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 22, Problem 117RQ

(a)

To determine

The tube length if heat exchanger has one-shell pass and one tube pass.

(a)

Expert Solution
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Explanation of Solution

Given:

The mass flow rate of hot steam (mh) is 47kg/s.

The mass flow rate of water (mc) is 66kg/s.

The heat capacity of water is cp,c=4.18kJ/kgK.

The heat capacity of hot stream is cp,h=3.5J/kgK.

The hot stream inlet temperature (Th,i) is 160°C.

The hot stream exit temperature (Th,o) is 100°C.

The water inlet temperature (Tc,i) is 10°C.

The average heat transfer coefficient (h) is 220W/m2K.

The dynamic viscosity of stream (μ) is 2.0mPas.

The density of stream (μ) is 950kg/m3.

Calculation:

Calculate the rate of heat transfer.

    Q˙=mhcp,h(Th,iTh,o)=(47kg/s)(3.5kJ/kgK)((160°C273)K(100°C273)K)=9870kW

Calculate the outlet temperature of cold water using energy balance.

    mccp,c(Tc,oTc,i)=Q˙[(66kg/s)(4.18kJ/kgK)×(Tc,o(10°C+273)K)]=9870kWTc,o=(318.77K273)°C=45.77°C

Calculate the temperature at the ends.

    ΔT1=(Th,iTc,o)=160°C45.77°C=114.23°CΔT2=(Th,oTc,i)=100°C10°C=90°C

Calculate the LMTD (log mean temperature difference).

    Tlm=ΔT1ΔT2lnΔT1ΔT2=(114.2°C+273)K(90°C+273)Kln114.2°C90°C=101.6K

Calculate the Reynolds number.

    Re=VDρμ=(m˙ntubeρπD24)Dρμ=(47kg/s100×950kg/m3π(2.5cm102m1cm)24)(2.5cm102m1cm)(950kg/m3)2.0mPas103kg/ms1mPas=11968.45

The value of Reynolds number is greater than 10000. Therefore, the flow is turbulent.

Calculate the value of Prandtl number.

    Pr=μcpk=(0.002kg/ms)(3500J/kgK)0.50W/mK=14

Calculate the Nusselt number for turbulent flow.

    Nu=0.023Re0.8Pr0.3=0.023(11968.45)0.8(14)0.3=92.89

Calculate the heat transfer coefficient on inner surface.

    hi=0.50W/mK(2.5cm102m1cm)92.89=1858W/m2K

Calculate the overall heat transfer coefficient.

    U=11hi+1ho=111858W/m2K+14000W/m2K=1268.69W/m2K

Calculate the tube length.

    L=Q˙UntubeπDΔTlm=9870kW1000W1kW(1268.69W/m2K)(100π)(2.5cm102m1cm)(101.6K)=9.75m

Thus, the tube length if heat exchanger has one-shell pass and one tube pass is 9.75m.

(b)

To determine

The tube length if heat exchanger has one-shell pass and four tube pass.

(b)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Calculate the number of tubes per pass.

    ntube=ntube4=1004=25

Calculate the Reynolds number.

    Re=VDρμ=(m˙ntubeρπD24)Dρμ=(47kg/s25×950kg/m3π(2.5cm102m1cm)24)(2.5cm102m1cm)(950kg/m3)2.0mPas103kg/ms1mPas=47873.8

The value of Reynolds number is greater than 10000. Therefore, the flow is turbulent.

Calculate the Nusselt number for turbulent flow.

    Nu=0.023Re0.8Pr0.3=0.023(47873.8)0.8(14)0.3=281.60

Calculate the heat transfer coefficient on inner surface.

    hi=0.50W/mK(2.5cm102m1cm)281.60=5632.15W/m2K

Calculate the overall heat transfer coefficient.

    U=11hi+1ho=115632.15W/m2K+14000W/m2K=2338.89W/m2K

Calculate the two temperature ratio.

    P=t2t1T1t1=100°C160°C10°C160°C=0.4

    R=T1T2t2t1=10°C45.8°C100°C160°C=0.60

Refer figure-22-19, “Correction factor F charts for common shell-and-tube and cross-flow heat exchangers” in book corresponding to calculated value of P and R to obtain the value of correction factor.

    F=0.96

Calculate the tube length.

    L=Q˙UFntubeπDΔTlm=9870kW1000W1kW(2338.89W/m2K)(0.96)(25π)(2.5cm102m1cm)(101.6K)=5.51m

Thus, the tube length if heat exchanger has one-shell pass and four tube pass is 5.51m.

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Chapter 22 Solutions

Fundamentals of Thermal-Fluid Sciences

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