Mechanical Knowledge

Thermodynamics  Chapter: First law of Thermodynamics  Question 2. An insulated pressure vessel divided into two parts. One part of vessel is occupiedby an ideal gas at a pressure P1, volume V1 & temp T1.The other part occupied by the same gas at a pressure P2, volume V2, & temp T2. The partition is removed and the two parts are mixed. Then show that the final pressure P3 & final T3 is   Solution:  Same ideal gas  PV = mRT   ...................................................................(1) Mass Conservation : m1+m2 = m3      ......................................(2) Energy conservation: m1 u1 + m2 u2 = m3 u3                                       m1 Cv T1 + m2 Cv T2 = m3 Cv T3                                 m1 T1 + m2 T2 = m3 T3 Use PV =  mRT,                 P1 V1/ R + P2 V2/R = P3 V3/R "R" Is canceled out with each other, so P1 V1 + P2 V2 + = P3 V3 P3 = P1 V1 + P2 V2/V3                                                            (:: V3= V1+V2 ) P3 = P1 V1+ P2

THERMODYNAMICS CHAPTER: FIRST LAW OF THERMODYNAMICS Question 1. Air at initial temp of 300K & initial volume of 0.002 m^3 is contain in a piston cylinder arrangement. Initially a spring is attached to the piston having a spring constant of 10 kN/m is touching the piston but no force exerting on it. Now heat is added to the air & it expended slowly to occupy a final volume of 0.03 m^3 and the area of piston is 0.02 m^2. The atmospheric pressure assume to be 100KPa. Then determine the final pressure inside the cylinder. Solution  T1 = 300K V1 = 0.002 m^3  V2  = 0.03 m^3 Spring constant (K) = 10 KN/m A = 0.02 m^2 Patm = 100 KPa  Psp = pressure applied by spring Patm = pressure applied by atmosphere P1 = inside pressure of piston at position one  P2 = inside pressure of piston at position two   P1 =  Patm + Psp1   ……………..............................(0)                                                   ( Psp1 = pressure of  spring at position one  is zero =0) P1 = Patm = 100 KPa ………

THERMODYNAMICS  CHAPTER: ZEROTH LAW OF THERMODYNAMICS Question 3. What does a temperature difference of 10 on °C scale corresponding toon a Fahrenheit scale. Solution:-                            Δtc = 10              (:: Δ= d/dt) We know     tf = 9/5tc + 32   ....................(0) Applying Δ on both sides of equation (0),                            Δtf  = 9/5Δtc   ....................(1)                                                                      (:: here 32 is zero because it is constant   Put Δtc = 10  in equation (1)                     (:: Δ this use for differentiation  Δtf  = 9/5(10) Δtf = 18°F *at any temperature difference in °C find by Δtf = 1.8(Δtc) equation