Calculating Maximum Theoretical Performances of Heat Engines and Heat Pumps

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Remember, energy is always conserved and the total entropy changes of all thermal reservoirs must be greater than 0 (IE the sum of all entropy increases must be greater than the sum of all entropy decreases).

Performance is measured by comparing what we want to what we pay.

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Remarks about sign conventions used in this presentation

• Although heat losses from a system are usually represented as negative numbers, and heat gains of a system by positive numbers, in this presentation, heat and entropy gains and losses of heat resevoirs will be stated in terms of positive numbers to simplify calculations

Heat Engines

Q_hot > 0, Q_cold > 0, by convention

Q_cold + W = Q_hot, by conservation of energy

Q_cold/T_cold > Q_hot/T_hot, by the second law of thermodynamics (IE the law of entropy, entropy lost by hot resivoir is less than entropy gained by cold resevoir: ΔS_cold > ΔS_hot, where ΔS_hot = Q_hot/T_hot and ΔS_cold = Q_cold/T_cold)

Heat Engines (continued)

efficiency, e = W/Q_hot (IE what we want divided by what we pay)

T_hot(Q_cold/T_cold) > Q_hot

T_hot((Q_hot - W)/T_cold) > Q_hot

T_hot/T_cold > Q_hot/(Q_hot – W)

T_cold/T_hot < (Q_hot – W)/Q_hot

T_cold/T_hot < 1 – e

e < 1 – T_cold/T_hot,

and since 0 < T_cold/T_hot < 1 (because T_hot > T_cold > 0), e < 1

Work Driven Heat Pumps

Q_hot > 0, Q_cold > 0, by convention

Q_hot = Q_cold + W, by conservation of energy

Q_cold/T_cold < Q_hot/T_hot, by the law of entropy (IE the law of entropy, entropy gained by hot resivoir is greater than entropy lost by cold resevoir: ΔS_cold < ΔS_hot, where ΔS_hot = Q_hot/T_hot and ΔS_cold = Q_cold/T_cold)

Heating Performance for Work driven Heat Pumps

coefficient of performance, cop, or c_heating = Q_hot/W (IE what we want divided by what we pay)

T_hot(Q_cold/T_cold) < Q_hot

T_hot(Q_hot - W)/T_cold) < Q_hot

(T_hot/T_cold)Q_hot – (T_hot/T_cold\)W < Q_hot

(T_hot/T_cold)Q_hot – Q_hot < (T_hot/T_cold)W

Q_hot(T_hot/T_cold - 1) < (T_hot/T_cold)W

(T_hot/T_cold - 1)/(T_hot/T_cold) < W/Q_hot

1 - T_cold/T_hot < 1/c_heating

So, 1/(1 – T_cold/T_hot) > c_heating, note the left side of the inequality is the same as 1/e_max, where e_max is the maximum efficiency of a heat engine with identical hot and cold resevoirs

Cooling Performance for Work driven Heat Pumps

for cooling, cop, or c_cooling = Q_cold/W (IE what we want divided by what we pay)

T_hot(Q_cold/T_cold) < Q_hot

T_hot(Q_cold/T_cold) < Q_cold + W

(T_hot/T_cold)Q_cold - Q_cold < W

(T_hot/T_cold - 1)Q_cold < W

(T_hot/T_cold – 1) < W/Q_cold, or 1/c_cooling

multiply both sides by T_cold/T_hot

1- T_cold/T_hot < (T_cold/T_hot)/c_cooling

(T_cold/T_hot)/(1- T_cold/T_hot) > c_cooling; note, the left side of the inequality is the same as (1 – e_max)/e_max, where e_max is the maximum efficiency of a heat engine with identical hot and cold resevoirs

multiply numerator and denominator by (T_hot/T_cold)

1/(T_hot/T_cold – 1) > c_cooling

multiply numerator and denominator by T_cold

T_cold/(T_hot – T_cold) > c_cooling

Remarks on Heat Engines and Work Driven heat Pumps

recall, 1- T_cold/T_hot = e_max for a heat engine

c_{heating, max} = 1/e_max, c_{cooling, max} = (1 – e_max)/e_max,

and 0 < e_max < 1, since T_hot > T_cold > 0,

Therefore, 0 < 1 – e_max < 1

so, c_{heating, max} > c_{cooling, max}, as expected

Heat Driven Heat Pumps

Q_hot > 0, Q_cold > 0, Q_warm > 0, by convention

Q_hot/T_hot + Q_cold/T_cold < Q_warm/T_warm, by the law of entropy (IE the sum of entropies lost by the hot and cold resevoirs is less than the entropy gained by the warm resevoir: ΔS_hot + ΔS_cold < ΔS_warm, where ΔS_hot = Q_hot/T_hot, ΔS_cold = Q_cold/T_cold, and ΔS_warm = Q_warm/T_warm)

Q_hot + Q_cold= Q_warm, by conservation of energy, since heat is the only energy input and output in this model

Heating Performance for Heat Driven Heat Pumps

For heating, coefficient of performance, cop, or c_heating = Q_warm/Q_hot (IE what we want divided by what we pay)

Q_hot < (Q_warm/T_warm – Q_cold/T_cold)T_hot

Q_warm >T_warm(Q_hot/T_hot + Q_cold/T_cold)

Q_hot = Q_warm - Q_cold < (Q_warm/T_warm – Q_cold/T_cold)T_hot

Q_warm - Q_cold - Q_warmT_hot/T_warm < -Q_coldT_hot/T_cold

Q_warm - Q_warmT_hot/T_warm < Q_cold - Q_coldT_hot/T_cold

Q_warm(1-T_hot/T_warm) < Q_cold(1 – T_hot/T_cold)

Q_cold/Q_warm < (1-T_hot/T_warm)/(1- T_hot/T_cold)�Q_cold/Q_warm = (Q_warm- Q_hot)/Q_warm < (1-T_hot/T_warm)/(1- T_hot/T_cold)

1 - 1/c_heating < (1-T_hot/T_warm)/(1-T_hot/T_cold)

1 - (1-T_hot/T_warm)/(1-T_hot/T_cold) < 1/c_heating

c_heating < 1/(1 - (1-T_hot/T_warm)/(1-T_hot/T_cold))

Cooling Performance for Heat Driven Heat Pumps

For cooling, cop, c_cooling = Q_cold/Q_hot (IE what we want divided by what we pay)

From previous slide: Q_cold/Q_warm < (1-T_hot/T_warm)/(1- T_hot/T_cold)

Q_cold/Q_warm = (Q_warm- Q_hot)/Q_warm < (1-T_hot/T_warm)/(1- T_hot/T_cold)

Q_cold/(Q_hot + Q_cold) < (1-T_hot/T_warm)/(1- T_hot/T_cold)

(Q_hot + Q_cold)/Q_cold > (1- T_hot/T_cold)/(1-T_hot/T_warm)

1/c_cooling + 1 > (1- T_hot/T_cold)/(1-T_hot/T_warm)

1/c_cooling > (1- T_hot/T_cold)/(1-T_hot/T_warm) – 1

c_cooling < 1/((1- T_hot/T_cold)/(1-T_hot/T_warm) – 1)

Final Remarks

let (1-T_hot/T_warm)/(1-T_hot/T_cold) = r, then c_heating < (1/(1-r)) and c_cooling < 1/(1/r – 1), and

c_{heating, max}/c_{cooling, max} = (1/(1-r))/(1/(1/r – 1))

(1/(1-r))/(1/(1/r – 1)) = (1/r – 1)/(1-r), and 0 < r < 1 (because T_hot > T_warm > T_cold > 0)

(1/r-1)(1-r)r = r(1/r – 1 – 1 + r)

r(1/r – 1 – 1 + r) = r – 2r + r²

r – 2r + r² = (r-1)², or (1- r)², so

((1/r-1)/(1-r))((r(1-r))/(r(1-r))) = (1- r)²/(r(1- r)²) or 1/r

and 0 < r < 1, since T_hot > T_warm > T_cold > 0, and therefore

c_{heating, max}/c_{cooling, max} = 1/r, which is greater than one,

So, again, c_{heating, max} > c_{cooling, max}, which is expected.

Why is the maximum possible heating performance of heat pumps better than the maximum possible cooling performance?

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For a practical example of a heat driven heat pump, go to this link.

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In reality, real heat engines and heat pumps rarely if ever come close to meeting these theoretical limits to performance.

This is because, in addition to frictional and other losses, heat addition and removal to or from the working fluid from or to the heat reservoirs is not even close to isothermal (taking place at constant temperature) for a practical heat engine or heat pump. For heat to flow from or to the heat reservoir to or from the heat engine or heat pump at a useful rate, there usually must be substantial temperature differences between the heat reservoirs and working fluid for the heat engine or heat pump.