School of Engineering

EM203/208 (January 2013 Semester) - Assignment 1

1. (a) Differentiate between a car radiator and a can of soft drink in terms of thermodynamic

systems (i.e. open or closed systems). Explain your answer.

Answer: The radiator should be analyzed as an open system since mass is crossing the

boundaries of the system. A can of soft drink should be analyzed as a closed system since no

mass is crossing the boundaries of the system.

(b) A closed system is also called a control mass and an open system is also called a control

volume. Explain Why.

Answer: Mass cannot cross the boundary of a closed system, hence the mass in a closed

system is fixed at a constant quantity (i.e. controlled). Thus the closed system is also referred

as a control mass.

Mass can cross the boundary of an open system, but the boundary of an open system is clearly

specified and defined the volume subjected to the energy analysis. Thus the open system is

also referred as a control volume.

(c) What is the state postulate?

Answer: Refer to notes.

(d) Is the state of superheated steam in a closed system completely specified by the

temperature and the pressure? Explain your answer.

Answer: Yes. Temperature and pressure are two independent intensive properties of

superheated steam, and since superheated steam can be regarded as a simple compressible

system, these two properties completely can be used to completely specify the state of the

superheated system.

(e) Is the state of a saturated liquid vapour mixture in a closed system completely specified by

the temperature and the pressure? Explain your answer.

Answer: No. Temperature and pressure of saturated steam are interdependent, that is, they

are not two independent intensive properties of a saturated liquid vapour mixture. One other

intensive property which is independent of temperature or pressure is required (e.g. quality,

average specific volume, average enthalpy).

2. A 1.8-m3 rigid tank contains steam at 220°C. One third (1/3) of the volume is in the liquid

phase and the rest is in the vapor form. Determine (a) the pressure of the steam, (b) the quality

of the saturated mixture, and (c) the density of the mixture.

Solution:

(a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The

pressure of the steam is the saturation pressure at the given temperature. Thus,

P = Tsat@220°C = 2320 kPa

(b) The total mass and the quality are determined as

(c) The density is determined from

3. A closed, rigid container of volume 0.5 m3 is placed on a hot plate. Initially, the container

holds a two-phase mixture of saturated liquid water and saturated water vapor at p1 = 1 bar

with a quality of 0.5. After heating, the pressure in the container is p2 = 1.5 bar. Indicate the

initial and final states on a T–v diagram, and determine

(a) the temperature, in C, at each state.

(b) the mass of vapor present at each state, in kg.

(c) If heating continues, determine the pressure, in bar, when the container holds only

saturated vapor.

Solution:

Two independent properties are required to fix states 1 and 2.

At the initial state, the pressure and quality are known. As these are independent, the state is

fixed. State 1 is shown on the T–v diagram below (in the two-phase region).

The specific volume at state 1 is found using the given quality. That is

v1=v f +x (v g−v f )

From Table A-5, at p1 = 1 bar (100 kPa), vf = 0.001043 m3/kg and vg = 1.6941 m3/kg. Thus

v1=0 . 001043+0 . 5 (1 .6941−0 .001043 )=0.8475 m3/kg.

At state 2, the pressure is known. One other property required to fix the state. This property is

the specific volume v2 which can be determined since the specific volume is constant in a

closed system with a rigid (fixed) boundary. Hence v2 = v1 = 0.8475 m3/kg.

Since p2 = 1.5 bar (150 kPa), Table A-5 gives vf,2 = 0.001053 and vg,2 = 1.1594 m3/kg. Since

vf,2 < v2 < vg,2

state 2 must be in the two-phase region as well, as shown on the T–v diagram above.

(a) Since states 1 and 2 are in the liquid–vapour mixture region, the temperatures correspond

to the saturation temperatures for the given pressures (100 kPa and 150 kPa). Table A-5 gives

T1 = 99.63°C and T2 = 111.4°C

(b) To find the mass of water vapor present, we use the volume and the specific volume to

first find the total mass, m. That is

m=Vv

= 0 . 5 m3

0. 8475 m3 /kg=0 . 59 kg

Then, with the given value of quality, the mass of vapor at state 1 is

mg,1 = x1m = 0.5(0.59 kg) = 0.295 kg

The mass of vapor at state 2 is found similarly using the quality x2. So first we determine x2,

x2=v−v f , 2

vg ,2−v f ,2

=0. 731

Then,

mg,2 = 0.731(0.59 kg) = 0.431 kg

(c) If heating continued, state 3 would be on the saturated vapor line, as shown on the T–v

diagram above. Thus the pressure would be the corresponding saturation pressure. That is, it

would be the pressure corresponding to a saturated vapor with a specific volume of 0.8475

kg/m3. Interpolating in Table A-5, we obtain

P3−200

225−200= 0 . 8475−0 . 88578

0 . 79329−0 . 88578 → P3=210 kPa =2 . 1 bar

4. A rigid tank of 1 m3 contains nitrogen gas at 600 kPa, 400K. By mistake, someone lets 0.5

kg of the gas flow out. If the final temperature is 375 K, what is the final pressure?

Solution

m= PVRT

=600 kPa × 1 m3

0 . 2968kPa⋅m3

kg⋅K×400 K

=5 .054 kg

m2 = m – 0.5 kg = 5.054 – 0.5 kg = 4.554 kg

P2=m2 RT 2

V=

4 . 554 kg ×0 . 2968kPa⋅m3

kg⋅K×375 K

1 m3=506 .9 kPa

5. Complete the following table for H2O.

T, C P, kPa v, m3/kg u, kJ/kg x Phase description

240 300 0.780446 2713.32 - Superheated vapor

133.52 300 0.5 2196.34 0.825 Liquid-vapor mixture

419 50 6.385 3000 - Superheated vapor

100 101.42 1.0036 1671.26 0.6 Liquid-vapor mixture

160 618.23 0.30680 2567.8 1 Saturated vapor

(a) Check Table A-5, Tsat at 300 kPa is133.52C. Since actual temperature is higher, we have

superheated phase. From Table A-6 in the section for 0.3 MPa (300 kPa), we obtain for a

temperature of 240C, using interpolation:

v−0.716430.79645−0.71643

=240−200250−200

→ v=0.780446 m3/kg

u−2651.02728.9−2651.0

=240−200250−200

→ u=2713.32 kJ /kg

(b) Check Table A-5, at 300 kPa, vf and vg are 0.001073 and 0.60582 m3/kg respectively.

Since actual specific volume is between vf and vg, we have a saturated liquid-vapor mixture

phase. Hence, the temperature is Tsat@300kPa = 133.52 C. To determine u, we have to determine

the quality first.

v=v f +x ( vg−v f ) → x=v−v f

vg−v f

= 0.5−0.0010730.60582−0.001073

=0.825

Then u = uf + xufg = 561.11 +(0.825)(1982.1) = 2196.34 kJ/kg

(c) Check Table A-5, at 50 kPa, uf and ug are 340.49 and 2483.2 kJ/kg respectively. Since

actual u is higher than ug, we have a superheated phase. From Table A-6 in the section for

0.05 MPa (50 kPa), we obtain for u = 3000 kJ/kg, using interpolation:

v−6.20947.1338−6.2094

= 3000−2968.93132.6−2968.9

→ v=6.385 m3/kg

T−400500−400

= 3000−2968.93132.6−2968.9

→T=419C

(d) Since the quality is between 0 and 1, we have a saturated liquid-vapor mixture. Then the

pressure is the saturated pressure at From Table A-4, we obtain Psat@100C = 101.42 kPa. We

also obtain vf and vg to be 0.001043 and 1.6720 m3/kg and uf and ufg to be 419.06 and 2087.0

kJ/kg respectively. Hence

v = vf + x vfg = 0.001043 + (0.6)(1.67200.001043) = 1.0036 m3/kg, and

u = uf + x ufg = 419.06 + (0.6)(2087.0) = 1419.8 kJ/kg , and

(e) Check Table A-4, at 160C, uf and ug are 674.79 and 2567.8 kJ/kg respectively. Since

actual u = ug, we have a saturated vapour phase. Thus the pressure is Psat@100C = 618.23 kPa

and v is vg= 0.30680 m3/kg. Quality x equals 1.