Problem 44 A closed system of mass of 10 kg undergoes a process during which there is energy transfer by work from the system of 0147 kj per kg, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s The specific internal energy decreases by 5 kj/kg and the acceleration of gravity is constant at 97 m/s Determine the heat transfer for the process, in kj KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity FIND: Determine the heat transfer for the process SCHEMATIC AND GIVEN DATA: V 1 = 15 m/s Δu = -5 kj/kg W/m = + 0147 kj/kg 50 m ENGINEERING MODEL: (1) The system is a closed system () The acceleration of gravity is constant z m = 10 kg g = 97 m/s V = 30 m/s ANALYSIS: ΔU + ΔPE + ΔKE = Q - W Q = ΔU + ΔPE + ΔKE - W W = m [W/m] = 10 kg [-0147 kj/kg] = -147 kj ΔU = mδu = 10 kg [- 5 kj/kg] = -50 kj ΔKE = m (V V 1 ) = 10 kg [(30 m s ) (15 m s ) ] ΔPE = mg(z z 1 ) = (10 kg) (97 m/s )(-50 m) Q = (-50) + (-485) + (338) (-147) = -50 kj (out) 1 N 1 kg m s 1 N 1 kg m s 1 kj 1 kj 10 3 N m 10 3 N m = +338 kj = - 485 kj
Problem 57 A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three processes: Process 1-: Constant pressure, p = 14 bar, V 1 = 008 m 3, W 1 = 105 kj Process -3: Compression with pv = constant, U 3 = U Process 3-1: Constant volume, U 1 U 3 = -64 kj There are no significant changes in kinetic or potential energy (a) Sketch the cycle on a p-v diagram (b) Calculate the net work for the cycle, in kj (c) Calculate the heat transfer for process 1-, in kj KNOWN: A gas undergoes a cycle consisting of three processes FIND: Sketch the cycle on a p-v diagram and determine the net work for the cycle and the heat transfer for process 1- SCHEMATIC AND GIVEN DATA: Process 1-: Constant pressure, p = 14 bar, V 1 = 008 m 3, W 1 = 105 kj Process -3: Compression with pv = constant, U 3 = U Gas Process 3-1: Constant volume, U 1 U 3 = -64 kj ENGINEERING MODEL: (1) The gas is a closed system () Kinetic and potential energy effects are negligible (3) The compression from state to 3 is a polytropic process ANALYSIS: (a) Since W 1 > 0, the process is an expansion Thus p 3 1 V
Problem 57 (Continued) 0 (b) The net work for the cycle is W cycle = W 1 +W 3 + W 31 W 1 = 105 kj, so we need W 3 V 3 V 3 W 3 = pdv = const V dv = (p V V )ln ( V 3 ) = (p V V V )ln ( V 1 ) (*) V where V 3 = V 1 has been incorporated But, we still need to evaluate V For Process 1- at constant pressure or V W 1 = pdv = p(v V 1 V 1 ) V = W 1 p + V 1 = (105 kj) N m (14 bar) 103 1 kj 1 bar + 008 m 3 = 0103 m 3 10 5 N m Thus, with Eq (*) W 3 = (14 bar)(0103 m 3 )ln ( 008 ) N m 0103 105 1 kj 1 bar 10 3 = -1878 kj N m Thus W cycle = 105 kj + (-1878 kj) + 0 = -88 kj 0 0 (c) To get Q 1, we apply the energy balance to process 1-: ΔKE + ΔPE + (U U 1 ) = Q 1 W 1 With U = U 3, Q 1 = (U 3 U 1 ) + W 1 = (+64 kj) + (105 kj) = 369 kj
PROBLEM 37 For H O, determine the specific volume at the indicated state, in m 3 /kg Locate the states on a sketch of the T-v diagram (a) T = 400 o C, p = 0 MPa (b) T = 40 o C, p = 0 MPa (c) T = 40 o C, p = MPa
38 For water (H O), locate each of the following states on sketches of the T-v, p-v, and phase diagrams (a) T = 300 o F, p = 0 lbf/in (b) T = 300 o F, p = 90 lbf/in (c) T = 300 o F, v = 5 ft 3 /lb KNOWN: H O at specified states FIND: Locate each specified state on sketches of the T-v, p-v, and phase diagrams SCHEMATIC AND GIVEN DATA: State (a) State (b) State (c) T = 300 o F T = 300 o F T = 300 o F p = 0 lbf/in p = 90 lbf/in v = 5 ft 3 /lb ENGINEERING MODEL: 1 States (a), (b), and (c) are equilibrium states ANALYSIS: State (a): Since p < p sat (6698 lbf/in from Table A-E at T = 300 o F), state (a) is superheated vapor State (b): Since p > p sat (6698 lbf/in from Table A-E at T = 300 o F), state (b) is compressed liquid State (c): From Table A-E at T = 300 o F, v f = 001745 ft 3 /lb and v g = 647 ft 3 /lb Since v f < v < v g, state (c) is a two-phase liquid-vapor mixture T T-v Phase Diagram p = 90 lbf/in p = 6698 lbf/in (Table A-E) 3031 o F (Table A-3E) 300 o F (b) f (c) g p = 0 lbf/in (a) Table A-E at T = 300 o F v f = 001745 ft 3 /lb v g = 647 ft 3 /lb v
p p-v Phase Diagram 90 lbf/in (b) 6698 lbf/in (c) 0 lbf/in (a) T = 300 o F v p Phase Diagram Critical Liquid point 90 lbf/in (b) 6698 lbf/in (c) Since (c) is a two-phase 0 lbf/in (a) Solid liquid-vapor state, it is not fixed by T = 300 o F and p = 6698 lbf/in Vapor 300 o F T
PROBLEM 311 Determine the volume, in ft 3, of lb of a two-phase liquid-vapor mixture of Refrigerant 134a at 40 o F with a quality of 0% What is the pressure, in lbf/in? saturated vapor R-134a T = 40 o F x = 0 m = lb saturated liquid First, find the specific volume using Eq 3 and data from Table A-10E at 40 o F v = vf + x (vf - v g ) = 00151 + (0) (09470 0151) = 08948 ft 3 /lb Now V = v m = (01994 ft 3 /lb) ( lb) = 0579 ft 3 p x= 0 f g p = 49738 lbf/in (Table A-10E) T = 40 o F v
PROBLEM 318 A closed, rigid tank contains a two-phase liquid-vapor mixture of Refrigerant- initially at -0 o C with a quality of 5036% Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar Determine the final temperature, in o C If the final state is in the superheated vapor region, at what temperature, in o C, does the tank contain only saturated vapor? State 1 State R- T 1 = -0 o C x 1 = 05036 V constant m constant Therefore, v = v1 R- p = 6 bar T 6 bar 0 o C (v g = 00470 m 3 /kg) x 1 = 05036 1-0 o C v First, using data from Table A-7 and Eq 3, we can determine v 1 as follows v 1 = v f1 + x 1 (v g1 v f1 ) = 0747 x 10-3 + (05036)(0096-0747 x 10-3 ) = 00470 m 3 /kg Since v = v 1, State is in the superheated vapor region (v > v g@6bar ) Thus, interpolating at 6 bar with v = 00470 m 3 /kg in Table A-9 we get T 4375 o C Since State is superheated vapor, the tank contains only saturated vapor at the condition where v g = 00470 m 3 /kg Referring to Table A-7, this occurs at T = 0 o C
PROBLEM 330
PROBLEM 333 4 The volume is constant
340 A piston-cylinder assembly contains water, initially a saturated vapor at 00 o C The water is cooled at constant temperature to saturated liquid Kinetic and potential energy effects are negligible (a) For the water as a closed system, determine the work per unit mass of water, in kj/kg (b) If the energy transfer by heat for the process is 100 kj, determine the mass of the water, in kg KNOWN: A piston-cylinder assembly contains water, which is cooled at constant temperature FIND: For the water as the system, determine the work per unit mass For a specified energy transfer by heat, determine the mass of the water SCHEMATIC AND GIVEN DATA: Q = 100 kj State 1 State Water (Saturated Vapor) Moving Piston Water (Saturated Liquid) Moving Piston T = 00 o C T = 00 o C T p sat (00 o C) = 1554 bar 00 o C 1 v ENGINEERING MODEL: 1 Water in the piston-cylinder assembly is a closed system Volume change is the only work mode 3 Kinetic and potential energy effects are ignored ANALYSIS: (a) Since the process occurs at constant temperature in the two-phase liquid-vapor region, the pressure is also constant Thus, the work per unit mass is obtained using W 1 v pdv mp( v 1 )
W m p( v 1 ) v From Table A-4 at T = 00 o C: v 1 = v g1 = 0174 m 3 /kg, v = v f = 00011565 m 3 /kg, and p = 1554 bar Substituting values yields W m 5 N 3 10 m m 1kJ ( 1554 bar)(00011565 0174) 3 kg 1bar 10 N m = 196 kj/kg The negative sign associated with the work indicates work is into the system, as expected (b) An energy balance reduces to read U + KE + PE = Q W, or Q = U + W Expressing the right-hand side of the equation in terms of mass gives Solving for mass gives W Q m( u u1) m m Q m W ( u u1 ) m From Table A-4 at T = 00 o C: u 1 = u g1 = 5953 kj/kg, u = u f = 85065 kj/kg Since heat transfer is from the system, Q = 100 kj Substituting values yields m ( 100 kj) kj kg 85065 5953 ( 196 ) = 0618 kg kj kg
PROBLEM 35 A piston-cylinder assembly contains lb of water, initially at 100 lbf/in and 400 o F The water undergoes two processes in series: a constant pressure process followed by a constant volume process At the of the constant volume process, the temperature is 300 o F and the water is a twophase liquid-vapor mixture with a quality of 60% Neglect kinetic and potential energy effects (a) Sketch T-v and p-v diagrams showing the key states and the processes (b) Determine the work and heat transfer for each process, all in Btu T KNOWN: Water contained in a pistoncylinder assembly undergoes two processes in series FIND: Sketch the T-v and p-v diagrams and for each process determine Q and W SCHEMATIC AND GIVEN DATA: Water m = lb p 3 1 x 3 = 06 100 lbf/in 400 o F 300 o F v ENGINEERING MODEL: 1 The `water is a closed system Volume change is the only work mode 3 Process 1- occurs at constant pressure and Process -3 occurs at constant volume 4 Kinetic and potential energy effects can be neglected ANALYSIS: First, we fix each state State 1 is in the superheated vapor region From Table A-4E; v 1 = 4934 ft 3 /lb and u 1 = 1136 Btu/lb With T 3 = 300 o F and x 3 = 06, we can evaluate v 3 and u 3 using data from Table A-E at 300 o F as follows T 3 100 lbf/in 1 400 o F 300 o F x 3 = 06 v v 3 = v f3 +x 3 (v g3 v f3 ) = 001745 + (06)(647 001745) = 389 ft 3 /lb u 3 = u f3 + x 3 (u g3 u f3 ) = 695 + (06)(11000 695) = 7678 Btu/lb Note that v = v 3 = 389 ft 3 /lb, and from Table A-3E we see that v < v g (100 lbf/in ) Thus x = v v f v g v f = 389 001774 4434 001774 = 08768
PROBLEM 35 (CONTINUED) PAGE and u = u f + x (u g u f ) = 983 + (08768)(11058 983) = 10063 Btu/lb Now, for Process 1- the pressure is constant Thus 1 W 1 = pdv = mp 1 (v v 1 ) = ( lb) (100 lbf ft3 ) (389 4934) = 3865 Btu (in) An energy balance reduces to give in lb in 144 1 Btu 1 ft 778 ft lbf Q 1 = m(u u 1 ) + W 1 = ( lb)(10063 1136) Btu/lb + ( 3865 Btu) = 985 Btu (out) Now, for Process -3, the volume is constant, so W 3 = 0 And, the energy balance reduces to give Q 3 = m(u 3 u ) = ( lb)(7678 10063)But/lb = 477 Btu (out)