Diese Präsentation wurde erfolgreich gemeldet.
Wir verwenden Ihre LinkedIn Profilangaben und Informationen zu Ihren Aktivitäten, um Anzeigen zu personalisieren und Ihnen relevantere Inhalte anzuzeigen. Sie können Ihre Anzeigeneinstellungen jederzeit ändern.

007 isometric process

7.382 Aufrufe

Veröffentlicht am

Veröffentlicht in: Technologie, Business
  • Loggen Sie sich ein, um Kommentare anzuzeigen.

007 isometric process

  1. 1. LECTURE UNIT 0072. CONSTANT VOLUME or ISOCHORIC PROCESS (V = c) A. CONSIDERING A CLOSED OR NON-FLOW SYSTEM ILLUSTRATION: (Ideal gas confined in a Rigid Container) ma Q P-v diagram T-s diagram P T 2 2 V=c P=c 1 1 v s Equations: 1. PVT Relationships a. General Gas Law Equation (Equation of State) PV = mRT P1V1 P2V2 b. = = mR = constant T1 T2 P1 T1 = P2 T2 2. Work Non-Flow, WNF 2 General Formula: WNF = P dV (Derivative of a constant is zero) 1 WNF = 0 3. Heat Energy, Q Using NFEE; Q= U + WNF Q = mCv T 4. Change in Total Internal Energy, U U = mCv(T2 - T1) = mCv( T) 5. Change in Total Enthalpy, H H = mCp(T2 - T1) = mCp( T) “Talent does you no good unless it is recognized by someone else.”
  2. 2. 6. Change in Total Entropy, S From; dQ = Tds dQ dS = 2 T 2 dT dS = mCv 1 1 T S2 - S2 = mCv (ln T2 - ln T1) T2 P2 S = mCv ln = mCv ln T1 P1 B. CONSIDERING AN OPEN OR STEADY-FLOW SYSTEM ILLUSTRATION: WSF PE1 PE2 KE1 TS KE2 U1 (P = C) U2 Wf1 Wf2 Q [Ein = Eout] PE1 + KE1 + U1 + Wf1 + Q = PE2 + KE2 + U2 + Wf2 + WSF PE1 + KE1 + U1 + Wf1 + Q = PE2 + KE2 + U2 + Wf2 + WSF Q = (PE1 - PE2) + (KE2 - KE1) + (U2 - U1) + (Wf2 - Wf1) + WSF Q= PE + KE + U+ Wf + WSF When: PE ˜ 0 (z1 ˜ z2) ˜ ˜ KE ˜ 0 (v1 ˜ v2) ˜ Q= U+ Wf + WSF But: Q = U WSF = - Wf WSF = - (P2V2 - P1V1) Since: V1 = V2 = V WSF = V(P1 - P2) = mR (T1 - T2)Fluid Flow Continuity Equation(Law of Conservation of Mass) . . D2 m 1 = m2 t2 ρ 2 . D1 Q2 = V2 . . m 1 = m2 t1 ρ1 . Q1 = V1 “The best leader brings out the best in those he has stewardship over.”
  3. 3. For frictionless flow: [Mass in = Mass out] . . m 1 = m2 . . ρ V1 = ρ V2 ρ1 A1 1 = ρ2 A 2 2 A1 1 A2 2 = v1 v2 If temperature remains constant: (t1 = t2), (ρ1 = ρ2) . V = Q = A1 1 = A2 2 Q=A If temperature is not constant: t1 = t2 ρ1 = 1 ρ2 = 1 v v Q = CdA1 1= CdA2 2 Where: Cd = coefficient of discharge = velocity of the fluid A = area normal to the flow Pumps Machine which is used to add energy to a liquid on order to transfer the liquid from one point to another point of higher energy level. Types: 1. Reciprocating 2. Centrifugal . . 3. Rotary m 1 = m2 t2 BP EP t1 . . WP m 1 = m2 2 . WSF = - V dP 1 2 . = -V dP 1 . P2 = -V P P1 . = - V (P2 - P1) Since: P2 - P1 = γ h . = - V [(P1 + γwh) - P1 . WSF = - V γ wh WP = γw Q TDH Where: γw = specific weight of the liquid, lbf/ft3, kN/m3 if water = 9.81 kN/m3 Q = pump capacity or volume flow or rate of discharge, ft3/sec, m3/s TDH = total dynamic head or the head handled by the pump, ft, m WP = water power, ft-lbf/sec, hp, kW “A good leader remains focused. Controlling your destination is better than being controlled by it.”
  4. 4. PROBLEM SET:1. A closed rigid container has a volume of 1 m3 and holds air at 344.8 kPa and 273 K. Heat is added until the temperature is 600 K. Determine the heat added and the final pressure. [ ]2. A scuba tank contains 1.5 kg of air. The air in the tank initially at 15oC. The tank is left near an engine exhaust line, and the tank’s pressure doubles. Determine (a) the final temperature (b) the change in internal energy (c) the heat added.[ ]3. A reversible nonflow, constant volume process decreases the internal energy by 315 kJ for 2.5 kg of gas for which R = 430 J/kg.K and k=1.35. For the process, determine: (a) the work [ ] (b) the heat [ ] (c) the change in total entropy [ ] The initial temperature is 205oC.4. Air is heated at constant volume from 20oC to 150oC. Determine the change in specific entropy for this isometric process. [0. ]5. A pump system under a steady flow operation delivers water at 25oC from an inlet pressure of 100 kPa to an exit pressure of 1200 kPa. Determine the work done per unit mass. [ ]6. Helium is heated at a constant volume from 350 kPaa and 38oC to 220oC. Determine the change in specific entropy. [ ]7. 40 kg of air at 345 kPa is contained in a rigid vessel at a temperature of 20oC, heat is then added until the temperature reaches 320oC. Determine (a) the final pressure (b) the work done (c) the heat. [ ]8. An ideal gas with a molecular mass of 36 is heated at a constant volume from 250oC to 485oC. If the change in internal energy is found to be 180 kJ/kg, determine (a) the change in enthalpy (b) the change in entropy. [ ]9. A rigid insulated tank (V=C, Q=0) contains 0.05 lbm of air initially at 80oF. An electrical resistance coil in the air within the tank receives 0.50 watt-hours of energy. Neglecting any energy storage within the coil itself, determine the final temperature of the gas, in oF. Note: 1 watt-hour = 3.413 Btu. [ ]10. A closed rigid container has a volume of 40 ft3 and holds helium at 55 psia and 35oF. Heat is added until the temperature is 625oF. Determine the heat added and the final pressure. “Learning is a treasure that accompanies its owner everywhere.”
  5. 5. Fans A machine used to apply power to a gas in order to cause movement of the gas. Types: 1. Propeller 2. Tubeaxial 3. Vaneaxial AP = γA Q H Where: γw = specific weight of the gas, lbf/ft3, kN/m3 if air = 1.2 kg/m3 at 21oC and 101.325 kPa Q = fan capacity or volume flow or rate of discharge, ft3/sec, m3/s H = head handled by the fan, ft, m AP = air power, ft-lbf/sec, hp, kW

×