EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Chapter 14.7, Problem 81P

a)

To determine

The rate of heat transfer of air stream.

a)

Expert Solution
Check Mark

Answer to Problem 81P

The rate of heat transfer of air stream is 250.9Btu/min.

Explanation of Solution

Write the expression to obtain the volume flow rate (V˙1).

V˙1=V1A1=V1(πD24) (I)

Here, diameter is D, and velocity of air is V1.

Write the expression to obtain the mass flow rate of dry air through the cooling section (m˙a).

m˙1=V˙1v1 (II)

Here, volume flow rate of air entering the cooling section is v1.

Apply the water mass balance equation to the combined cooling and dehumidification section.

m˙w,i=m˙w,em˙a1ω1=m˙a2ω2+m˙wm˙w=m˙a1(ω2ω1) (III)

Here, initial and final mass flow rate of dry air is m˙a1 and m˙a2, specific humidity of air entering and leaving the cooling coils is ω1 and ω2, and mass flow rate of vapor is m˙w.

Apply the water energy balance equation to the combined cooling and dehumidification section.

E˙inE˙out=ΔE˙system(steady)E˙inE˙out=0E˙in=E˙out

m˙ihi=Q˙out+m˙eheQ˙out=m˙a1h1(m˙a2h2+m˙whw)Q˙out=m˙a1(h1h2)m˙whw (IV)

Here, the amount of energy rate required for cooling coils is E˙in, amount of rate of energy rejected from cooling coils is E˙out, change in the energy rate of a system is ΔE˙system, and rate of heat removal from air is Q˙out.

Conclusion:

Refer Table A-4E, “Saturated water – Temperature table”, write the value of saturation pressure of liquid (Pv) as 0.69904 psia at a temperature of 90°F.

Refer Table A-5E, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 90°F.

Tdp=Tsat@Pv=Tsat@0.69904 psia=74.2°F

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at inlet temperature (T1=90°F) and relative humidity (ϕ=60%), write the value of enthalpy at state 1 (h1) as 41.8Btu/lbmdry air, specific humidity of air entering the cooling section (ω1) as 0.0183lbmH2O/lbmdryair, and specific volume (v1) as 14.26ft3/lbmdryair.

Refer Fig A-31E, “Psychrometric chart at 1 atm total pressure”, at exit temperature (T2=70°F) and relative humidity (ϕ2=100%), write the value of enthalpy at state 2 (h2) as 34.1Btu/lbmdry air, and specific humidity of air leaving the cooling section (ω2) as 0.0158lbmH2O/lbmdryair and specific volume (v2) as 13.68ft3/lbmdryair.

Refer Table A-4E, “Saturated water – Temperature table”, write the value of enthalpy at saturation liquid (hf@70°Fhw) as 38.08Btu/lbm at a temperature of 70°F.

Substitute 1 ft for D and 600ft/min for V1 in Equation (I).

V˙1=600ft/min(π(1ft)24)=471ft3/min

Substitute 471ft3/min for V˙1, and 14.26ft3/lbmdryair for v1 in Equation (II).

m˙a1=471ft3/min14.26ft3/lbmdryair=33.0lbm/min

Substitute 33.0lbm/min for m˙a, 0.0158lbmH2O/lbmdryair for ω2, and 0.0183lbmH2O/lbmdryair for ω1 in Equation (III).

m˙w=33.0lbm/min(0.0158lbmH2O/lbmdryair0.0183lbmH2O/lbmdryair)=0.083lbm/min

Substitute 38.08Btu/lbm for hw, 0.083lbm/min for m˙w, 33.0lbm/min for m˙a1, 41.8Btu/lbmdry air for h1, and 34.1Btu/lbmdry air for h2 in Equation (IV).

Q˙out=33.0lbm/min(41.8Btu/lbmdry air34.1Btu/lbmdry air)(0.083lbm/min)(38.08Btu/lbm)=250.9Btu/min

Thus, rate of heat transfer of air stream is 250.9Btu/min.

b)

To determine

The mass flow rate of cooling water.

b)

Expert Solution
Check Mark

Answer to Problem 81P

The mass flow rate of cooling water is 17.9lbm/min.

Explanation of Solution

Write the expression to obtain the mass flow rate of cooling water (m˙coolingwater).

m˙coolingwater=Q˙wcpΔT (V)

Here, specific heat of water is cp, rate of heat transfer of water is Q˙w, and change in temperature of air is ΔT.

Conclusion:

Substitute 1.0Btu/lbm°F for cp, 14°F for ΔT, and 250.9Btu/min for Q˙w in Equation (V).

m˙coolingwater=250.9Btu/min(1.0Btu/lbm°F)(14°F)=17.9lbm/min

Thus, the mass flow rate of cooling water is 17.9lbm/min.

c)

To determine

The exit velocity of the airstream.

c)

Expert Solution
Check Mark

Answer to Problem 81P

The exit velocity is 576ft/min.

Explanation of Solution

Apply the conservation of mass of dry air to calculate exit velocity (V2).

m˙a1=m˙a2V1Av1=V2Av2V2=v2v1V1 (VI)

Conclusion:

Substitute 13.68ft3/lbmdryair for v2, 14.26ft3/lbmdryair for v1, and 600ft/min for V1 in Equation (VI).

V2=(13.68ft3/lbmdryair)(14.26ft3/lbmdryair)(600ft/min)=576ft/min

Thus, the exit velocity is 576ft/min.

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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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