Chapter 1,2,3

Fundamental Numerical Values

First Bohr Electron Radius............... a 0= 0.5292 X 10-10 mSpeed of light in a vacuum c0 = 2.9979 X 108 m/sElectronic Charge e = 1.6022 X 10-19 CGravitational Acceleration g = 9.81 m/s2

Plank's Constant h = 6.6262 X 10-34 J.sBoltzmann Constant k = 1.3806 X 10-23 J/KElectron Rest Mass me = 9.1096 X 10-31 kgClassical Electron Radius ro = 2.8179 X 10-15 mElectron Volt 1 eV = 1.6022 X 10-19 JAvagadro's Number Na = 6.023 X 1026 particles/kmol

Conversion Factors

Mass: Density:

1 kg = 2.2046 lbm 1 lbm/ft3 = 16.019 kg/m3

1 lbm = 0.45359 kg 1 kg/m3 = 0.062428 lbm/ft3

Area: Volume:

1 m2 = 10.764 ft2 1 m3 = 35.315 ft3

1 ft2 = 0.092903 m2 1 ft3 = 0.028317m 2

1 in2 = 6.4516 E-04 m2 1 gal = 0.13368 ft3

1 yd2 = 0.83613 m2 1 liter = 1000 cm3

Velocity: Force:

1 m/s = 3.281 ft/s 1 N = 0.22481 lbf 1 ft/s = 0.3048 m/s 1 lbf = 4.4482 N

Acceleration: Time:

1 m/s2 = 3.28 ft/s2 1 year = 3.16 E07 s 1 ft/s2 = 0.3048 m/s2 1 day = 1.44 E03 min = 8.64 E04 s

Pressure:

1 Pa = 1 E-05 bar = 1.4504 E-04 psia = 9.8692 E-6 atm = 0.020886 lbf/ft2

1 kPa = 0.14504 psia 1 bar = 1 E05 Pa = 0.98692 atm = 14.504 psia = 2088.6 lbf/ft2

1 psia = 144 lbf/ft2 = 6894.8 Pa = 6.8948 E-02 bar = 0.068046 atm 1 atm = 101.325 kPa = 14.696 psia = 1.0133 bar = 2116.2 lbf/ft2

Energy:

1 J = 1 N*m = 1 kg*m/s2

1 kJ = 1 kW*s = 0.94783 Btu = 0.23885 kcal = 737.56 ft*lbf 1 Btu = 1.0550 kJ = 0.25200 kcal = 778.16 ft*lbf 1 kcal = 4.1868 kJ = 3.9684 Btu = 3088.0 ft*lbf 1 kWh = 3.60 E03 kJ = 2655.2 E03 ft*lbf = 3412.2 Btu = 859.86 kcal/h 1 ft*lbf = 1.2851 E-03 Btu = 1.3558 E-03 kJ

Power:

1 W = 1 J/s 1 W = 3.4122 Btu/h = 0.85987 kcal/h = 1.34102 E-03 hp = 0.73756 ft*lbf/s 1 Btu/h = 0.29307 W = 0.25200 kcal/h = 3.930 E-04 hp = 0.21616 ft*lbf/s 1 kcal/h = 1.1630 W = 3.9683 Btu/h = 1.5595 E-03 hp = 0.85778 ft*lbf/s 1 hp = 550 ft*lbf/s = 2544.5 Btu/h = 745.7 W 1 ft*lbf/s = 4.6262 Btu/h = 1.3558 W = 1.8182 E-03 hp 1 ton(Cooling Capacity)= 12,000 Btu/h = 3.5168 kW

Energy Rate per Unit Area:

1 W/m2 = 0.31700 Btu/(h*ft2) = 0.85986 kcal/(h*m2) 1 Btu/(h*ft2) = 3.1546W/m2 = 2.7125 kcal/(h*m2) 1 kcal/(h*m2) = 0.36867 Btu/(h*ft2) = 1.1630 W/m2

Entropy:

1 kJ/K = 0.52657 Btu/oR = 0.23885 kcal/K 1 Btu/oR = 1.8991 kJ/K = 0.45359 kcal/K 1 kcal/K = 4,1868 kJ/K = 2.2047 Btu/oR

Temperature:

T,K = 5/9(T,oR) = 5/9(T,oF + 459.67) = T,oC + 273.15 T,oR = 9/5(T,K) = 9/5(T,oC + 273.15) = T,oF + 459.67 T,oF = 9/5(T,oC + 32) T,oC = 5/9(T,oF - 32)

Specific Energy, Specific Enthalpy:

1 kJ/kg = 0.42992 Btu/lbm = 0.23885 kcal/kg = 334.55 ft*lbf/lbm 1 Btu/lbm = 2.3260 kJ/kg = 0.55556 kcal/kg = 778.16 ft*lbf/lbm 1 kcal/kg = 4.1868 kJ/kg = 1.8000 Btu/lbm = 1400.7 ft*lbf/lbm

___

Universal Gas Constant R:

= 8.31441 kPa*m3/(kmol*K) = 8.31441 kJ/(kmol*K) = 0.0820568 liter*atm/(gmol*K) = 0.0820568 m3*atm/(kmol*K) = 1.98586 Btu/(lbmol*oR) = 0.730235 ft3*atm/(lbmol*oR) = 1545.3 ft*lbf/(lbmol*oR) = 10.73 psia*ft3/(lbmol*oR)

TUTORIAL 1

Example 1-1

Problem Statement:

A spacecraft weighting 273,000 N leaves Earth with 900,000 N of fuel on board and flies to a planet where the acceleration due to gravity is 4 m/s2. During the flight to the planet, 2/3 of the fuel is consumed. How much thrust does the rocket need to insure lift off from the planet?SOLUTION:

Total mass on planet = 27,840 + 30,590 = 58,430 kgTo lift off:Thrust = Weight

HB 1.8s5/12/2001

Example 1-2

Problem Statement:

An external force of 225 N is acting vertically upwards on a body whose mass is 5 kg. In the absence of friction, is the body rising or falling, and what is the acceleration? The body is initially at rest, and the acceleration due to gravity is 9.8 m/s2.SOLUTION:

Example 1-3

Problem Statement:

A kilogram mass is "weighed" with a beam balance at a point where g = 9.7 m/s2. What reading is expected? If it is weighed with a spring scale that reads correctly for standard gravity, what scale reading is expected?SOLUTION:

The beam balance reads in proportion to mass, and will balance with 1 kg. If calibrated with "weights" based on Earth's gravity, it will give:

The spring scale would read:

Example 1-4

Problem Statement:

What is the weight of the H2O occupying a volume of 0.85 m3 if v = 0.021 m3/kg? The acceleration due to gravity is 9.807 m/s2.SOLUTION:

Example 1-5

Problem Statement:

A mass of 5 lbm is moving at a speed of 10 ft/s at a height of 30 ft above a reference plane. What are the values of kinetic and potential energy of the mass, and what is the weight of the mass?

SOLUTIONThe kinetic energy is

The potential energy is

The weight is

COMMENT: Watch the units carefully! All terms must be converted to the same units, usually either Btu or ft*lbf for energy in the USCS.

Example 1-6

Problem Statement:

The exit pressure from a steam turbine is measured to be 12 inches of Hg vacuum. If Patm = 101.3 kPa, calculate the absolute pressure of the H2O.SOLUTION:

Example 1-7

Problem Statement:

A cylinder encloses a gas with a piston as shown. The area of the piston is 0.01 m2. Take the atmospheric pressure to be 0.101 MPa and the local gravitational acceleration as 9.8 m/s2. If the piston supports a mass of 50 kg (including the mass of the piston), what is the gas pressure? Will the gas pressure change if the gas volume beneath the piston is doubled?

SOLUTION:By a force balance on the piston,

Even if the gas volume is doubled, the gas pressure is the same because the force balance contains no function of volume. The force on the gas exerted by the piston weight plus atmospheric pressure is constant regardless of piston position and, therefore, gas volume.Example 1-8

Problem Statement:

At the beginning of a process, the thermometer reads 15 °F. You want to stop the experiment when the absolute temperature has doubled. What is the final temperature in degrees Celsius?SOLUTION:

Example 1-9

Problem Statement:

Determine the SI values for the following energy-related quantities: 1 Btu/(ft2-hr), 13,000 Btu/lbm, and 50,000 Btu/h.SOLUTION:

Example 2-1

Problem Statement:

The numerical values of the properties x, y, and z are given. Linearly interpolate and obtain the specified property.a) Given:

x y

110 1.2074

115 1.031

Find: y for x = 112.3

b) Given:x = 10 x = 50

y z y z

150 19.513 150 3.8894

200 21.820 200 4.3561

Find: z for x = 27 and y = 150c) Given: Same data as in b). Find: z for x = 32 and y = 181SOLUTION:

a)

b) c) We must effectively create new entries for z at 181. Thus, at x=10

at x = 50

Now, finding the values at x = 32 yields:

COMMENTS: These are the types of interpolation that are required to obtain properties from many tables.Example 2-2

PROBLEM STATEMENT:

Indicate whether the following states for water are in the liquid, saturation, or superheated region. Specify the quality of the states in the saturation region:

State P (kPa) T (°C) v (m3/kg)

1 1700 200 --

2 1200 -- 0.0010

3 -- 75 3.0

4 500 202 --

5 350 -- 0.005

Solution: State 1


1 1700 200 --

In order to determine the state of the water, first find the saturation temperature at P = 1700 kPa = 17 bar. From the steam tables:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) uf(kJ/kg) ug(kJ/kg)

15.0 198.3 0.001154 0.1318 843.16 2594.5

20.0 212.4 0.001177 0.09963 906.44 2600.3

Using linear interpolation:

Since T = 200°C < Tsat = 204.35°C, water is in the liquid region.

Solution: State 2


2 1200 -- 0.0010

In order to determine the state of the water, examine the specific volume columns in the saturated water table at P = 1200 kPa = 12 Bar:


10.0 179.9 0.001127 0.1944 761.68 2583.6

15.0 198.3 0.001154 0.1318 843.16 2594.5


Since v = 0.0010 m3/kg < vl = 0.001138 m3/kg, water is in the liquid region

Solution: State 3


3 -- 75 3.0

In order to determine the state of the water, examine the specific volume columns in the saturated water table at T = 75oC:

T (°C) P (bar) vf (m3/kg) vg (m3/kg) uf(kJ/kg) ug(kJ/kg)

75 0.3858 0.0010259 4.131 313.90 2475.9

Since vf = 0.001059 < v = 3.0 < vg = 4.131 m3/kg, water is in the saturation region.

Using linear interpolation to determine the quality:

Solution: State 4


4 500 202 --

In order to determine the state of the water, first find the saturation temperature at P = 500 kPa = 5 Bar. From the steam tables:


5.0 151.9 0.0010926 0.3749 639.68 2561.2

Since T = 202°C > Tsat = 151.9°C, water is in the superheat region.

Solution: State 5


5 350 -- 0.005

In order to determine the state of the water, examine the specific volume columns in the saturated water table at P = 350 kPa = 3.5 Bar:


3.50 138.9 0.0010786 0.5243 583.95 2546.9

Since vf = 0.0010786 < v = 0.005 < vg = 0.5243 m3/kg, water is in the saturation region.

Using linear interpolation to determine the quality:

Example 2-3

Problem Statement:

Given: water at P = 0.08 MPa and h = 1000 kJ/kg.Find: T and vSOLUTION:From Table 11s, at P = 0.08 MPa; hf = 391.71 kJ/kg, hg = 2665.3 kJ/kghf < h < hg, so fluid is a saturated mixture. At the given P, T = Tsat = 93.5 °C.

Example 2-4

PROBLEM STATEMENT:

A sealed volume of 2.0 cm3 contains saturated water. The container is heated from an initial pressure of 0.10 MPa until the vessel contains a single phase. Determine the final state of the water if the vessel contains:

(a) 0.10 gram and

(b) 1.0 gram.

Solution: Part a

Problem State P (kPa) T (oC) v (m3/kg)

(a) 1 100 99.63 ?

2 ? ? ?

The states are determined by the following steps:

The saturated water tables at 100 kPa = 1 bar are:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) hf(kJ/kg) hfg(kJ/kg) hg(kJ/kg)

1.00 99.63 0.0010432 1.694 417.46 2258.0 2675.5

The quality is then:

since v is constant,

For the final state a saturated vapor, the saturated water tables near this point are:


90.0 303.4 0.0014178 0.02048 1363.3 1378.9 2742.1

100. 311.1 0.0014524 0.01803 1407.6 1317.1 2724.7

By interpolation:

Solution: Part b

Problem State P (kPa) T (oC) V (m3/kg)

(b) 1 100 99.63 ?

2 ? ? ?

The states are determined by the following steps:

The saturated water tables at 100 kPa = 1 bar are:


1.00 99.63 0.0010432 1.694 417.46 2258.0 2675.5

The quality is then:

since v is constant,

For the final state a saturated liquid, the saturated water tables near this point are:


190. 361.5 0.0019243 0.006657 1776.5 688.0 2464.5

200. 365.8 0.002036 0.005834 1826.3 583.4 2409.7

By interpolation:

Comments:

The specific volume at the critical point for water is 0.00311 m3/kg. This value shows directly that (a) is a saturated vapor (va > vcr) and that (b) is a saturated liquid (vb <vcr).

Example 2-5

Problem Statement:

Find the change in value of enthalpy h for water (steam) between an initial state where P = 100 kPa and v = 1.6500 m3/kg and a final state where P = 3000 kPa and T = 525°C.

Solution:

State 1 is saturated (check Table 11s) v lies between vf and vg at P = 100 kPa.

x = (1.6500 - 0.001043) / 1.6933 = 0.9738

Using the known x at state 1,

At state 2, Table 11s shows Tsat = 233.89°C at Psat = 3000 kPa, so state 2 is superheated. Go to Table 12s. For P = 3000, T = 525°C,

so

Example 2-6

Problem Statement:

Given: water at P = 10.0 MPa and u = 418.96 kJ/kg.Find: T and hSOLUTION:From Table 11s, at P = 10 MPa; uf = 1392.8 kJ/kg, ug = 2544.2 kJ/kg

u < uf, so it's in the compressed liquid region. From Table 13s,T = 100.7 °C, h = 429.35 kJ/kg.The incompressible liquid approximation gives, for uf(T) = 418.96 kJ/kg, T = 100 °C (Table 11s), and h = uf + Pvf = 418.96 kJ/kg + (10 * 103 kPa * 0.001043 m3/kg) = 429.39 kJ/kg.

Example 2-7

PROBLEM STATEMENT:

Given: Water at P = 15 MPa and v = .001006 m3/kg.

Find: T and h

SOLUTION (Saturated Water Tables):

StateP (MPa) T (oC) v (m3/kg) h (kJ/kg)

1 15 ? 0.001006 ? Examine the Saturated Water table at P = 15 MPa = 150 bar:

P (bar)T (°C) vf

(m3/kg)vg

(m3/kg) hf

(kJ/kg) hg

(kJ/kg)

150 342.2 0.0016581 0.010304 1610.5 2610.5

At Psat = 15 Mpa, vf = 0.0016581 m3/kg

Therefore => v = 0.001006 < vf = 0.0016581, so water is in the compressed liquid state.

Now, to find T & h, assume that the specific volume is equal to the saturated liquid specific volume at the state temperature:

v = 0.001006 m3/kg vf

Examining the Saturated Water tables again to determine where v = vf:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) hf(kJ/kg) hg(kJ/kg)

0.04 28.96 0.0010040 34.800 121.46 2554.4

0.06 36.16 0.0010064 23.739 151.53 2567.4

Linear interpolation gives:

and

Using linear interpolation to obtain hf and Psat:

Now,

SOLUTION (Compressed Liquid Tables):

State P (MPa) T (oC) v (m3/kg) h (kJ/kg) 1 15 ? 0.001006 ?

Examine the Compressed Liquid table at P = 15 MPa = 150 bar:

P = 150 bars = 15.0 MPaTsat = 342.24 oC

T (oC) v (m3/kg) u (kJ/kg) h (kJ/kg)

40 1.0013 165.76 180.78

80 1.0222 331.48 346.81


Comments:In this case, the incompressible liquid approximation breaks down badly. The results from the compressed liquid tables are more accurate.

Example 2-8

PROBLEM STATEMENT:

Find the following properties for steam from the steam tables. Show all interpolation calculations.

(a) u(T=400°C, P=1500 kPa)

(b) h(T=200°C, P=2000 kPa)

(c) T(h=2100 kJ/kg, P=6900 kPa)

(d) x(P=500 kPa, h=2000 kJ/kg)

(e) h(u=2000 kJ/kg, x=0.65)

(f) h(u=3100 kJ/kg, v=0.1 m3/kg)

Solution: State a State T (oC) P (kPa) u (kJ/kg)

a 400 1500 ?In this case, the temperature is higher than the critical temperature for water (373.98oC) but the pressure is well below the critical pressure of 22056 kPa for water. Examining the Saturated Water tables:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) uf(kJ/kg)

ug(kJ/kg)

15.0 198.3 0.0011539 0.1318 843.16 2594.5

Since the temperature (400oC) is higher than the saturation temperature at the state pressure, the state lies in the superheat region. The specific internal energy can be found in the superheat tables:

P = 15.0 bars = 1500 kPaTsat = 198.32 oC

T (oC) v (m3/kg) u (kJ/kg) h (kJ/kg) s

(kJ/kgK)

360 0.1899 2884.4 3169.2 7.1363

400 0.2030 2951.3 3255.8 7.2690

440 0.2160 3018.5 3342.5 7.3940

The specific internal energy is 2951.3 kJ/kg.

Solution: State b State T (oC) P (kPa) h (kJ/kg)

b 200 2000 ?Examining the Saturated Water tables:

T (°C) P (bar) vf (m3/kg)

vg (m3/kg)

uf(kJ/kg)

ug(kJ/kg) hf(kJ/kg) hg(kJ/

kg)

200 15.54 0.001565

0.1274 850.65 2595.3 852.45 2793.2

shows the given pressure (20.0 bar) to be above the saturation pressure at the given temperature, so this state is subcooled. The enthalpy of a subcooled liquid can be determined by noting that u for a subcooled liquid is chiefly a function of temperature, so that the change in h between the saturated liquid state and the subcooled liquid state is:

or, since v is constant for an incompressible liquid and usubcooled(T) = usat(T),

Solution: State c State T (oC) P (kPa) h (kJ/kg)

c ? 6900 2100For this state, checking the Saturated Water (pressure) tables:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) hf(kJ/kg)

hg(kJ/kg)

60.0 275.6 0.0013187 0.03244 1213.4 2784.3

70.0 285.9 0.0013513 0.02737 1267.0 2772.1

shows it to be in the saturation region, with a temperature between 275.62°C (for Psat = 6000 kPa) and 285.86°C (for Psat = 7000 kPa). Using linear interpolation gives:

Solution: State d State x P (kPa) h (kJ/kg)

d ? 500 2000Again using the Saturated Water (pressure) table:

P (bar) T (°C) vf (m3/kg) vg (m3/kg) hf(kJ/kg)

hfg(kJ/kg)

hg(kJ/kg)

4.50 147.9 0.0010882 0.4140 623.25 2120.7 2743.9

5.00 151.9 0.0010926 0.3749 640.23 2108.5 2748.7

6.00 158.9 0.0011006 0.3157 670.56 2086.3 2756.8

we find that the given h does lie in the saturation region. Thus, from the definition x we find

Solution: State e State x u (kJ/kg) h (kJ/kg)

e 0.65 2000 ?This problem requires an iterative solution. One way is to assume a value of Tsat (or Psat), and using the corresponding saturation tables, calculate the value of u that corresponds to the given value of x = 0.65. The selected T (or P) is then adjusted until the computed u value agrees with that given. Then the value of h at that T (or P) is computed for the given x. Such an iterative solution is outlined below.

Trial T Guess (°C) x u(kJ/kg) =uf+xufg

1 200 0.65 1984.7

2 210 0.65 2003.11

3 205 0.65 1993.91

4 208.3 0.65 2000.0

Where the Saturated Water table is shown here:

T (°C)

P (bar) uf (kJ/kg) ug (kJ/kg) hf (kJ/kg) hg (kJ/kg)

200 15.54 850.65 2595.3 852.45 2793.2

210 19.06 895.53 2599.5 897.76 2798.5

Linear interpolation between T = 200°C and 210°C can be carried out to obtain u at T = 205°C. The values at trial 4 can also be obtained through linear interpolation:

Now, h is computed for Tsat = 208.3°C (based on interpolation from the trials 2 and 3) at x = 0.65 to give:

Solution: State f State v m3/kg u (kJ/kg) h (kJ/kg)

f 0.1 3100 ?Since u > ug for all conditions in the Saturated Water tables, this set of values lies in the superheat region. Again, we note that u is a weak function of P, so we can search the superheat tables for a temperature with a value of u close to the value of 3100 kJ/kg. At moderate pressures, a temperature of about 500°C appears about right (check pressures in the range 600 to 8000 kPa to see that this is so). Now the pressures are scanned to see which pressure gives v near 0.1 m3/kg at T = 500°C. It appears that at 500°C, the pressure is between 3000 and 4000 kPa. We note that we are very close to the required state. We can construct the table below for states that bracket the given state:

State P(kPa) T(°C) v(m3/kg) u(kJ/kg) h(kJ/kg)

1 3000 450 0.1079 3020.5 3344.1

2 3000 500 0.1162 3108.1 3456.6

3 4000 450 0.08002 3010.3 3330.4

4 4000 500 0.08642 3099.7 3445.4

The table entries have P, T values that bracket both the required v and the required u. The figure below shows the situation on a plot of u vs. v. The constant P and constant T lines are superimposed on the diagram, assuming a linear relation over the region between the points.

We now must perform a double interpolation to find that T, P and h that correspond to the given v,u at state 5. This is done as follows: First, the values of u, h, and P are found at states 6 and 7, where T and v are known. Any of these properties can be found from the relations

where z is the property to be determined. This linear interpolation results in the table.

State P(kPa) T(°C) v(m3/kg) u(kJ/kg) h(kJ/kg)

6 3544 500 0.1 3103.5 3450.5

7 3283 450 0.1 3017.6 3340.2

Now, with states 6 and 7 known, linear interpolation can be used to define state 5 using

This results in h5 = 3446.0 kJ/kg at T = 498.0°C.

Comments:

Obviously, problems involving property determinations of the type necessary in parts (e) and (f) require tedious calculations and provide room for considerable error. Further, the results obtained from the table entries depend on linear interpolations among the available states in the table. More complete tables allow better interpolation and therefore better accuracy.

Example 2-9

Problem Statement:

Steam is at T = 650°C and P = 2000 kPa. Find the change in the enthalpy of the steam whena) T is held constant and P is changed to 40 MPab) P is held constant and T is changed to 800°C.SOLUTION:a) From Table 12s, h(T = 650°C, P = 2000 kPa) = 3803.2 kJ/kg, and h(T = 650°C, P = 40 MPa) = 3516.3 kJ/kg. Thus, h2-h1 = 3516.3 - 3803.2 = -286.9 kJ/kgb) From Table 12s, h(T = 800°C, P = 2000 kPa) = 4150.9 kJ/kg, so h2-h1 = 4150.9 - 3803.2 = 347.7 kJ/kgExample 2-10

Problem Statement:

Helium occupies a rigid container at the given conditions. Determine the specific volume in cubic meters per kilogram and the mass in kilograms.

T = 20°C; P = 0.1 MPa; V = 60 m3

SOLUTION:Assuming ideal gas behavior

The value of R is from

Example 2-11

Problem Statement:

In a rigid tank 1.0 m3 of O2 is heated from an initial pressure of 0.25 MPa and a temperature of 50°C to a final temperature of 325°C. Determine the mass of the gas and its final pressure.SOLUTION:Assuming ideal gas behavior

Example 2-11

Problem Statement:

In a rigid tank 1.0 m3 of O2 is heated from an initial pressure of 0.25 MPa and a temperature of 50°C to a final temperature of 325°C. Determine the mass of the gas and its final pressure.SOLUTION:Assuming ideal gas behavior

Example 2-12

Problem Statement:

Air occupies a rigid container at the conditions shown below. Determine the temperature in Kelvins and the total volume of the container.P = 0.1 MPa; density = 1/v = 1.00 kg/m3 ; m = 453.6 kg

SOLUTION:Assuming ideal gas behavior

Example 2-13

Problem Statement:

Can the following fluids in the specified states be treated as ideal gases? a) Air at 0.1 MPa and 20°C b) Air at 13 MPa and 900°C c) Methane at 2 MPa and 1000°C d) Water at 0.1 MPa and 20°C e) Water at 0.01 MPa and 30°C f) Ammonia at 12 MPa and 450 KSOLUTIONS:The critical point data is given in Table 2.

Substance Tcr (K) Pcr (MPa)

Air 132.5 3.77

Methane 190.4 4.60

Water 647.1 22.06

Ammonia 405.5 11.35

To use the compressibility charts, the reduced temperature and pressure must be found, Tr = T/Tcr and Pr = P/Pcr.

Problem Tr Pr z

a) 2.211 0.027 1.0

b) 8.853 3.448 1.05

c) 6.689 0.434 1.005

d) 0.453 0.005 0.0007

e) 0.468 0.0005 0.00007

f) 1.110 1.057 0.71

It is not possible to find z of Parts (d) and (e) from the compressibility chart because this state lies outside the range of plotted values and below the saturation curve indicated by the long dashed curve. Hence, we use the compressibility factor law. The two states of (d) and (e) are in the sub-cooled region (from steam table). The specific volume is almost constant.

z is close to 1 for an ideal gas.Thus:a) b) c): ideal gas d) e) f): non ideal behaviorExample 2-14

Problem Statement:

Nitrogen gas is originally at P = 200 atm, T = 252.4 K. It is cooled at constant volume to T = 189.3 K. What is the pressure at the lower temperature?SOLUTION:From Table 2, Pcr = 33.5 atm, Tcr = 126.2 K for nitrogen. At State 1, Pr,1 = 200/33.5 = 5.97 and Tr,1 = 252.4/126.2 = 2. According to compressibility factor chart, vr' = 0.34. Following the constant vr' line until it intersects with the line at Tr,2 = 189.3/126.2 = 1.5 give Pr,2 = 3.55. Thus P2 = 3.55 x 33.5 = 119 atm. Since the chart shows z far from 1.0 for this process, it is not appropriate to use the ideal gas law or the gas tables.Example 2-15

PROBLEM STATEMENT:

5 kg of methane is enclosed in a rigid container at a pressure of 90 bar and a temperature of 250 K. Find the volume of the container using the generalized compressibility chart.

Given & Find

Given: Substance: methane (1 kg)

T = 250 K

P = 90 bar

Find: Volume (m3)

Assumptions

1. It will be shown that methane in these conditions cannot be assumed an ideal gas

Solution

The critical temperature and pressure of methane are, respectively

The molecular weight is

The reduced temperature is

The reduced pressure is

Using the generalized compressibility chart

Z = 0.71

The compressibility factor is defined as:

Solving for V:

Substituting values:

Example 2-16

PROBLEM STATEMENT:

Carbon dioxide is contained in a piston cylinder device. Weights are placed on the piston such the pressure inside the cylinder is 300 bar and the temperature is 150oC. The initial volume is 10 liters. Heat is transferred to the gas until the temperature reaches 310oC. Using the generalized compressibility chart, determine the final volume.

Given & Find

Given: Carbon dioxide contained in a piston cylinder device

T1 = 150oC

P1 = 300 bar V1 = 10 liters T2 = 310oC

Find: V2 (in liters)

Assumptions

1. It will be shown that CO2 in these conditions cannot be assumed an ideal gas

Solution

The critical temperature and pressure of carbon dioxide are, respectively:

The molecular weight is:

State 1

Using the generalized compressibility chart,

Z = 0.74

The compressibility factor is defined as:

The mass contained in the cylinder is:


State 2

Since the process is constant pressure,

P2 = P1 = 300 bar

Pr2 = Pr1 = 4.06

The reduced temperature at State 2 is:

So that the compressibility factor at State 2 is:

Z2 = 0.95

The volume at State 2 can then be calculated from:

Example 2-17

PROBLEM STATEMENT

Calculate the change in enthalpy (kJ/kg) for nitrogen between T1 = 25°C and T2 = 650°C using the ideal gas tables. Compare the result to that found by using the polynomial equations for cp and to that found by using the mean value of cp for the temperature range between 25 and 650°C.

Given and Find

Given: Nitrogen, T1 = 25°C, T2 = 650°CFind: h2 - h1 (kJ/kg)

Assumptions: Nitrogen behaves as an ideal gas

Solution Using Ideal Gas Tables

The ideal gas table for nitrogen (Table 6s) gives the following values for enthalpy:

h1(25°C) = 309.99 kJ/kg h2(650°C) = 987.52 kJ/kg

and therefore the enthalpy change can be calculated directly: h2 - h1 = 677.53 kJ/kg

Solution Using Polynomial Function for cp(T)

The change in enthalpy for a simple compressible substance can be calculated from:

so that

The function cp(T) can be obtained from Table 3s as a polynomial of the form

Note temperatures must be entered in K, not °C. Substituting the polynomial function into the integral yields:

For nitrogen, the coefficients of the polynomial are:a=28.90b= -0.157x10-2c=0.8081x10-5d=-2.873x10-9

Substituting M=28.01 kg/kmol, T1 = 25 + 273 = 298 K, and T2 = 650 + 273 = 923 K:

h2 - h1 = 678.12 kJ/kg

Solution Using Mean cp

From Table 4s, the mean cp for the temperature range between 25 and 650 °C is (by interpolation) 1.0838 kJ/kg-K. Taking this value as a constant specific heat:

Substituting values yields:

h2 - h1 = 1.0838 kJ/kg-K x (923-298) K = 677.34 kJ/kg

Note that, since we are taking a difference in temperatures in this equation, it would have been OK to use °C instead of K for T1 and T2.

COMMENTS: All three methods in this case yield results that agree within a fraction of a percent. The assumption of constant specific heat, using the mean value of cp over the temperature range, is quite accurate so long as cp varies linearly with T over the range of interest (reasonably valid for diatomic gases). For gases with more than two atoms in the molecule, this assumption begins to break down and thus assuming constant cp would result in greater error.

Problem 3-1

PROBLEM STATEMENT:

A piston-cylinder device contains 50 kg of water at 150 kPa and 25oC. The cross-sectional area of the piston is 0.1 m2. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the piston reaches a linear spring whose spring constant is 100 kN/m. More heat is transferred to the water until the piston rises 20 cm more.

Determine:

The final pressure and temperature (P3, T3) The work done during this process (1W3) Also show the process on a P-V diagram.

Given & Find

Given: Substance: Water Vapor (50 kg) Piston Area: 0.1 m2

Spring Constant: 100 kN/m State 1: T = 25oC, P = 150 kPa Note: The initial state is a subcooled liquid (T < Tsat at P) State 2: V = 0.2 m3

Piston travels: 2x3 = 20 cm Find: The final pressure and temperature (P3, T3) The work done during this process (1W3) Also show the process on a P-V diagram.

Assumptions

1. Water at State 1 can be treated as a saturated liquid 2. Piston is frictionless 3. Neglect Kinetic and Potential Energy

Final Pressure

Between states 1 and 2 the piston is not in contact with the spring. Therefore,

or

The process from state 1 to state 2 is a constant pressure process.

The piston comes into contact with the spring at state 2. The force balance now becomes:

or

Final Temperature

The temperature can be determined from the steam tables.

thus

V1 = m * v1 = 50*0.0010 = 0.050 m 3

V3 = V2 + 2x3*Ap = 0.20+0.2*0.1 = 0.22 m 3

v2 = 0.0040 < vg @ 150 kPa => State 2 is in the saturation region.

Thus

v3 = 0.0044 < vg @ 350 kPa => State 3 is also in the saturation region.

Thus

P (kPa) V (m3) v (m3/kg) T(oC) 1 150 0.05 0.0010 25 2 150 0.20 0.0040 111.4 3 350 0.22 0.0044 138.9

Work

The pressure remains constant during process 1-2 and changes linearly (a straight line) during process 2-3. Then the boundary work during this process is simply the total area under the process curve,

Wb =

=27.5 kJ

PV Diagram

Comments

The above is based on the principle of using the simple concept that work = area under the process path on a P-V diagram. The work can also be calculated from the expression for an expansion or compression process:

From state 1 to state 2 P is constant, therefore

where

From state 2 to state 3 the compression of the spring adds force and the work integral becomes

and

Now,

The total boundary work is thus

Wb = 1W2 + 2W3

or

Example 3-2

PROBLEM STATEMENT:

Air with mass of 1 kg and initially at P = 101.3 kPa and T = 300K is contained within a cylinder. The cup on top of the piston is then filled at a constant rate until it contains 50 kg of water, thus compressing the gas. The compression proceeds slowly, so that heat transfer to the surroundings maintains the temperature of the air at 300K. Assume air is an ideal gas.

(a) Draw a P-V diagram of the process. Label the initial state 1 and the final state 2.(b) Calculate the work done on the gas during the process.

Given & Find

Given: Air contained in a piston-cylinder device is an ideal gas

T1 = 300 K

P1 = 101.3 kPa

T2 = 300 K

Find: The work done on the air as it is compressed

Assumptions

1. Air behaves as an ideal gas 2. Isothermal process

Solution

Part a:

A force balance on the piston yields:

At the initial state, the mass of water mwater = 0 and the pressure of the air in the cylinder is due to the combined effects of the atmospheric pressure acting on the piston area and the weight of the piston. Since these remain constant throughout the process, they can be combined into one pressure Po. A general expression fot the pressure of the air in the cylinder is then:

The pressure at State 2 can then be calculated from:

And since the temperature is known:

A property table can be constructed:

State T (K) P (kPa) v (m3/kg) 1 300 101.3 0.850 2 300 116.9 0.736

And the P-v diagram is:

Note that while this appears to be a straight line, it is in fact a curve.

Part b:

For an ideal gas undergoing a constant temperature process, the following is true:

which is equivalent to a polytropic process (Pvn = const.) with n = 1. This provides the P-v relationship that is necessary to calculate the work using the relationship:

Substituting into the above equation yields:

Using the values calculated in part a:

Comments: Note that this is not an adiabatic process. An energy balane for this process yields Q-W = U2 - U1 = 0 so that Q = W. In other words, all of the energy that is put into the gas through compression is rejected as heat transfer.

Example 3-3

PROBLEM STATEMENT:

Find the work done, in kilojoules, by an ideal gas in going from state A to state C along the path shown in the P-V diagram.

Given & Find

Given: An Ideal Gas undergoes a constant temperature process followed by a constant pressure process. VA = 3 m3

VB = 2 m3

PB = 100 kPa VC = 1 m3

PC = 100 kPaFind: The work done on the air as it is compressed

Assumptions

1. The substance behaves as an ideal gas

Solution

The process from A to B is constant temperature. Since the substance in question is an ideal gas, we can say:

and

The work for this process is then:

Substituting values gives:

The process from B to C is constant presure. The work for this process is expressed as:

Substituting values results in:

The total work for the process from State A to C is:

Example 3-4

PROBLEM STATEMENT:

A weightless and frictionless piston moves 10 cm. against an ideal gas (that is, a gas with properties that obey Pv=RT) contained in a cylinder that has an inside diameter of 15 cm. A spring is also within the cylinder, positioned as shown. The spring exerts no force on the piston in the initial position, but is touching the piston. The spring has a spring

constant of 2 x 104 N/m. During the piston motion, the gas in the cylinder is maintained at 1 atm.

(a) How much work was done by the piston? (b) What fraction of the work is done in compressing the gas?(c) What fraction of the total work done by the piston is done in the first 5 cm of the piston travel?

Given & Find

Given: An Ideal Gas undergoes a constant pressure process.

Piston diameter = 0.15 m

Piston displacement = 0.1 m

Spring constant = 2 x 104 N/m

Find: (a) How much work was done by the piston? (b) What fraction of the work is done to compress the gas?

(c) What fraction of the total work done by the piston is done in the first 5 cm of the piston travel?

(d) What is the heat transfer associated with this process?

Assumptions

1. Neglect changes in kinetic and potential energy 2. The temperature change is small enough that specific heats can be treated as

constant

The total work is the combined work done against the air and the spring:

Now, with pressure constant, F = kx and let dS = dX, the work can be expressed as:

The first term can be rearranged using the relationship V = A x. Since we are considering work done by the piston, x will be considered positive in the downward direction. Taking the initial position of the piston to be x1 = 0 gives:

b) The fraction of the work that was done compressing the gas is:

c) The fraction of the total work done in the first 5 cm of compression is:

d) For this analysis it is convenient to consider the air in the cylinder as the system. With

assumptions 1 and 2, the heat transfer associated with this process can be calculated by first examining the 1st Law:

In this case, the work term is only the work done on the air. Using the Ideal Gas Law:

and substituting this into the energy balance:

Since pressure is constant, and with V = mv, this can be rewritten as:


Note: The heat transfer out of the system is greater than the work input due to the temperature drop that was necessary to maintain a constant pressure. The temperature drop can be calculated from:

Example 3-5

Problem Statement:

The piston/cylinder device shown contains one kilogram of water at a temperature of 200°C and pressure of 0.8 MPa. The water undergoes an isobaric process until it becomes a saturated liquid. Find the total work done in the process. Is it work done on or by the system? Show the process on a T-v diagram.

SOLUTION:

Example 3-6

Problem Statement:

A piston of mass equal to 900 kg and area equal to 9.8 x 10-3 m2 is supported against stops (see figure) by water at the saturated vapor state and 20 MPa (state 1). The water is cooled until half the container is filled with liquid and half is vapor by volume (state 3). Determine the work per unit mass for the process (w13,out). Draw the process on the P-v diagram.

SOLUTION:

State 1 is specified by P1 = 20 MPa and x1 = 1.0. The final state (3) is indirectly specified by the volume (which will yield quality when the other independent property is determined). The process starts at the saturated vapor state and could either be a constant volume process directly to state 3 or the process could be constant volume to an intermediate state (2) and constant pressure process to state 3.

The pressure where the piston leaves the stop, state 2, is given by

and the specific volume is v2 = vf3 = vg1 = 0.005874 m3/kg. Interpolating from Table 12s yields

To determine if this is the final state, the fraction of vapor is found at state 2 as

which is greater than the required 50 percent.

Thus this is not the final state since more liquid is in state 3. The final state is given by P3 = 1001 kPa and Vg/V = 0.5. The volume fraction is put into quality form by

Solving for x gives

so that state 3 is x3 = 0.00576 and v3 = 0.002240 m3/kg. The P-v diagram and the state table are now complete as indicated.

State P (kPa) v (m3/kg) x

1 20,000 (0.005874) 1.0

2 1,001 0.005874 (0.02459)

3 1,001 (0.002240) 0.00577

The work is evaluated as

Example 3-7

Problem Statement:

A piston encloses 1 kg of saturated water which is initially 10.0 percent liquid on a mass basis. The initial pressure is 1.0 MPa. The water is heated in the system shown in the figure until the water is at the saturated vapor state. The volume at the top stops is 0.19 m3. Determine the work done by the water and the final temperature and pressure.

SOLUTION:

The specific volume at state 1 is

and the value at state 2 is

State 2 (0.19 m3/kg and 1.0 MPa) is in the saturation region so that the process is not completed. The piston now rests against the stops and is heated at constant volume until the final state (saturated vapor) is reached. State 3 is given by v3 = 0.19 m3/kg and x3 = 1.0.Thus, by interpolation from Table 11s.

and the work done by the water is W12 = +15 kJ.The state variables are summarized in the table below

State P(MPa) V(m3) T(°C) v(m3/kg) x

1 1.0 (0.175) (179.9) (0.1750) 0.9

2 1.0 (0.19) (179.9) 0.19 --

3 (1025) (0.19) (181) 0.19 1.0

Example 3-8

Problem Statement:

A piston-cylinder device with one set of stops has dimensions as shown. The piston is free to move. The H2O inside is initially a saturated vapor at 270°C. It is cooled until half the mass is a liquid. The piston requires a pressure of 5.4987 MPa to support it. Draw and label the processes on a T-u diagram. Determine: a) the initial and final mass of H2O, and b) the total work done (kJ).

SOLUTION:

Example 3-9

Problem Statement:

Water (14 kg) is contained in the piston-cylinder device. The H2O is heated from an initial temperature of 150°C to a final temperature and pressure of 450°C and 0.6 MPa. a) Plot and label the process (T-v plot). b) What is the total work done (kJ)? c) What is the change in internal energy?

SOLUTION:

a)

b)

Example 3-10

Problem Statement:

The cylinder shown contains 0.1 kg of water at 0.150 MPa. The volume initially enclosed is 0.005 m3. The spring touches the piston but exerts no force at this initial state. The piston rises owing to a heat transfer to the water until the water is in the saturated vapor state. The spring constant is 150 kN/m and the piston area is 0.02 m2. Determine the final temperature and pressure in the water and evaluate the change in internal energy.

SOLUTION:

First we will need to calculate our initial state variables

where Patm + mg/Agc is constant and given by

From the tables,

P (MPa) vg (m3/kg)

2.0 0.09959

2.25 0.08872

Taking a linear relation for this saturation data gives

The spring relation and the saturation data must intersect at State 2, so

Solving for v2 yields v2 = 0.0994 m3/kg and P2 = 2.003 MPa

and u2 - u1 = 2599.52 - 553.7 = 2045.8 kJ/kg or, equivalently,

The state variables are summarized in the table below

State T (°C) P (MPa)M v (m3/kg) x u (kJ/kg)

1 (111.4) 0.15 0.05 (0.04225) (533.73)

2 (212.49) 2.003 (0.09943) 1.0 (2599.52)

http://www.wiley.com/college/moran/CL_0471129313_S/user/examples/tut3_ex/ex4_12/tut3eq17.gif

Chapter 1,2,3

Documents

Transcript of Chapter 1,2,3