TITLE.PM5

98 ENGINEERING THERMODYNAMICS Dharm \M-therm/th3-2.p65 With the increase in pressure (a) boiling point of water increases and e ...

PROPERTIES OF PURE SUBSTANCES 99 Dharm \M-therm/th3-2.p65 In throttling process (a)h 12 = h 2 (b)h 1 = h 2 (c)h 1 = h 2 + h T ...

100 ENGINEERING THERMODYNAMICS Dharm \M-therm/th3-2.p65 Steam at 150 bar has an enthalpy of 3309 kJ/kg, find the temperature, t ...

4 First Law of Thermodynamics 4.1. Internal energy. 4.2. Law of conservation of energy. 4.3. First law of thermodynamics. 4.4. A ...

102 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 “When a system undergoes a thermodynamic cycle then the net heat supplied ...

FIRST LAW OF THERMODYNAMICS 103 dharm M-therm/th4-1.pm5 by heat transfer. The experiments show : (i) A definite quantity of work ...

104 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 p V 1 L 2 M N Fig. 4.2. Energy—a property of system. The processes L and ...

FIRST LAW OF THERMODYNAMICS 105 dharm M-therm/th4-1.pm5 Fig. 4.3. A PPM 1. Fig. 4.4. The converse of PMM 1. — The converse of th ...

106 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 The kilogram-mole is defined as a quantity of a gas equivalent to M kg of ...

FIRST LAW OF THERMODYNAMICS 107 dharm M-therm/th4-1.pm5 Specific heat at constant volume, cv and, Specific heat at constant pres ...

108 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 In a constant pressure process, the work done by the fluid, W = p(V 2 – V ...

FIRST LAW OF THERMODYNAMICS 109 dharm M-therm/th4-1.pm5 = cvT + RT [Q pv = RT] = (cv + R)T = cpT [Q cp = cv + R] i.e., h = cpT a ...

110 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 Considering mass of the working substance unity and applying first law of ...

FIRST LAW OF THERMODYNAMICS 111 dharm M-therm/th4-1.pm5 = (u 2 + pv 2 ) – (u 1 + pv 1 ) = h 2 – h 1 [Q h = u + pv] or Q = h 2 – ...

112 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 The constant C can either be written as p 1 v 1 or as p 2 v 2 , since p 1 ...

FIRST LAW OF THERMODYNAMICS 113 dharm M-therm/th4-1.pm5 Dividing both sides by T, we get cv dTT + Rdvv = 0 Integrating cv loge T ...

114 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 The work done is given by the shaded area, and this area can be evaluated ...

FIRST LAW OF THERMODYNAMICS 115 dharm M-therm/th4-1.pm5 From Eqn. (4.31), pv 11 γ = pv 22 γ or p p v v 2 1 1 2 =FHG IKJ γ ...(4. ...

116 ENGINEERING THERMODYNAMICS dharm M-therm/th4-1.pm5 i.e., Work done, W pv p v = n − − 11 2 2 1 ...(4.39) or W RT T = n − − () ...

FIRST LAW OF THERMODYNAMICS 117 dharm M-therm/th4-1.pm5 This is illustrated on a p-v diagram in Fig. 4.9. (i) State 1 to state A ...