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MECH 213 - 9.17.07

MECH 213 - 9.17.07 - Chapter 4 65E o A rigid tank contains...

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MECH 213 – September 17, 2007 Know for Test: Know modes of heat transfer not heat transfer equations Total energy = blah blah blah Memorize all equations Ideal gas law give p, v, T c p , c v give h, u real pure substance – properties stay the same across the phases Δh = ∫ T1 T2 c p bar dT o c p bar = a + bT + cT 2 + dT 3 o use gas tables o Δhbar = hbar(T 2 ) – hbar(T 1 ) o Δh = Δhbar/M o MOST ACCURATE Assume c p,avg at (T 1 + T 2 )/2 or c p,avg = [c p (T 1 ) + c p (T 2 )]/2 o Δh = c p,avg ΔT o SECOND MOST ACCURATE Assume constant c p at T 1 o Δh = c p ΔT Δu = ∫ T1 T2 c v dT = u 2 (T 2 ) – u 1 (T 1 ) o From gas tables o c v,avg = (c v1 + c v2 )/2 o Δu = c v,avg ΔT or o Δu = c v1 ΔT h = u + pv o dh = du + d(pv) o but for Ideal Gas pv = RT o dh = du + R dt o c p = (δh/δT) p o c p dT = c v dT + RdT o c p – c v = R o c p bar – c v bar = R u universal gas constant = R*Molar mass o k ≡ c p /c v > 1.0 monatomic k = 1.667 air k = 1.4 Incompressible Substance o v = constant, ρ = constant, not f(p) o c p = c v c (specific heat) o Δu = cΔT o dh = du + pdv + vdp
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o Δh = Δu + vΔp Constant T: Δh = vΔp Constant p: Δh = Δu
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Unformatted text preview: Chapter 4 # 65E o A rigid tank contains 20lbm of air at 50 psia and 80°F. The air is now heated until its pressure doubles. Determine (a) the volume of the tank and (b) the amount of heat transfer Closed system Substance: Assuming that air acts as an ideal gas Process: constant volume First law for a stationary system • (Q in – Q out ) + (w in – w out ) • Q net,in – w net,out = mΔu • w b = ∫pdv = 0 o shaft work = 0 no shaft o electrical work = 0 no electrical connections Q in = mΔu Q in = mc v ΔT p 2 v 2 /T 2 = p 1 v 1 /T 1 p 2 /T 2 = p 1 /T 1 2p 1 /T 2 = p 1 /T 1 T 2 = 1080 R = 620°F Q in = mΔu Q in = mc v ΔT Q in = 20 lbm (.175) (540 btu/lbm) Q in = 1890 AIR TABLE • At 80 = 540 R u 1 = 92.04 • At 1080R u 2 = 186.93 • Δu = 94.89 • ΔU = mΔu = 1897.8 Btu • Q = Δu = 1897.8 Btu o pV = mRT...
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