¡¶¹¤³ÌÈÈÁ¦Ñ§¡·(µÚÎå°æ)µÚ3ÕÂÁ·Ï°Ìâ.. ÏÂÔØ±¾ÎÄ

C W?0 D W?Q??U

2£®Ñ¹Æø»úѹËõÆøÌåËùºÄÀíÂÛÖṦΪ£ß£ß£ß¡£

A ?pdv

122B ?d(pv)

12C ?pdv?p1v1?p2v2

13£®?w??cvdTÊÊÓÃÌõ¼þΪ£ß£ß£ß¡£

A ÀíÏëÆøÌå¿ÉÄæ¹ý³Ì C Èκι¤Öʶ¨Èݹý³Ì

B ÀíÏëÆøÌå¾øÈȹý³Ì£» D Èκι¤ÖʾøÈȹý³Ì

4£®q?cv?T??RdvÊÊÓÃÓڣߣߣߡ£

A ÀíÏëÆøÌå¿ÉÄæ¹ý³Ì C ÀíÏëÆøÌåÒ»Çйý³Ì

B Ò»ÇÐÆøÌå¿ÉÄæ¹ý³Ì D ÀíÏëÆøÌå×¼¾²Ì¬¹ý³Ì

5£®dq?dh??wtÖ»ÊÊÓÃÓڣߣߣߡ£

A ÀíÏëÆøÌå¿ÉÄæ¹ý³Ì C ÀíÏëÆøÌåÈκιý³Ì

B Èκι¤ÖÊÈκιý³Ì D Èκι¤ÖÊ¿ÉÄæ¹ý³Ì

6£®ÖüÓÐ¿ÕÆøµÄ¾øÈȸÕÐÔÃܱÕÈÝÆ÷ÖУ¬°²×°Óеç¼ÓÈÈË¿£¬Í¨µçºó£¬ÈçÈ¡¿ÕÆøÎªÏµÍ³£¬Ôò¹ý³ÌÖеÄÄÜÁ¿¹ØÏµÓÐ______

A Q>0 , ¦¤U>0 , W>0 C Q>0 , ¦¤U>0 , W=0

Èý¡¢Ìî¿Õ

1£®ÈÈÁ¿ÓëÅòÕ͹¦¶¼ÊÇ Á¿£¬ÈÈÄÜͨ¹ý ²î¶ø´«µÝ £¬ÅòÕ͹¦Í¨¹ý ´«µÝ ¡£

2£®±Õ¿Úϵͳq??hÊÊÓÃÓÚ ¹ý³Ì£¬¿ª¿Úϵͳq??hÊÊÓÃÓÚ ¡£

3£®ÄÜÁ¿·½³Ìʽq??h+wsÊÊÓõÄÌõ¼þÊÇ ¡£

4£®¹«Ê½?q??wÊÊÓÃÓÚÀíÏëÆøÌåµÄ ¹ý³Ì¡£

B Q=0 , ¦¤U>0 , W<0 D Q=0 , ¦¤U=0 , W=0

5£®¹«Ê½?q?dhÊÊÓÃÓÚÈÎºÎÆøÌåµÄ ¹ý³Ì¡£ 6£®¹«Ê½?ws??dhÊÊÓÃÓÚ ¹ý³Ì¡£

ËÄ¡¢Ãû´Ê½âÊÍ ÈÈÁ¦Ñ§µÚÒ»¶¨ÂÉ ìÊ

ϵͳµÄ´¢´æÄÜ ¼¼Êõ¹¦ ÎÈ̬ÎÈÁ÷

Îå¡¢¼ÆËãÌâ

1£®lkg¿ÕÆø´Ó³õ̬p1=5bar£¬T1=340K¡£ÔÚ±Õ¿ÚϵͳÖнøÐпÉÄæ¾øÈÈÅòÕÍ£¬ÆäÈÝ»ý±äΪԭÀ´µÄ2±¶(v2?2v1)ÇóÖÕ̬ѹÁ¦¡¢Î¶ȡ¢ÄÚÄÜ¡¢ìʵı仯¼°ÅòÕ͹¦¡£

2£®Ñ¹Æø»ú²úÉúѹÁ¦Îª6bar£¬Á÷Á¿Îª20kg/sµÄѹËõ¿ÕÆø£¬ÒÑÖªÑ¹Æø»ú½ø¿Ú״̬p1=1bar£¬t1=20¡æ£¬ÈçΪ²»¿ÉÄæ¾øÈÈѹËõ£¬Êµ¼ÊÏûºÄ¹¦ÊÇÀíÂÛÖṦµÄ1.15±¶£¬ÇóÑ¹Æø»ú³ö¿ÚζÈt2¼°Êµ¼ÊÏûºÄ¹¦ÂÊP¡£

3£®ÆøÌå´Óp1=1bar£¬v1=0.3m3ѹËõµ½p2=4bar£¬Ñ¹Ëõ¹ý³ÌÖÐά³ÖÏÂÁйØÏµp=av+bÆäÖÐa=-15bar/m3£¬ÊÔ¼ÆËã¹ý³ÌÖÐËùÐèµÄ¹¦,²¢½«¹ý³Ì±íʾÔÚP-vͼÉÏ¡£