2. Two moist air streams flow into a modified cooling tower to cool liquid water

Question

2. Two moist air streams flow into a modified cooling tower to cool liquid water from 120F to 75F. The first moist air stream has a relative humidity of 60% and enters the cooling tower at 80F. The second moist air stream enters the cooling tower at 40F with a relative humidity of 40% and a dry air mass flow rate that is 1/4 that of the 80F air stream. The air exiting the cooling tower exits with a relative humidity of 90% at 110F. Assuming that no makeup water is supplied to the return water stream and that atmospheric pressure is 14.7 lb/in2, determine

a. the amount of water evaporated from the incoming stream, in lb/mass dry air.

b. the mass flow rate of the hot water stream, in lb/mass dry air. (Hint: let the %u201Cmass of dry air%u201D for the units be the mass of air in the 80F stream)

3. If all of the same conditions hold (including the mass evaporation rate that you found) from #2, determine the rate of air flow for the 80F stream (in lb/hr) if a 7 hp fan is added to homogenize the air mixture within the cooling tower.

Explanation / Answer

w = 0.622 *Pv / (P - Pv)

At 80 F, we have sat. pressure = 0.507 psi

At 40 F, we have sat. pressure = 0.122 psi

At 110 F, we have sa.t pressure = 1.28 psi

Pv1 = 0.6*0.507 = 0.3042 psi

Pv2 = 0.4*0.122 = 0.0488 psi

Pv3 = 0.9*1.28 = 1.152 psi

w1 = 0.622*0.3042 / (14.7 - 0.3042) = 0.0131

w2 = 0.622*0.0488 / (14.7 - 0.0488) = 0.0021

w3 = 0.622*1.152 / (14.7 - 1.152) = 0.0529

a)

Given, m_a2 = 1/4*m_a1

For dry air, mass conservation yields,

m_a1 + m_a2 = m_a3

m_a1 + 1/4*m_a1 = m_a3

5/4*m_a1 = m_a3

For water, mass conservation yields,

m_in + m_v1 + m_v2 = m_out + m_v3

m_v1 = w1*m_a1

m_v2 = w2*m_a2 = w2*1/4*m_a1

m_v3 = w3*m_a3 = w3*5/4*m_a1

Thus, m_in + m_a1 [w1 + w2 /4] = m_out + w3 *5/4*m_a1

m_in - m_out = (w3 *5/4 - w2* 1/4 - w1) *m_a1

If dry air mass = m_a1 = 1 lb we get

m_in - m_out = (w3 *5/4 - w2* 1/4 - w1)

m_in - m_out = (0.0529 *5/4 - 0.0021* 1/4 - 0.0131)

m_in - m_out = 0.0524 lb / lb of dry air

b)

Energy rate balance yields:

0 = (m_a1*h_a1 + m_v1*h_v1) + (m_a2*h_a2 + m_v2*h_v2) - (m_a3*h_a3 + m_v3*h_v3) + m_in*h_in - m_out*h_out

0 = (m_a1*h_a1 + w1*m_a1*h_v1) + (1/4*m_a1*h_a2 + w2*m_a2*h_v2) - (5/4*m_a1*h_a3 + w3*m_a3*h_v3) + m_in*h_in - m_out*h_out

0 = (m_a1*h_a1 + w1*m_a1*h_v1) + (1/4*m_a1*h_a2 + w2*1/4*m_a1*h_v2) - (5/4*m_a1*h_a3 + w3*5/4*m_a1*h_v3) + m_in*h_in - m_out*h_out

0 = m_a1*[(h_a1 + w1*h_v1) + (1/4)*(h_a2 + w2*h_v2) - (5/4)*(h_a3 + w3*h_v3)] + m_in*h_in - m_out*h_out

Evaluating the enthalpies of the water vapor as the saturated vapor values at the respective temperatures and the enthalpy of

each liquid stream as the saturated liquid enthalpy at the respective temperature, the energy rate equation becomes

At 80 F, we have h_v1 = sat.vapor enthalpy = 1100 btu/lb

At 40 F, we have h_v2 = sat.vapor enthalpy = 1080 Btu/lb

At 110 F, we have h_v3 = sat.vapor enthalpy = 1110 Btu/lb

At 120 F, we have h_in = sat.liquid enthalpy = 88 Btu/lb

At 75 F, we have h_out = sat.liquid enthalpy = 43.1 Btu/lb

From air property tables,

At 80 F, we have h_a1 = 129 Btu/lb

At 40 F, we have h_a2 = 119.5 Btu/lb

At 110 F, we have h_a3 = 136 Btu/lb

0 = m_a1*[(129 + 0.0131*1100) + (1/4)*(119.5 + 0.0021*1080) - (5/4)*(136 + 0.0529*1110)] + m_in*88 - m_out*43.1

0 = -m_a1*69.547 + m_in*88 - m_out*43.1

For m_a1 = 1 we get,

m_in*88 - m_out*43.1 = 69.547

From part (a), m_in - m_out = 0.0524

Solving both these, m_in = 1.499 lb / lb of dry air.

m_out = 1.446 lb/lb of dry air.

c)

Not sure how to do part c. Request you to rate 4-stars for my efforts.

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