A chilled-water heat-exchange unit is designed to cool 5 m3/s5 \mathrm{~m}^{3} / \mathrm{s} of air at 100 kPa100 \mathrm{~kPa} and 30C30^{\circ} \mathrm{C} to 100 kPa100 \mathrm{~kPa} and 18C18^{\circ} \mathrm{C} by using water at 8C8^{\circ} \mathrm{C}. Find the maximum water outlet temperature when the mass flow rate of the water is 2 kg/s2 \mathrm{~kg} / \mathrm{s}.


Answered 1 year ago
Answered 1 year ago
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Given data:

P1=100  kPaP_1=100 \; \text{kPa} v˙1=5  m3s\dot{v}_1=5 \; \frac{\text{m}^3}{\text s} T1=30  C303.15  KT_1=30^{\circ} \; \text C \to 303.15 \; \text K

P2=100  kPaP_2=100 \; \text{kPa} T2=18  CT_2=18^{\circ} \; \text C

Tw=8  CT_w=8 ^{\circ} \; \text C

m˙w=2  kgs\dot{m}_w=2\; \frac{\text{kg}}{\text s} cw=4.18  kJkg    Kc_w=4.18 \; \frac{\text{kJ}}{\text{kg \; K}}\\


We have a steady flow process. We will use TABLES to find missing properties. We will show how to apply First Law and Law of Conservation of mass in the equations. We must be careful not to mistake what our control volume is.

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