This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 3.6 Power and Refrigeration Cycles AIM: Show common ways of assembling some of the components previously discussed. Power Cycle • We can extract energy in the form of work from a fluid that has energy in the form of heat. Pump Boiler Turbine Condenser Qb Wt Qc Wp 1 2 3 4 NOTE: Boiler and condenser are heat exchangers. NOTE: There is a pressure loss in the turbine. Need a pump to compensate for this. 3–41 NOTE: The thermal efficiency , symbol η (eta), is defined as η = work we get out heat we put in (3.103) = d W p d t + d W t d t d Q b d t . (3.104) NOTE: This formula will depend on how the terms are defined. In our sign convention, W t will be negative and W p positive. NOTE: We will see (second law of thermodynamics) that there is an upper limit to the thermal efficiency. EXAMPLE: (Schaum) • Steam leaves the boiler at P 2,3 = 4000kPa and T 3 = 600 ◦ C • It exits the turbine at P 1,4 = 20kPa as saturated steam ( x = 1 ). • It exits the condensor as saturated water ( x = ). QUESTION: What is the thermal efficiency, ignoring pressure loss in boiler and condenser. ANSWER: We need to calculate d W t d t , d W p d t and d Q b d t . We choose a mass of 1kg and mass flux of ˙ m = 1 kg s . ‹ The pump: We know the enthalpy of saturated water from the steam tables (Schaum p. 355) H spec 1 = 251.4 kJ kg . • The pump increases the pressure from P 1 to P 2 : d W p d t = ˙ m P 1 P 2 ρ (3.105) 3–42 = 1 kg s " 20kPa 4000kPa 1000 kg m 3 # = 3.98 kJ s . (3.106) • We can now calculate H spec 2 because we know in general that d W p d t = ˙ m h H spec 1 H spec 2 i . (3.107) • For ˙ m = 1 kg s , we have H spec 2 = d W p d t + H spec 1 (3.108) = 3.98 kJ kg + 251.4 kJ kg (3.109) = 255.38 kJ kg . (3.110) › The boiler: From the steam tables, we find H spec 3 = 3674 kJ kg . Then the boiler heating rate d Q b d t = ˙ m h H spec 2 H spec 3 i (3.111) = 1 kg s 255.38 kJ kg 3674 kJ kg = 3418.6 kJ s . (3.112) 3–43 fi The turbine : From the steam tables, we find H spec 4 = 2610 kJ kg . d W t d t = ˙ m h H spec 3 H spec 4 i (3.113) = 1 kg s 3674 kJ kg 2610 kJ kg = 1064 kJ s . (3.114) • Now, finally, we can calculate η = d W p d t + d W t d t d Q b d t (3.115) = 3.98 kJ s + 1064 kJ s 3418.6 kJ s (3.116) = 0.31 . (3.117) 3–44 Refrigeration Cycle • We can transfer heat from one place to another. Evaporator Expansion Valve Condenser Compressor Wcmp Qe Qc NOTE: Two names for the same cycle, depending on the main purpose of the application: ‹ cool something (ice cream) down: refrigeration cycle › heat something (building) up: heat pump QUESTION: How do we measure efficiency in this case? ANSWER: As the aim here is to transfer heat, not perform work, we do not use η in this case....
View
Full
Document
This note was uploaded on 12/05/2010 for the course CHBE 251 taught by Professor Scotty during the Winter '09 term at UBC.
 Winter '09
 scotty

Click to edit the document details