Udara dikompres secara tunak oleh kompresor $8 \text{ kW}$ dari kondisi-1: $P_1 = 100 \text{ kPa}$, $T_1 = 17^\circ C$ menjadi kondisi-2: $P_2 = 600 \text{ kPa}$, $T_2 = 167^\circ C$ pada laju aliran massa $2,1 \text{ kg/menit}$. Perubahan energi kinetik dan energi potensial dapat diabaikan. Tentukan peningkatan eksergi pada air (asumsi suhu lingkungan = $17^\circ C$).
Identifikasi Data
Kita ambil komponen dan ubah ke bentuk standar
- $C_{p, air} = 1,005 \text{ kJ/kg} \cdot \text{K}$
- $R = 0,287 \text{ kJ/kg} \cdot \text{K}$
- $W_{compressor} = 8 \text{ kW}$
- $\dot{m} = 0,035 \text{ kg/s}$
- $T_{environment} = 290,15 \text{ K}$
- $T_1 = 290,15 \text{ K}$
- $T_2 = 440,15 \text{ K}$
Perubahan Entropi Spesifik
$$\Delta s = s_2 – s_1$$
$$\Delta s = C_p \ln\left(\frac{T_2}{T_1} \right) – R \ln\left(\frac{P_2}{P_1}\right)$$
$$\Delta s = 1,005 \ln\left(\frac{440,15}{290,15} \right) – 0,287 \ln\left(\frac{600}{100}\right)$$
$$\Delta s = 1,005 \ln(1,5169) – 0,287 \ln(6)$$
$$\Delta s = 1,005 \cdot 0,4166 – 0,287 \cdot 1,7917$$
$$\Delta s = 0,4186 – 0,5142$$
$$\Delta s = -0,0956 \text{ kJ/kg} \cdot \text{K}$$
Perubahan Entalpi Spesifik
$$\Delta h = h_2 – h_1$$
$$\Delta h = C_p(T_2 – T_1)$$
$$\Delta h = 1,005(440,15 – 290,15)$$
$$\Delta h = 1,005(150)$$
$$\Delta h = 150,75 \text{ kJ/kg}$$
Peningkatan Eksergi Spesifik
$$\Delta \psi = \psi_2 – \psi_1$$
$$\Delta \psi = (h_2 – h_1) – T_{environment}(s_2 – s_1)$$
$$\Delta \psi = 150,75 – 290,15(-0,0956)$$
$$\Delta \psi = 150,75 – (-27,7383)$$
$$\Delta \psi = 178,4883 \text{ kJ/kg}$$
Laju Peningkatan Eksergi
$$\dot{\Delta X} = \dot{m} \times \Delta psi$$
$$\dot{\Delta X} = 0,035 \times 178,4883$$
$$\dot{\Delta X} = 6,247 \text{ kW}$$
