Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 6, Problem 169RQ

(a)

To determine

The time taken to attain the building’s average temperature of 24°C.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The volume of the building (v) is 400 m3.

The local atmospheric pressure (P1) is 95 kPa.

The initial temperature of the (T1) building is 14°C.

The heat lost to the surroundings (Q˙out) is 450 kJ/min.

The work input to the fan (Wfan,in) is 250W.

The temperature rise (T2T1) is 5°C

Calculation:

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of air (R) is 0.287kPam3/kgK.

Calculate the mass of air (m) that circulates inside the building.

  m=P1νRT1

  m=(95kPa)(400m3)(0.287kPam3/kgK)(14°C)=38000kPam3(0.287kPam3/kgK)(14+273)K=38000kPam382.369kPam3/kg=461.3386kg

Calculate the mass flow rate (m˙).

  m˙=mΔt

  m˙=461.3386kgΔt461.3kgΔt

Here, the change in time or time interval is Δt.

Consider the entire building as system and the air circulates the in the building itself. There is no leakage to the surrounding.

The air flows at steady state through one inlet and one exit system (pipe and duct flow). Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the energy balance equation.

  EinEout=ΔEsystem{[Qin+Win+min(h+ke+pe)in][Qout+Wout+mout(h+ke+pe)out]}=[m2(u+ke+pe)2m1(u+ke+pe)1]system        (I)

Here, the heat transfer is Q, the work transfer is W, the enthalpy is h, the internal energy is u, the kinetic energy is ke, the potential energy is pe and the change in net energy of the system is ΔEsystem; the suffixes 1 and 2 indicates the inlet and outlet of the system.

In this system two work inputs are involved namely, the work input to the electric heater (We,.in)- used to heat the air, the work input to the fan (Wfan,in)- used to circulate the air. There is an heat loss from the building (Q˙2). Neglect work transfer at the outlet, kinetic and potential energies.

The Equations (I) reduced as follows.

  {[0+(We,in+Wfan,in)+m(hin+0+0)][Qout+0+m(hout+0+0)]}=mu2mu1We,in+Wfan,in+mhinQoutmhout=m(u2u1)        (II)

Here, there is no mass leakage from the building to the surrounding. The mass of air circulates in the building itself. Hence, inlet and exit enthalpies are neglected.

The change in internal energy is expresses as follow.

  u2u1=cv(T2T1)

Here, the specific heat at constant volume is cv, the exit temperature is T2 and the inlet temperature is T1.

Neglect the inlet and exit enthalpies and substitute cv(T2T1) for u2u1 in

Equation (II).

  We,in+Wfan,in+m(0)Qoutm(0)=mcv(T2T1)We,in+Wfan,inQout=mcv(T2T1)        (III)

Express the Equation (III) with respect to change of time and rearrange it to obtain W˙e,in as follows.

  W˙e,in+W˙fan,inQ˙out=m˙cv(T2T1)        (IV)

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant volume (cv) of air is 0.718kJ/kgK.

Substitute m˙=461.3kgΔt, cv=0.718kJ/kgK, T2=24°C, T1=14°C, W˙e,in=30kW, W˙fan,in=250W and Q˙out=450kJ/min in Equation (IV).

  30kW+250W450kJ/min=(461.3kgΔt)(0.718kJ/kgK)(24°C14°C){30kW+(250W×1kW1000W)(450kJ/min×1min60s×1kW1kJ/s)}={(461.3kgΔt)(0.718kJ/kgK)[(24+273)K(14+273)K]}30kW+0.25kW7.5kW=(461.3kgΔt)7.18kJ/kg22.75kW=1Δt(3312.134kJ)

  Δt=3312.134kJ22.75kW×1kJ/s1kWΔt=3312.134kJ22.75kJ/s=145.588s146s

Thus, the time taken to attain the building’s average temperature of 24°C is 146s.

(b)

To determine

The average mass flow rate of air through the duct.

(b)

Expert Solution
Check Mark

Explanation of Solution

Consider the heating duct with fan and heater only as the system. The air passes through in it steadily.

The system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

The heating duct is an adiabatic duct. Hence, there is no heat loss.

The Equations (II) reduced as follows.

  [0+(We,in+Wfan,in)+m(hin+0+0)][0+0+m(hout+0+0)]=0We,in+Wfan,in+mhinmhout=0We,in+Wfan,in=mhoutmhinWe,in+Wfan,in=m(houthin)        (V)

Express the Equation (V) with respect to change of time as follows.

  W˙e,in+W˙fan,in=m˙(houthin)        (VI)

The change in enthalpy is expresses as follow.

  houthin=cp(ToutTin)=cp(T2T1)

Here, the specific heat at constant pressure is cp, the outlet temperature is T2 and the inlet temperature is T1.

Substitute cp(T2T1) for houthin in Equation (VIII) and rearrange it to obtain m˙.

  W˙e,in+W˙fan,in=m˙[cp(T2T1)]W˙e,in+W˙fan,in=m˙cp(T2T1)m˙=W˙e,in+W˙fan,incp(T2T1)        (VII)

Refer Table A-2, “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure (cp) of air is 1.005kJ/kg°C.

Substitute W˙e,in=30kW, W˙fan,in=250W, cp=1.005kJ/kg°C, and (T2T1)=5°C in Equation (VII).

  m˙=30kW+250W(1.005kJ/kg°C)(5°C)=(30kW×1kJ/s1kW)+(250W×1kJ/s1000W)(40kg/min×1min60s)(1.005kJ/kg°C)=30.25kJ/s5.025kJ/kg=6.0199kg/s

  6.02kg/s

Thus, The average mass flow rate of air through the duct is 6.02kg/s.

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 6 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

Ch. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - A house is maintained at 1 atm and 24°C, and warm...Ch. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - The kinetic energy of a fluid increases as it is...Ch. 6 - Prob. 23PCh. 6 - Air enters a nozzle steadily at 50 psia, 140°F,...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Air at 13 psia and 65°F enters an adiabatic...Ch. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Steam at 4 MPa and 400°C enters a nozzle steadily...Ch. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Helium is to be compressed from 105 kPa and 295 K...Ch. 6 - Carbon dioxide enters an adiabatic compressor at...Ch. 6 - Air is compressed from 14.7 psia and 60°F to a...Ch. 6 - Prob. 47PCh. 6 - An adiabatic gas turbine expands air at 1300 kPa...Ch. 6 - Steam flows steadily into a turbine with a mass...Ch. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Refrigerant-134a is throttled from the saturated...Ch. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - A thin-walled double-pipe counter-flow heat...Ch. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - The components of an electronic system dissipating...Ch. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - A house has an electric heating system that...Ch. 6 - Prob. 95PCh. 6 - Refrigerant-134a enters the condenser of a...Ch. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 100PCh. 6 - Air enters the duct of an air-conditioning system...Ch. 6 - Prob. 102PCh. 6 - A rigid, insulated tank that is initially...Ch. 6 - Prob. 105PCh. 6 - Prob. 106PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - An air-conditioning system is to be filled from a...Ch. 6 - Prob. 111PCh. 6 - A 0.06-m3 rigid tank initially contains...Ch. 6 - A 0.3-m3 rigid tank is filled with saturated...Ch. 6 - Prob. 114PCh. 6 - A 0.3-m3 rigid tank initially contains...Ch. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - The air in a 6-m × 5-m × 4-m hospital room is to...Ch. 6 - Prob. 124RQCh. 6 - Prob. 125RQCh. 6 - Prob. 126RQCh. 6 - Prob. 127RQCh. 6 - Prob. 128RQCh. 6 - Prob. 129RQCh. 6 - Prob. 130RQCh. 6 - Prob. 131RQCh. 6 - Prob. 132RQCh. 6 - Steam enters a nozzle with a low velocity at 150°C...Ch. 6 - Prob. 134RQCh. 6 - Prob. 135RQCh. 6 - Prob. 136RQCh. 6 - In large steam power plants, the feedwater is...Ch. 6 - Prob. 138RQCh. 6 - Prob. 139RQCh. 6 - Prob. 140RQCh. 6 - Prob. 141RQCh. 6 - Prob. 142RQCh. 6 - Prob. 143RQCh. 6 - Prob. 144RQCh. 6 - Prob. 145RQCh. 6 - Prob. 146RQCh. 6 - Repeat Prob. 6–146 for a copper wire ( = 8950...Ch. 6 - Prob. 148RQCh. 6 - Prob. 149RQCh. 6 - Prob. 150RQCh. 6 - Prob. 151RQCh. 6 - Prob. 152RQCh. 6 - Prob. 153RQCh. 6 - An adiabatic air compressor is to be powered by a...Ch. 6 - Prob. 156RQCh. 6 - Prob. 157RQCh. 6 - Prob. 158RQCh. 6 - Prob. 159RQCh. 6 - Prob. 160RQCh. 6 - Prob. 161RQCh. 6 - Prob. 162RQCh. 6 - Prob. 163RQCh. 6 - Prob. 164RQCh. 6 - Prob. 166RQCh. 6 - Prob. 167RQCh. 6 - Prob. 168RQCh. 6 - Prob. 169RQ
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
First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=NyOYW07-L5g;License: Standard youtube license