Question

0.5kg of air is compressed reversibly and adiabatically from 80kpa, 60°c to 0.4Mpa and is then expanded at constant pressure to the original volume. Calculate the heat transfer and work transfer for the whole path.
Take R = 0.247kj/kg.k, Cp = 1.005kj/kg.k, Cv = 0.718kj/kg.k

Answer 1

Question #79126

0.5kg of air is compressed reversibly and adiabatically from 80kPa, $60^{\circ}\mathrm{C}$ to 0.4Mpa and is then expanded at constant pressure to the original volume. Calculate the heat transfer and work transfer for the whole path.

Take $R = 0.247\mathrm{kJ / kg.K}$, $C_p = 1.005\mathrm{kJ / kg.K}$, $C_v = 0.718\mathrm{kJ / kg.K}$

Answer:

Assume the air is an ideal gas. Thus, from the ideal gas equation we have:

$$pV = mRT,$$

where $p$, $V$ and $T$ are pressure, volume and absolute temperature, respectively,

$m = 0.5\mathrm{kg}$ – the mass of the air (which remains constant),

$R = 0.247\mathrm{kJ / kg.K}$ – the gas constant of air.

From (1) we can find the initial volume of the processed air ($T_0 = 60^{\circ}\mathrm{C} + 273\mathrm{K} = 333\mathrm{K}$):

$$V_0 = \frac{mRT_0}{p_0} = \frac{0.5 \cdot 0.247 \cdot 333}{80} = 0.514\mathrm{m^3}.$$

Since the compression is reversible and adiabatic, the heat transfer is zero ($Q_c = 0$), and the transferred work is given by:

$$W_c = -\alpha mRT_0 \left(\left(\frac{p_1}{p_0}\right)^{\frac{\gamma - 1}{\gamma}} - 1\right),$$

where

$$\gamma = \frac{C_p}{C_V},$$

the adiabatic index,

$\alpha = 2.5$ – the number of degrees of freedom (which is 5 for the air) divided by two.

Substitute into (3) and (2):

$$\gamma = \frac{1.005}{0.718} = 1.400,$$ $$W_c = -2.5 \cdot 0.5 \cdot 0.247 \cdot 333 \left(\left(\frac{400}{80}\right)^{\frac{1.4 - 1}{1.4}} - 1\right) = -60.00\mathrm{kJ}.$$

The final volume of the air could be determined from the equation of an adiabatic process:

$$p_0 V_0^{\gamma} = p_1 V_1^{\gamma},$$ $$V_1 = V_0 \left(\frac{p_0}{p_1}\right)^{\frac{1}{\gamma}} = 0.514 \cdot \left(\frac{80}{400}\right)^{\frac{1}{1.4}} = 0.163\mathrm{m^3}.$$

The temperature at the end of compression could be defined from (1) as follow:

$$T _ {1} = \frac {p _ {1} V _ {1}}{m R} = \frac {4 0 0 \cdot 0 . 1 6 3}{0 . 5 \cdot 0 . 2 4 7} = 5 2 7. 3 \mathrm {K}.$$

For the second process, which is isobaric expansion, the heat and the work transfer, respectively, are given by:

$$Q = m C _ {p} \Delta T,$$ $$W = p \Delta V,$$

where $\Delta T$ and $\Delta V$ are change in temperature and volume, respectively.

From (1), the final temperature of the air is:

$$T _ {2} = \frac {p _ {2} V _ {2}}{m R} = \frac {4 0 0 \cdot 0 . 5 1 4}{0 . 5 \cdot 0 . 2 4 7} = 1 6 6 4. 8 \mathrm {K}.$$

Substitute into (5) and (6):

$$Q _ {e} = 0. 5 \cdot 1. 0 0 5 \cdot (1 6 6 4. 8 - 5 2 7. 3) = 5 7 1. 7 \mathrm {k J},$$ $$W _ {e} = 4 0 0 \cdot (0. 5 1 4 - 0. 1 6 3) = 1 4 0. 5 \mathrm {k J}.$$

The total heat transfer is:

$$Q = Q _ {c} + Q _ {e} = 5 7 1. 7 \mathrm {k J}.$$

The total work transfer is:

$$W = - 6 0 + 1 4 0. 5 = 8 0. 5 \mathrm {k J}.$$