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Handbook of air conditioning and refrigeration / Shan K

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15.54 CHAPTER FIFTEEN<br />

Using an <strong>air</strong>-side heat-transfer coefficient h o � 13.4 Btu/h�ft 2 �°F as in Example 15.1, from Eq.<br />

(10.21), the heat-transfer coefficient for the wet surface is<br />

Then<br />

3. On the water side,<br />

As in Example 15.1, B � 11.4 � 1.15 � 13.1, from Eq. (15.43), for 1 ft 2 <strong>of</strong> outer surface area at<br />

the dry-wet boundary,<br />

From Eq. (15.44),<br />

hwet � m�<br />

ho � 2.84 � 13.4 � 38.06 Btu/h�ft<br />

cpa 2�°F (216 W/m2 ��C)<br />

Re D � �v wD i<br />

� w<br />

According to the above simultaneous equations,<br />

The dry-wet boundary can therefore be determined as shown in Fig. 15.31.<br />

4. From Eq. (15.45), the sensible cooling capacity <strong>of</strong> the dry part <strong>of</strong> the dry-wet coil is<br />

Q cs � 60V˙ a� ac pa(T ae � T ab)<br />

The average temperature <strong>of</strong> the <strong>air</strong>stream in the dry part is<br />

�<br />

Pr � � wc pw<br />

k w<br />

h i � 0.023 Re D 0.8 Pr 0.4 � 0.023 � (12,020) 0.8 (9.12) 0.4<br />

� 838.6 Btu/h�ft 2 �°F<br />

h wet� s(T ab � T sb) � h i<br />

B (T sb � T wb)<br />

38.06 � 0.76 (T ab � 60) � 838.6<br />

13.1 (60 � T wb)<br />

V˙ a� ac pa(T ae � T ab) � 8.33V˙ galc pw(T wl � T wb)<br />

11,000 � 0.075 � 0.243(80 � T ab) � 8.33 � 74 � 1(55 � T wb)<br />

T ab � 74.8�F <strong>and</strong> T wb � 53.3�F (11.83�C)<br />

� 11,000 � 60 � 0.075 � 0.243 � (80 � 74.8) � 62,548 Btu/h (18,327 W)<br />

T ad, m � T ae � T ab<br />

2<br />

The average temperature <strong>of</strong> the water stream corresponding to the dry part <strong>of</strong> the dry-wet coil is<br />

T wd,m � T wl � T wb<br />

2<br />

62.4 � 4 � 3600 � 0.496<br />

3.09 � 12<br />

�<br />

�<br />

3.09 � 1<br />

0.339<br />

80 � 74.8<br />

2<br />

� 9.12<br />

� 77.4�F<br />

55 � 53.3<br />

� � 54.15�F<br />

2<br />

� 12,020

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