Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 5, Problem 111RQ
To determine

The average temperature of air in 45 min.

Expert Solution & Answer
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Explanation of Solution

Given:

The size of the room (V) is (3-m × 4–m × 6-m =72m3).

The air pressure in the room remains constant (p1) at 100 kPa.

The initial temperature of the room (T1) is 7°C.

The volume of the radiator (V1) is 15 L.

The pressure of superheated vapor in radiator (P1)rad  is 200 kPa

The temperature of superheated vapor (T1)rad  is 200°C.

The work input for the fan (W˙fan,in) is 120-W.

The final pressure of the steam (P2)  is 100 kPa

The change in time (Δt) is 45 min.

Calculation:

Draw the free body diagram of the well-insulated room as shown in figure (1).

Fundamentals of Thermal-Fluid Sciences, Chapter 5, Problem 111RQ

Refer the Table (A-4 through A-6), obtain the value of specific volume, specific internal energy at the initial and the final states.

At initial pressure and temperature of 200kPa and 200°C.

  The initial specific internal energy (u1) is 2654.6kJ/kg.

  The initial specific volume (v1) is 1.08049m3/kg,

At final pressure of 100kPa.

The specific volume of saturated liquid (vf) is 0.01043.

The specific internal energy of saturated liquid (uf) is 417.40.

The specific volume of saturated vapour (vg) is 1.6941m3/kg

The specific internal energy of vapour mixture (ufg) is 2088.2kJ/kg.

Calculate the final quality at the final state.

  x2=v1vfvfgx2=(1.08049m3/kg)(0.01043m3/kg)(1.693057m3/kg)=0.6376

Calculate the final specific internal energy of a well-insulated room.

  u2=uf+x2ufgu2=417.40+0.6376(2088kJ/kg)=1748.7kJ/kg

Write the expression for total mass of a water in the radiator.

  m=V1v1

  m=0.015m31.08049m3/kg=0.01388kg

Write expression for the volume of the well-insulated room.

  V=lbh        (VI)

  V=3m×4m×6m=72m3

Write the expression of mass of air in a well-insulated room.

  mair=P1V1RT1

  mair=(100kPa)(72m3)(0.2870kPam3/kgK)(200K)=89.60kg

Write the expression for amount of fan work done in 45 min.

  Wfan,in=W˙fan,inΔt        (VIII)

  Wfan,in=(0.120kJ/s)(40min)=(0.120kJ/s)(40min×(60s1min))=324kJ

Write the expression for the energy balance equation.

  EinEout=ΔEsystem        (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Substitute 0 for Ein, Qout for Eout, and m(u1u2) for ΔEsystem in Equation (I).

  Qout=ΔU=m(u2u1)=m(u1u2)

  Qout=(0.01388kg)(2654.61748.7)kJ/kg=12.58kJ

For well-insulated room the energy balance equation.\

Substitute Qin+Wfan,in for Ein, Wb,out for Eout, and ΔU for ΔEsystem in Equation (I).

  Qin+Wfan,inWb,out=ΔUQin+Wfan,in=ΔHmcP(T2T1)        (II)

Express the change in internal energy and boundary work into the constant pressure expansion and compression in Equation (IX).

  (Q˙in+W˙fan,in)Δt=mcP,avg(T2T1)        (III)

Here, the rate of heat transfer entering is Q˙in, the rate of work of fan entering is W˙fan,in, change in temperature is Δt, the mass is m, the constant average pressure is cP,avg, the initial temperature is T1, and the final temperature is T2.

Substitute Q˙in=12.58kJ, W˙fan,in=324kJ, m=89.60kg, cP,avg=1.005kJ/kg°C, and T1=7°C in Equation (III).

  (12.58kJ)+(324kJ)=(89.60kg)(1.005kJ/kg°C)(T27°C)(336.58kJ)=(90.048kJ°C)(T27°C)T2=10.7°C

Therefore, the well-insulated room temperature rises from 7°Cto 10.7°C in 45 min time.

Thus, the average temperature of air in 45 min is 10.7°C_.

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Chapter 5 Solutions

Fundamentals of Thermal-Fluid Sciences

Ch. 5 - Prob. 11PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - During an expansion process, the pressure of a gas...Ch. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - 0.75-kg water that is initially at 0.5 MPa and 30...Ch. 5 - Prob. 22PCh. 5 - A piston–cylinder device contains 50 kg of water...Ch. 5 - Reconsider Prob. 5–23. Using an appropriate...Ch. 5 - Prob. 25PCh. 5 - A closed system undergoes a process in which there...Ch. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - A fixed mass of saturated water vapor at 400 kPa...Ch. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 36PCh. 5 - A 40-L electrical radiator containing heating oil...Ch. 5 - Prob. 38PCh. 5 - Saturated R-134a vapor at 100°F is condensed at...Ch. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Prob. 46PCh. 5 - Prob. 47PCh. 5 - Prob. 48PCh. 5 - Prob. 49PCh. 5 - Prob. 50PCh. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53PCh. 5 - Prob. 54PCh. 5 - Is it possible to compress an ideal gas...Ch. 5 - Prob. 56PCh. 5 - Prob. 57PCh. 5 - A rigid tank contains 10 lbm of air at 30 psia and...Ch. 5 - Prob. 59PCh. 5 - Prob. 60PCh. 5 - Prob. 61PCh. 5 - Prob. 62PCh. 5 - Prob. 63PCh. 5 - Prob. 64PCh. 5 - Prob. 65PCh. 5 - Prob. 66PCh. 5 - Prob. 67PCh. 5 - Air is contained in a variable-load...Ch. 5 - A mass of 15 kg of air in a piston–cylinder device...Ch. 5 - Prob. 70PCh. 5 - Prob. 72PCh. 5 - Prob. 73PCh. 5 - Air is contained in a cylinder device fitted with...Ch. 5 - Air is contained in a piston–cylinder device at...Ch. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 81PCh. 5 - Prob. 82PCh. 5 - Prob. 83PCh. 5 - Prob. 85PCh. 5 - Prob. 86PCh. 5 - Repeat Prob. 5–86 for aluminum balls. 5-86. In a...Ch. 5 - Prob. 88RQCh. 5 - Prob. 89RQCh. 5 - Air in the amount of 2 lbm is contained in a...Ch. 5 - Air is expanded in a polytropic process with n =...Ch. 5 - Nitrogen at 100 kPa and 25°C in a rigid vessel is...Ch. 5 - A well-insulated rigid vessel contains 3 kg of...Ch. 5 - In order to cool 1 ton of water at 20°C in an...Ch. 5 - Prob. 95RQCh. 5 - Prob. 96RQCh. 5 - Saturated water vapor at 200°C is condensed to a...Ch. 5 - A piston–cylinder device contains 0.8 kg of an...Ch. 5 - A piston–cylinder device contains helium gas...Ch. 5 - Prob. 100RQCh. 5 - Prob. 101RQCh. 5 - Prob. 102RQCh. 5 - Prob. 103RQCh. 5 - Prob. 104RQCh. 5 - Prob. 105RQCh. 5 - Prob. 106RQCh. 5 - A 68-kg man whose average body temperature is 39°C...Ch. 5 - An insulated rigid tank initially contains 1.4-kg...Ch. 5 - Prob. 109RQCh. 5 - Prob. 111RQCh. 5 - Prob. 112RQCh. 5 - Prob. 114RQCh. 5 - Prob. 115RQCh. 5 - An insulated piston–cylinder device initially...Ch. 5 - Prob. 118RQCh. 5 - Prob. 119RQCh. 5 - Prob. 120RQ
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