EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220100257056
Author: CENGEL
Publisher: YUZU
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Chapter 14.7, Problem 110P

a)

To determine

The volume flow rate of air into the cooling tower.

a)

Expert Solution
Check Mark

Answer to Problem 110P

The volume flow rate of air into the cooling tower is 11.2 m3/s.

Explanation of Solution

Apply the dry air mass balance on the cooling tower.

m˙a1=m˙a2=m˙a

Here, mass flow rate of air at inlet and outlet is m˙a1andm˙a2, and mass flow rate of liquid water is m˙a.

Apply the water mass balance on the cooling tower.

m˙3+m˙a1ω1=m˙4+m˙a2ω2m˙3m˙4=m˙a(ω2ω1)m˙3m˙4=m˙makeup (I)

Here, mass flow rate of water at state 3 and 4 is m˙3 and m˙4, and mass flow rate of required makeup water is m˙makeup.

Apply the energy balance on the cooling tower.

inm˙h=outm˙hm˙a1h1+m˙3h3=m˙a1h1+m˙4h4m˙3h3=m˙a(h2h1)+(m˙3m˙makeup)h4m˙a=m˙3(h3h4)(h2h1)(ω2ω1)h4 (II)

Write the expression to obtain the volume flow rate of air into the cooling tower (ν˙1).

ν˙1=m˙av1 (III)

Write the expression to obtain the vapor pressure at inlet conditions (Pv1).

Pv1=ϕ1Pg1=ϕ1Psat@30°C (IV)

Here, saturation pressure of water at 30°C is Pg1 and relative humidity at state 1 is ϕ1.

Write the expression to obtain the atmospheric pressure of an ideal gas mixture (P1).

P1=Pa1+Pv1 (V)

Here, dry air partial pressure at state 1 is Pa1.

Write the expression to obtain the specific volume of duct (v1).

v1=RaT1Pa1 (VI)

Here, inlet temperature is T1 and gas constant of air is Ra.

Write the expression to obtain the specific humidity (ω1) of air incoming.

ω1=0.622Pv1P1Pv1 (VII)

Here, total pressure at state 1 is P1.

Write the expression to obtain the enthalpy at state 1 (h1).

h1=cpT1+ω1hg1 (VIII)

Here, specific heat of air is cp, initial temperature is T1, and initial condition of enthalpy at saturation vapor is hg1.

Write the expression to obtain the vapor pressure at second inlet conditions (Pv2).

Pv2=ϕ2Pg2=ϕ2Psat@20°C (IX)

Here, saturation pressure of water at 20°C is Pg2 and relative humidity at state 2 is ϕ2.

Write the expression to obtain the specific humidity (ω2) of air incoming from second duct.

ω2=0.622Pv2P2Pv2 (X)

Here, total pressure at state 2 is P2.

Write the expression to obtain the enthalpy at state 2 (h2).

h2=cpT2+ω2hg2 (XI)

Here, initial condition of enthalpy at saturation vapor at state 2 is hg2.

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 20°C.

Psat@20°C=2.3392 kPahg1=2,537.4kJ/kg

Substitute 2.3392 kPa for Psat@30°C and 0.7 for ϕ1 in Equation (IV).

Pv1=(0.7)(2.3392kPa)=1.637kPa

Rewrite Equation (V) and substitute 96 kPa for P1 and 1.637 kPa for Pv1.

Pa1=P1Pv1=96kPa1.637kPa=94.363kPa

Convert the unit of T1 from °C to K.

T1=20°C=(20+273)K=293K

Refer Table A-2, “Ideal gas specific heats of various common gases”, obtain the value of Ra for air as 0.287kPam3/kgK and cp for air as 1.005kJ/kg°C.

Substitute 0.287kPam3/kgK for Ra, 293 K for T1, and 94.363 kPa for Pa1 in Equation (VI).

v1=(0.287kPam3/kgK)(293K)(94.363kPa)=0.891m3/kgdryair

Substitute 1.637 kPa for Pv1 and 96 kPa for P1 in Equation (VII).

ω1=0.622(1.637kPa)96kPa1.637kPa=0.0108kgH2O/kgdryair

Substitute 1.005kJ/kg°C for cp, 20°C for T1, 0.0108kgH2O/kgdryair for ω1, and 2,537.4kJ/kg for hg1 in Equation (VIII).

h1=(1.005kJ/kg°C)(20°C)+[(0.0108kgH2O/kgdryair)(2,537.4kJ/kg)]=47.5kJ/kgdryair

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 35°C.

Psat@35°C=5.6291 kPahg2=2,564.6kJ/kg

Substitute 5.6291 kPa for Psat@35°C and 1.00 for ϕ2 in Equation (IX).

Pv2=(1.00)(5.6291kPa)=5.6291kPa

Substitute 5.6291 kPa for Pv2 and 96 kPa for P2 in Equation (X).

ω2=0.622(5.6291kPa)96kPa5.6291kPa=0.0387kgH2O/kgdryair

Substitute 1.005kJ/kg°C for cp, 35°C for T2, 0.0387kgH2O/kgdryair for ω2, and 2,564.6kJ/kg for hg2 in Equation (XI).

h2=(1.005kJ/kg°C)(35°C)+[(0.0387kgH2O/kgdryair)(2,564.6kJ/kg)]=134.4kJ/kgdryair

Refer Table A-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid (hf@40°Ch3) as 167.53kJ/kgH2O at a temperature of 40°C.

Refer Table A-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid (hf@30°Ch4) as 125.74kJ/kgH2O at a temperature of 30°C.

Substitute 25kg/s for m˙3, 125.74kJ/kgH2O for h4, 167.53kJ/kgH2O for h3, 47.5kJ/kgdryair for h1, 134.4kJ/kgdryair for h2, 0.0108kgH2O/kgdryair for ω1, and 0.0387kgH2O/kgdryair for ω2 in Equation (II).

m˙a=25kg/s(167.53kJ/kgH2O125.74kJ/kgH2O)(134.4kJ/kgdryair47.5kJ/kgdryair)[(0.0387lbmH2O/lbmdryair0.0108lbmH2O/lbmdryair)125.74kJ/kgH2O]=12.53 kg/s

Substitute 12.53 kg/s for m˙a and 13.76ft3/lbmdryair for v1 in Equation (III).

ν˙1=12.53 kg/s(0.891m3/kgdryair)=11.2 m3/s

Thus, the volume flow rate of air into the cooling tower is 11.2 m3/s.

b)

To determine

The mass flow rate of required makeup water.

b)

Expert Solution
Check Mark

Answer to Problem 110P

The mass flow rate of required makeup water is 0.35 kg/s.

Explanation of Solution

Write the expression to obtain the mass flow rate of required makeup water (m˙makeup) from Equation (I).

m˙makeup=m˙a(ω2ω1) (XII)

Conclusion:

Substitute 12.53 kg/s for m˙a, 0.0108kgH2O/kgdryair for ω1 and 0.0387kgH2O/kgdryair for ω2 in Equation (XII).

m˙makeup=12.53 kg/s(0.0387kgH2O/kgdryair0.0108kgH2O/kgdryair)=0.35 kg/s

Thus, the mass flow rate of required makeup water is 0.35 kg/s.

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Chapter 14 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

Ch. 14.7 - A tank contains 15 kg of dry air and 0.17 kg of...Ch. 14.7 - Prob. 12PCh. 14.7 - Prob. 13PCh. 14.7 - 14–13 A room contains air at 20°C and 98 kPa at a...Ch. 14.7 - A room contains air at 85F and 13.5 psia at a...Ch. 14.7 - An 8-m3 tank contains saturated air at 30C, 105...Ch. 14.7 - Prob. 17PCh. 14.7 - Prob. 18PCh. 14.7 - Prob. 19PCh. 14.7 - Andy and Wendy both wear glasses. On a cold winter...Ch. 14.7 - In summer, the outer surface of a glass filled...Ch. 14.7 - In some climates, cleaning the ice off the...Ch. 14.7 - Prob. 23PCh. 14.7 - Prob. 24PCh. 14.7 - Prob. 25PCh. 14.7 - Prob. 26PCh. 14.7 - A thirsty woman opens the refrigerator and picks...Ch. 14.7 - Prob. 28PCh. 14.7 - The air in a room has a dry-bulb temperature of...Ch. 14.7 - Prob. 31PCh. 14.7 - Prob. 32PCh. 14.7 - How do constant-enthalpy and...Ch. 14.7 - At what states on the psychrometric chart are the...Ch. 14.7 - How is the dew-point temperature at a specified...Ch. 14.7 - Can the enthalpy values determined from a...Ch. 14.7 - Prob. 37PCh. 14.7 - Prob. 39PCh. 14.7 - Prob. 41PCh. 14.7 - Prob. 42PCh. 14.7 - Prob. 43PCh. 14.7 - Prob. 44PCh. 14.7 - What does a modern air-conditioning system do...Ch. 14.7 - How does the human body respond to (a) hot...Ch. 14.7 - Prob. 47PCh. 14.7 - How does the air motion in the vicinity of the...Ch. 14.7 - Consider a tennis match in cold weather where both...Ch. 14.7 - Prob. 50PCh. 14.7 - Prob. 51PCh. 14.7 - Prob. 52PCh. 14.7 - What is metabolism? What is the range of metabolic...Ch. 14.7 - What is sensible heat? How is the sensible heat...Ch. 14.7 - Prob. 55PCh. 14.7 - Prob. 56PCh. 14.7 - Prob. 57PCh. 14.7 - Prob. 58PCh. 14.7 - Repeat Prob. 1459 for an infiltration rate of 1.8...Ch. 14.7 - An average person produces 0.25 kg of moisture...Ch. 14.7 - An average (1.82 kg or 4.0 lbm) chicken has a...Ch. 14.7 - How do relative and specific humidities change...Ch. 14.7 - Prob. 63PCh. 14.7 - Prob. 64PCh. 14.7 - Prob. 65PCh. 14.7 - Humid air at 40 psia, 50F, and 90 percent relative...Ch. 14.7 - Air enters a 30-cm-diameter cooling section at 1...Ch. 14.7 - Prob. 68PCh. 14.7 - Prob. 69PCh. 14.7 - Why is heated air sometimes humidified?Ch. 14.7 - Air at 1 atm, 15C, and 60 percent relative...Ch. 14.7 - Prob. 72PCh. 14.7 - An air-conditioning system operates at a total...Ch. 14.7 - Prob. 74PCh. 14.7 - Why is cooled air sometimes reheated in summer...Ch. 14.7 - Prob. 76PCh. 14.7 - Prob. 77PCh. 14.7 - Air enters a 40-cm-diameter cooling section at 1...Ch. 14.7 - Repeat Prob. 1479 for a total pressure of 88 kPa...Ch. 14.7 - Prob. 81PCh. 14.7 - Prob. 83PCh. 14.7 - Prob. 84PCh. 14.7 - Prob. 85PCh. 14.7 - Atmospheric air at 1 atm, 32C, and 95 percent...Ch. 14.7 - Prob. 88PCh. 14.7 - Prob. 89PCh. 14.7 - Does an evaporation process have to involve heat...Ch. 14.7 - Prob. 93PCh. 14.7 - Prob. 94PCh. 14.7 - Air at 1 atm, 20C, and 70 percent relative...Ch. 14.7 - Two unsaturated airstreams are mixed...Ch. 14.7 - Consider the adiabatic mixing of two airstreams....Ch. 14.7 - Prob. 98PCh. 14.7 - Two airstreams are mixed steadily and...Ch. 14.7 - A stream of warm air with a dry-bulb temperature...Ch. 14.7 - Prob. 104PCh. 14.7 - How does a natural-draft wet cooling tower work?Ch. 14.7 - What is a spray pond? How does its performance...Ch. 14.7 - The cooling water from the condenser of a power...Ch. 14.7 - Prob. 108PCh. 14.7 - A wet cooling tower is to cool 60 kg/s of water...Ch. 14.7 - Prob. 110PCh. 14.7 - Prob. 111PCh. 14.7 - Prob. 112PCh. 14.7 - Prob. 113RPCh. 14.7 - Prob. 114RPCh. 14.7 - Prob. 115RPCh. 14.7 - Prob. 116RPCh. 14.7 - Prob. 117RPCh. 14.7 - Prob. 118RPCh. 14.7 - Prob. 119RPCh. 14.7 - Prob. 120RPCh. 14.7 - 14–121 The relative humidity inside dacha of Prob....Ch. 14.7 - Prob. 122RPCh. 14.7 - Prob. 124RPCh. 14.7 - 14–126E Air at 15 psia, 60°F, and 70 percent...Ch. 14.7 - Prob. 127RPCh. 14.7 - Air enters a cooling section at 97 kPa, 35C, and...Ch. 14.7 - Prob. 129RPCh. 14.7 - Humid air at 101.3 kPa, 36C dry bulb and 65...Ch. 14.7 - 14–131 Air enters an air-conditioning system that...Ch. 14.7 - Prob. 132RPCh. 14.7 - Prob. 133RPCh. 14.7 - Conditioned air at 13C and 90 percent relative...Ch. 14.7 - Prob. 138RPCh. 14.7 - A room is filled with saturated moist air at 25C...Ch. 14.7 - Prob. 141FEPCh. 14.7 - A 40-m3 room contains air at 30C and a total...Ch. 14.7 - Prob. 143FEPCh. 14.7 - The air in a house is at 25C and 65 percent...Ch. 14.7 - On the psychrometric chart, a cooling and...Ch. 14.7 - On the psychrometric chart, a heating and...Ch. 14.7 - An airstream at a specified temperature and...Ch. 14.7 - Prob. 148FEPCh. 14.7 - Air at a total pressure of 90 kPa, 15C, and 75...
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