The work done during the process can be calculated using the formula for work done during an expansion or compression process that follows a polytropic process, which is given by:

W = (P2*V2 - P1*V1) / (n-1)

Where:
P1 = Initial pressure
V1 = Initial specific volume
P2 = Final pressure
V2 = Final specific volume
n = polytropic index

Given:
P1 = 3 bar = 3 * 10^5 Pa (since 1 bar = 10^5 Pa)
V1 = 0.18 m^3/kg
P2 = 0.6 bar = 0.6 * 10^5 Pa
The process follows the law pv^2 = c, which means n = 2.

We need to find V2. We can use the polytropic process equation:

P1 * V1^n = P2 * V2^n

Substituting the given values:

3 * 10^5 * (0.18)^2 = 0.6 * 10^5 * V2^2

Solving for V2, we get V2 = 0.3 m^3/kg.

Now we can substitute all the values into the work done equation:

W = (0.6 * 10^5 * 0.3 - 3 * 10^5 * 0.18) / (2-1)
W = -27,000 J/kg

The negative sign indicates that work is done by the system (expansion).

Question

The work done during the process can be calculated using the formula for work done during an expansion or compression process that follows a polytropic process, which is given by:

W = (P2*V2 - P1*V1) / (n-1)

Where:
P1 = Initial pressure
V1 = Initial specific volume
P2 = Final pressure
V2 = Final specific volume
n = polytropic index

Given:
P1 = 3 bar = 3 * 10^5 Pa (since 1 bar = 10^5 Pa)
V1 = 0.18 m^3/kg
P2 = 0.6 bar = 0.6 * 10^5 Pa
The process follows the law pv^2 = c, which means n = 2.

We need to find V2. We can use the polytropic process equation:

P1 * V1^n = P2 * V2^n

Substituting the given values:

3 * 10^5 * (0.18)^2 = 0.6 * 10^5 * V2^2

Solving for V2, we get V2 = 0.3 m^3/kg.

Now we can substitute all the values into the work done equation:

W = (0.6 * 10^5 * 0.3 - 3 * 10^5 * 0.18) / (2-1)
W = -27,000 J/kg

The negative sign indicates that work is done by the system (expansion).

Knowee AI · Accepted Answer