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First Law for Transient Open Systems - Thermodynamics - Solved Quiz, Exercises of Thermodynamics

Assam Agricultural UniversityThermodynamics

Some of the topic in thermodynamics are: Property tables and ideal gases, First law for closed and open (steady and unsteady) systems, Entropy and maximum work calculations, Isentropic efficiencies, Cycle calculations (Rankine, refrigeration, air standard) with mass flow rate ratios. This quiz is about: First Law for Transient Open Systems, Pressure, Inlet Conditions, Kinetic and Potential Energy, Specific Volumes, First Law and Mass Balance, Heat Transfer

Typology: Exercises

2013/2014

Uploaded on 02/01/2014

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Solution First Law for Transient Open Systems
An evaporator in a home refrigerator has a
volume of 0.8 ft3, and operates at a constant
pressure of 20 psia. The inlet to the
evaporator is R-134a at a pressure of 20 psia
and a quality of 40%. The evaporator exit is
saturated R-134a vapor (quality = 100%) at a
pressure of 20 psia. Initially the evaporator
contains 0.4 lbm of R-134a. Compute the heat
transfer if 0.01 lbm of R-134a enters the
evaporator (at the inlet conditions) and the
amount of saturated R-134a vapor leaving the
evaporator outlet is 0.03 lbm.
The system boundary, defined by the dashed line
in the figure, has one inlet and one outlet. In general, we have an unsteady problem because
more mass enters the system than leaves it. The general first law equation for unsteady open
systems is shown below.
inl et i
i
ii
outlet i
i
ii
u
system
gz
V
hmgz
V
hm
WQgz
V
umgz
V
um
22
22
22
1
2
1
2
2
2
We see that there is no mechanism for useful work in this system so we set Wu = 0 and make the
usual assumption that kinetic and potential energy terms are zero. This gives the following
expression for the first law for our system with one inlet and one outlet.
outoutinin
system hmhmQumumumum 11221122
Solving this equation for the heat transfer gives:
ininoutout hmhmumumQ 1122
. We
can simplify the general the mass balance equation for this problem where there is only one
outlet and one inlet. This gives the following result.
outin
outlet i
inlet i
system mmmmmmmm 1212
We are given that m1 = 0.4 lbm, min = 0.01 lbm, and mout = 0.03 lbm. Thus we can calculate the
final mass remaining in the evaporator, m2, from the mass-balance equation.
mmmmmoutin lblblblblbmmmm 38.003.001.04.0
12
.
From the data that Pin = Pout = 20 psia, xin =40%, and the outlet state is saturated vapor, we can
find the values of hin = hf(20 psia a) + 0.4hfg(20 psia) = 75.14 Btu/lbm + 0.4(91.29 Btu/lbm) =
111.66 Btu/lbm. Since the outlet is saturated vapor we have hout = hg(20 psia) = 166.42 Btu/lbm.
Inflow: 0.01 lbm at
20 psia, x = 40%
Condenser
V = 100 ft3
P = 2 psia =
P1 = P2
m1 = 50 lbm
Q
Outflow
0.03 lbm
20 psia.
sat vap,
Evaporator
V = 0.8 ft3
Pin = Pout = P1 =
P2 = 20 psia
m1 = 0.4 lbm
pf2

Partial preview of the text

Download First Law for Transient Open Systems - Thermodynamics - Solved Quiz and more Exercises Thermodynamics in PDF only on Docsity!

Solution First Law for Transient Open Systems

An evaporator in a home refrigerator has a

volume of 0.8 ft^3 , and operates at a constant

pressure of 20 psia. The inlet to the

evaporator is R-134a at a pressure of 20 psia

and a quality of 40%. The evaporator exit is

saturated R-134a vapor (quality = 100%) at a

pressure of 20 psia. Initially the evaporator

contains 0.4 lbm of R-134a. Compute the heat

transfer if 0.01 lbm of R-134a enters the

evaporator (at the inlet conditions) and the

amount of saturated R-134a vapor leaving the

evaporator outlet is 0.03 lbm.

The system boundary, defined by the dashed line

in the figure, has one inlet and one outlet. In general, we have an unsteady problem because

more mass enters the system than leaves it. The general first law equation for unsteady open

systems is shown below.

inlet

i

i i i outlet

i

i i i

u

system

gz

V

gz m h

V

m h

gz Q W

V

gz m u

V

m u

2 2

1

2

1

2

2

2

We see that there is no mechanism for useful work in this system so we set Wu = 0 and make the

usual assumption that kinetic and potential energy terms are zero. This gives the following

expression for the first law for our system with one inlet and one outlet.

m 2 u 2  m 1 u 1 system m 2 u 2 m 1 u 1 Qminhinmouthout

Solving this equation for the heat transfer gives: Q  m 2 u 2 m 1 u 1 mout houtminhin. We

can simplify the general the mass balance equation for this problem where there is only one

outlet and one inlet. This gives the following result.

  in out

outlet

i inlet

m 2  m 1 system  mim  m 2 m 1 m m

We are given that m 1 = 0.4 lbm, min = 0.01 lbm, and mout = 0.03 lbm. Thus we can calculate the

final mass remaining in the evaporator, m 2 , from the mass-balance equation.

m 2 m 1 min mout 0. 4 lbm 0. 01 lbm 0. 03 lbm 0. 38 lbmlbm.

From the data that Pin = Pout = 20 psia, xin =40%, and the outlet state is saturated vapor, we can

find the values of hin = hf(20 psia a) + 0.4hfg(20 psia) = 75.14 Btu/lbm + 0.4(91.29 Btu/lbm) =

111.66 Btu/lbm. Since the outlet is saturated vapor we have hout = hg(20 psia) = 166.42 Btu/lbm.

Inflow: 0.01 lbm at 20 psia, x = 40%

Condenser V = 100 ft^3 P = 2 psia = P 1 = P 2 m 1 = 50 lbm

Q

Outflow

0.03 lbm 20 psia. sat vap,

Evaporator V = 0.8 ft^3 Pin = Pout = P 1 = P 2 = 20 psia m 1 = 0.4 lbm

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We can use the initial and final mass to compute the initial and final specific volumes, v 1 and v 2 ;

we can use these specific volumes and the given condenser pressure, P 1 = P 2 = 20 psia to find

the initial and final internal energy for the condenser.

m m m lbm

ft

lb

ft v lb

ft

lb

ft v

3 3

2

3 3

1

From the table we see that both specific volumes are between the saturated liquid and saturated

vapor at 20 psia, so all data are in the mixed region. We have to use the quality to compute the

internal energy from the equations v = vf + xvfg and u = uf + xufg. Eliminating x from these two

equations gives the following equation for the enthalpy as a function of the specific volume, using

the data for the given initial and final pressure of 20 psia.

m m

m

m m

m

fg fg

f f

m m

m

m m

m

fg fg

f f

lb

Btu

lb

Btu

lb

ft

lb

ft

lb

ft

lb

Btu u v

v v u u

lb

Btu

lb

Btu

lb

ft

lb

ft

lb

ft

lb

Btu u v

v v u u

3

3 3

3

3 3

2 2

1 1

We now have all the information required to compute the heat transfer from the combination of

the first law and mass balance derived above,

 

      

m

m m

m m

m

m

out out in in m

lb

Btu lb lb

Btu lb lb

Btu lb

lb

Btu Q mu mu m h m h lb

2 2 11 0.^38

.

Q = 2.383 Btu

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