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Entropy. Thermodynamics Professor Lee Carkner Lecture 13. PAL # 12 Carnot. Engine powering a heat pump Find work needed for heat pump COP HP = 1 /(1-T L /T H ) = 1/(1 – 275/295) = COP HP = Q H /W in W = Q/COP = 62000/14.75 = Total engine work is twice that needed for heat pump
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Entropy Thermodynamics Professor Lee Carkner Lecture 13
PAL # 12 Carnot • Engine powering a heat pump • Find work needed for heat pump • COPHP = 1 /(1-TL/TH) = 1/(1 – 275/295) = • COPHP = QH/Win • W = Q/COP = 62000/14.75 = • Total engine work is twice that needed for heat pump • Wengine = 2WHP = • hth = 1 – (TL/TH) = 1 – (293/1073) = 0.727 • hth = W/QH • QH = W/hth = 8406/0.727 =
Clausius Inequality ∫ dQ/T = 0 ∫ dQ/T ≤ 0 • Valid for all cycles
Entropy • The integral is a state quantity called the entropy Ds = ∫ dQ/T • Entropy is a tabulated property of a system like v, h or u DS = Q/T
TS Diagram • Can plot a Temperature-Entropy Diagram • The total heat is the integral of TdS, or the area under the process line on a TS diagram • Similar to W being the area in a Pv diagram
Mollier Diagram • The vertical (enthalpy) distance gives a measure of the work • The horizontal (entropy) distance gives a measure of the irreversibilities (and thus inefficiencies)
Entropy Rules • Processes must proceed in the direction that increases entropy • Entropy is not conserved • Rate of entropy generations tells us the degree of irreversibility and thus efficiency
Finding Entropy Change • Entropy in tables assign zero to some arbitrary point • For mixtures of vapor and liquid: • We will normally assume the entropy of a compressed liquid is the same as a saturated liquid
Isentropic Process • e.g. well-insulated and frictionless • Can use to determine the properties of the initial and final states of the system • Find s for state 1 and you know that state 2 must have a P and T to give the same s
Disorder • The higher the entropy the more random the distribution of molecules • High entropy → high disorder → low quality → low efficiency
The 3rd Law • Molecule motions decrease with decreasing temperature • Only one configuration means absolute order
Gibbs Equation • but we already have relationships for Q • Solving for Tds TdS – PdV = dU • Called the Gibbs equation • Substituting into the Gibbs equation: Tds = dh - vdP
Entropy Relations • We can know write integrable equations ds = dh/T + vdP/T • We still need: • e.g., du = cvdT • e.g., Pv = RT
Incompressible Entropy • For solids and liquids the volume does not change much • dv = 0 • We can integrate the entropy equation: DS = ∫ c dT/T = c ln T2/T1
Next Time • Read: 7.7-7.13 • Homework: Ch 7, P: 46, 54, 66, 93