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 6, Problem 106P

(a)

To determine

The final pressure in the tank.

(a)

Expert Solution
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Explanation of Solution

Given:

The initial volume of the rigid tank (ν1) is 3-ft3.

The initial temperature of the tank (T1) is 300°F.

The pressure of the vapor in the valve (Pin) is 200 psia.

The temperature of the vapor in the valve (Tin) is 400°F.

The surroundings temperature in the tank (T2) is 300°F.

Calculation:

At the final observation, the valve is closed and the tank composed with one-half water and vapor at the temperature of 300°F. The temperature of the tank is kept constant.

Hence, the pressure (P2) of mixture in the tank at final state is equal to the saturation pressure (Psat) of that mixture.

  P2=Psat@300°F

Refer Table A-4E, “Saturated water-Temperature table”.

The saturation pressure corresponding to the temperature of 300°F is 67.028psia.

Thus, the pressure of the mixture in the tank at the final state is 67.028psia.

(b)

To determine

The amount of steam entered in the tank.

(b)

Expert Solution
Check Mark

Explanation of Solution

It is given that the tank consist of one-half of the volume of the tank is occupied by liquid water.

  νf,2=1.5ft3νg,2=1.5ft3

At the initial state (1):

The tank consist of saturated water vapor (g).

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the initial specific volume (v1=vg) corresponding to the temperature of 300°F.

  vg=v1=6.4663ft3/lbm

At the final state (2):

The tank consist of mixture of vapor (g) and liquid (f).

Refer Table A-4E, “Saturated water-Temperature table” at to the temperature of 300°F.

The final fluid specific volume (vf,2) is 0.01745ft3/lbm.

The final gaseous specific volume (vg,2) is 6.4663ft3/lbm.

Calculate the mass of steam (m1) at initial.

  m1=ν1vg2

  m1=3ft36.4663ft3/lbm=0.4639lbm

At final state, the tank consist of mixture of vapor (g) and liquid (f).

Calculate mass of steam (m2) at final.

  m2=mf,2+mg,2=νf,2vf,2+νg,2vg,2

  m2=1.5ft30.01745ft3/lbm+1.5ft36.4663ft3/lbm=85.9599lbm+0.2319lbm=86.1919lbm

Write the equation of mass balance.

  minme=Δmsystem        (I)

Here, the inlet mass is min, the exit mass is me and the change in mass of the system is Δmsystem.

The change in mass of the system for the control volume is expressed as,

  Δmsystem=(m2m1)cv

Here, the suffixes 1 and 2 indicate the initial and final states of the system.

Consider the given rigid tank as the control volume.

At final state, the valve is close and the steam is not allowed to exit, i.e. me=0.

Rewrite the Equation (I) as follows.

  min0=(m2m1)cvmin=m2m1        (II)

  min=86.1919lbm0.4639lbm=85.7279lbm

Thus, the amount of steam entered in the tank is 85.7279lbm.

(c)

To determine

The amount of the heat transfer.

(c)

Expert Solution
Check Mark

Explanation of Solution

At the line (while entering the tank):

The supply line consist of superheated water vapor.

Refer Table A-6E, “Superheated water”.

Obtain the line enthalpy (hin) corresponding to the pressure of 200psia and the temperature of 400°F.

  hin=1210.9Btu/lbm

At the initial state (1):

The tank consist of saturated water vapor (g).

Refer Table A-4E, “Saturated water-Temperature table”.

Obtain the initial internal energy (u1=ug) corresponding to the temperature of 300°F.

  ug=u1=1099.8Btu/lbm

At the final state (2):

The tank consist of mixture of vapor (g) and liquid (f).

Refer Table A-4E, “Saturated water-Temperature table “at the temperature of 300°F.

The final fluid internal energies (uf,2) is 269.51Btu/lbm.

The final gaseous internal energies (ug,2) is 1099.8Btu/lbm.

At the final state, the tank is composed of vapor and liquid. Hence, the final state energy is expressed as follows.

  m2u2=mf,2uf,2+mg,2ug,2=νf,2vf,2uf,2+νg,2vg,2ug,2=85.9599uf+0.2319ug

  m2u2=85.9599(269.51Btu/lbm)+0.2319(1099.8Btu/lbm)=23167.0527Btu+255.0436Btu=23422.0963Btu

Write the energy balance equation.

  EinEout=ΔEsystem{[Qin+Win+min(h+ke+pe)in][Qe+We+me(h+ke+pe)e]}=[m2(u+ke+pe)2m1(u+ke+pe)1]system        (V)

Here, the heat transfer is Q, the work transfer is W, the enthalpy is h, the internal energy is u, the kinetic energy is ke, the potential energy is pe and the change in net energy of the system is ΔEsystem; the suffixes 1 and 2 indicates the inlet and outlet of the system.

Since the tank is not insulated, the heat transfer occurs through the tank wall. In control volume, there is no work transfer, i.e. (Win=We=0). Neglect the kinetic and potential energy changes i.e. (Δke=Δpe=0). There is no exit for the tank, the exit mass is neglected i.e. (me=0).

The Equation (V) reduced as follows.

  Qin+minhin=m2u2m1u1Qin=m2u2m1u1minhin

  Qin=[23422.0963Btu/lbm(0.4639lbm)(1099.8Btu/lbm)(85.7279lbm)(1210.9Btu/lbm)]=(23422.0963510.1972103807.9141)Btu=80896.0151Btu

Here, the negative sign indicates that the heat transfer occurs from the tank to the surrounding.

  Qout=80896.0151Btu

Thus, the amount of the heat transfer is 80896.0151Btu.

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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