TITLE.PM5

118 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 Fig. 4.10. Free expansion. ∴ For a free expansion of a perfect gas, cvT 1 ...

FIRST LAW OF THERMODYNAMICS 119 dharm M-therm/th4-2.pm5 Example 4.1. In an internal combustion engine, during the compression st ...

120 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 Temperature before compression = 40°C or 313 K Temperature after compress ...

FIRST LAW OF THERMODYNAMICS 121 dharm M-therm/th4-2.pm5 = 0.105(0.20 – 0.40) MJ = – 21 kJ [Q 1 MJ = 10^3 kJ] Substituting this v ...

122 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 (i)For isothermal process : W1–2 = pdv. 1 2 z = p^1 v^1 loge^ p p 1 2 F H ...

FIRST LAW OF THERMODYNAMICS 123 dharm M-therm/th4-2.pm5 Expansion stroke. Process 2-1 : Fig. 4.15 Work done by air on the piston ...

124 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 Work input to the paddle wheel = 9000 kJ Heat transferred to the surround ...

FIRST LAW OF THERMODYNAMICS 125 dharm M-therm/th4-2.pm5 Here Q = Heat leaving the boundary. (i) When the stone is about to enter ...

126 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 We have, Ql–q–m = (Um – Ul) + Wl–q–m 168 = (Um – Ul) + 64 ∴ Um – Ul = 104 ...

FIRST LAW OF THERMODYNAMICS 127 dharm M-therm/th4-2.pm5 ∴ 6400 = ∆ U + 8000 or ∆ U = – 1600 J = – 1.6 kJ. (Ans.) The –ve sign in ...

128 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 The completed table is given below : Process Q(kJ/min) W(kJ/min) ∆ E(kJ/m ...

FIRST LAW OF THERMODYNAMICS 129 dharm M-therm/th4-2.pm5 Let the system flow be in kg/s. ∴ zdQ = 840 m& kJ/s zdW = 1200 – 6 = ...

130 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 Solution. Heat received by the system, Q = 50 kJ Change in volume ∆ V = 0 ...

FIRST LAW OF THERMODYNAMICS 131 dharm M-therm/th4-2.pm5 (i) Applying the first law energy equation, Q = ∆ U + W 45 = ∆ U + (– 5 ...

132 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 ∴ ∆U = 3.64 (100 × 10^3 × 0.906 – 500 × 10^3 × 0.25) J [Q 1 Pa = 1 N/m^2 ...

FIRST LAW OF THERMODYNAMICS 133 dharm M-therm/th4-2.pm5 (ii) The work done per kg of fluid is given by Wpdv v v =z 1 (^2) = p(v ...

134 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 The change in internal energy of the fluid during the process U 2 – U 1 = ...

FIRST LAW OF THERMODYNAMICS 135 dharm M-therm/th4-2.pm5 l Constant volume Constant pressure n m V p Fig. 4.20 ∴ Un = 195 – 77 = ...

136 ENGINEERING THERMODYNAMICS dharm M-therm/th4-2.pm5 QW==+=+=+=paddle ∆∆∆∆ ∆U pV U () (pV U pV)∆H Hence paddle work is equal t ...

FIRST LAW OF THERMODYNAMICS 137 dharm /M-therm/Th4-3.pm5 Also γ= = c c p v 1 0 714. = 1.4 ∴ V 2 = 0.2 × 410 102 10 5 5 1 ×^1 × F ...