Availability & Irreversibility
By: S K Mondal
Chapter 8
⎛T ⎞ WRII = Q2 ⎜ 0 − 1 ⎟ ⎝ T2 ⎠ ⎛T ⎞ = mL ⎜ 0 − 1 ⎟ ⎝ T2 ⎠ ⎛ T ⎞ = 1000 × 335 ⎜ 0 − 1 ⎟ ⎝ 273 ⎠ ⎛ T ⎞ = 335000 ⎜ 0 − 1 ⎟ kJ ⎝ 273 ⎠ (iii) WR required for 0º C ice to –10 º C ice. When temperature is T if dT temperature decreases dQ 2 = – mc p ice dT ∴ T0 ⎞ − 1⎟ ⎝T ⎠
∴
dWR = − m c p ice dT ⎛⎜
∴
WRII = m c p ice
273
1 4.187 kJ/kg c p,water = 2 2 4.187 ⎡ 273 ⎤ = 1000 × − 10 ⎥ T0 ln ⎢ 2 ⎣ 263 ⎦ 273 ⎡ ⎤ − 10 ⎥ kJ = 2093.5 ⎢T0 ln 263 ⎣ ⎦
∴
Total work required
∴
273 ⎡ ⎤ ⎢⎣T0 ln 263 − (273 − 263) ⎥⎦
=
c p,ice
∴
Solution:
263
⎞ − 1 ⎟ dT = m c p ice ⎠
Let
∴
Q8.9
⎛ T0
∫ ⎜⎝ T
WR = (i) + (ii) + (iii) = [1529.2 T0 – 418740] kJ WR and T0 has linear relationship 15 + 30 T0 = º C = 22.5ºC = 295.5 K 2 WR = 33138.6 kJ = 33.139 MJ
A pressure vessel has a volume of 1 m3 and contains air at 1.4 MPa, 175°C. The air is cooled to 25°C by heat transfer to the surroundings at 25°C. Calculate the availability in the initial and final states and the irreversibility of this process. Take p0 = 100 kPa. (Ans. 135 kJ/kg, 114.6 kJ/kg, 222 kJ) Given Ti = 175ºC = 448 K Tf = 25ºC = 298 K Vf = 1 m3 Vi = 1 m3 pi = 1.4 MPa = 1400 kPa p f = 931.25 kPa Calculated Data: p 0 = 101.325 kPa,
T0 = 298 K
c p = 1.005 kJ/kg – K, cV = 0.718 kJ/kg – K; R = 0.287 kJ/kg – K ∴
Mass of air (m) =
pi Vi 1400 × 1 = 10.8885 kg = RTi 0.287 × 448 Page 103 of 265