Inlet:Flow regime

Option = subsonicMass and momentum

Option = Mass Flow rate

Mass flow rate = where:  

= mass flow rate, kg/s

C = discharge coefficient, dimensionless

A = discharge hole cross-sectional area, m²

k = cp/cv of the gas

cp = specific heat of the gas at constant pressure

cv = specific heat of the gas at constant volume

ρ = real gas density at P and T, kg/m³

P = absolute upstream pressure of the gas, Pa

M = the gas molecular mass, kg/kmole    (also known as the molecular weight)

R = Universal gas law constant = 8314.5 (N·m) / (kmole·K)

T = absolute upstream temperature of the gas, K

Z = the gas compressibility factor at P and T, dimensionless

Ratio of Specific Heats for some common gases: Reference: http://www.engineeringtoolbox.com/specific-heat-ratio-d_608.html

Gas Ratio of Specific Heats

Acetylene 1.30

Air, Standard 1.40

Ammonia 1.32

Argon 1.66

Benzene 1.12

N-butane 1.18

Iso-butane 1.19

Carbon Dioxide 1.28

Carbon Disulphide

1.21

Carbon Monoxide

1.40

Chlorine 1.33

Ethane 1.18

Ethyl alcohol 1.13

Ethyl chloride 1.19

Ethylene 1.24

Helium 1.66

N-heptane 1.05

Hexane 1.06

Hydrochloric acid

1.41

Hydrogen 1.41

Hydrogen chloride

1.41

Hydrogen sulphide

1.32

Methane 1.32

Methyl alcohol 1.20

Methyl butane 1.08

Methyl chloride 1.20

Natural Gas (Methane)

1.32

Nitric oxide 1.40

Nitrogen 1.40

Nitrous oxide 1.31

N-octane 1.05

Oxygen 1.40

N-pentane 1.08

Iso-pentane 1.08

Propane 1.12

R-11 1.14

R-12 1.14

R-22 1.18

R-114 1.09

R-123 1.10

R-134a 1.20

Steam (water) k

1.33

Sulphur dioxide 1.26

Toulene 1.09

Reference: http://www.engineeringtoolbox.com/saturated-

steam-properties-d_457.html

Absolutepressure

Boiling pointSpecific volume

(steam)Density (steam)

Specific enthalpy of liquid water (sensible heat)

Specific enthalpy of steam

(total heat)

Latent heat of vaporization

Specific heat

(bar) (oC) (m3/kg) (kg/m3) (kJ/kg) (kcal/kg) (kJ/kg) (kcal/kg) (kJ/kg) (kcal/kg) (kJ/kg.K)

0.02 17.51 67.006 0.015 73.45 17.54 2533.64 605.15 2460.19 587.61 1.8644

0.03 24.10 45.667 0.022 101.00 24.12 2545.64 608.02 2444.65 583.89 1.8694

0.04 28.98 34.802 0.029 121.41 29.00 2554.51 610.13 2433.10 581.14 1.8736

0.05 32.90 28.194 0.035 137.77 32.91 2561.59 611.83 2423.82 578.92 1.8774

0.06 36.18 23.741 0.042 151.50 36.19 2567.51 613.24 2416.01 577.05 1.8808

0.07 39.02 20.531 0.049 163.38 39.02 2572.62 614.46 2409.24 575.44 1.8840

0.08 41.53 18.105 0.055 173.87 41.53 2577.11 615.53 2403.25 574.01 1.8871

0.09 43.79 16.204 0.062 183.28 43.78 2581.14 616.49 2397.85 572.72 1.8899

0.1 45.83 14.675 0.068 191.84 45.82 2584.78 617.36 2392.94 571.54 1.8927

0.2 60.09 7.650 0.131 251.46 60.06 2609.86 623.35 2358.40 563.30 1.9156

0.3 69.13 5.229 0.191 289.31 69.10 2625.43 627.07 2336.13 557.97 1.9343

0.4 75.89 3.993 0.250 317.65 75.87 2636.88 629.81 2319.23 553.94 1.9506

0.5 81.35 3.240 0.309 340.57 81.34 2645.99 631.98 2305.42 550.64 1.9654

0.6 85.95 2.732 0.366 359.93 85.97 2653.57 633.79 2293.64 547.83 1.9790

0.7 89.96 2.365 0.423 376.77 89.99 2660.07 635.35 2283.30 545.36 1.9919

0.8 93.51 2.087 0.479 391.73 93.56 2665.77 636.71 2274.05 543.15 2.0040

0.9 96.71 1.869 0.535 405.21 96.78 2670.85 637.92 2265.65 541.14 2.0156

1 99.63 1.694 0.590 417.51 99.72 2675.43 639.02 2257.92 539.30 2.0267

1.1 102.32 1.549 0.645 428.84 102.43 2679.61 640.01 2250.76 537.59 2.0373

1.2 104.81 1.428 0.700 439.36 104.94 2683.44 640.93 2244.08 535.99 2.0476

1.3 107.13 1.325 0.755 449.19 107.29 2686.98 641.77 2237.79 534.49 2.0576

1.4 109.32 1.236 0.809 458.42 109.49 2690.28 642.56 2231.86 533.07 2.0673

1.5 111.37 1.159 0.863 467.13 111.57 2693.36 643.30 2226.23 531.73 2.0768

1.5 111.37 1.159 0.863 467.13 111.57 2693.36 643.30 2226.23 531.73 2.0768

1.6 113.32 1.091 0.916 475.38 113.54 2696.25 643.99 2220.87 530.45 2.0860

1.7 115.17 1.031 0.970 483.22 115.42 2698.97 644.64 2215.75 529.22 2.0950

1.8 116.93 0.977 1.023 490.70 117.20 2701.54 645.25 2210.84 528.05 2.1037

1.9 118.62 0.929 1.076 497.85 118.91 2703.98 645.83 2206.13 526.92 2.1124

2 120.23 0.885 1.129 504.71 120.55 2706.29 646.39 2201.59 525.84 2.1208

2.2 123.27 0.810 1.235 517.63 123.63 2710.60 647.42 2192.98 523.78 2.1372

2.4 126.09 0.746 1.340 529.64 126.50 2714.55 648.36 2184.91 521.86 2.1531

2.6 128.73 0.693 1.444 540.88 129.19 2718.17 649.22 2177.30 520.04 2.1685

2.8 131.20 0.646 1.548 551.45 131.71 2721.54 650.03 2170.08 518.32 2.1835

3 133.54 0.606 1.651 561.44 134.10 2724.66 650.77 2163.22 516.68 2.1981

3.5 138.87 0.524 1.908 584.28 139.55 2731.63 652.44 2147.35 512.89 2.2331

4 143.63 0.462 2.163 604.68 144.43 2737.63 653.87 2132.95 509.45 2.2664

4.5 147.92 0.414 2.417 623.17 148.84 2742.88 655.13 2119.71 506.29 2.2983

5 151.85 0.375 2.669 640.12 152.89 2747.54 656.24 2107.42 503.35 2.3289

5.5 155.47 0.342 2.920 655.81 156.64 2751.70 657.23 2095.90 500.60 2.3585

6 158.84 0.315 3.170 670.43 160.13 2755.46 658.13 2085.03 498.00 2.3873

6.5 161.99 0.292 3.419 684.14 163.40 2758.87 658.94 2074.73 495.54 2.4152

7 164.96 0.273 3.667 697.07 166.49 2761.98 659.69 2064.92 493.20 2.4424

7.5 167.76 0.255 3.915 709.30 169.41 2764.84 660.37 2055.53 490.96 2.4690

8 170.42 0.240 4.162 720.94 172.19 2767.46 661.00 2046.53 488.80 2.4951

8.5 172.94 0.227 4.409 732.03 174.84 2769.89 661.58 2037.86 486.73 2.5206

9 175.36 0.215 4.655 742.64 177.38 2772.13 662.11 2029.49 484.74 2.5456

9.5 177.67 0.204 4.901 752.82 179.81 2774.22 662.61 2021.40 482.80 2.5702

10 179.88 0.194 5.147 762.60 182.14 2776.16 663.07 2013.56 480.93 2.5944

11 184.06 0.177 5.638 781.11 186.57 2779.66 663.91 1998.55 477.35 2.6418

12 187.96 0.163 6.127 798.42 190.70 2782.73 664.64 1984.31 473.94 2.6878

13 191.60 0.151 6.617 814.68 194.58 2785.42 665.29 1970.73 470.70 2.7327

14 195.04 0.141 7.106 830.05 198.26 2787.79 665.85 1957.73 467.60 2.7767

15 198.28 0.132 7.596 844.64 201.74 2789.88 666.35 1945.24 464.61 2.8197

16 201.37 0.124 8.085 858.54 205.06 2791.73 666.79 1933.19 461.74 2.8620

17 204.30 0.117 8.575 871.82 208.23 2793.37 667.18 1921.55 458.95 2.9036

18 207.11 0.110 9.065 884.55 211.27 2794.81 667.53 1910.27 456.26 2.9445

19 209.79 0.105 9.556 896.78 214.19 2796.09 667.83 1899.31 453.64 2.9849

20 212.37 0.100 10.047 908.56 217.01 2797.21 668.10 1888.65 451.10 3.0248

21 214.85 0.095 10.539 919.93 219.72 2798.18 668.33 1878.25 448.61 3.0643

22 217.24 0.091 11.032 930.92 222.35 2799.03 668.54 1868.11 446.19 3.1034

23 219.55 0.087 11.525 941.57 224.89 2799.77 668.71 1858.20 443.82 3.1421

24 221.78 0.083 12.020 951.90 227.36 2800.39 668.86 1848.49 441.50 3.1805

25 223.94 0.080 12.515 961.93 229.75 2800.91 668.99 1838.98 439.23 3.2187

26 226.03 0.077 13.012 971.69 232.08 2801.35 669.09 1829.66 437.01 3.2567

27 228.06 0.074 13.509 981.19 234.35 2801.69 669.17 1820.50 434.82 3.2944

28 230.04 0.071 14.008 990.46 236.57 2801.96 669.24 1811.50 432.67 3.3320

29 231.96 0.069 14.508 999.50 238.73 2802.15 669.28 1802.65 430.56 3.3695

30 233.84 0.067 15.009 1008.33 240.84 2802.27 669.31 1793.94 428.48 3.4069

Coefficient of Discharge (C)

Contents

i. Theory of Discharge from an Orifice ii. Experiments

iii. References

Theory of Discharge from an Orifice

The Roman engineer Frontinus, who was in charge of the water supply under Augustus, used short pipes of graduated sizes to meter water delivered to different users. This was purely empirical, since the effects of pressure, or "head," and orifice size were not known quantitatively until Torricelli, in 1643, showed that the velocity of efflux was given by Vi = √2gh. We still calculate the

velocity from Bernoulli's principle, that h + p/ρg + V2/2, is a constant along a streamline in irrotational flow, which is equivalent to the conservation of energy.

We'll consider here the case of zero initial velocity, as at the surface of a liquid in a container with an orifice in the side. We assume that a streamline starts at the surface, a distance h above the orifice, and neglect the pressure on the surface of the liquid, since it would cancel out anyway. The streamline then leads somehow to the orifice, and out into the jet that issues from it. We choose the point at which the streamlines are parallel a short distance from the orifice, and find that the velocity there is Vi = √2gh, as given by Torricelli's theorem.

A jet surrounded only by air (or another fluid of small density) is called a free jet, and is acted upon by gravity. A jet surrounded by fluid is called a submerged jet. If the fluid is the same as that of the jet, then buoyancy eliminates the effect of gravity on it. A submerged jet is also subject to much greater friction at its boundary. We shall consider here only free jets of water, and neglect the viscosity of water, which is small, but finite.

A cross section of a circular orifice of diameter Do is shown. The thickness of the wall is assumed small compared to the diameter of the orifice. Because of the convergence of the streamlines approaching the orifice, the cross section of the jet decreases slightly until the pressure is equalized over the cross-section, and the velocity profile is nearly rectangular. This point of minimum area is called the vena contracta. Beyond the vena contracta, friction with the fluid outside the jet (air) slows it down, and the cross section increases perforce. This divergence is usually quite small, and the jet is nearly cylindrical with a constant velocity. The jet is held together by surface tension, of course, which has a stronger effect the smaller the diameter of the jet.

The area A of the vena contracta is smaller than the area Ao of the orifice because the velocity is higher there (converging streamlines). For a sharp-edged, or "ideal" circular orifice, A/Ao = Cc = π/(π + 2) = 0.611. Cc is called the coefficient of

contraction. For a sharp orifice, is usually estimated to be 0.62, a figure that can be used if the exact value is not known. For an orifice that resembles a short tube, Cc = 1, but then there are turbulence losses that affect the discharge.

The average velocity V is defined so that it gives the correct rate of discharge when it is assumed constant over the vena contracta, or Q = VA. Then, we can write V = CvVi, where Cv is the coefficient of velocity. The coefficient of velocity is usually quite high, between 0.95 and 0.99. Combining the results of this paragraph and the preceding one, the discharge Q = VA = CvViCcAo = CdAoVi. Cd, the coefficient of discharge, allows us to use the ideal velocity and the orifice area in calculating the discharge.

Experiments

Our apparatus consists of a tomato juice can with the top removed, and a hole near the bottom. With this can, a scale, and a timing device, we can measure the coefficients of discharge and velocity, and from them the coefficient of contraction, for the orifice. In this small-scale experiment, the influence of

surface tension, adhesion and capillarity will be greater than for most practical cases. Nevertheless, we shall obtain very instructive results at small expense and in little time.

The first experiment, to measure Cd, is performed by measuring the time required for the container to empty between levels h1 and h2 through the orifice. To find the rate at which the level h decreases, we equate the rate at which fluid is leaving the reservoir to the discharge of the

orifice: Acdh/dt = - CdAo√2gh, or dh/dt = -Kh1/2, where K = √(2g)Cd(Ao/Ac), or dh/h1/2 = -Kdt. Ac is the area of the cross-section of the can. Integrating and substituting the limits, we find 2(√h1 - √h2) = KT, where T is the time required.

The corrugations in the can make convenient reference points for the liquid level. For my experiment, Ao = 0.09932 cm2, Ac = 84.95 cm2, h1 = 10.5 cm, and h2 = 1.5 cm. Using these numbers, K = 0.0518Cd, and the difference in the square roots of the heights is 2.239, so T = 86.53/Cd seconds. The time was measured with my HP-48G calculator as 126.3 s. The result was Cd = 0.685.

The second experiment measures Cv. Water was allowed to run from the tap into the reservoir, keeping h constant at 16 cm. The height of the orifice was y = 10.0 cm, and the horizontal distance was 22.8 cm. Since x = vt and y = gt2/2, v = x√(g/2y). From this equation, V = 160 cm/s. By Torricelli's theorem, Vi = √(2gh) = 177 cm/s. Therefore, the coefficient of velocity was Cv = 160/177 = 0.90. Finally, the coefficient of contraction was 0.685/0.90 = 0.76. These figures are quite reasonable, if somewhat different from those for larger scales. Orifices are considered small if they have diameters less than 2.5 in and heads less that 3 ft, so these experiments are indeed small-scale, where effects that can be neglected on larger scales may play a role. These effects generally act to increase the coefficient of contraction, so our value is not out of line.

Other experiments and demonstrations suggest themselves. The discharge coefficient could also be found by keeping the head constant and measuring the water discharged in a known time interval. Orifices at different heights could be made to flow simultaneously, demonstrating the increase in velocity with head. Three holes would be appropriate. They should not be in the same vertical line, so that each jet would be independent. A square orifice could be considered. Care should be taken to make the orifice as accurate as possible. A short tube could be soldered at the hole. In particular, a Borda's mouthpiece, which is a short tube of length about equal to the radius of the orifice that projects into the reservoir. The ideal coefficient of contraction for a Borda's mouthpiece is 0.5. Flow separation should occur at the inner edge. These suggestions would require a little more preparation, but the experiments would still be very inexpensive and easy.

References

R. L. Daugherty and J. B. Franzini, Fluid Mechanics, 6th ed. (New York: McGraw-Hill, 1965). pp. 338-349.

The molecular weight of a substance, also called molecular mass, is the mass of one molecule of that substance, relative to the unified atomic mass unit u equal to 1/12 the mass of one atom of carbon-12.

Gas or Vapor Molecular Weight

Acetylene, C2H2 26.04

Air 28.966

Ammonia (R-717) 17.02

Argon, Ar 39.948

Benzene 78.11

N-Butane, C4H10 58.12

Iso-Butane (2-Metyl propane)

58.12

Butadiene 54.09

1-Butene 56.108

cis -2-Butene 56.108

trans-2-Butene 56.108

Isobutene 56.108

Carbon Dioxide, CO2 44.01

Carbon Disulphide 76.13

Carbon Monoxide, CO 28.011

Chlorine 70.906

Cyclohexane 84.16

Deuterium 2.014

Ethane, C2H6 30.070

Ethyl Alcohol 46.07

Ethyl Chloride 64.515

Ethylene, C2H4 28.054

Fluorine 37.996

Helium, He 4.02

N-Heptane 100.20

Hexane 86.17

Hydrochloric Acid 36.47

Hydrogen, H2 2.016

Hydrogen Chloride 36.461

Hydrogen Sulfide 34.076

Hydroxyl, OH 17.01

Krypton 83.80

Methane, CH4 16.044

Methyl Alcohol 32.04

Methyl Butane 72.15

Methyl Chloride 50.488

Natural Gas 19.00

Neon, Ne 20.179

Nitric Oxide, NO2 30.006

Nitrogen, N2 28.0134

Nitrous Oxide 44.012

N-Octane 114.22

Oxygen, O2 31.9988

Ozone 47.998

N-Pentane 72.15

Iso-Pentane 72.15

Propane, C3H8 44.097

Propylene 42.08

R-11 137.37

R-12 120.92

R-22 86.48

R-114 170.93

R-123 152.93

R-134a 102.03

R-611 60.05

Sulfur 32.02

Sulfur Dioxide (Sulphur Dioxide)

64.06

Sulfuric Oxide 48.1

Toluene 92.13

Xenon 131.30

Water Vapor - Steam, H2O

18.02

Reference: http://www.engineeringtoolbox.com/molecular-

weight-gas-vapor-d_1156.html

Exhaust gas velocity

As the gas enters a nozzle, it is traveling at subsonic velocities. As the throat contracts down the gas is forced to accelerate until

at the nozzle throat, where the cross-sectional area is the smallest, the linear velocity becomes sonic. From the throat the cross-

sectional area then increases, the gas expands and the linear velocity becomes progressively more supersonic.

The linear velocity of the exiting exhaust gases can be calculated using the following equation: [1] [2] [3]

where:  

Ve = Exhaust velocity at nozzle exit, m/s

T = absolute temperature of inlet gas, K

R = Universal gas law constant = 8314.5 J/(kmol·K)

M = the gas molecular mass, kg/kmol    (also known as the molecular weight)

k = cp/cv = isentropic expansion factor

cp = specific heat of the gas at constant pressure

cv = specific heat of the gas at constant volume

Pe = absolute pressure of exhaust gas at nozzle exit, Pa

P = absolute pressure of inlet gas, PaHow much pressure does a column of water 47 feet high exert?A column of water 2.31 feet high exerts a pressure of 1 pound per square inch.

Best Answer - Chosen by Voters

Each foot of water exerts a pressure of .433 psi so 47 feet of water column would exert .433 X 47 = 20.351 psi. 

This assumes you have pure water and not salt water.

One can argue about the density of water depending on your reference conditions but in general it is 62.2 to 62.4 pounds per cubic foot. This leads to a pressure per foot of height of 62.2/144 = .43194 psi to 62.4/144 = .4333 psi.

Reference: Engineering reference book.

Static Fluid Pressure

The pressure exerted by a static fluid depends only upon the depth of the fluid, the density of the fluid,

and the acceleration of gravity.

The pressure in a static fluid arises from the weight of the fluid and is given by the expression

Pstatic fluid = ρgh where

ρ = m/V = fluid densityg = acceleration of gravityh = depth of fluid

The pressure from the weight of a column of liquid of area A and height h is

The most remarkable thing about this expression is what it does not include. The fluid pressure at a given depth does not depend upon the total mass or total volume of the liquid. The above pressure expression is easy to see for the straight, unobstructed column, but not obvious for the cases of different geometry which are shown.

Because of the ease of visualizing a column height of a known liquid, it has become common practice to state all kinds of pressures in column height units, like mmHg or cm H2O, etc. Pressures are often measured by manometers in terms of a liquid column height.

Index

fluid pressure

calculation

Pressure concepts

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HyperPhysics*****  Mechanics   *****  Fluids R Nave

Fluid Pressure Calculation

Discussion

Fluid column height in the relationship

is often used for the measurement of pressure. After entering the relevant data, any one of the highlighted quantities below can be calculated by clicking on it. 

Pressure difference = density x g x height

If the fluid density is 

ρ =  gm/cm3 =  kg/m3

and the column height is 

h =   m =   x 10^   m

h =   ftthen the pressure difference is

ΔP =  kPa

ΔP =  lb/in2

ΔP =  mmHg=  inches Hg

ΔP =  atmos 

ΔP =  inches water=  cm water

Note that this static fluid pressure is dependent on density and depth only; it is independent of total mass, weight, volume, etc. of the fluid.

Reference: http://hyperphysics.phy-astr.gsu.edu/hbase/pflu.html

Rocket Nozzle Design: Optimizing Expansion for Maximum Thrust

A rocket engine is a device in which propellants are burned in a combustion chamber and the resulting high pressure gases are expanded through a specially shaped nozzle to produce thrust. The function of the nozzle is to convert the chemical-thermal energy generated in the combustion chamber into kinetic energy. The nozzle converts the slow moving, high pressure, high temperature gas in the combustion chamber into high velocity gas of lower pressure and temperature. Gas velocities from 2 to 4.5 kilometers per second can be obtained in rocket nozzles. The nozzles which perform this feat are called DeLaval nozzles (after the inventor) and consist of a convergent and divergent section. The minimum flow area between the convergent and divergent section is called the nozzle throat. The flow area at the end of the divergent section is called the nozzle exit area.

Hot exhaust gases expand in the diverging section of the nozzle. The pressure of these gases will decrease as energy is used to accelerate the gas to high velocity. The nozzle is usually made long enough (or the exit area great enough) such that the pressure in the combustion chamber is reduced at the nozzle exit to the pressure existing outside the nozzle. It is under this condition that thrust is maximum and the nozzle is said to be adapted, also called optimum or correct expansion. To understand this we must examine the basic thrust equation:

F = q × Ve + (Pe - Pa) × Ae

where F = Thrust q = Propellant mass flow rate Ve = Velocity of exhaust gases Pe = Pressure at nozzle exit Pa = Ambient pressure Ae = Area of nozzle exit

The product qVe is called the momentum, or velocity, thrust and the product (Pe-Pa)Ae is called the pressure thrust. As we have seen, Ve and Pe are inversely proportional, that is, as one increases the other decreases. If a nozzle is under-extended we have Pe>Pa and Ve is small. For an over-extended nozzle we have Pe<Pa and Ve is large. Thus, momentum thrust and pressure thrust are inversely proportional and, as we shall see, maximum thrust occurs when Pe=Pa.

Let us now consider an example. Assume we have a rocket engine equipped with an extendible nozzle. The engine is test fired in an

environment with a constant ambient pressure. During the burn, the nozzle is extended from its fully retracted position to its fully extended position. At some point between fully retracted and fully extended Pe=Pa (see figure below).

As we extend the nozzle, the momentum thrust increases as Ve increases. At the same time the pressure thrust decreases as Pe decreases. The increase in momentum thrust is greater than the decrease in pressure thrust, thus the total thrust of the engine increases as we approach the condition Pe=Pa. As we continue to extend to nozzle the situation changes slightly. Now the pressure thrust changes in magnitude more rapidly than the momentum thrust, thus the total thrust begins to decrease.

Let's now apply some numbers to our example and run through the calculations to prove that this is true. Assume our rocket engine operates under the following conditions:

q = Propellant mass flow rate = 100 kg/s k = Specific heat ratio = 1.20 M = Exhaust gas molecular weight = 24 Tc = Combustion chamber temperature = 3600 K Pc = Combustion chamber pressure = 5 MPa Pa = Ambient pressure = 0.05 MPa

If the nozzle is properly adapted to the operating conditions we have Pe=Pa, or Pe=0.05 MPa.

The gas pressure and temperature at the nozzle throat is less than in the combustion chamber due to the loss of thermal energy in accelerating the gas to the local speed of sound at the throat. Therefore, we calculate the pressure and temperature at the nozzle throat,

Pt = Pc × [1 + (k - 1) / 2]-k/(k-1)

Pt = 5 × [1 + (1.20 - 1) / 2]-1.20/(1.20-1)

Pt = 2.82 MPa = 2.82x106 N/m2

Tt = Tc × [1 / (1 + (k - 1) / 2)] Tt = 3,600 × [1 / (1 + (1.20 - 1) / 2)] Tt = 3,273 K

The area at the nozzle throat is given by

At = (q / Pt) × SQRT[ (R' × Tt) / (M × k) ] At = (100 / 2.82x106) × SQRT[ (8,314 × 3,273) / (24 × 1.20) ] At = 0.0345 m2

The hot gases must now be expanded in the diverging section of the nozzle to obtain maximum thrust. The Mach number at the nozzle exit is given by

Nm2 = (2 / (k - 1)) × [(Pc / Pa)(k-1)/k - 1] Nm2 = (2 / (1.20 - 1)) × [(5 / 0.05)(1.20-1)/1.20 - 1] Nm2 = 11.54 Nm = (11.54)1/2 = 3.40

The nozzle exit area corresponding to the exit Mach number is given by

Ae = (At / Nm) × [(1 + (k - 1) / 2 × Nm2)/((k + 1) / 2)](k+1)/(2(k-1))

Ae = (0.0345 / 3.40) × [(1 + (1.20 - 1) / 2 × 11.54)/((1.20 + 1) / 2)](1.20+1)/(2(1.20-1))

Ae = 0.409 m2

The velocity of the exhaust gases at the nozzle exit is given by

Ve = SQRT[ (2 × k / (k - 1)) × (R' × Tc / M) × (1 - (Pe / Pc)(k-1)/k) ] Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.05 / 5)(1.20-1)/1.20) ] Ve = 2,832 m/s

Finally, we calculate the thrust,

F = q × Ve + (Pe - Pa) × Ae F = 100 × 2,832 + (0.05x106 - 0.05x106) × 0.409 F = 283,200 N

Let's now consider what happens when the nozzle is under-extended, that is Pe>Pa. If we assume Pe=Pa × 2, we have

Pe = 0.05 × 2 = 0.10 MPa At = 0.0345 m2

Nm2 = (2 / (1.20 - 1)) × [(5 / 0.10)(1.20-1)/1.20 - 1] Nm2 = 9.19 Nm = (9.19)1/2 = 3.03

Ae = (0.0345 / 3.03) × [(1 + (1.20 - 1) / 2 × 9.19)/((1.20 + 1) / 2)](1.20+1)/(2(1.20-1))

Ae = 0.243 m2

Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.10 / 5)(1.20-1)/1.20) ] Ve = 2,677 m/s

F = 100 × 2,677 + (0.10x106 - 0.05x106) × 0.243 F = 279,850 N

Now we consider the over-extended condition, that is Pe<Pa. If we assume Pe=Pa / 2, we have

Pe = 0.05 / 2 = 0.025 MPa At = 0.0345 m2

Nm2 = (2 / (1.20 - 1)) × [(5 / 0.025)(1.20-1)/1.20 - 1] Nm2 = 14.18 Nm = (14.18)1/2 = 3.77

Ae = (0.0345 / 3.77) × [(1 + (1.20 - 1) / 2 × 14.18)/((1.20 + 1) / 2)](1.20+1)/(2(1.20-1))

Ae = 0.696 m2

Ve = SQRT[ (2 × 1.20 / (1.20 - 1)) × (8,314 × 3,600 / 24) × (1 - (0.025 / 5)(1.20-1)/1.20) ] Ve = 2,963 m/s

F = 100 × 2,963 + (0.025x106 - 0.05x106) × 0.696 F = 278,900 N

We see that both the under-extended and over-extended nozzles produce thrusts less than that produced when the condition Pe=Pa is satisfied. When we plot a graph of total thrust versus the ratio Pa/Pe we obtain the following:

As can be easily seen, thrust is maximum when Pa/Pe=1, or when Pe=Pa.

Reference: http://www.braeunig.us/space/sup1.htm