Outline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples

Size: px

Start display at page:

Download "Outline. Example. Solution. Property evaluation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids Examples"

Caren Rogers
6 years ago
Views:

1 Outline Property ealuation examples Specific heat Internal energy, enthalpy, and specific heats of solids and liquids s A piston-cylinder deice initially contains 0.5m of saturated water apor at 00kPa. At this state, the piston is resting on stops, and the mass of the piston is such that a pressure of 00kPa is required to moe it. Heat is now slowly transferred to the steam until the olume doubles. Show the process on a P- diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done during this process, and (c) the total heat transfer. Solution First recognize that we are dealing with a step process Process -: Constant olume heating Process -: Constant pressure heating Let s apply the st law to entire process: Q W = E = U = U U = m( u u ) Q W = m( u u) o find Q we need to find W and internal energy change (property ealuation)

2 Property ealuation 00kPa Solution P = 00kPa ( a) = = m / kg sat. apor u = u = 59.5 kj / kg m = V / = 0.5 / = kg = / = / =.775 / V m m kg P = 00kPa = C, u = 8.8 kj / kg Solution Work ealuation ( b) W = 0 (constant olume) W = PdV = P( V V ) = 00( 0.5) = 50kJ How does P ary with V? Heat ealuation ( c) Q = W + m( u u ) = + 50 kj + ( kg)( ) kj / kg Q = 875.0kJ A 0kg mass of superheated refrigerant- 4a at 0.8MPa and 40C is cooled at constant pressure until it exists as a compressed liquid at 0C. (a) Show the process on a - diagram with respect to saturation lines (b) Determine the change in olume (c) Find the change in total internal energy

3 R-4a 40C 0.8MPa P = 0.8MPa u = 5. kj / kg ; = m / kg = 40C P = 0.8MPa u u = kj / kg ; = m / kg = 0C V = m( ) = (0 kg)( ) m / kg = 0.6m A- A-0 U = m( u u ) = (0 kg)( ) kj / kg = 75.kJ Superheated water apor cools at constant olume until the temperature drops to 50F. At the final state, the pressure, the quality, and the enthalpy are to be determined. P = 80 psia =.04 ft / lbm able A-4E = 500F ( a) = ft / lbm; =.86 ft / lbm F < < SLVM P = P = 9.8 psia f g f ( b) x = = = 0.9 fg f fg HO 80psia 500F ( c) h = h + x h = = 45.7 Btu / lbm HO 00C A rigid essel that contains a SLVM is heated until it reaches the critical state. he mass of the liquid water and the olume occupied by the liquid at the initial state is to be determined. = = = cr m / kg (A-) m = V / = 0.5m/ m / kg = 58.48kg = m / kg; =.679 m / kg (A-) C f x = = = fg f t m = ( x ) m = ( 0.006)(58.48) = 58.8kg V = m = (58.8 kg)( m / kg) = 0.65m f f f

4 Superheated steam in a piston-cylinder deice is cooled at constant pressure until half of the mass condenses. he final temperature and the olume change are to be determined, and the process should be shown on a - diagram. x = 0.5 = = 79.9C P =.0MPa P =.0MPa = m / kg & = f + x fg = 00C x = 0.5 = ( ) = m / kg ( ) V = m = 0.8 kg( ) m / kg = 0.8m HO 00C MPa Specific Heats Related to materials ability to store energy It is defined as energy required to raise temperature of a unit mass of a substance by one degree HEA m = kg = C C 5 kj / kg C Amount will depend on exact process Let s consider two important processes: Constant olume Constant pressure Constant olume and constant pressure specific heats Constant olume specific heat, C Constant pressure specific heat, C p () V = const m = kg = C C. kj / kg C () P = const m = kg = C C 5. kj / kg C HEA HEA 4

5 More on specific heats Express specific heats in terms of other thermodynamic properties Consider fixed mass in stationary closed system undergoing a constant-olume process (no expansion or compression) Conseration of energy principle says... e e = e or δ e δ e = du in out system in out δ e δ e = C d C d = du in out u h C = can also derie CP = at constant olume P Obserations Equations are property relations and so are independent of type of process, i.e. good for any substance undergoing any process (but does depend on states, i.e., P) For constant olume/pressure process C /C p happen to represent amount of energy transferred to raise temperature of unit mass one degree Constant olume/pressure specific heats relate changes in internal energy u/enthalpy h to changes in temperature SI units are kj/kg-c or kj/kg-k; numbers equialent since change in temperature in C scale same as in K scale Incompressible substance First, define incompressible substance as one whose specific olume (density) is constant Second, notice that specific olume of solids and liquids essentially remains constant during process i.e. see saturate liquid line on P- or - diagram Hence, solids and liquids can be modeled (approximated) as incompressible substances his implies that energy associated with olume change (what we will call boundary energy) is negligible compared to other forms of energy such as heat interaction or non-qe forms of work (tbd) 5

6 Specific heats and energy for incompressible substances It can be shown that for solids and liquids: CP = C = C ypical alues for common solids/liquids in A-9 Also specific heat for solids/liquids function of temperature alone (as for ideal gases - tbd) du = C d = C( ) d u = u u = C( ) d C ( ) a dh = du + dp + Pd = du + dp h = u + P C + P a Enthalpy changes for solids and liquids For solids nd term is small For liquids, two special cases: Constant pressure processes (as in heaters) P = h = u C 0 a Constant temperature processes (as in pumps) = 0 h = P h h = ( P P ) h h + ( P P P, Determine enthalpy of liquid water at 00C and 5MPa (a) by using compressed liquid tables, and (b) by approximating it as saturated liquid, and (c) using correction gien aboe. P > P = 0.kPa CL P = 5MPa ( a) h = 40.8 kj / kg (A-5) = 00C ( b) h = h = kj / kg (A-) (.6%) ( c) h = h + ( P P P, kj = kj / kg m/ kg(5,000 0.) kpa kpa m = kj / kg (%) 6

7 Summary Know what substance you are dealing with, e.g. pure substances, ideal gas For pure substances such as water, R-4a, etc. use tables Know where you are relatie to the wet dome e.g. fix the state, clearly describe it, draw P- or - diagram and label your states In general, always look for properties that may be constant during a process, e.g. olume for rigid tank Read question carefully and think before you act! Summary Specific heats proide property relations between changes in enthalpy/internal energy and temperature for different materials Internal energy and enthalpy change for solid/liquid related to temperature change ia single specific heat alue (aerage alue) For constant temperature case, change in enthalpy related to change in pressure ia specific olume 7

GAS. Outline. Experiments. Device for in-class thought experiments to prove 1 st law. First law of thermodynamics Closed systems (no mass flow)

GAS. Outline. Experiments. Device for in-class thought experiments to prove 1 st law. First law of thermodynamics Closed systems (no mass flow) Outline First law of thermodynamics Closed systems (no mass flow) Device for in-class thought experiments to prove 1 st law Rubber stops GAS Features: Quasi-equlibrium expansion/compression Constant volume