Problem 4.23 - Problem 4.23 Difficulty: 4] Given: Data on...

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Problem 4.23 Difficulty: 4] Given: Data on flow into and out of cooling tower Find: Volume and mass flow rate of cool water; mass flow rate of moist and dry air Solution: Basic equation CS ρ V A () 0 = and at each inlet/exit QV A = Assumptions: 1) Uniform flow 2) Incompressible flow Given data: m warm 2.5 10 5 lb hr = D6 i n = V5 ft s = ρ moist 0.065 lb ft 3 = At the cool water exit Q cool VA = Q cool 5 ft s π 4 × 0.5 ft 2 × = Q cool 0.982 ft 3 s = Q cool 441 gpm = The mass flow rate is m cool ρ Q cool = m cool 1.94 slug ft 3 0.982 × ft 3 s = m cool 1.91 slug s = m cool 2.21 10 5 × lb hr = NOTE: Software does not allow dots over terms, so m represents mass flow rate, not mass! For the water flow we need CS ρ V A 0 = to balance the water flow We have m warm m cool + m v + 0 = m v m warm m cool = m v 29341 lb hr = This is the mass flow rate of water vapor. To obtain air flow rates, from psychrometrics x m v m air = where x is the relative humidity. It is also known (try Googling "density of moist air") that
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Problem 4.23 - Problem 4.23 Difficulty: 4] Given: Data on...

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