Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 9, Problem 9.101P
To determine

(a)

The value of mass flow for isothermal flow.

Expert Solution
Check Mark

Answer to Problem 9.101P

The value of mass flow for isothermal flow is 6.731 kg/h

Explanation of Solution

Given Information:

The value of pressure at entrance, p1=102000 Pa

The value of pressure at entrance, p2=100000 Pa

The value of friction coefficient f=0.028

The diameter of the tube, D = 0.01 m

The length of the tube, L = 3.0 m

Concept Used:

(m˙A)=p12p22RT[ fLD+2ln p 1 p 2 ]m˙=massflowrateinkg/s

Calculation:

( m ˙A)2=(102000)2(100000)2(287)(293)[ (0.028)3 0.01+2ln 102 100]=404000000(84091)(8.4+0.0396)=569.26(m˙A)=23.86m˙=23.86×A=23.86×(π4(0.01)2)=0.001874kg/s=6.731kg/h.

Conclusion:

The value of mass flow for isothermal flow is 6.731 kg/h.

To determine

(b)

The value of mass flow for adiabatic flow.

Expert Solution
Check Mark

Answer to Problem 9.101P

The value of mass flow for adiabatic flow is 6.79 kg/h

Explanation of Solution

Given Information:

The value of temperature at entrance, T0 = 293 K

The value of pressure at entrance, p1 = 102000 Pa

The diameter of tube,D=0.01 m

The value of k for adiabatic flow = 1.4

Concept Used:

p1=ρ1RT1ρ1=densityofairatentranceThevelocityofairatentrance,a0=kRT0fromequation9.75:V1V2=p2p1fromequation9.74:V12=a02(1( V 1 V 2 )2)kfL/D+(k+1)ln( V 2 V 1)m˙=ρ1(π/4×D2)(V1)

Calculation:

p1=RT1ρ1=p1RT=102000287(293)=1.2129kg/m3Thevelocityofairatentrance,a0=1.4×287×293=11772.4=343.11m/sfromequation9.75:V1V2=100000102000=0.9803fromequation9.74:V12=(343.11)2(1(0.9803)2)(1.4)(0.028×3/0.01)+(1.4+1)ln(1.02)=117724.47×0.039(1.4×8.4)+(2.4×0.0198)=4591.2511.76+0.04752=388.84V1=19.72m/sm˙=1.219(π/4×0.012)(19.72)m˙=0.001887kg/s=6.79kg/h

Conclusion:

The value of mass flow for adiabatic flow is 6.79 kg/h.

To determine

(c)

The value of mass flow for incompressible flow.

Expert Solution
Check Mark

Answer to Problem 9.101P

The value of mass flow for incompressible flow is 6.81 kg/h

Explanation of Solution

Given Information:

The value of temperature at entrance, T0 = 293 K

The value of pressure at entrance, p1 = 102000 Pa

The value of pressure at exit, p2 = 10000 Pa

The change in pressure = p1-p2 = 2000 Pa

The diameter of tube, D=0.01 m

Concept Used:

p1=ρ1RT1ρ1=densityofairatentranceThevelocityofairatentrance,V12=Δp(fLD)(ρ1/2)m˙incompressible=ρ1(π/4×D2)(V1)

Calculation:

ρ1=p1RT=102000287(293)=1.2129kg/m3ρ1=densityofairatentranceThevelocityofairatentrance,V12=2000(0.028×3×0.01)(1.2129/2)=20005.12=390.625V1=19.76m/sm˙incompressible=(1.219((π/4×0.012)(19.76)=0.001892kg/s=6.81kg/h

Conclusion:

The value of mass flow for incompressible flow is 6.81 kg/h.

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