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 136RQ

(a)

To determine

The rate of heat loss from the water.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The Length of pipe is (l) is 15m.

The thermal conductivity (k) is 386W/mK.

The inner diameter (d1) is 4cm.

The inner diameter (d2) is 4.6cm.

The temperature of air (T) is 10°C.

The temperature of hot water (T) is 90°C.

The speed of water (V) is 1.2m/s.

The emissivity (ε) is 0.7.

The convection heat transfer coefficient (h) is 12W/m2K.

Calculation:

Refer Table A-22 “Properties of air at 1atm pressure”.

Obtain the following properties of air corresponding to the temperature of 90°C as follows:

k=0.675W/mKPr=1.96ρ=965.3kg/m3cp=4206J/kgKv=0.326×106m2/s

Calculate the mass flow rate of water (m) using the relation.

    m=ρAV=(965.3kg/m3)(π4(0.04m)2)(1.2m/s)=1.456kg/s

Calculate the Reynolds number (Re) using the relation.

  Re=Vdv=(1.2m/s)(4cm×1m100cm)0.326×106m2/s=147240

Calculate the Nusselt number (Nu) using the relation.

    Nu=hDk=0.125fRePr1/3=0.125×0.032×147240×1.961/3=737.1

Calculate the heat transfer coefficient (h) using the relation.

    h=kDNu=0.675W/mK(4cm×1m100cm)(737.1)=12440W/m2K

Calculate the rate of heat transfer by convection (Qc) using the relation.

    Q=hA(TsT)=hπd2l(TT)=(12W/m2K)(π×4.6cm×1m100cm)(15m)((90°C+273)K(10°C+273)K)=2081W

Calculate the rate of heat transfer by radiation (Qr) using the relation.

    Qr=εAσ(T4T4)=[(0.7)(π4.6cm×1m100cm)(15m)(5.67×108W/m2K4)×((90°C+273K)4(10°C+273K)4)]=942W

Calculate the total heat transfer (Q) using the relation.

    Q=Qc+Qr=2081W+942W=3023W

Thus, the rate of heat loss from the water is 3023W.

(b)

To determine

The temperature at which the water leaves the basement.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given:

The Length of pipe is (l) is 15m.

The thermal conductivity (k) is 386W/mK.

The inner diameter (d1) is 4cm.

The inner diameter (d2) is 4.6cm.

The temperature of air (T) is 10°C.

The temperature of hot water (T) is 90°C.

The speed of water (V) is 1.2m/s.

The emissivity (ε) is 0.7.

The convection heat transfer coefficient (h) is 12W/m2K.

Calculation:

Calculate the temperature at which water leaves the basement (Te) using the relation

  Q=mcp(TTe)Te=TQmcp=(90°C+273)K3023W(1.456kg/s)(4206J/kgK)=(362.5K273)°C=89.5°C

Calculate the temperature at which the water (ΔTp) leaves the basement.

    ΔTP=Q(ln(d2/d1))2πkL=3023W×((ln4.64)2π(386W/mK)(15m))=0.012K

The value of temperature difference is same in term of °C and K therefore, the value of ΔTp can be taken as 0.012°C.

Thus, the temperature at which the water leaves the basement is 0.012°C.

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Chapter 19 Solutions

Fundamentals of Thermal-Fluid Sciences

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