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ISOBARIC PROCESS OF VAPORS

RB Astillero

Asst. Prof. I

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THE ENERGY EQUATION

The General Energy Equation:

P1 + K1 + U1 + Wf1 + Q = P2 + K2 + U2 + Wf2 + Wn

Simple Energy Equation:

Q = U2 – U1 + Wn

The Enthalpy:

h = u + pv

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THE CONSTANT PRESURE PROCESS

The pv plane The Ts planeThe Tv plane

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Defining the condition of the substance

• Point 1 is in the liquid region and the condition of the substance is “subcooled” or “compressed” liquid. It is generally defined by giving its pressure and temperature.

• Point f lies on the saturation curve and is therefore saturated liquid at a given pressure or temperature.

• Point m is in the wet region, is a mixture of liquid and vapor. A quality x and a pressure or temperature generally defines the condition of the substance. Thus, for the enthalpy at point m,

hm = hfm + xhfgm

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Defining the condition of the substance

Point g is on the saturation vapor curve. A pressure or temperature defines the state or condition of the substance.

Point 2 in the superheat region is generally, but not necessarily, defined by giving its temperature and pressure.

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NONFLOW WORK, Wn

The work of a reversible nonflow constant pressure process.

Wn = is the area on the pv plane under the constant pressure process.

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NONFLOW WORK, Wn

Wn =

At p = c,

Wn = = p(v2 – v1) [unit mass]

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STEADY FLOW WORK, Ws

For steady flow process, the work Ws is

Ws = h1 – h2 - K + Q

And h1 = hf1 + x1hfg1

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HEAT TRANSFERRED

From the simple energy equation,

Q = u2 – u1 + Wn

For steady flow and nonflow processes

Q = u2 – u1 + p(v2 – v1)

= u2 – u1 + pv2 – pv1

= u2 – u1 + p2v2 – p1v1

= (u2 + p2v2) – (u1 + p1v1)

= h2 – h1

Where h1 = hf1 + x1hfg1

u2 – u1 = (h2 – p2v2) – (h1 – p1v1)

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EXAMPLE 1:

Steam with a specific volume of 0.09596 m3/kg undergoes a constant pressure process at 1.70 MPa until the specific volume becomes 0.13796 m3/kg. What are (a) the final temperature, (b) u, (c) W, (d) s, and (e) Q?

Solution:

Point 1:

Steam at a pressure of 1.70 MPa and with a specific volume of 0.09596 m3/kg is a wet-mixture, the specific volume of saturated steam at 1.70 mPa is 0.11673 m3/kg.

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Solution Example 1:

@ p1 = 1.70 mPa

vf1 = 0.0011634 uf1 = 870.09 hf1 = 872.06 sf1 = 2.3718

vfg1 = 0.11557 ufg1 = 1727.2 hfg1 = 1923.6 sfg1 = 4.0282

vg1 = 0.11673 ug1 = 2597.3 hg1 = 2795.7 sg1 = 6.400

= = 0.8203

u1 = uf1 + x1ufg1 = 870.09 + (0.8203)(1727.2) = 2286.9 kJ/kg

h1 = hf1+ x1hfg1 = 872. 06 + (0.8203)(1923.6) = 2450.0 kJ/kg

s1 = sf1+x1sfg1 = 2.3718+(0.8203)(4.0282) = 5.6761 kJ/kg•K

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Solution Example 1:

Point 2:

At p2 = 1.70 MPa and a specific volume of 0.13796 m3/kg, the steam is superheated.

From table 3:

0.13621 260 2707.7 2939.3 6.6849

0.13796 t2 u2 h2 s2

0.13944 270 2725 2963.0 6.7290

x1 = 0.542; t2 = 260 +5.42 = 265.4C

x2 = 9.86; u2 = 2707.7 + 9.86 = 2717.6 kJ/kg

0.00

175

0.00

323

x 110

x 2

18.2 x 3

23.7 x 4

0.04

41

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Solution Example 1:x3 = 12.84; h2 = 2939.3 + 12.84 = 2952.1 kJ/kg

x4 = 0.02389; s2 = 6.6849 + 0.02389 = 6.7088 kJ/kg•K

Therefore:

(a) t 2 = 265.4C (ans)

(b) u = u2 – u1 = 2717.6 – 2286.9 = 430.7 kJ/kg (ans)

(c) W = p(v2 – v1) = 1700(0.13796–09596)=71.4 kJ/kg (ans)

(d) s = s2 – s1 = 6.7088 – 5.6761 = 1.0327 kJ/kg•K (ans)

(e) Q = h2 – h1 = 2952.1 – 2450.0 = 502.1 kJ/kg (ans)

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pv and Ts diagrams for Example 1

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EXAMPLE 2:

Steam with an enthalpy of 2843.5 kJ/kg undergoes a constant pressure process at 0.9 MPa until the enthalpy becomes 2056.1 kJ/kg. What are (a) the initial temperature or quality, (b) u, (c) W, (d) s, and (e) Q

Solution:

Point 1:

Steam with an enthalpy of 2843.5 kJ/kg at a pressure of 0.90 MPa is a superheated vapor.

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Solution Example 2:

@ p1 = 0.90 Mpa

2833.6 200 0.2303 2626.3 6.7522

2843.5 t1 v1 u1 s1

2856.9 210 0.2364 2644.2 6.8008

x1 = 4.2; t1 = 200 + 4.2 = 204.2 C

x2 = 0.00259; v1 = 0.2303+0.00259 = 0.2329 m3/kg

9.9

23.3 x 1 10

x 2

17.9x 3

0.00

61 x 4

0.04

86

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Solution Example 2:

x3 = 7.6; u1 = 2626.3 + 7.6 = 2633.9 kJ/kg

x4 = 0.0206; s1 = 6.7522 + 0.0206 = 6.7728 kJ/kg•K

Point 2:

Steam with an enthalpy of 2056.1 kJ/kg at a pressure of 0.90 MPa is a wet mixture.

@ p2 = 0.90 MPa

vf2 = 0.0011212 hf2 = 742.83 sf2 = 2.09046 uf2 = 741.83

vfg2 = 0.2139 hfg2 = 2031.1 sfg2 = 4.5280 ufg2 = 1838.6

vg2 = 0.215 vg2 = 2773.91 sg2 = 6.6226 ug2 = 2580.5

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Solution Example 2:

== 0.6466v2 = vf2 + x2vfg2 = 0.00112112 + (0.6466)(0.2139) = 0.1394 m3/kg

u2 = uf2 + x2ufg2 = 741.83 + (0.6466)(1838.6) = 1930.7 kJ/kg

s2 = sf2 + x2sfg2 = 2.0946 + (0.6466)(4.5280) = 5.0224 kJ/kg•K

Therefore:

(a) t1 = 204.2 C (ans)

( )b u = u2 – u1 = 1930.7 – 2633.9 = -703.2 kJ/kg (ans)

(c) W = p(v2 – v1) = 900(0.1394 – 0.2329) = -84.15 kJ/kg (ans)

( )d s = s2 – s1 = 5.0224 – 6.7728 = -1.7505 kJ/kg•K (ans)

(e) Q = h2 – h1 = 2056.1 – 2843.5 = -787.4 kJ/kg (ans)

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pv and Ts Diagrams for Example 2p

v

12

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EXERCISES:

1. A piston-cylinder containing steam at 700 kPa and 250C undergoes a constant pressure process until the quality is 70%. Determine per kilogram (a) the work done, (b) the heat transferred, (c) the change of internal energy, and (d) the change of enthalpy.

Ans. (a) – 101.4 kJ/kg; (b) -810 kJ/kg; (c) -708 kJ/kg;

(d) – 810 kJ/kg

2. Steam at 3.1 MPa and with 74.28C SH rejects 1973.36 kJ/kg of steam heat at constant pressure, determine (a) the final temperature and (b) the change in specific entropy.

Ans. (a) 235.7C; (b) -3.8572 kJ/kg•K

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EXERCISES:

3. With 3% moisture, 2.5 kg of steam has an enthalpy of 6600 kJ. It is heated at constant pressure to a final condition of 80 superheat. Find (a) the pressure of the process and (b) the quantity of heat received by the steam.

Ans. (a) 0.22 MPa; (b) 575 kJ

4. Five kg of water vapor are contained at 150 kPa and 90% quality in a suitable enclosure. Calculate the heat which must be added in order to just produce a saturated vapor. What will the pressure be at the end of the heating process?

Ans. 1045.5 kJ, 168 kPa

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END OF THE LESSONDO THE ASSIGNMENT

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QUIZ#1

1. Steam at a temperature of 190C has a specific internal energy of 1250.5 kJ/kg. Determine the (a) pressure, (b) specific volume, (c) specific enthalpy, & (d) specific entropy of the steam.

2. Steam at a pressure of 1.40 MPa has a specific volume of 0.1602 m3/kg. What are the (a) temperature, (b) specific internal energy, (c) specific enthalpy, & (d) specific entropy of the steam?

3. Steam with a specific enthalpy of 2500.5 kJ/kg undergoes an isobaric process at 4000 kPa until the enthalpy becomes 2902.4 kJ/kg. Determine the (a) final temperature (t2), (b) v, (c) u, (d) Wn, (e) Q.

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QUIZ #1

1. Steam at a temperature of 135C has a specific internal energy of 2000.5 kJ/kg. Determine the (a) pressure, (b) specific volume, (c) specific enthalpy, & (d) specific entropy of the steam.

2. Steam at a pressure of 1.45 MPa has a specific volume of 0.17476 m3/kg. What are the (a) temperature, (b) specific internal energy, (c) specific enthalpy, & (d) specific entropy of the steam?

3. Steam with a specific enthalpy of 2915.3 kJ/kg undergoes an isobaric process at 1.50 MPa until the enthalpy becomes 2000.4 kJ/kg. Determine the (a) final temperature (t2), (b) v, (c) u, (d) Wn, (e) Q.