TITLE.PM5

138 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 (ii)Index of expansion, n : If the work done by the polytropic process i ...

FIRST LAW OF THERMODYNAMICS 139 dharm /M-therm/Th4-3.pm5 Also, bb+^1 = γ ...(Given) ∴ d(loge p) + γd (loge v) = 0 Integrating, w ...

140 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 The right hand side is a positive figure indicating the increase in ener ...

FIRST LAW OF THERMODYNAMICS 141 dharm /M-therm/Th4-3.pm5 (iv) Q = ∆ U + W Now, W pV pV n mR T T = n − − = − − 11 2 2 1 2 11 () = ...

142 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 pV = mRT ∴ m pV RT ==×× ×× 11 1 5 3 02 10 0 0 10 295 .015 .287 [Q R for ...

FIRST LAW OF THERMODYNAMICS 143 dharm /M-therm/Th4-3.pm5 or Taking log on both sides, we get loge (0.6357) = (γ – 1) loge (0.333 ...

144 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 In case of a polytropic process, T T p p n 2 n 1 2 1 (^131) 66 3 1 F HG ...

FIRST LAW OF THERMODYNAMICS 145 dharm /M-therm/Th4-3.pm5 ∴ Change in internal energy during constant volume process 2–3, U 3 – U ...

146 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 = (15 × 10^5 ) × 0.15 × loge (4) = 311916 J = 311.9 kJ Considering const ...

FIRST LAW OF THERMODYNAMICS 147 dharm /M-therm/Th4-3.pm5 5 = a + 0.02b ...(i) 1.5 = a + 0.08b ...(ii) From (i) and (ii) we get, ...

148 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 Solution. Insider diameter of the cylinder = 8 cm Stiffness of the sprin ...

FIRST LAW OF THERMODYNAMICS 149 dharm /M-therm/Th4-3.pm5 Vacuum dy Air Q y y 1 y 0 0 0 2 1 c p (Pressure) V (Volume) Fig. 4.28 F ...

150 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-3.pm5 = F − HG I KJ = F + HG I KJ F − HG I KJ A S ppA S (^222122) pp pp21 21 2 ...

FIRST LAW OF THERMODYNAMICS 151 dharm /M-therm/Th4-4.pm5 Fig. 4.30 shows a schematic flow process for an open system. An open sy ...

152 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 C = Velocity of fluid , Z = Height above datum, p = Pressure of the flui ...

FIRST LAW OF THERMODYNAMICS 153 dharm /M-therm/Th4-4.pm5 The internal energy is a function of temperature only and it is a point ...

154 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 i.e., vdp 1 2 z = C^2 2 in the case of a nozzle. If ∆ PE = 0 and ∆ KE = ...

FIRST LAW OF THERMODYNAMICS 155 dharm /M-therm/Th4-4.pm5 (ii)Steady flow constant volume process : W = – 1 Vdp 2 z = – V(p 2 – p ...

156 ENGINEERING THERMODYNAMICS dharm /M-therm/Th4-4.pm5 ∴ pv Z g C pv Z g C 111 2 222 2 ++ 22 F HG I KJ =++ F HG I KJ + W ...(4. ...

FIRST LAW OF THERMODYNAMICS 157 dharm /M-therm/Th4-4.pm5 Applying energy equation to the system. Here, Z 1 = Z 2 (i.e., ∆ Z = 0) ...