Specific work done by an adiabatic & reversible compressor on air

[Master1022]

Homework Statement

Air at 290 K is compressed in a steady-flow process between absolute pressures 1 bar and 10 bar. Calculate the work done per kg by an adiabatic & reversible compressor

Relevant Equations

$w = \frac{R}{1-n} (T_2 - T_1)$

Hi,

A quick question on a conundrum I seem to have encountered. My main question is: why is it wrong to use the formula above instead of the SFEE approach?

My approach:
Use the formula:
$$ w = \frac{R}{1-n} (T_2 - T_1) $$

From the data book, $R = 0.287$ kJ/kg K and $n = \gamma = 1.4$
For a reversible adiabatic process, we can find $T_2$:
$$ T_2 = T_1 \left( \frac{p_2}{p_1} \right) ^{\frac{\gamma - 1}{\gamma}} $$
$$ (290) \cdot (10)^{0.4/1.4} = 559.902 \text{K} $$
and thus:
$$ w = \frac{0.287}{1-1.4} (559.902 - 290) = (-) 193.65 \text{kJ/kg} $$

Answer book:
The answer book seems to use the SFEE and does: $$ w = - \Delta h = - c_p (T_2 - T_1) = (-) 270 \text{kJ/kg} $$
(The units were most likely kW/kg for the SFEE problem)

Trying to understand the differences:
- Earlier in the question, we did assume perfect gas behavior to derive a certain formula
- In a later part of this question (for an reversible isothermal compressor), we use a variant of this formula which yields the same answer as the book

The only reason I can think of that the above formula that I used is not applicable to a steady flow process. However, I don't really understand why this is the case?

Any help is greatly appreciated

 

[Chestermiller]

Are you familiar with the derivation of the SFEE?

 

[Master1022]

Are you familiar with the derivation of the SFEE?

Yes

Does the formula that I used only apply to closed systems, whereas we use the SFEE for open systems?

 

[Chestermiller]

Yes

Does the formula that I used only apply to closed systems, whereas we use the SFEE for open systems?

Yes. If you are familiar with the derivation of the SFEE (open system version of the first law of thermodynamics), please say in words how work is subdivided and accounted for in the SFEE, and describe the difference between shaft work and total work.

 

[Master1022]

Yes. If you are familiar with the derivation of the SFEE (open system version of the first law of thermodynamics), please say in words how work is subdivided and accounted for in the SFEE, and describe the difference between shaft work and total work.

IN the SFEE, $\dot q - \dot w_s = \dot m \Delta (h + c^2 / 2 + gz)$ where $w_s$ is the shaft work done BY the system (i.e. the work done for a moving component).

The total work done is split up into three components:
1. Shaft work
2. Work needed to push fluid into system $- \dot m v_1 p_1$
3. Work is done by the system on the surroundings $\dot m v_2 p_2$

Shaft work doesn't include components (2) and (3)

 

[Chestermiller]

IN the SFEE, $\dot q - \dot w_s = \dot m \Delta (h + c^2 / 2 + gz)$ where $w_s$ is the shaft work done BY the system (i.e. the work done for a moving component).

The total work done is split up into three components:
1. Shaft work
2. Work needed to push fluid into system $- \dot m v_1 p_1$
3. Work is done by the system on the surroundings $\dot m v_2 p_2$

Shaft work doesn't include components (2) and (3)

Perfect. So in the closed system problem, you are determining the total work $W=nC_v\Delta T$, and in the SFEE problem, you are determining the shaft work $W_s=nC_p\Delta T$ where n is the number of moles of gas in the closed system or the number of moles of gas that passes through the open system.

 

[Master1022]

Perfect. So in the closed system problem, you are determining the total work $W=nC_v\Delta T$, and in the SFEE problem, you are determining the shaft work $W_s=nC_p\Delta T$ where n is the number of moles of gas in the closed system or the number of moles of gas that passes through the open system.

Okay thank you for the clarification!