3000K Solutions to 2nd law problem sheet questions
Solutions to 2nd law problem sheet questions
Solutions to 2nd law problem sheet questions
1. (a) QL = 3kJ, irreversible, η = 70% < ηrev= (1-TL/TH) = 80%
(b) QL = 10kJ, impossible, increase to TH = 3000K
(c) W = 8kJ, reversible (η = W/QH = 1-TL/TH = 80%)
(d) QH = 10kJ, irreversible, COP = 5 < COPrev = TH/(TH –TL) = 6
(e) W = 1kJ, impossible, reduce to TH = 278K
2. Answers:
(a) 12. qL = 20kJ/kg, qH = 80kJ/kg, η = 75%
η = 75±0.83%, w = ηqH = 60±0.67kJ/kg, qL = (qH-w) = 20-/+ 0.67kJ/kg
(b) TL = 250K, COR = (qL/w) = 5
qH = 122.5J/kg ⇒ w = (qH - qL) = 22.5J/kg
(c) qL = 0.93kJ/kg ⇒ TH = 296K, COP = (qH/ w) = 14.3
qL = 1.81kJ/kg ⇒ qH = 1.88kJ/kg, COP = 26.9
3. (a) Q′f = 169W (with Q′rej = 294W)
(b) from given data COR = (100+100)/50 = 4 ⇒ for Q′rej = 250W Clausius = -0.109 < 0 ⇒
possible
4. (a) Heat extracted from environment Q′a = (Q′d - W′) = (1000 – 100)W = 900W
Clausius Inequality ∑Q′/T = (-1000W/293K) + (+900W/283K) = -0.176W/K < 0 ⇒
possible
(b) Clausius Inequality ∑Q′/T =(-500W/293K)+ (-500W/313K) + (+900W/283K) = -
0.095W/K < 0
⇒ possible in principle
5. (a) with x2 = 0.913, ∆s12 = +108J/kgK > 0 ⇒ irreversible (Δs > 0)
(b) with superheated T2 = 108°C, ∆s12 = +180J/kgK > 0 ⇒ irreversible
6. (i) For s2 = s1 = 6.796kJ/kgK (giving two known properties at state 2),
T2 = 444°C, h2 = 3301kJ/kg, hence from equation c2 = 447m/s
(ii) For h3 = h1 = 3410kJ/kg (giving two known properties at state 3),
T3 = 490°C, ∆s13 = +148J/kgK > 0 ⇒ irreversible
7. (a) Reading off chart:
saturated vapour (x=1) at ~160bar, 350°C
superheated vapour at ~160bar, 530°C
wet vapour (x=0.917) at 0.04bar, ~30°C (29°C)
(b) From differences of ordinates (H):
+79.5 MW, yes - entropy increases with heat addition
-103 MW, yes, entropy increases with work output
(c) For the two Qu 14 processes:
14.(a) At h1 = h2 ≈ 2610kJ/kg (2611kJ/kg from Tables), ∆s ≈ 0.105kJ/kgK > 0
14.(b) At h1 = h2 ≈ 2690kJ/kg (2693kJ/kg from Tables), ∆s ≈ 0.160kJ/kgK > 0
(d) h1 = h3 ≈3310kJ/kg (3073kJ/kg from Tables), h2 ≈3305kJ/kg, T2 ≈ 444°C, ∆s23 ≈
0.15kJ/kgK > 0
8. From Mollier Diagram:
(a) For an expansion expect pressure decrease and specific volume increase with enthalpy
decrease (work done). In this case Δs > 0 (entropy increase) ⇒ possible but irreversible.
(b) For heat addition must have both enthalpy and entropy increase (Δs > 0) ⇒ possible but
irreversible (all real heat transfers are irreversible).
(c) Cannot have isentropic heat transfer by definition (dq = ds/T = 0 for ds = 0) ⇒
impossible.
(d) Heat rejection is the only way to decrease entropy (by thermal entropy flux) so in this
case entropy increase ⇒ impossible.
(e) For throttling pressure must decrease as given, but in this case entropy also decreases (Δs
< 0) ⇒ impossible.
(f) For heat rejection enthalpy must decrease, as must entropy (by thermal entropy flux), ⇒
possible but irreversible (all real heat transfers are irreversible).
9. (a) η = Δhactual/Δhisentropic ≈ (3380 - 2320) / (3380 - 1925) ≈ 73%
(b) +20.23kJ/kg, 158.0kJ/kg, +1.19°C, -0.7%
(c) +20.10kJ/kg (-0.7% error), hdischarge ≈ hsuction + w = 157.9kJ/kg (-0.06% error)
10. 0.95x100kW = 95kW (with 100-95 = 5kW lost as heat to the surroundings)
gearbox: dQ′/T = (-5kW)/(333K) = -15W/K (< 0)
surroundings: dQ′/T = (+5kW)/(293K) = +17W/K
Temperature difference between gearbox and surroundings, heat loss from gearbox (cooling)
vs heating of surroundings (heat flux to surroundings).
Ta.ΔS′ = (293K).(-15+17)W/K = 586W < 5kW rate of heat loss. In principle some of heat
above ambient could be available to do useful work.
11. hot summer day:
from condenser dQ/T = -1J/(15+273+30)K = -0.315x10-3J/K
to ambient dQ/Ta = +1J/(273+30)K = +0.330x10-3J/K
irreversibility Ta.ΔS = (273+30)K.(-0.315+0.330)x10-3J/K = +4.55x10-3J
cold winter day:
from condenser dQ/T = -1J/(15+273+0)K = -0.347x10-3J/
to ambient dQ/Ta = +1J/(273+0)K = +0.366x10-3J/K
irreversibility Ta.ΔS = (273+0)K.(-0.347+0.366)x10-3J/K = +5.19x10-3J
Entropy flux from condensate –ve (cooling condensate), entropy flux to ambient +ve (heating
environment), latter greater than former due to heat exchange temperature difference ΔT =
15°C (difference would increase with ΔT). Cold winter day gives greater irreversibility,
because heat transfer at higher temperature gives lower irreversibility for same temperature
difference.