A wet cooling tower is to cool 60 kg/s of water from 40 to 33°C. Atmospheric air enters the tower at 1 atm with dry-and wet-bulb temperatures of 22 and 16°C, respectively, and leaves at 30°C with a relative humidity of 95 percent. Using the psychrometric chart, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water.

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Question:

A wet cooling tower is to cool 60 kg/s of water from 40 to 33°C. Atmospheric air enters the tower at 1 atm with dry-and wet-bulb temperatures of 22 and 16°C, respectively, and leaves at 30°C with a relative humidity of 95 percent. Using the psychrometric chart, determine (a) the volume flow rate of air into the cooling tower and (b) the mass flow rate of the required makeup water.

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Answer:32.06m³/s0.62kg/sExplanation:Using our psychometric chart, at T1 = 22°C and wet bulb = 16°C, we have h1 = 45.6kj/kgv1 = 0.85m³/kgw1 = 0.00923At T2 = 30°C, T3 = 40°C, T4 = 33°Ch2 = 96.1kj/kgw2 = 0.02579h3 = 168kj/kgh4 = 138kj/kgUsing the mass for rate balances for water and air, we have m'(a1) = m'(a2) = m'(a)m'(3) - m'(4) = m'(a) [w2 - w1] = m'(m)Now, we use the energy balance equation to solve for the needed mass, m'(a). Mass flow rate of water, m'3 = 60kg/s0 = m'(a)h2 + m'4h4 - m'(a)h1 - m'3h30 = m'(a) [h2 - h1] + [-m'(a) {w2-w1} + m'3]h4 - m'3h3m'(a) = [m'3(h3 - h4) ] / [ h2 - h1 - (w2 - w1)h4m'(a) = [60(168-138)] / 96.1 - 45.6 - (0.02579 - 0.00923)168m'(a) = 60*30 / 50.5 - 0.01656*168m'(a) = 1800 / 47.72m'(a) = 37.72kg/sVolume flow rate = m'(a) * v1Volume flow rate = 37.72 * 0.85Volume flow rate = 32.06m³/sMass flow rate of make up water = m'(m) = m'(a) [w2 - w1]m'(m) = 37.72 [0.02579 - 0.00923]m'(m) = 37.72 * 0.01656m'(m) = 0.62kg/s

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