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 160RQ
To determine

The heat transfer with the surrounding that is needed to maintain the temperature and pressure of R-134a constant.

Expert Solution & Answer
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Explanation of Solution

Given:

The bottle internal volume (v) of 0.0015 m3.

The initial mass of the refrigerant (m1) is 0.55 kg.

The temperature of the refrigerant (T1) is 26°C.

The final mass of the refrigerant (m2) is 0.15 kg.

Calculation:

At the initial state 1:

The rigid container consist of saturated mixture refrigerant at 26°C.

Refer Table A-11, “Saturated refrigerant-134a-Temperature table”.

Obtain the following corresponding to the temperature of 26°C.

  vf,1=0.0008312m3/kgvg,1=0.030008m3/kguf,1=87.26kJ/kgufg,1=156.89kJ/kg

The quality of the refrigerant at state 1 is expressed as follows.

  x1=v1vf,1vg,1vf,1

  x1=0.002727m3/kg0.0008312m3/kg0.030008m3/kg0.0008312m3/kg=0.00189580.0291768=0.064976

The internal energy of the refrigerant at state 1 is expressed as follows.

  u1=uf,1+x1ufg,1

  u1=87.26kJ/kg+(0.064976)(156.89kJ/kg)=87.26kJ/kg+10.194085kJ/kg=97.4541kJ/kg97.45kJ/kg

At the final state (2):

When the valve is opened, the vapor refrigerant only allowed to escape and the temperature is kept constant.

The final temperature of the refrigerant is also 26°C.

Refer Table A-12, “Saturated refrigerant-134a-Temperature table”.

Obtain the following corresponding to the temperature of 26°C.

  vf,2=0.0008312m3/kgvg,2=0.030008m3/kguf,2=87.26kJ/kgufg,2=156.89kJ/kg

The quality of the refrigerant at state 2 is expressed as follows.

  x2=v2vf,2vg,2vf,2

  x2=0.01m3/kg0.0008312m3/kg0.030008m3/kg0.0008312m3/kg=0.00916880.0291768=0.31425

The internal energy of the refrigerant at state 2 is expressed as follows.

  u2=uf,2+x2ufg,2

  u2=87.26kJ/kg+(0.31425)(156.89kJ/kg)=87.26kJ/kg+49.3027kJ/kg=136.5627kJ/kg136.56kJ/kg

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 subscripts 1 and 2 indicates the initial and final states of the system.

Consider the given rigid container as the control volume.

Initially the container is filled with liquid refrigerant and the valve is in closed position, further no other mass is allowed to enter the container. Hence, the inlet mass is neglected i.e. min=0.

Rewrite the Equation (I) as follows.

  0me=(m2m1)cvme=m1m2

  me=0.55kg0.15kg=0.40kg

Write the formula for initial specific volume (v1).

  v1=νm1

  v1=0.0015m30.55kg=0.002727m3/kg

Write the formula for final specific volume (v2).

  v2=νm2

  v2=0.0015m30.15kg=0.01m3/kg

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        (II)

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.

The process is maintained at isothermal at the open condition of valve, there is no heat transfer while the mass leaves the container .i.e. Qe=0. 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 mass inlet for the rigid container, the inlet mass is neglected i.e. (min=0).

The Equation (V) reduced as follows.

  Qinmehe=m2u2m1u1Qin=m2u2m1u1+mehe        (III)

Here, the temperature is kept constant until the final state and the vapor only exits the tank. Hence the exit enthalpy is expressed as follows.

  he=hg@26°C

Refer Table A-12, “Saturated refrigerant-134a-Temperature table”.

The exit enthalpy (he) corresponding to the temperature of 26°C is 264.73kJ/kg.

Substitute m2=0.15kg, u2=136.56kJ/kg, m1=0.55kg, u1=97.45kJ/kg, me=0.40kg, and he=264.73kJ/kg in Equation (III).

  Qin=[(0.15kg)(136.56kJ/kg)(0.55kg)(97.45kJ/kg)+(0.40kg)(264.73kJ/kg)]=(20.48453.5975+105.892)kJ=72.7785kJ72.8kJ

Thus, the heat transfer with the surrounding that is needed to maintain the temperature and pressure of R-134a constant is 72.8kJ.

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Chapter 6 Solutions

Fundamentals Of Thermal-fluid Sciences In Si Units

Ch. 6 - Prob. 11PCh. 6 - Prob. 12PCh. 6 - Prob. 13PCh. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - Prob. 16PCh. 6 - A house is maintained at 1 atm and 24°C, and warm...Ch. 6 - Prob. 18PCh. 6 - Prob. 19PCh. 6 - Prob. 20PCh. 6 - Prob. 21PCh. 6 - The kinetic energy of a fluid increases as it is...Ch. 6 - Prob. 23PCh. 6 - Air enters a nozzle steadily at 50 psia, 140°F,...Ch. 6 - Prob. 25PCh. 6 - Prob. 26PCh. 6 - Air at 600 kPa and 500 K enters an adiabatic...Ch. 6 - Prob. 28PCh. 6 - Prob. 29PCh. 6 - Air at 13 psia and 65°F enters an adiabatic...Ch. 6 - Prob. 31PCh. 6 - Prob. 32PCh. 6 - Prob. 33PCh. 6 - Steam at 4 MPa and 400°C enters a nozzle steadily...Ch. 6 - Prob. 35PCh. 6 - Prob. 36PCh. 6 - Prob. 37PCh. 6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Helium is to be compressed from 105 kPa and 295 K...Ch. 6 - Carbon dioxide enters an adiabatic compressor at...Ch. 6 - Air is compressed from 14.7 psia and 60°F to a...Ch. 6 - Prob. 47PCh. 6 - An adiabatic gas turbine expands air at 1300 kPa...Ch. 6 - Steam flows steadily into a turbine with a mass...Ch. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Refrigerant-134a is throttled from the saturated...Ch. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - A thin-walled double-pipe counter-flow heat...Ch. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - The components of an electronic system dissipating...Ch. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - Prob. 93PCh. 6 - A house has an electric heating system that...Ch. 6 - Prob. 95PCh. 6 - Refrigerant-134a enters the condenser of a...Ch. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 100PCh. 6 - Air enters the duct of an air-conditioning system...Ch. 6 - Prob. 102PCh. 6 - A rigid, insulated tank that is initially...Ch. 6 - Prob. 105PCh. 6 - Prob. 106PCh. 6 - Prob. 107PCh. 6 - Prob. 108PCh. 6 - Prob. 109PCh. 6 - An air-conditioning system is to be filled from a...Ch. 6 - Prob. 111PCh. 6 - A 0.06-m3 rigid tank initially contains...Ch. 6 - A 0.3-m3 rigid tank is filled with saturated...Ch. 6 - Prob. 114PCh. 6 - A 0.3-m3 rigid tank initially contains...Ch. 6 - Prob. 116PCh. 6 - Prob. 117PCh. 6 - An insulated 40-ft3 rigid tank contains air at 50...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - A vertical piston–cylinder device initially...Ch. 6 - The air in a 6-m × 5-m × 4-m hospital room is to...Ch. 6 - Prob. 124RQCh. 6 - Prob. 125RQCh. 6 - Prob. 126RQCh. 6 - Prob. 127RQCh. 6 - Prob. 128RQCh. 6 - Prob. 129RQCh. 6 - Prob. 130RQCh. 6 - Prob. 131RQCh. 6 - Prob. 132RQCh. 6 - Steam enters a nozzle with a low velocity at 150°C...Ch. 6 - Prob. 134RQCh. 6 - Prob. 135RQCh. 6 - Prob. 136RQCh. 6 - In large steam power plants, the feedwater is...Ch. 6 - Prob. 138RQCh. 6 - Prob. 139RQCh. 6 - Prob. 140RQCh. 6 - Prob. 141RQCh. 6 - Prob. 142RQCh. 6 - Prob. 143RQCh. 6 - Prob. 144RQCh. 6 - Prob. 145RQCh. 6 - Prob. 146RQCh. 6 - Repeat Prob. 6–146 for a copper wire ( = 8950...Ch. 6 - Prob. 148RQCh. 6 - Prob. 149RQCh. 6 - Prob. 150RQCh. 6 - Prob. 151RQCh. 6 - Prob. 152RQCh. 6 - Prob. 153RQCh. 6 - An adiabatic air compressor is to be powered by a...Ch. 6 - Prob. 156RQCh. 6 - Prob. 157RQCh. 6 - Prob. 158RQCh. 6 - Prob. 159RQCh. 6 - Prob. 160RQCh. 6 - Prob. 161RQCh. 6 - Prob. 162RQCh. 6 - Prob. 163RQCh. 6 - Prob. 164RQCh. 6 - Prob. 166RQCh. 6 - Prob. 167RQCh. 6 - Prob. 168RQCh. 6 - Prob. 169RQ
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