Question

Esporti 90 Edit PDF Create PC Comment A cylinder/piston arrangement contains saturated water at 105°C with a volume of 0.1 L (state 1). The system is heated, causing the piston to rise and encounter a linear spring, as shown in figure. At this point (state 2) the volume is 15 L, the piston diameter is 334.7 mm, and the spring constant is 115 N/mm. The heating continues, so the piston compresses the spring until all the water becomes saturated vapor (state 3). Neglect the weight of the piston. a) Find the pressure and the mass of the water at the initial state (state 1). (5 Points) b) Find the quality of the water at the state 2. (5 Points) c) Find the pressure and temperature of the water at the state 3. (15 Points) d) Sketch the p-v diagram for the process. Explain what happens to the p-v diagram if we increase the stiffness of the spring. (10 HO Points) Combine 0 Organize Redact Protect Compress PD Fill & Sign

Answer 1

Solution:- The given problem has been solved using the standard formulas and principles used in the subject of Thermodynamics. Standard thermodynamic tables for the saturated water have been used to determine the various properties of the water and saturated vapour.

Solution- Given data :- Ta = 105°C V = 0.1 L V2 15 L piton diameter, dp=334.7mm or dp = 334.7 x 103 m constant, k=115N/mm. Spring (a) Pressure and the mass of the water at the initiat state (state 1) (Prandm) Refer the thermodynamic tables for saturated wonter find that corresponding T = 105°C ие. to And P₂ = 120.8 kPa

sling m and Vi Vf@ 105°c Droo 10473 Now the mass of the water at initial state can be given by : Vi v where V = o.lx10-3 m3 (As 1L = 10-3 m3) so, 0,1x103 oooolo47 ma m. -0.0955 kg And (6) queility of water at state 2 (x2) As we have V2 = ISL 15x10-3 m3 Also or Vz vi V₂ Y U2= V V₂ Vi ( 22 23

v 3 V 0.00 1047 15803 0.1%15? Uz = 0.15705 m?leg As we Steite 2 is achieved from State 1 at a constant previure process SO, P = P2 = 120.8 kPa frou steam tables we house حملہ Vf 2 = Vf @ 120.8kpan 0.0olour m? Vaga = Vfg@ 120.8kpa = 1.41831 m² Ikg milkg Also we know, V = Vaat Xa Vega or 0.15705 = 0.001047 + X(1.418 31) X₂ = 0.1570S -0.Do 1047 1.4183) x2 = 0.11 Aus

(6) Pressure and temperature of the water at the state 3 (Pz and Tz) Now the pressure out state 3 can be by. P₂ = P2 tk (vse vz_) A where Ap = Cross-sectional area givein ml 2 of the piston Ixldp) 2 4 IX (334.7x103) 2 = 0.087984 m2 and KE 115 N 115 KN mm o, P₂ = 120.8 + 115 (0.087984) [cate (0.0955)(1-015705)

At state 3 we know that the condition of the substance is saturated vapour. After trying various values Vg (as is) find 466 kPa And Vog@ 466 kPa 0.400sl m3 And the correspon temperature T3 = 149,2 °C] Dry am and 3 m² kg corresponding saturation d P-v diangisan ram for the process Pq 3 PA Ź P-v diagram P, = P₂ V₂ V3 V-

1-2 Process constant pressure process. 2-3 process Pressure and volume increases water is frosthes hearted we increase the stiffner ness (k) of the spring, then and V increase ou d cal T १३