What Is the Flow Work for Isothermal Process?

Yes the work done by the system will be zero

No, it's not zero and you're not correct!

In polytropic compression non-flow work = 1/(r-1)*P₁*V₁*([P₂/P₁]**[(r-1)/r] - 1).

Flow work is r*non-flow work. Then putting (r - 1)/r = n, flow work = 1/n*P₁*V₁*([P₂/P₁]**n - 1). For adiabatic compression r = γ = 1.41. For practical compressor duties r usually taken as 1.3, giving n = 0.3.

In isothermal compression r = 1, n = 0, and non-flow work and flow work are the same, and = P₁*V₁*ln[P₂/P₁].

You can show using de l'Hopital's theorem that the limit of 1/n*P₁*V₁*([P₂/P₁]**n - 1) as n → 0 is P₁*V₁*ln[P₂/P₁].

Codey

Perhaps I should have added that to calculate power (implied by your ....rate of flow work for isothermal) the V₁ in the equations is flow, not volume.

Codey

You can put it also in an other way. Temperature is constant if the output and input of energy are equal. For instance if you heat an expanding gas you make a work and at same time heat it temperature stays constant. If you compress a gas and cool it the effect is the same.

Well, yes, but what has that to do with the OP's question?

Codey