- #1

S. Moger

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I am preparing for a re-exam, this is a problem from the exam I took, but I can't see what I did wrong and why.

Case: A turbine that is producing work dW is powered by compressed air (treated as a diatomic ideal gas).

Known quantities:

[itex]P_0, P_f, T_0[/itex]

Wanted quantity:

[itex]dW[/itex] per mole air.

It is also known that the process is adiabatic, so [itex]dQ = 0[/itex], and that the flow is stationary.

By the first law of thermodynamics, and the adiabatic property:

[itex]\Delta U = dQ + dW = dW[/itex].

The energy content of a diatomic ideal gas is given by:

[itex]U = \frac{f}{2} nRT[/itex], where f = 5 (the amount of quadratic degrees of freedom)

Thus, ΔU should equal the change in energy content of the air before and after the turbine:

[itex]\Delta U = U_0 - U_f =\frac{f}{2} nR (T_0 - T_f) = dW [/itex].

To compute the unknown [itex]T_f[/itex] we again use the fact that the process is adiabatic, so the following should hold

[itex]P_0^{-\frac{2}{f+2}} T_0 = P_f^{-\frac{2}{f+2}} T_f \iff T_f = T_0 (\frac{P_f}{P_0})^{\frac{2}{f+2}}[/itex].

Inserting this result into the prior equation gives

[itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

However, the solution sheet states that the term f/2=5/2 should be 7/2 (rest unchanged). I can't see why. They use a different technique as well, which I don't understand.

Stationary flow implies that [itex] H_0 = W + H_f[/itex] by the first law of thermodynamics. The enthalpy [itex]H=C_P T[/itex]. So, [itex]W=C_P (T_0 - T_f)[/itex]. Furthermore,

[itex]T_f = T_0 (\frac{P_f}{P_0})^{1-1/\gamma}[/itex] and

[itex]C_P = 7/2 nR[/itex] by the properties of diatomic ideal gases.

Finally,

[itex]W/n = \frac{7}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{1-1/\gamma})[/itex]

which is not what I get:

[itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

Gamma is defined as (f+2)/f. Also observe that their notation is W instead of dW ( which here is not meant to be read as a change in work, but a quantity of work ).

## Homework Statement

Case: A turbine that is producing work dW is powered by compressed air (treated as a diatomic ideal gas).

Known quantities:

[itex]P_0, P_f, T_0[/itex]

Wanted quantity:

[itex]dW[/itex] per mole air.

It is also known that the process is adiabatic, so [itex]dQ = 0[/itex], and that the flow is stationary.

## Homework Equations

## The Attempt at a Solution

By the first law of thermodynamics, and the adiabatic property:

[itex]\Delta U = dQ + dW = dW[/itex].

The energy content of a diatomic ideal gas is given by:

[itex]U = \frac{f}{2} nRT[/itex], where f = 5 (the amount of quadratic degrees of freedom)

Thus, ΔU should equal the change in energy content of the air before and after the turbine:

[itex]\Delta U = U_0 - U_f =\frac{f}{2} nR (T_0 - T_f) = dW [/itex].

To compute the unknown [itex]T_f[/itex] we again use the fact that the process is adiabatic, so the following should hold

[itex]P_0^{-\frac{2}{f+2}} T_0 = P_f^{-\frac{2}{f+2}} T_f \iff T_f = T_0 (\frac{P_f}{P_0})^{\frac{2}{f+2}}[/itex].

Inserting this result into the prior equation gives

[itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

**_________________________________**However, the solution sheet states that the term f/2=5/2 should be 7/2 (rest unchanged). I can't see why. They use a different technique as well, which I don't understand.

**The correct solution:**Stationary flow implies that [itex] H_0 = W + H_f[/itex] by the first law of thermodynamics. The enthalpy [itex]H=C_P T[/itex]. So, [itex]W=C_P (T_0 - T_f)[/itex]. Furthermore,

[itex]T_f = T_0 (\frac{P_f}{P_0})^{1-1/\gamma}[/itex] and

[itex]C_P = 7/2 nR[/itex] by the properties of diatomic ideal gases.

Finally,

[itex]W/n = \frac{7}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{1-1/\gamma})[/itex]

**_________________________________**which is not what I get:

[itex]\Delta U = \frac{f}{2} R T_0 ( 1 - (\frac{P_f}{P_0})^{\frac{2}{f+2}}) = dW[/itex] per mole (with n=1).

Gamma is defined as (f+2)/f. Also observe that their notation is W instead of dW ( which here is not meant to be read as a change in work, but a quantity of work ).

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