Unformatted text preview: Ph2b Spring 2007 HW 9 Solutions Winter 2008 Problem 1, K&K 8-1 (10 points) (a) The reversible heat pump is simply a Carnot engine with all flows reversed, so that its efficiency is still given by the Carnot efficiency, ! #! "C $ h l . (1) !h For a reversible process, Qh Q ! $ % h $ % l $ l , or Ql $ l Qh . !h !l !h The work required to drive the heat pump is then (2) & !' W W $ Qh # Ql $ (1 # l ) Qh $ "C Qh , or $ "C . Qh * !h + (3) If the heat pump is not reversible, the entropy leaving the heat pump will be greater than the entropy entering it. Then % h , % l , and more work is required, & !' W $ Qh # Ql , (1 # l ) Qh $ "C Qh . * !h +
(b) The work generated by the Carnot engine is (4) & !' W $ (1 # l ) Qhh . * ! hh + (5) The work consumed by the reversible heat pump was found in (a), and it is equal to the work generated by the Carnot engine, & !l ' & !l ' (1 # ) Qhh $ ( 1 # ) Qh . * ! hh + * !h +
From this, we get (6) ! -! # ! . !' Qhh & ! l ' & $ (1 # ) ( 1 # l ) $ hh h l ! 0.18 ! h -! hh # ! l . Qh * ! h + * ! hh +
#1 (7) for the temperatures given. (c) See figure on the next page. Problem 2, K&K 8-2 (10 points) (a) This is the same as the figure in problem 1 above. (b) From energy and entropy conservation (assuming reversible operation), Q Q Q Qh $ Qhh / Ql and % h $ h $ % hh / % l $ hh / l . !h ! hh ! l equation, we get 0 $ -1 # ! h / ! hh . Qhh / -1 # ! h / ! l . Ql Ql / Qhh $ -1 # ! h / ! hh . / -! h / ! l # 1. $ -! l / ! hh . -! hh # ! h . / -! h # ! l . . (9) (8) Multiplying the second of these equations by ! h and subtracting if from the first ...
View Full Document