ph2b_hw9_soln

# ph2b_hw9_soln - Ph2b Spring 2007 HW 9 Solutions Winter 2008...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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

{[ snackBarMessage ]}

Ask a homework question - tutors are online