EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
bartleby

Videos

Question
Book Icon
Chapter 14.7, Problem 132RP

a)

To determine

The rate of dehumidification.

a)

Expert Solution
Check Mark

Answer to Problem 132RP

The mass flow rate of condensate water is 0.01861kg/min.

Explanation of Solution

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

Pv1=ϕ1Pg1=ϕ1Psat@30°C (I)

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 (II)

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

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

v1=RaT1Pa1 (III)

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 (IV)

Here, total pressure at state 1 is P1.

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

h1=cpT1+ω1hg1 (V)

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 (VI)

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 (VII)

Here, total pressure at state 2 is P2.

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

h2=cpT2+ω2hg2 (VIII)

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

Write the expression to obtain the mass flow rate of dry air (m˙a1) in each stream.

m˙a1=V˙1v1 (IX)

Here, volume flow rate is V˙1.

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) (X)

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

Conclusion:

Refer Table A-4, “Saturated water – Temperature table”, obtain the properties of water at a temperature of 30°C.

Psat@30°C=4.247 kPahg1=2,555.6kJ/kg

Substitute 4.247 kPa for Psat@30°C and 0.7 for ϕ1 in Equation (I).

Pv1=(0.7)(4.247kPa)=2.973kPa

Refer Table A-5, “Saturated water – Pressure table”, obtain the properties of water at a pressure of 2.973kPa.

Tdp=23.9°C

Rewrite Equation (II) and substitute 90 kPa for P1 and 2.973 kPa for Pv1.

Pa1=P1Pv1=90kPa2.973kPa=87.03kPa

Convert the unit of T1 from °C to K.

T1=30°C=(30+273)K=303K

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, 303 K for T1, and 87.03 kPa for Pa1 in Equation (III).

v1=(0.287kPam3/kgK)(303K)(87.03kPa)=0.9992m3/kgdryair

Substitute 2.973 kPa for Pv1 and 90 kPa for P1 in Equation (IV).

ω1=0.622(2.973kPa)90kPa2.973kPa=0.02125kgH2O/kgdryair

Substitute 1.005kJ/kg°C for cp, 30°C for T1, 0.02125kgH2O/kgdryair for ω1, and 2,555.6kJ/kg for hg1 in Equation (V).

h1=(1.005kJ/kg°C)(30°C)+[(0.02125kgH2O/kgdryair)(2,555.6kJ/kg)]=84.45kJ/kgdryair

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

Psat@20°C=2.3392 kPahg2=2,537.4kJ/kg

Substitute 2.3392 kPa for Psat@20°C and 1.00 for ϕ2 in Equation (VI).

Pv2=(1.00)(2.3392kPa)=2.3392kPa

Substitute 2.3392 kPa for Pv2 and 95 kPa for P2 in Equation (VII).

ω2=0.622(2.3392kPa)95kPa2.3392kPa=0.01660kgH2O/kgdryair

Substitute 1.005kJ/kg°C for cp, 20°C for T2, 0.01660kgH2O/kgdryair for ω2, and 2,537.4kJ/kg for hg2 in Equation (VIII).

h2=(1.005kJ/kg°C)(20°C)+[(0.01660kgH2O/kgdryair)(2,537.4kJ/kg)]=62.22kJ/kgdryair

Substitute 4m3/min for V˙1 and 0.9992m3/kgdryair for v1 in Equation (IX).

m˙a1=4m3/min0.9992m3/kgdryair=4.003kg/min

Substitute 4.003kg/min for m˙a1, 0.01660kgH2O/kgdryair for ω2, and 0.02125kgH2O/kgdryair for ω1 in Equation (X).

m˙w=4.003kg/min(0.02125kgH2O/kgdryair0.01660kgH2O/kgdryair)=0.01861kg/min

Thus, the mass flow rate of condensate water is 0.01861kg/min.

b)

To determine

The rate of heat transfer.

b)

Expert Solution
Check Mark

Answer to Problem 132RP

The rate of heat transfer of air stream is 87.44kJ/min.

Explanation of Solution

Apply the 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˙a1(h1h2)m˙whw (XI)

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-4, “Saturated water – Temperature table”, obtain the value of enthalpy at saturation liquid (hf@20°Chw2) as 83.915kJ/kgdry air at a temperature of 20°C.

Substitute 83.915kJ/kgdry air for hw, 0.01861kg/min for m˙w, 4.003kg/min for m˙a1, 84.45kJ/kgdry air for h1, and 62.22kJ/kgdry air for h2 in Equation (XI).

Q˙out=(4.003kg/min(84.45kJ/kgdry air62.22kJ/kgdry air)(0.01861kg/min)(83.915kJ/kgdry air))=87.44kJ/min

Thus, rate of heat transfer of air stream is 87.44kJ/min.

c)

To determine

The mass flow rate of the refrigerant.

c)

Expert Solution
Check Mark

Answer to Problem 132RP

The mass flow rate of the refrigerant is 0.620kg/min.

Explanation of Solution

Write the expression to obtain the inlet enthalpy of the refrigerant (h3).

h3=hf+x3hfg (XII)

Here, enthalpy evaporation is hfg, enthalpy saturation liquid is hf, and dryness fraction of refrigerant is x3.

Write the expression to obtain the mass flow rate of a refrigerant (m˙R).

m˙R=Q˙Rh4h3 (XIII)

Here, heat lost by the air is Q˙R.

Conclusion:

Refer Table A-12, “Saturated refrigerant 134a – Pressure table”, obtain the properties of refrigerant 134a at a pressure of 700 kPa.

hf=88.82kJ/kghfg=176.26kJ/kg

Substitute 88.82kJ/kg for hf, 0.2 for x3, and 176.26kJ/kg for hfg in Equation (XII).

h3=88.82kJ/kg+0.2(176.26kJ/kg)=124.07kJ/kg

Refer Table A-12, “Saturated refrigerant 134a – Pressure table”, obtain the value of (hg=h4) as 265.08kJ/kg at a pressure of 700 kPa.

Substitute 87.44kJ/min for Q˙R, 124.07kJ/kg for h3, and 265.08kJ/kg for h4 in Equation (XIII).

m˙R=87.44kJ/min265.08kJ/kg124.07kJ/kg=0.620kg/min

Thus, the mass flow rate of the refrigerant is 0.620kg/min.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
The Refrigeration Cycle Explained - The Four Major Components; Author: HVAC Know It All;https://www.youtube.com/watch?v=zfciSvOZDUY;License: Standard YouTube License, CC-BY