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 19, Problem 102P

(a)

To determine

The exit temperature of the hot air leaving the basement.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The length (L) of the duct is 12m.

The cross section area (Ac) of the duct is 20cm×20cm.

The average velocity (V) of the hot air in the duct is 4m/s

The temperature (T1) of the hot air leaving the furnace is 60°C.

The emissivity (ε) of the outer surface of duct is 0.3.

The temperature (T0) of the cold air is 10°C.

Calculation:

Refer to the table A22 “Properties of air at 1atm pressure.”

Obtain the following properties of air corresponding to the bulk mean temperature of 50°C as follows:

cp=1007J/kgK

k=0.02735W/mK

ν=1.798×105m2/s

ρ=1.092kg/m3

Pr=0.7228

Calculate the cross sectional area of the duct by using the relation.

    As=4aL=4(0.2m)(12m)=9.6m2

Calculate the hydrodynamic diameter by using the relation.

    Dh=4AcP=4(0.2m×0.2m)4(0.2m)=0.2m

Calculate the Reynolds number by using the relation.

    Re=VDhν=(4m/s)(20cm×1m100cm)1.798×105m2/s=44494

The Reynolds number is greater than 10000 so the flow in the tube is turbulent.

Calculate the length of hydrodynamic layer by using the relation.

    Lh=10Dh=10(0.2m)=2m

Calculate the length of thermal boundary layer by using the relation.

    Lt=10Dh=10(0.2m)=2m

The length of hydrodynamic layer and thermal boundary layer is less than the length of duct.

Calculate the heat transfer coefficient by using the relation.

    Nu=0.023Re0.8Pr0.4hDk=0.023(44494)0.8(0.7228)0.4h=(105.68)(0.02735W/mK0.2m)=14.45W/m2K

Calculate the mass flow rate by using the relation.

    m˙=ρAcV=(1.092kg/m3)((20cm×1m100cm)×(20cm×1m100cm))(4m/s)=0.1747kg/s

Calculate the heat transfer due to convection by using the relation.

    Qcon=h1As(ΔTlm)=h1As[(T2T1)ln(TsT2TsT1)]=(14.45W/m2K)(9.6m2)[(T260°C)ln(TsT2Ts60°C)]=138.72W/K[(T260°C)ln(TsT2Ts60°C)]

Calculate the heat transfer due to convection and radiation by using the relation.

    Qcon+rad=h2As(TsTo)+εAsσ(Ts4To4)=[(10W/m2K)(9.6m2)(Ts10°C)+(0.3)(9.6m2)(5.67×108W/m2K4)[(Ts)4(10°C+273)4K4]]

Calculate the total heat transfer by using the relation.

    Qt=m˙cp(T2T1)=(0.1747kg/s)(1007J/kgK)(60°CT1)

In the steady operation heat transfer from the hot air to duct must be equal to heat transfer due to convection and radiation, which also equal to energy loss of the hot air in the duct.

    Qcon=Qcon+rad=Qt

From above condition

Q=2622W

T2=45.1°C

Ts=33.3°C

Thus, the exit temperature of the hot air leaving the basement is 45.1°C.

(b)

To determine

The rate of heat loss from the hot air in the duct.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the total heat transfer by using the relation.

    Qt=m˙cp(T2T1)=(0.1747kg/s)(1007J/kgK)((60°C+273)K(45.1°C+273)K)=2621.25W

Thus, the rate of heat loss from the hot air in the duct is 2621.25W.

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Chapter 19 Solutions

Fundamentals of Thermal-Fluid Sciences

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