ME Thermodynamics I = = = 98.3% 1

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Lionel Ball
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1 HW-08 (25 points) i) : a) 1 Since ν f < ν < ν g we conclude the state is a Saturated Liquid-Vapor Mixture (SLVM) 1, from the saturation tables we obtain p bar. 1 Calculating the quality, x: x ν ν f ν g ν f % 1 1

From the compressed liquid tables we obtain: ν 1.

2 b) 1 c) 1 Since P > P sat@100 o C bar, we conclude the state is Compressed Liquid (CL) 1. From the compressed liquid tables we obtain: ν m 3 /kg 1 Since T > T sat,p100bar 311 o C we conclude the state is Superheated Vapor (SHV) 1. From the superheated vapor tables: Interpolation for ν: 2

p 100 bar p 100 bar T 480 485 520 ν 0.03163 ν 0.03397 ν 0.03163 485 480 0.03397 0.03163 520 480 ν 0.03163 + (0.03397 0.03163) ν 0.

3 p 100 bar p 100 bar T ν ν ν ν ( ) ν m 3 /kg 1 ( ) d) 1 Since 0 < x < 1 we conclude the state is Saturated Liquid-Vapor Mixture (SLVM) 1. Calculating the specific volume: ν ν f + x(ν g ν f ) ( ) 2.55 m 3 /kg 1 3

ii) Given: A closed, rigid tank that contains a two-phase liquid-vapor mixture of Refrigerant R134 that receives energy transfer through heat. State properties are provided.

4 ii) Given: A closed, rigid tank that contains a two-phase liquid-vapor mixture of Refrigerant R134 that receives energy transfer through heat. State properties are provided. Find: The final temperature in o C. If the final state is SHV, find the temperature at which the tank contains saturated vapor only. Locate the states on a T-ν diagram indicating lines of constant p. EFD: 2, 1 for the boundary and 1 for the heat transfer Assumptions: None Basic Equations: None : a) Using the saturation tables we obtain: ν 1 ν f + x(ν g ν f ) ( ) m 3 /kg 1 4

5 Since the tank is closed (mconstant) and rigid (Vconstant) we conclude that ν 1 ν 2 constant, from the saturated table we can conclude that ν 2 > ν g@4 bar 1, i.e, state 2 is Superheated Vapor (SHV). Interpolation for T 2 : p 4 bar p 4 bar ν T 100 T T T ( ) T o C 2 ( ) b) From the saturation tables we can find that the temperature at which only saturated vapor will be present inside the tank is, the at which ν g (T ) m 3 /kg: Interpolation for T : p bar p bar ν T 0 T -4 T ( ) T 0 + ( 4 0) T 1.80 o C 2 5

6 c) 4, 2 points for locating the states and 2 for the process line 6

7 HW-09 (25 points) i) Given: R 134a contained in a closed, rigid tank is cooled until it becomes saturated vapor. State data is provided. Find: For the refrigerant, determine the initial and final pressures, each in bar, and the heat transfer, in kj/kg.. EFD: 1 Assumptions: 1 Neglect KE and P E. Rigid tank, W b 0 Basic Equations: dm dt i ṁ i o ṁo For a closed system: mconstant de dt Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o For a closed system: Q W E U + KE + P E 1 : 7

8 Since the tank is closed (mconstant) and rigid (Vconstant) we conclude that ν 1 ν 2 constant 1. Since state 2 is saturated vapor, T 2 20 o C we retrieve: p bar 1 ν 2 ν g m 3 /kg 1 u 2 u g kj/kg 1 Since ν 1 ν 2 > ν g,100 o C, we conclude state 1 is superheated vapor. From the superheated tables: Interpolation for p 1 : T 100 o C T 100 o C ν p 7 p 1 8 p ( ) p (8 7) p bar 1 Interpolation for u 1 : T 100 o C T 100 o C ν u 310 u u u ( ) u kj/kg 1 ( )

9 From the first law analysis, we obtain: U Q W Q U Q m u 2 u kj/kg 1 9

ii) Given: Water contained in a piston-cylinder assembly undergoes two processes in series Find: Evaluate Q and W for each process: sketch the two processes in series on a p-ν diagram EFD: 1

10 ii) Given: Water contained in a piston-cylinder assembly undergoes two processes in series Find: Evaluate Q and W for each process: sketch the two processes in series on a p-ν diagram EFD: 1 Assumptions: 1 Frictionless piston Neglect kinetic and potential energy changes. Quasi-equilibrium Basic Equations: W pdv dm dt i ṁ i o ṁo For a closed system: mconstant de dt Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o For a closed system: Q W E U + KE + P E 1 10

: a) 3 b) Since T 1 > T sat@10bar we conclude state 1 is superheated vapor. From the superheated vapor tables we retrieve: ν 1 0.35411 m 3 /kg 1 u 1 3125 kj/kg 1 We know ν 2 ν 1 /2 0.177 m 3 /kg.

11 : a) 3 b) Since T 1 > T sat@10bar we conclude state 1 is superheated vapor. From the superheated vapor tables we retrieve: ν m 3 /kg 1 u kj/kg 1 We know ν 2 ν 1 / m 3 /kg. Since ν f@10bar < ν 2 < ν g@10bar we conclude state 2 is SLVM. Calculating the quality of the state 2 and the internal energy, u 2 : x 2 ν ν f ν g ν f % 1 u 2 u f + x(u g u f ) ( ) 2419 kj/kg 1 11

12 Using the definition of work: W pdv p(v 2 V 1 ) m p(ν 2 ν 1 ) 0.1 kg N ( )m3 m2 kg J 17.7 kj 1 Applying the first law to the closed system: U Q W Q 1 2 U + W ( ) + ( 17.7) 88.3 kj 1 For state 3 it is known that ν 3 ν 2, since ν f@25 o C < ν 3 < ν g@25 o C we conclude state 3 is SLVM. Calculating the quality at state 3 and the internal energy, u 3 : x 3 ν ν f ν g ν f % 1 u 3 u f + x(u g u f ) ( ) kj/kg 1 Since there is no volume change associated with process 2-3, W From the first law we obtain: U Q W Q 2 3 U + W ( ) 230 kj 1 12

HW-10 (25 points) Given: Water contained in a piston-cylinder assembly, initially a two-phase liquid-vapor mixture undergoes two processes in series. State data is provided.

13 HW-10 (25 points) Given: Water contained in a piston-cylinder assembly, initially a two-phase liquid-vapor mixture undergoes two processes in series. State data is provided. Find: Show the two processes of the water in series on a T-ν diagram. For the overall process of the water evaluate the work and heat transfer, each in kj/kg. EFD: 2, 1 for the boundary and 1 for the work interaction Assumptions: 2 Frictionless piston. Quasi-equilibrium. Neglect kinetic and potential energy changes. Basic Equations: W pdv 1 dm dt i ṁ i o ṁo For a closed system: mconstant de dt Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 13

For a closed system: Q W E U + KE + P E 1 : (a) 5 (b) From the saturation tables we calculate ν 1 : ν 1 ν f + x(ν g ν f ) 0.0010527 + 0.2(1.1593 0.

14 For a closed system: Q W E U + KE + P E 1 : (a) 5 (b) From the saturation tables we calculate ν 1 : ν 1 ν f + x(ν g ν f ) ( ) m 3 /kg 2 From the definition of specific volume the following can be obtained: ν 1 AL 1 m 1 ; L m ν 2 A(L 1 + L 2 ) ; L m m 2 14

15 Solving for ν 2 in terms of ν 1 : ν 2 ν 1(L 1 + L 2 ) L ( ) ν m 3 /kg 2 From the definition of boundary work: W 1 3 W pdv 2 1 pdv W 1 3 m p(ν 2 ν 1 ) pdv pdv w N ( )m3 m2 kg J/kg 20.9 kj/kg 2 From the saturation tables we obtain: u 1 u f + x(u g u f ) ( ) kj/kg 2 Interpolation for u 3 : p 6 bar p 5 bar ν u u u u ( ) u kj/kg 2 ( )

16 Applying the first law to the whole process: U Q W q 1 3 u w 1 3 ( ) kj/kg kj/kg 2 16

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity.

KNOWN: Data are provided for a closed system undergoing a process involving work, heat transfer, change in elevation, and change in velocity. 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