJ/kg. During the process, there is a heat transfer from

To find the work and final specific volume of the steam in the piston-cylinder assembly, we will use the first law of thermodynamics and the ideal gas equation. The first law of thermodynamics can be written as:

ΔQ – ΔW = ΔU

Where ΔQ is the heat transfer, ΔW is the work done, and ΔU is the change in internal energy. In this case, the heat transfer is given as 361.76 kJ.

Since the steam undergoes a polytropic process, we can use the specific heat ratio (γ) to relate the pressure and volume during the process. For an ideal gas, γ is the ratio of the specific heat at constant pressure (cp) to the specific heat at constant volume (cv). It can be expressed as:

γ = cp / cv

For steam, the specific heat ratio γ is typically around 1.3.

During the polytropic process, the relationship between pressure and specific volume can be expressed as:

p * v^n = constant

Where p is the pressure, v is the specific volume, and n is the polytropic index. In this case, we are given that n = 2.

To determine the work done, we can integrate the pressure-volume relationship over the process. Since the process is between two fixed states, we can write the work as:

W = ∫ (p * dv)

To perform the integration, we need to express pressure (p) as a function of specific volume (v). Rearranging the pressure-volume relationship, we get:

p = (constant / v^n)

Rearranging the equation for the polytropic process, we have:

p1 * v1^n = p2 * v2^n

Substituting the given values, we have:

(3.45 MPa) * (0.106 m^3/kg)^2 = p2 * v2^2

Solving for p2, we find:

p2 = (3.45 MPa) * (0.106 m^3/kg)^2 / (v2^2)

To solve for the final specific volume v2, we can use the fact that u is a function of temperature (T) and specific volume (v). The change in internal energy ΔU can be expressed as:

ΔU = m * (u2 – u1)

Substituting the given values, we have:

ΔU = (0.54 kg) * (2303.9 kJ/kg – 3171.1 kJ/kg)

Now, let’s solve for the work done:

ΔQ – ΔW = ΔU

361.76 kJ – ΔW = (0.54 kg) * (2303.9 kJ/kg – 3171.1 kJ/kg)

ΔW = 361.76 kJ – (0.54 kg) * (2303.9 kJ/kg – 3171.1 kJ/kg)

Finally, to find the final specific volume v2, we can rearrange the equation for the polytropic process and substitute the known values:

(3.45 MPa) * (0.106 m^3/kg)^2 = p2 * v2^2

Substituting the value of p2 obtained earlier, we have:

(3.45 MPa) * (0.106 m^3/kg)^2 = [(3.45 MPa) * (0.106 m^3/kg)^2 / (v2^2)] * v2^2

By canceling out the terms, we find:

1 = 1

This indicates that the equation is valid for any value of v2, as long as it satisfies the given pressure and specific volume conditions.

In conclusion, the work done during the polytropic process is ΔW = 361.76 kJ, and the final specific volume is indeterminate due to the equation being valid for any value of v2 that satisfies the given conditions.