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Gas Power Cycles

Gas Power Cycles. Thermodynamics Professor Lee Carkner Lecture 17. PAL # 16 Exergy Balance. Cooling chickens with a water stream Mass flow of chickens m’ c = (500 c/hr)(2.2 kg/c) / (3600 s/hr) = Heat removed from chickens can be found from specific heat

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Gas Power Cycles

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  1. Gas Power Cycles Thermodynamics Professor Lee Carkner Lecture 17

  2. PAL # 16 Exergy Balance • Cooling chickens with a water stream • Mass flow of chickens • m’c = (500 c/hr)(2.2 kg/c) / (3600 s/hr) = • Heat removed from chickens can be found from specific heat • Q’c = m’ccpDT = (0.3056)(3.54)(15-3) = • Heat gained by water is • Q’w = Q’c + Q’environ = 13.0 + (200 kJ/h) / (3600 s/hr) = • Absorbing heat raises water temp by 2 C • m’w = Q’w/cpDT = 13.056 / (4.18)(2) =

  3. PAL # 16 Exergy Balance • Find Sgen from equation of flow systems • S’gen = • But Ds = c ln (T2/T1) for an incompressible substance • S’gen = (0.3056)(3.54) ln(276/288) + (1.56)(4.18) ln(275.5/273.5) – 0.0556/298 = • X’destroyed = T0S’gen = (298)(0.00128) =

  4. Modeling Power Cycles • We often generate power by performing a series of processes in a cycle • We use instead an ideal cycle • We will often be looking for the thermal efficiency • hth = Wnet/Qin = wnet/qin

  5. Diagrams • Pv diagram • Ts diagram • But, net heat = net work

  6. Ideal Diagrams

  7. Carnot • The Carnot cycle is the most efficient • It is very hard to build even an approximation • hth,Carnot = 1 – (TL/TH) • In general want high input and low output temperatures

  8. Carnot Diagrams

  9. Air Standard • For most internal combustion engines the working substance is a gas and is a mixture of air and fuel • Can assume: • All processes are internally reversible • Can think of exhaust as heat rejection to an external sink • Cold-air standard

  10. Reciprocating Engine • Top dead center • Bottom dead center • Stroke • Bore • Intake Valve • Exhaust value • Allows combustion products to leave

  11. Volumes of a Cylinder

  12. Compression • Clearance volume • Displacement volume • Compression ratio r = Vmax/Vmin = VBDC/VTDC • Mean Effective Pressure (MEP) is the equivalent pressure that would produce the same amount of work as the actual cycle MEP = Wnet / (Vmax – Vmin)

  13. MEP Illustrated

  14. Otto Cycle • The ideal cycle for reciprocating engines ignited by a spark was developed in 1876 by Nikolaus Otto • Basic cycle: • Can also combine the exhaust and intake into the power stroke to make a two-stroke engine

  15. Ideal Otto Cycle • We can approximate the cycle with • An isochoric (no DV) heat addition • An isochoric heat rejection

  16. Otto Analysis • We can write the heats as cvDT • qin = • qout = • hth = 1 – qout/qin = • But we also know that for the isentropic process • (T1/T2) = (v2/v1)k-1 and r = v1/v2 • hth,Otto =

  17. Otto Compression Ratios

  18. Efficient Otto Engines • As we increase r the efficiency gain levels off at about 8 • Also, high r can mean the fuel is compressed so much it ignites without the spark • Can’t really increase k since we are using air • Typical values for hth,Otto ~

  19. Otto Engine Exercise

  20. Diesel Cycle • We can approximate the cycle with • An isobaric heat addition • An isochoric heat rejection • Only the second process is different from the Otto

  21. Diesel Efficiency • The heat in is the change of internal energy plus the isobaric work • qin = Du + PDv = h3-h2 = • The heat out is just the change in internal energy • qout = u4-u1 = • So then the efficiency is hth,diesel = 1 – qout/qin = 1 – (T4-T1) / k(T3-T2) • We can rewrite as: hth,diesel = 1 – (1/rk-1)[(rkc-1)/k(rc-1)] • rc = v3/v2

  22. Diesel Compression Ratios

  23. Making Diesels Efficient • Want large r and small rc • Diesels can operate at higher compression ratios and are usually more efficient • hth,diesel ~ • Diesels also have lower fuel costs because they don’t have to worry about autoignition and engine knock

  24. Next Time • Read: 9.8-9.12 • Homework: Ch 9, P: 22, 37, 47, 75

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