Thermodynamics filled in class notes_Part_11

# Thermodynamics filled in class notes_Part_11 - 29 2.3...

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Unformatted text preview: 29 2.3. RANKINE T 3 2 1 4 s Figure 2.4: T − s plane for Rankine cycle. • thermal eﬃciency • back work ratio • mass ﬂow rate of steam ˙ • rate of heat transfer Qin to the ﬂuid in the boiler ˙ • rate of heat transfer Qout in the condenser • mass ﬂow rate of condenser cooling water if the cooling water enters at 15 ◦ C and exits at 35 ◦ C . Use the steam tables to ﬁx the state. At the turbine inlet, one has P3 = 8.0 M P a, and x3 = 1 (saturated steam). This gives two properties to ﬁx the state, so that h3 = 2758 kJ , kg s3 = 5.7432 kJ . kg K (2.104) State 4 has P4 = 0.008 M P a and s4 = s3 = 5.7432 kJ/kg/K , so the state is ﬁxed. From the saturation tables, it is found then that 5.7432 s4 − sf = x4 = sg − sf kJ kg K − 0.5926 7.6361 kJ kg K kJ kg K = 0.6745. (2.105) Note the quality is 0 ≤ x4 ≤ 1, as it must be. The enthalpy is then h4 = hf + x4 hf g = 173.88 kJ kg + (0.6745) 2403.1 kJ kg = 1794.8 kJ . kg (2.106) State 1 is saturated liquid at 0.008 M P a, so x1 = 0, P1 = 0.008 M P a. One then gets h1 = hf = 173.88 kJ/kg, v1 = vf = 0.0010084 m3 /kg . Now state 2 is ﬁxed by the boiler pressure and s2 = s1 . But this requires use of the sparse compressed liquid tables. Alternatively, the pump work is easily approximated by assuming an incomCC BY-NC-ND. 18 November 2011, J. M. Powers. 30 CHAPTER 2. CYCLE ANALYSIS pressible ﬂuid so that h2 = h2 ˙ W = h1 + v1 (P2 − P1 ), m ˙ m3 kJ + 0.0010084 173.88 kg kg = (2.107) h1 + The net power is (8000 kP a − 8 kP a) = 181.94 kJ . kg ˙ ˙ ˙ Wcycle = Wt + Wp . (2.108) (2.109) Now the ﬁrst law for the turbine and pump give ˙ Wt = h3 − h4 , m ˙ ˙ Wp = h1 − h2 . m ˙ (2.110) The energy input that is paid for is ˙ Qin = h3 − h2 . m ˙ (2.111) ˙ ˙ (h3 − h4 ) + (h1 − h2 ) Wt + Wp = , ˙ in h3 − h2 Q (2.112) The thermal eﬃciency is then found by η = = − 1794.8 kJ kg kJ kg + 173.88 kJ kg 2758 2758 = kJ kg − 181.94 kJ kg − 181.94 kJ kg , 0.371. (2.113) (2.114) By deﬁnition the back work ratio bwr is the ratio of pump work to turbine work: = ˙ Wp , ˙ Wt (2.115) = bwr h1 − h2 , h3 − h4 (2.116) 173.88 = kJ kg 2758 = kJ kg − 181.94 − 1794.8 kJ kg , kJ kg (2.117) 0.00837. (2.118) The desired mass ﬂow can be determined since we know the desired net power. Thus m ˙ = ˙ Wcycle , (h3 − h4 ) + (h1 − h2 ) = 2758 kJ kg kg , s kg 104.697 s − 1794.8 (2.119) 100 × 103 kW kJ kg + 173.88 kJ kg − 181.94 = 104.697 = kJ kg , (2.120) (2.121) 3600 s hr CC BY-NC-ND. 18 November 2011, J. M. Powers. = 3.769 × 105 kg . hr (2.122) ...
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