homework

Name:

HW #07 Due: 03/29/2012

ES 346

Solution Set

5-6 1 of 24

Problem: A steady-flow compressor is used to compress helium from 15 psia and 70°F at the inlet to 200 psia and 600°F at the outlet. The outlet area and velocity are 0.01 ft2 and 100 ft/s, respectively, and the inlet velocity is 50 ft/s. Determine the mass flow rate and the inlet area. Solution: Assume that flow through the compressor is steady and that helium can be treated as an ideal gas under these conditions.

200 psia 600°F 0.01 ft2

From Tbl A-1E: R = 2.6809 psia⋅ft3/lbm⋅°R

Compressor

The mass flow rate is:

m& =

A2V2

v2

=

15 psia 70°F 50 ft/s

A2V2 P2 RT2

(

)

(0.01 ft 2 ) 100 ft (200 psia) s = = 70.38 × 10 −3 lbm 3 s   psia ⋅ ft (600 + 460 )°R  2.6809  lbm ⋅ °R  

Ans

Therefore, the inlet area is:

A1 =

m& v 1 m& RT1 = V1 V1 P1

(70.38 × 10 =

-3

3  lbm  2.6809 psia ⋅ ft (70 + 460 )°R s lbm ⋅ °R  = 0.1333 ft 2 50 ft (15 psia) s

)

(

)

hw07soln.docx

Ans

HW #07 Due: 03/29/2012

Name:

Solution Set

ES 346

5-11 2 of 24

Problem: A cyclone separator like that in Fig. P5–11 is used to remove fine solid particles, such as fly ash, that are suspended in a gas stream. In the flue-gas system of an electrical power plant, the weight fraction of fly ash in the exhaust gases is approximately 0.001. Determine the mass flow rates at the two outlets (flue gas and fly ash) when 10 kg/s of flue gas and ash mixture enters this unit. Also determine the amount of fly ash collected per year.

Solution: Assume flow through the separator is steady. Since the ash particles cannot be converted into gas and vice-versa, the mass flow rate of ash into the control volume must equal that going out, and the mass flow rate of flue gas into the control volume must equal that going out. Use Conservation of Mass to solve this problem. m& ash = y ash m& in = (0.001)10 kg  = 0.01 kg s  s  Ans

m& flue gas = m& in − m& ash = (10 − 0.01) kg

s

= 9.99 kg

s

The amount of fly ash collected per year is:

mash = m& ash ∆t =  0.01 kg s 

 365day  24hr  3600s  = 315.4 × 10 3 kg yr  yr  day  hr 

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-15 3 of 24

Problem: Refrigerant-134a enters a 28-cm diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat as it flows and leaves the pipe at 180 kPa and 40°C. Determine (a) the volume flow rate of the refrigerant at the inlet, (b) the mass flow rate of the refrigerant, and (c) the velocity and volume flow rate at the exit. Solution: A diagram of the system is: Q

R-134a 200 kPa 20°C 5 m/s

180 kPa 40°C

From Tbl A-13:

P1 = 200 kPa  3 v 1 = 0.1142 m kg T1 = 20°C 

P2 = 180 kPa  3 v 2 = 0.1374 m kg T2 = 40°C 

The volume flow rate at the inlet and the mass flow rate can be determined by:

V&1 = AcV1 = m& =

1

v1

πD 2 4

π (0.28 m) 2

V1 =

4

(5 m s ) = 0.3079 m

2

AcV1 =

3

s

Ans 2

1 πD 1 π (0.28 m) V1 = 5 m = 2.696 kg s s 3 v1 4 4  0.1142 m    kg  

(

)

Noting that mass flow rate is constant, the volume flow rate and the velocity at the exit of the pipe are: 3 3 V&2 = m& v 2 =  2.696 kg s  0.1374 m kg  = 0.3704 m s



V2 =

V&2 Ac



0.3705 m =



3

s = 6.016 m

π (0.28 m) 2

Ans

s

4

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-32 4 of 24

Problem: The stators in a gas turbine are designed to increase the kinetic energy of the gas passing through them adiabatically. Air enters a set of these nozzles at 300 psia and 700°F with a velocity of 80 ft/s and exits at 250 psia and 645°F. Calculate the velocity at the exit of the nozzles. Solution: Assume that this is a steady-flow process since there is no change with time and that we can treat air is an ideal gas with constant specific heats. Also, assume that the potential energy changes are negligible, there are no work interactions, and that the nozzle is adiabatic. From Tbl A-2E(b): (700 + 645)°F ⇒ c = 0.253 Btu Tavg = p lbm ⋅ ° R 2 Using the Conservation of Mass:

m& 1 = m& 2 = m& The energy balance for the steady-flow system (nozzle) can be expressed in the rate form as:

E& in − E& out 14 24 3 Rate of net energy transfer by heat, work, and mass

=

0 (steady)

∆E& system 144244 3

=0

Rate of change in internal, kinetic, potential, etc. energies

E& in = E& out   V1 2  V22    & & m h1 + = m h2 + 2  2   V2 V2 h1 + 1 = h2 + 2 2 2

  

300 psia 700°F 80 ft/s

AIR

250 psia 645°F

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-32 5 of 24

Solving for exit velocity:

[

V2 = V1 2 + 2(h1 − h2 )

]

0.5

[

= V1 2 + 2c p (T1 − T2 )

]

0.5

  25,037 ft 2  2   s 2  =  80 ft + 2 0.253 Btu (700 − 645)°R   s lbm ⋅ °R Btu   lbm   = 838.6 ft s

(

) (

0.5

)

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-34 6 of 24

Problem: Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and 200 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm2, determine the velocity and the volume flow rate of the steam at the nozzle exit.

Solution: Assume this is a steady-flow process since there is no change with time and that the potential energy change is negligible. Also, assume that there are no work interactions.

400°C 800 kPa 10 m/s

Q

The steam in the control volume of the nozzle is the system. The energy balance for this steady-flow system can be expressed in the rate form as:


=

0 (steady)

∆E& system 144244 3

300°C 200 kPa

STEAM

=0


E& in = E& out  V1 2  & m h1 + 2 

  V2  = m&  h2 + 2  + Q& out 2    V1 2 V22 Q& out = h2 + + h1 + 2 2 m&

or

since W& ≅ ∆pe ≅ 0)

From Tbl A-6:

P1 = 800 kPa v 1 = 0.38429 m 3 /kg  T1 = 400°C  h1 = 3267.7 kJ/kg P2 = 200 kPa v 2 = 1.31623 m 3 /kg  T2 = 300°C  h2 = 3072.1 kJ/kg

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-34 7 of 24

Calculating the mass flow rate of the steam:

m& =

1

v1

A1V1 =

1 3 0.38429 m

(0.08 m 2 )(10 m ) = 2.082 kg s s s

(10 m s )  2

⇒ 3267.7 kJ = 3072.1 kJ

kg

+

1 kJ/kg   2 2   1000 m /s 

2

25 kJ V22  1 kJ/kg  s + + kg 2  1000 m 2 /s 2  2.082 kg

⇒ V2 = 606.0 m

s Ans

s

The volume flow rate at the exit of the nozzle is: 3 3 V&2 = m& v 2 =  2.082 kg s 1.31623 m kg  = 2.74 m s





Ans



hw07soln.docx

Name:

HW #07 Due: 03/29/2012

Solution Set

ES 346

5-47 8 of 24

Problem: Refrigerant-134a enters a compressor at 100 kPa and –24°C with a flow rate of 1.35 m3/min and leaves at 800 kPa and 60°C. Determine the mass flow rate of R134a and the power input to the compressor. Solution: Assume that this is a steady-flow process since there is no change with time. Also assume that the kinetic and potential energy changes are negligible. Take the compressor as the system. Noting that one fluid stream enters and leaves the compressor, the energy balance for this steady-flow system can be expressed in the rate form as: E& in − E& out 14 24 3

=

Rate of net energy transfer by heat, work, and mass

0 (steady)

∆E& system 1442443

=0


E& in = E& out W& in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0) W& in = m& ( h2 − h2 )

800 kPa 60°C Compressor

100 kPa -24°C 1.35 m3/min

From Tbl A -13:

P1 = 100 kPa  h1 = 232.42 kJ/kg  T1 = −24°C  v1 = 0.19019 m 3 /kg P2 = 800 kPa  h2 = 296.81 kJ/kg T2 = 60°C  Find the mass flow rate:

(

)

3  min  1.35 m   & min V1 60 s   m& = = = 0.1183 kg s v1 0.19019 m 3 /kg

Ans

Substituting into the energy balance:  kW ⋅ s  W& in = m& ( h2 − h1 ) = (0.1183 kg/s)(296. 81 − 232.42)kJ/kg  = 7.62kW  kJ 

hw07soln.docx

Ans

HW #07 Due: 03/29/2012

Name:

Solution Set

ES 346

5-49 9 of 24

Problem: Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 6 MPa, 400°C, and 80 m/s, and the exit conditions are 40 kPa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 20 kg/s. Determine (a) the change in kinetic energy, (b) the power output, and (c) the turbine inlet area.

Solution: Assume that this is a steady-flow process since there is no change with time. Assume that potential energy changes are negligible. Since the device is adiabatic, the heat transfer is negligible. From the steam tables (Tbl A-4 thru A-6):

P1 = 6 MPa v 1 = 0.04742 m 3 /kg  T1 = 400°C  h1 = 3178.3 kJ/kg P2 = 40 kPa   h2 = h f + x 2 h fg = 317.62kJ/kg + (0.92)(2318.4kJ/kg ) = 2450.5 kJ/kg x 2 = 0.92  (a) The change in kinetic energy is determined from:

∆ke =

V22 − V1 2 (50 m/s )2 − (80 m/s) 2 = 2 2

 1 kJ/kg  2 2  1000 m /s

  = − 1.95 kJ kg 

Ans (a)

hw07soln.docx

Name:

HW #07 Due: 03/29/2012 (b)

5-49

Solution Set 10 of 24

ES 346

There is only one inlet and one exit, therefore m& 1 = m& 2 = m& . Take the turbine as the system. The energy balance for this steady-flow system can be expressed in the rate form as:

E& in − E& out 14 24 3

=


0 (steady)

∆E& system 144244 3

=0


E& in = E& out  V2 m&  h1 + 1 2 

 &  V2 & ≅ ∆pe ≅ 0)  = Wout + m&  h2 + 2  (since Q 2   

 V 2 − V1 2 W& out = − m&  h2 − h1 + 2 2 

  

The power output of the turbine is determined by substitution to be:  kW ⋅ s  3 W& out = −(20 kg/s )( 2450.5 − 3178.3 − 1.95)kJ/kg  = 14.60 × 10 kW = 14.6MW  kJ 

(c)

Ans (b)

Finally, the inlet area of the turbine is determined from the mass flow rate relation:

m& =

1

v1

⇒ A1 =

A1V1 m& v 1 (20 kg/s )(0.04742 m 3 /kg ) = = 0.0119m 2 = 119cm 2 V1 80 m/s

Ans (c)

hw07soln.docx

HW #07 Due: 03/29/2012

Name:

ES 346

5-59


Problem: Steam enters a steady-flow turbine with a mass flow rate of 20 kg/s at 600°C, 5 MPa, and a negligible velocity. The steam expands in the turbine to a saturated vapor at 500 kPa where 10 percent of the steam is removed for some other use. The remainder of the steam continues to expand to the turbine exit where the pressure is 10 kPa and quality is 85 percent. If the turbine is adiabatic, determine the rate of work done by the steam during this process.

Solution: Assume that this is a steady-flow process since there is no change with time. Also assume that the kinetic and potential energy changes are negligible. The turbine is adiabatic so the heat transfer is negligible. From the steam tables (Tbl A-5 and A-6): P1 = 5 MPa   h1 = 3666.9 kJ/kg T1 = 600°C  P2 = 0.5 MPa   h2 = 2748.1 kJ/kg Sat.Vap.  P3 = 10 kPa  h3 = h f + xh fg  x 2 = 0.85  = 191.81kJ/kg + (0.85)(2392.1kJ/kg) = 2225.1 kJ/kg

Take the entire turbine, including the connection part between the two stages, as the system. Noting that one fluid stream enters the turbine and two fluid streams leave, the energy balance for this steady-flow system can be expressed in rate form as:

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

ES 346


=

0 (steady)

∆E& system 1442443

5-59


=0


E& in = E& out m& 1 h1 = m& 2 h2 + m& 3 h3 + W& out W& out = m& 1 (h1 − 0.1h2 − 0.9h3 )

Substituting our values, the power output of the turbine is: W& out = m& 1 (h1 − 0.1h2 − 0.9h3 )  kW ⋅ s  = (20 kg/s)[3666.9kJ/kg − (0.1)(2748.1kJ/kg ) − (0.9 )(2225.1kJ/kg )]    kJ  = 27.79 × 10 3 kW = 27.79 MW

hw07soln.docx

Ans

Name:

HW #07 Due: 03/29/2012

5-77


ES 346

Problem: Water at 65°F and 20 psia is heated in a chamber by mixing it with saturated water vapor at 20 psia. If both streams enter the mixing chamber at the same mass flow rate, determine the temperature and the quality of the exiting stream. Solution: Assume this is a steady-flow process since there is no change with time and that kinetic and potential energy changes are negligible. Also, assume that there are no work interactions and that the device is adiabatic and thus heat transfer is negligible. From Tbl A-5E thru A-6E: h1 ≅ hf @ 65°F = 33.08 Btu/lbm h2 = hg @ 20 psia = 1156.2 Btu/lbm This is an adiabatic mixing process as shown in the diagram. The mass balance is: T1 = 65°F H2O (P = 20 psia) T3, x3

0 (steady) m& in − m& out = ∆m& system =0

m& in = m& out m& 1 + m& 2 = m& 3 = 2m& m& 1 = m& 2 = m&

Energy balance is: E& in − E& out = 14 24 3 Rate of net energy transfer by heat, work, and mass

Sat. vapor · · m2 = m1 0 (steady)

∆E& system 1442443

=0


E& in = E& out m& 1 h1 + m& 2 h2 = m& 3 h3 (since Q& = W& = ∆ke ≅ ∆pe ≅ 0)

⇒ m& h1 + m& h2 = 2m& h3 ⇒ h3 =

(h1 + h2 ) 2

Substituting our known values:

h3 =

(33.08 + 1156.2) Btu lbm 2

= 594.6 Btu

lbm

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

ES 346

5-77


From Tbl A-5: h f = 196.27 Btu lbm   P3 = 20 psia ⇒ h fg = 959.93 Btu lbm  T3 = Tsat = 227.92° F  x3 =

h3 − h f h fg

Ans

(594.6 − 196.27 ) Btu lbm = = 0.415 (1156.2 − 196.27 ) Btu lbm

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-84


ES 346

Problem: A thin-walled double-pipe counter-flow heat exchanger is used to cool oil (cp = 2.20 kJ/kg·°C) from 150 to 40°C at a rate of 2 kg/s by water (cp = 4.18 kJ/kg·°C) that enters at 22°C at a rate of 1.5 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of water.

Solution: Assume steady operating conditions exist and that the heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Also, assume that changes in the kinetic and potential energies of fluid streams are negligible and that the fluid properties are constant. We take the oil tubes as the system. The energy balance for this steady-flow system can be expressed in the rate form as:


=

0 (steady)

∆E& system 144244 3

Hot oil 150°C 2 kg/s

=0

Cold water 22°C 1.5 kg/s


E& in = E& out m& h1 = Q& out + m& h2 (since ∆ke ≅ ∆pe ≅ 0) Q& out = m& c p (T1 − T2 )

40°C

The rate of heat transfer from the oil is:

Q& = [m& c p (Tin − Tout )]oil

(150°C − 40°C) kW ⋅ s  = 484.0kW =  2 kg  2.2 kJ s  kg.°C    kJ 

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-84


ES 346

Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is:

Q& = [m& c p (Tout − Tin )] water ⇒ Tout = Tin +

Q& m& water c p

484 kJ

= 22°C +

s = 99.2°C 1.5 kg  4.18 kJ  s  kg.°C  

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-99


ES 346

Problem: The fan on a personal computer draws 0.5 ft3/s of air at 14.7 psia and 70°F through the box containing the CPU and other components. Air leaves at 14.7 psia and 80°F. Calculate the electrical power, in kW, dissipated by the PC components.

Solution: Assume this is a steady-flow process since there is no change with time and that the kinetic and potential energy changes are negligible. Also, assume that all the heat dissipated by the circuits is picked up by the air drawn by the fan. Finally, assume that air can be assumed to be an ideal gas with constant properties under the conditions of this problem. From Tbl A-1E & A-2Ea: R = 0.3704 psia·ft3/lbm·°R cp = 0.240 Btu/lbm·°F The system can be modeled as shown in the diagram. Consider the air to be the system. The energy balance for this steady-flow system can be expressed in the rate form as:


=

0 (steady)

∆E& system 144244 3

. We,in

=0


E& in = E& out m& h1 + W& e,in = m& h2 W& e,in = m& (h2 − h1 ) W& e,in = m& c p (T2 − T1 )

14.7 psia, 70°F 0.5 ft3/s

Air

14.7 psia 80°F

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-99


ES 346

The inlet specific volume and the mass flow rate of air are:   psia ⋅ ft 3 (70 + 460)°R  0.3704 lbm ⋅ °R  3 RT1  v1 = = = 13.35 ft lbm P1 14.7 psia 3 0.5 ft V&1 s = 0.03745 lbm m& = = s v 1 13.35 ft 3 lbm

Substituting into the energy balance:

(

W& e,out = 0.03745 lbm

s

)(

0.240 Btu

  3600 s 1 kW    (80 − 70)° R    lbm ⋅ °R  3412.14 Btu  hr   hr 

)

= 94.83 × 10 −3 kW

Ans

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-110


ES 346

Problem: The hot-water needs of a household are to be met by heating water at 55°F to 180°F by a parabolic solar collector at a rate of 4 lbm/s. Water flows through a 1.25-in diameter thin aluminum tube whose outer surface is black-anodized in order to maximize its solar absorption ability. The centerline of the tube coincides with the focal line of the collector, and a glass sleeve is placed outside the tube to minimize the heat losses. If solar energy is transferred to water at a net rate of 400 Btu/h per ft length of the tube, determine the required length of the parabolic collector to meet the hot-water requirements of this house. Solution: Assume steady operating conditions exist and that heat loss from the tube is negligible (as stated) so that the entire solar energy incident on the tube is transferred to the water. Also, assume that the kinetic and potential energy changes are negligible. From Tbl A-2E: cp = 1.00 Btu/lbm.°F Consider the thin aluminum tube to be the system. The energy balance for this steady-flow system can be expressed in the rate form as: E& in − E& out 14 24 3

0 (steady)

∆E& system 144244 3

=



E& in = E& out

=0

2

1 Water

Q& in + m& h1 = m& h2 (since ∆ke ≅ ∆pe ≅ 0) 55°F 4 lbm/s Q& in = m& water c p (T2 − T1 )

180°F

The total rate of heat transfer to the water flowing through the tube is: Q& total = m& c p (Te − Ti )

(

= 4 lbm

s

)(1.00 Btu lbm.°F)(180 − 55)°F = 500 Btu/s = 1,800,000 Btu/h

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

ES 346

5-110


We can calculate the collector length required from the information given in the problem:

L=

Btu Q& total 1,800,000 hr = 4500 ft = & Btu Q 400 hr ⋅ ft

Ans

Note that I used specific heat to calculate the solution rather than using the hf values from the steam tables. The example problem from class suggests that for liquid water at relatively low temperatures (such as this problem) little error should result. Let’s check this. From Tbl A-4E: h f @ 55° F = 23.07 Btu h f @ 180° F

(

⇒ Q& = 4 lbm

s

lbm = 148.04 Btu lbm

)(148.04 − 23.07) Btu lbm = 499.9 Btu s = 1,800,000 Btu hr

No significant difference.

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

5-125


ES 346

Problem: An air-conditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R-134a at 24°C. The valve connecting this container to the air-conditioning system is now opened until the mass in the container is 0.25 kg, at which time the valve is closed. During this time, only liquid R-134a flows from the container. Presuming that the process is isothermal while the valve is open, determine the final quality of the R-134a in the container and the total heat transfer.

Solution: Assume this is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remains constant. Also assume that the kinetic and potential energies are negligible and that there are no work interactions involved. Consider the refrigerant in the tank as the system. Mass balance: min − mout = ∆msystem

A-C line

− me = m2 − m1 me = m1 − m2

Energy balance: E in − E out 14 24 3 Net energy transfer by heat, work, and mass

=

∆E system 1 424 3 Change in internal, kinetic, potential, etc. energies

Liquid R-134a 5 kg 24°C

Qin − me he = m2 u 2 − m1u1 Qin = m2 u 2 − m1u1 + me he

Combining: Qin = m2 u 2 − m1u1 + (m1 − m2 )he

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

ES 346

5-125


From Tbl A-11: 3

v 1 = 0.0008261 m kg

T1 = 24°C  kJ  u1 = 84.44 kg x=0  he = 84.98 kJ kg

The volume of the tank is:

V = m1v 1 = (5 kg) 0.0008261 m kg  = 0.004131 m 3   3

We can now calculate the final specific volume:

v2 =

V m2

=

3 0.004131 m 3 = 0.01652 m kg 0.25 kg

Returning to Tbl A-11:

v2 −v f 0.01652 − 0.0008261 = = 0.5061  x2 =  v g − v f 0.031834 − 0.0008261 3  v 2 = 0.01652 m kg   u 2 = u f + x 2 u fg = 84.44 kJ kg + (0.5061)158.65 kJ kg  = 164.73 kJ kg   T2 = 24°C

Ans

Substituting into the energy balance: Qin = m2 u 2 − m1u1 + (m1 − m2 )he = (0.25 kg )164.73 kJ  − (5 kg ) 84.44 kJ  + (4.75 kg ) 84.98 kJ  kg  kg  kg     = 22.64kJ

hw07soln.docx

Ans

HW #07 Due: 03/29/2012

Name:

ES 346

5-139


Problem: The air in an insulated, rigid compressed-air tank whose volume is 0.5 m3 is initially at 4000 kPa and 20°C. Enough air is now released from the tank to reduce the pressure to 2000 kPa. Following this release, what is the temperature of the remaining air in the tank?

Solution: Assume this is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process. Also assume that air is an ideal gas with constant specific heats under the conditions of this problem, that the kinetic and potential energies are negligible, and that there are no work interactions involved. From Tbl A-1 & A-2a: R = 0.287 kPa·m3/kg·K cp = 1.005 kJ/kg·K cv = 0.718 kJ/kg·K Consider the air in the tank as the system. Mass balance:

Air 4000 kPa 20°C, 0.5 m3

min − mout = ∆msystem − me = m2 − m1 me = m1 − m2

hw07soln.docx

Name:

HW #07 Due: 03/29/2012

ES 346

5-139


Energy balance: E in − E out 14 24 3

=

Net energy transfer by heat, work, and mass

∆E system 1 424 3 Change in internal, kinetic, potential, etc. energies

− me he = m2 u 2 − m1u1 0 = m2 u 2 − m1u1 + me he 0 = m2 cv T2 − m1cv T1 + me c p Te

Combining: 0 = m 2 cv T2 − m1cv T1 + (m1 − m 2 )c p Te

The initial and final masses are:

m1 =

P1V (4000 kPa )(0.5 m 3 ) = = 23.78 kg RT1  0.287 kPa ⋅ m 3  (20 + 273) K  kg ⋅ K  

m2 =

P2V (2000 kPa )(0.5 m 3 ) 3484 = = kg RT2  0.287 kPa ⋅ m 3 T  2 T  kg ⋅ K  2 

The temperature of air leaving the tank changes from the initial temperature in the tank to the final temperature during the discharging process. We assume that the temperature of the air leaving the tank is the average of initial and final temperatures in the tank. Substituting into the energy balance equation: 0 = m2 cv T2 − m1cv T1 + (m1 − m2 )c p Te  3484   T − (23.78kg ) 0.718 kJ (293K ) kg  0.718 kJ 0 =   kg ⋅ K  2 kg ⋅ K    T  2   3484    293 + T2  kg 1.005 kJ +  23.78 −  kg ⋅ K  T2   2   ⇒ T2 = 241K = −32°C

Ans

hw07soln.docx

homework

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