Fundamentals Of Thermal-fluid Sciences In Si Units
Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 6, Problem 129RQ
To determine

The velocity of the nitrogen at the pipe’s inlet and outlet.

Expert Solution & Answer
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Explanation of Solution

Given:

The initial pressure (P1) is 100 psia.

The initial temperature (T1) is 120°F.

The final pressure (P2) is 50 psia.

The final temperature (T2) is 70°F.

Calculation:

Write the formula for mass flow rate of nitrogen gas at initial and final.

  m˙1=A1V1P1RT1        (I)

Write the formula for mass flow rate of nitrogen gas at initial and final.

  m˙2=A2V2P2RT2        (II)

Here, the cross sectional area of the pipe is A, the velocity is V, the gas constant is R and the temperature is T; the subscripts 1 and 2 indicates the inlet and outlet condition of nitrogen gas.

Here, the inlet and exit mass flow rates are equal and  the diameter of the pipe is constant (A1=A2).

Equate the Equations (I) and (II) to obtain the exit velocity V2 .

  A1V1P1RT1=A2V2P2RT2V2=A1V1P1RT1×RT2A2P2=A1P1T2A2P2T1V1=P1T2P2T1V1        (III)

Refer Table A-2E (a), “Ideal-gas specific heats of various common gases”.

The specific heat of nitrogen gas is 0.248Btu/lbmR.

Substitute P1=100psia, T2=70°F, P2=50psia, and T1=120°F in Equation (III).

  V2=(100psia)(70°F)(50psia)(120°F)V1=(100psia)(70+460)R(50psia)(120+460)RV1=5300029000=1.8276V1        (IV)

Consider the nitrogen gas flows at steady state through one inlet and one exit system (pipe and duct flow). Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the energy rate balance equation for one inlet and one outlet system.

  [Q˙1+W˙1+m˙(h1+V122+gz1)][Q˙2+W˙2+m˙(h2+V222+gz2)]=ΔE˙system        (V)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The nitrogen gas flows at steady state through the system. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

Neglect the heat transfer (adiabatic process), work transfer and potential energy changes i.e. W˙1=W˙2=0,Q˙1=Q˙2=0,ΔPE=0.

The Equations (I) reduced as follows.

  [0+0+m˙(h1+V122+0)][0+0+m˙(h2+V222+0)]=0m˙(h1+V122)=m˙(h2+V222)h1+V122=h2+V222V122V222=h2h1        (VI)

The change in enthalpy is expresses as follow.

  h2h1=cp(T2T1)

Here, the specific heat is cp, the exit temperature is T2 and the inlet temperature is T1.

Substitute cp(T2T1) for h2h1 in Equation (II) and rearrange it to obtain V1.

  V122=cp(T2T1)+V222V12=2[cp(T2T1)+V222]V12=2cp(T2T1)+V22        (VII)

Substitute cp=0.248Btu/lbmR, T2=70°F, T1=120°F and V2=1.8276V1 in Equation (VII).

  V12=2(0.248Btu/lbmR)(70°F120°F)+(1.8276V1)2V12=0.496Btu/lbmR[(70+460)R(120+460)R]+3.34V12V12=0.496Btu/lbmR(50R)+3.34V12V12=24.8Btu/lbm+3.34V12

  V123.34V12=24.8Btu/lbm2.34V12=24.8Btu/lbmV1=24.8Btu/lbm2.34×25037ft2/s21Btu/lbmV1=515.1208ft/s

  515ft/s

Thus, the velocity of the nitrogen at the pipe’s inlet is 515ft/s.

Substitute V1=515ft/s in Equation (IV).

  V2=1.8276(515ft/s)=941.4347ft/s941ft/s

Thus, the velocity of the nitrogen at the pipe outlet is 941ft/s.

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Chapter 6 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

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