BAB IV (CONT’D)

1. BAB IV (CONT’D) AGUS HARYANTO AGRENG DEPT. 06 MARET 2008

2. SPECIFIC HEATS • The specific heat (C) is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. The unit of C is J/kg.oC or J/kg.K • Two kinds of specific heats: specific heat at constant volume,Cv and specific heat at constant pressure, Cp. • Cv : energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. Similarly for Cp. • Cv : energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant.

3. Cv and Cp • Cvand Cp(values given are for helium gas).

4. Cv • Consider a fixed mass in a stationary closed system undergoing a process @ V constant (no expansion or compression work). • The conservation of energy principle: ein – eout = esystem(per unit mass) ein – eout = du (differential) • From the definition of Cv, left hand side must be equal to Cv dT, so: Cv dT = du (at constant V) or

5. Cp • Now, consider a constant-pressure expansion or compression process: • The conservation of energy principle: ein – eout = esystem(per unit mass) ein – eout = dh (differential, Ex. 4-5) • From the definition of Cp, left hand side must be equal to Cp dT, so: Cv dT = dh (at constant V) or

6. Defining equation Cv and Cp

7. INTERNAL ENERGY, ENTHALPY, and SPECIFIC HEATS OF IDEAL GASES • Remind for an ideal gas: Pv = RT • For ideal gas, the internal energy is a function of the temperature only: u = u(T) • Using enthalpy definition and EOS of ideal gas: • Since R constant and u = u(T), then enthalpy of ideal gas is also a function of temperature only: h = h(T)

8. Since u and h depend only on temperature for an ideal gas, then the specific heats Cv and Cp also depend, at most, on temperature only. Therefore, at a given temperature, u, h, Cv, and Cp of an ideal gas have fixed values regardless of the specific volume or pressure.

9. u, h, cv, and cp • The differential changes in the internal energy and enthalpy of an ideal gas can be expressed as: du = Cv(T) dT and dh = Cp(T) dT • The change in internal energy and enthalpy:

10. u, h, cv, and cp • Practically: • The average specific heats Cp,avgand Cv,avgareevaluated at the average temperature (T1 + T2)/2

11. Cp values • At low P, all real gases approach ideal-gas, and their specific heats depend on T only. • We need to have relations for Cvand Cpas functions of temperature.

12. NOTES • The u and h relations given previously are not restricted to any kind of process. They are valid for all processes. • The presence of the Cvin an equation should not lead one to believe that this equation is valid for a constant-volume process only. • The relation u = Cv,avg T is valid for any ideal gas undergoing any process. • Similar argument can be given for Cpand h.

14. Specific Heat Relations of Ideal Gases • From h = u + RT we have dh = du + RdT • From du = Cv(T) dT and dh = Cp(T) dT we have Cp(T) dT = Cv(T) dT + RdT • Divide by dT we have: Cp = Cv + R • Another ideal-gas property called the specific heat ratio k,defined as: k varies with T. For monatomic gases, k = 1.667. For diatomic gases (ex: air), k = 1.4 at room T.

15. SUMMARY: 3 ways to determine the internal energy and enthalpy changes of ideal gases: • Using the tabulated u and h data • using the Cvor Cprelations as a function of T and integrate them. • using average specific heats

16. EXAMPLE 4–7 Evaluation of u • Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) data from the air table (Table A–17), (b) the average specific heat value (Table A–2b).

17. SOLUtion (a) Using Table A-17: u1 = u@300 K = 214.07 kJ/kg u2 = u@600 K = 434.78 kJ/kg u = u2 – u1 = 1434.78 – 214.07 kJ/kg u= 220.71 kJ/kg (b) Average temperature = (T1 + T2)/2 = 450 K Cv,avg@450 K = 0.733 kJ/kg.K u = Cv,avg@450 K (T2 – T1) u = (0.733 kJ/kg.K)[(600 – 300) K] u = 220 kJ/kg

18. EXAMPLE 4–8 Heating of a Gas • An insulated rigid tank initially contains 1.5 lbm of helium at 80°F and 50 psia. A paddle wheel with a power rating of 0.02 hp is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas.

19. SKEtch

20. Solution (a) The amount of paddle-wheel work: Energy balanca equation:

21. (b) The final pressure is determined from the ideal-gas relation: where V1 = V2 so that:

22. EXAMPLE 4–9 Heating of a Gas • A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27°C. An electric heater within the device is turned on and is allowed to pass a current of 2 A for 5 min from a 120-V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen.

23. sketch

24. Slution • In this case we will have boundary work, Wb. Also, heat is lost (Qout) from the system and electrical work is done on the system (Wein) • Electrical work done on the nitrogen: • The mass of nitrogen is determined from

25. Solution (cont’d) • Energy balance (for a closed system undergoing a process at constant pressure, U  Wb + H): • From Table A–2a, Cp = 1.039 kJ/kg·K for N2, so

26. EXAMPLE 4–10 Heating of a Gas at Constant Pressure • A piston–cylinder device initially contains air at 150 kPa and 27°C. At this state, the piston is resting on a pair of stops, as shown in Fig. 4–32, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) the final temperature, (b) the work done by the air, and (c) the total heat transferred to the air.

27. sketch

28. Solution (cont’d) (a) The final temperature can be determined easily by using the ideal-gas relation between states 1 and 3 in the following form: (b) The work done could is the area under the process curve on a P-V diagram,

29. Solution (cont’d) (c) First, take energy balance:

30. INTERNAL ENERGY, ENTHALPY, ANDSPECIFIC HEATS OF incompressible • A substance whose specific volume (or density) is constant is called an incompressible. • Cp = Cv = C • Internal Energy Changes:

31. INTERNAL ENERGY, ENTHALPY, ANDSPECIFIC HEATS OF incompressible • Enthalpy Changes: • For solid v P = 0, and • For liquids: 1. Proses @ P constant 2. Proses @ T constant

32. For liquids • For a process between states 1 and 2, the last relation can be expressed: • By taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same T, the enthalpy of the compressed liquid can be expressed as

33. EXAMPLE 4–11 Enthalpy of Compressed Liquid • Determine the enthalpy of liquid water at 100°C and 15 MPa (a) by using compressed liquid tables, (b) by approximating it as a saturated liquid, and (c) by using the correction given by Eq. 4–38.

34. solution (a) From compressed liquid tables (Table A–7): (b) Approximating the compressed liquid as a saturated liquid at 100°C (error = 2.6%): (c) Using Equation 4-38 (error 1%):

35. EXAMPLE 4–12 Cooling of an Iron Block by Water • A 50-kg iron block at 80°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. Determine the temperature when thermal equilibrium is reached.

36. solution • The volume of a rigid tank is constant, and thus there is no boundary work. The energy balance on the system can be expressed as

37. Solution (cont’d) • The specific volume of liquid water at or about room temperature is 0.001 m3/kg. Then: • The specific heats of iron and liquid water (Table A–3) are Ciron = 0.45 kJ/kg·°Cand Cwater = 4.18 kJ/kg·°C. Substituting into energy equation: 0 =

38. EXAMPLE 4–13 Temperature Rise due to Slapping • If you ever slapped someone or got slapped yourself, you probably remember the burning sensation. Imagine you had the unfortunate occasion of being slapped by an angry person, which caused the temperature of the affected area of your face to rise by 1.8°C (ouch!). Assuming the slapping hand has a mass of 1.2 kg and about 0.150 kg of the tissue on the face and the hand is affected by the incident, estimate the velocity of the hand just before impact. Take the specific heat of the tissue to be 3.8 kJ/kg·°C.

39. sketch

40. analysis • This is a closed system since it involves a fixed amount of mass (no mass transfer). We observe that the kinetic energy of the hand decreases during the process, as evidenced by a decrease in velocity from initial value to zero, while the internal energy of the affected area increases, as evidenced by an increase in the temperature. There seems to be no significant energy transfer between the system and its surroundings during this process. Energy balance:

41. solution • NOTE: Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering.