Question

Air enters a compressor at 10⁵ Pa and 25°C having volume of 1.8 m³/kg and is compressed to 5 × 10⁵ Pa isothermally. Determine : 

(i) Work done ; 

(ii) Change in internal energy ; and 

(iii) Heat transferred.

Answer 1

Initial pressure of air, p₁ = 10⁵ Pa 

Initial temperature of air, T₁ = 25 + 273 = 298 K 

Final pressure of air, p₂ = 5 × 10⁵ Pa 

Final temperature of air, T₂ = T₁ = 298 K (isothermal process) 

Since, it is a closed steady state process, we can write down the first law of thermodynamics

Q = (u₂ – u₁) + W ......per kg

(i) For isothermal process :

p₁v₁ = p₂v₂ for isothermal process

(– ve sign indicates that the work is supplied to the air) 

∴ Work done on the air = 289.7 kJ/kg

(ii) Since temperature is constant,

∴ u₂ – u₁ = 0

∴ Change in internal energy = zero

(iii) Again, 

Q_1–2 = (u₂ – u₁) + W

= 0 + (– 289.7) = – 289.7 kJ

(– ve sign indicates that heat is lost from the system to the surroundings)

∴ Heat rejected = 289.7 kJ/kg.

Answer 2

nitial pressure of air,   P_{1}=10^{5}P1​=105Pa

Initial temperature of air,  T_{1} = 25 + 273 = 298T1​=25+273=298K

Final pressure of air,   P_{2} = 5 × 10^{5}P2​=5×105Pa

Final temperature of air,  T_{2} = T_{1} = 298T2​=T1​=298 K (isothermal process)

Since, it is a closed steady state process, we can write down the first law of thermodynamics as,

Q = (u_{2}-u_{1}) + WQ=(u2​−u1​)+W   ……per kg

(i) For isothermal process :

W_{1-2}=\int_{1}^{2} P.dv=P_{1}v_{1}\log_{e}\left ( \frac{P_{1}}{P_{2}} \right )W1−2​=∫12​P.dv=P1​v1​loge​(P2​P1​​)

As     P_{1}v_{1}=P_{2}v_{2}P1​v1​=P2​v2​ for isothermal process

\therefore∴          W_{1-2}=-10^{5}\times 1.8\log_{e}\left ( \frac{1\times 10^{5}}{5\times 10^{5}} \right )W1−2​=−105×1.8loge​(5×1051×105​)

= – 2.897 × 10^{5} = – 289.7 \frac{kJ}{kg}=–2.897×105=–289.7kgkJ​.

(– ve sign indicates that the work is supplied to the air)

\therefore∴     Work done on the air = 289.7 \frac{kJ}{kg}=289.7kgkJ​ .

(ii) Since temperature is constant,

\therefore∴       u_{2}-u_{1}=0u2​−u1​=0

\therefore∴       Change in internal energy = zero.

(iii) Again,     Q_{1-2}=\left ( u_{2}-u_{1} \right )+WQ1−2​=(u2​−u1​)+W

= 0 + (– 289.7) = – 289.7=0+(–289.7)=–289.7 kJ

(– ve sign indicates that heat is lost from the system to the surroundings)

\therefore∴      Heat rejected = 289.7\frac{kJ}{kg}kgkJ​.