N B11 = 0.801 x 4 x 2.75×10 22 = 8.80×10. The enrichment may be specified in terms of weight percent or weight fraction, ω i, of isotope i: The atomic number. Since (1-x) = (1 - 0.996) = 0.004, multiply by 100: nitrogen-15 is 0.4%. The abundance of the nitrogen-14 isotope is 99.6 percent, and the abundance of the nitrogen-15 isotope is 0.4 percent. Calculating Relative Abundance in Mass Spectroscopy.
8.1 Time and rotational frequency
The SI unit of time (actually time interval) is the second (s) and should be used in all technical calculations. When time relates to calendar cycles, the minute (min), hour (h), and day (d) might be necessary. For example, the kilometer per hour (km/h) is the usual unit for expressing vehicular speeds. Although there is no universally accepted symbol for the year, Ref. [4: ISO 80000-3] suggests the symbol a.
The rotational frequency n of a rotating body is defined to be the number of revolutions it makes in a time interval divided by that time interval [4: ISO 80000-3]. The SI unit of this quantity is thus the reciprocal second (s-1). However, as pointed out in Ref. [4: ISO 80000-3], the designations 'revolutions per second' (r/s) and 'revolutions per minute' (r/min) are widely used as units for rotational frequency in specifications on rotating machinery.
8.2 Volume
The SI unit of volume is the cubic meter (m3) and may be used to express the volume of any substance, whether solid, liquid, or gas. The liter (L) is a special name for the cubic decimeter (dm3), but the CGPM recommends that the liter not be used to give the results of high accuracy measurements of volumes [1, 2]. Also, it is not common practice to use the liter to express the volumes of solids nor to use multiples of the liter such as the kiloliter (kL) [see Sec. 6.2.8, and also Table 6, footnote (b)].
8.3 Weight
In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame [4: ISO 80000-4]. Thus the SI unit of the quantity weight defined in this way is the newton (N). When the reference frame is a celestial object, Earth for example, the weight of a body is commonly called the local force of gravity on the body.
Example: The local force of gravity on a copper sphere of mass 10 kg located on the surface of the Earth, which is its weight at that location, is approximately 98 N.
Note: The local force of gravity on a body, that is, its weight, consists of the resultant of all the gravitational forces acting on the body and the local centrifugal force due to the rotation of the celestial object. The effect of atmospheric buoyancy is usually excluded, and thus the weight of a body is generally the local force of gravity on the body in vacuum.
Example: The local force of gravity on a copper sphere of mass 10 kg located on the surface of the Earth, which is its weight at that location, is approximately 98 N.
Note: The local force of gravity on a body, that is, its weight, consists of the resultant of all the gravitational forces acting on the body and the local centrifugal force due to the rotation of the celestial object. The effect of atmospheric buoyancy is usually excluded, and thus the weight of a body is generally the local force of gravity on the body in vacuum.
In commercial and everyday use, and especially in common parlance, weight is usually used as a synonym for mass. Thus the SI unit of the quantity weight used in this sense is the kilogram (kg) and the verb 'to weigh' means 'to determine the mass of' or 'to have a mass of.'
Examples: the child's weight is 23 kg the briefcase weighs 6 kg Net wt. 227 g
Inasmuch as NIST is a scientific and technical organization, the word 'weight' used in the everyday sense (that is, to mean mass) should appear only occasionally in NIST publications; the word 'mass' should be used instead. In any case, in order to avoid confusion, whenever the word 'weight' is used, it should be made clear which meaning is intended.
Examples: the child's weight is 23 kg the briefcase weighs 6 kg Net wt. 227 g
Inasmuch as NIST is a scientific and technical organization, the word 'weight' used in the everyday sense (that is, to mean mass) should appear only occasionally in NIST publications; the word 'mass' should be used instead. In any case, in order to avoid confusion, whenever the word 'weight' is used, it should be made clear which meaning is intended.
8.4 Relative atomic mass and relative molecular mass
The terms atomic weight and molecular weight are obsolete and thus should be avoided. They have been replaced by the equivalent but preferred terms relative atomic mass, symbol Ar, and relative molecular mass, symbol Mr, respectively [4: ISO 31-8], which better reflect their definitions. Similar to atomic weight and molecular weight, relative atomic mass and relative molecular mass are quantities of dimension one and are expressed simply as numbers. The definitions of these quantities are as follows [4: ISO 31-8]:
Relative atomic mass (formerly atomic weight): ratio of the average mass per atom of an element to 1/12 of the mass of the atom of the nuclide 12C.
Relative molecular mass (formerly molecular weight): ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide 12C.
Examples:Ar(Si) = 28.0855, Mr(H2) = 2.0159, Ar(12C) = 12 exactly
Notes:
1. It follows from these definitions that if X denotes a specified atom or nuclide and B a specified molecule or entity (or more generally, a specified substance), then Ar(X) = m(X) / [m(12C) / 12] and Mr(B) = m(B) / [m(12C) / 12], where m(X) is the mass of X, m(B) is the mass of B, and m(12C) is the mass of an atom of the nuclide 12C. It should also be recognized that m(12C) / 12 = u, the unified atomic mass unit, which is approximately equal to 1.66 3 10-27 kg [see Table 7, footnote (d)].
2. It follows from the examples and note 1 that the respective average masses of Si, H2, and 12C are m(Si) = Ar(Si) u, m (H2) = Mr(H2) u, and m(12C) = Ar(12C) u.
3. In publications dealing with mass spectrometry, one often encounters statements such as 'the mass-to-charge ratio is 15.' What is usually meant in this case is that the ratio of the nucleon number (that is, mass number—see Sec. 10.4.2) of the ion to its number of charges is 15. Thus mass-to-charge ratio is a quantity of dimension one, even though it is commonly denoted by the symbol m / z. For example, the mass-to-charge ratio of the ion 12C71H7+ + is 91/2 = 45.5.
8.5 Temperature interval and temperature difference
As discussed in Sec. 4.2.1.1, Celsius temperature (t) is defined in terms of thermodynamic temperature (T) by the nist-equation t = T - T0, where T0 = 273.15 K by definition. This implies that the numerical value of a given temperature interval or temperature difference whose value is expressed in the unit degree Celsius (°C) is equal to the numerical value of the same interval or difference when its value is expressed in the unit kelvin (K); or in the notation of Sec. 7.1, note 2, {Δt }°C = {ΔT}K. Thus temperature intervals or temperature differences may be expressed in either the degree Celsius or the kelvin using the same numerical value.
Example: The difference in temperature between the freezing point of gallium and the triple point of water is Δt = 29.7546 °C = ΔT = 29.7546 K.
8.6 Amount of substance, concentration, molality, and the like
The following section discusses amount of substance, and the subsequent nine sections, which are based on Ref. [4: ISO 31-8] and which are succinctly summarized in Table 12, discuss quantities that are quotients involving amount of substance, volume, or mass. In the table and its associated sections, symbols for substances are shown as subscripts, for example, xB, nB, bB. However, it is generally preferable to place symbols for substances and their states in parentheses immediately after the quantity symbol, for example n(H2SO4). (For a detailed discussion of the use of the SI in physical chemistry, see the book cited in Ref.[6], note 3.)
8.6.1 Amount of substance
Quantity symbol: n (also v). SI unit: mole (mol).
Definition: See Sec. A.7.
Definition: See Sec. A.7.
Notes:
1. Amount of substance is one of the seven base quantities upon which the SI is founded (see Sec. 4.1 and Table 1).
2. In general, n(xB) = n(B) / x, where x is a number. Thus, for example, if the amount of substance of H2SO4 is 5 mol, the amount of substance of (1/3)H2SO4 is 15 mol: n[(1/3) H2 SO 4] = 3n(H2SO4).
1. Amount of substance is one of the seven base quantities upon which the SI is founded (see Sec. 4.1 and Table 1).
2. In general, n(xB) = n(B) / x, where x is a number. Thus, for example, if the amount of substance of H2SO4 is 5 mol, the amount of substance of (1/3)H2SO4 is 15 mol: n[(1/3) H2 SO 4] = 3n(H2SO4).
Example: The relative atomic mass of a fluorine atom is Ar(F) = 18.9984. The relative molecular mass of a fluorine molecule may therefore be taken as Mr(F2) = 2Ar(F) = 37.9968. The molar mass of F2 is then M(F2) = 37.9968 × 10-3 kg/mol = 37.9968 g/mol (see Sec. 8.6.4). The amount of substance of, for example, 100 g of F2 is then n(F2) = 100 g / (37.9968 g/mol) = 2.63 mol.
8.6.2 Mole fraction of B; amount-of-substance fraction of B
Quantity symbol:xB (also yB). SI unit: one (1) (amount-of-substance fraction is a quantity of dimension one).
Definition: ratio of the amount of substance of B to the amount of substance of the mixture: xB = nB/n.
Quantity in numerator | ||||
---|---|---|---|---|
Amount of substance Symbol: n SI unit: mol | Volume Symbol: V SI unit: m3 | Mass Symbol: m SI unit: kg | ||
Quantity in denominator | Amount of substance Symbol: n SI unit: mol | amount-of-substance fraction $$ x_{rm B} = frac{n_{rm B}}{n} $$ SI unit: mol/mol = 1 | molar volume $$V_{rm m} = frac{V}{n} $$ SI unit: m3/mol | molar mass $$ M = frac{m}{n} $$ SI unit: kg/mol |
Volume Symbol: V SI unit: m3 | amount-of-substance concentration $$ c_{rm B} = frac{n_{rm B}}{V} $$ SI unit: mol/m3 | volume fraction $$varphi_{rm B} = frac{x_{rm B} V_{rm m,B}^* }{Sigma x_{rm A} V_{rm m,A}^*}$$ SI unit: | mass density $$ rho = frac{m}{V}$$ SI unit: kg/m3 | |
Mass Symbol: m SI unit: kg | molality $$ b_{rm B} = frac{n_{rm B}}{m_{rm A}}$$ SI unit: mol/kg | specific volume $$v = frac{V}{m}$$ SI unit: m3/kg | mass fraction $$v = frac{V}{m}$$ | |
Adapted from Canadian Metric Practice Guide (see Ref. [8], note 3; the book cited in Ref. [8], note 5, may also be consulted). |
Notes:
1. This quantity is commonly called 'mole fraction of B' but this Guide prefers the name 'amount of- substance fraction of B,' because it does not contain the name of the unit mole (compare kilogram fraction to mass fraction).
2. For a mixture composed of substances A, B, C, . . . , nA + nB + nC + .. $$equiv sum_{rm A} n_{rm A}$$
3. A related quantity is amount-of-substance ratio of B (commonly called 'mole ratio of solute B'), symbol rB. It is the ratio of the amount of substance of B to the amount of substance of the solvent substance: rB = nB/nS. For a single solute C in a solvent substance (a one-solute solution), rC = xC/(1 - xC). This follows from the relations n = nC + nS, xC = nC / n, and rC = nC / nS, where the solvent substance S can itself be a mixture.
8.6.3 Molar volume
Quantity symbol: Vm. SI unit: cubic meter per mole (m3/mol).
Definition: volume of a substance divided by its amount of substance: Vm = V/n.
Definition: volume of a substance divided by its amount of substance: Vm = V/n.
Notes:
1. The word 'molar' means 'divided by amount of substance.'
2. For a mixture, this term is often called 'mean molar volume.' Disk cleaner 1 6 1.
3. The amagat should not be used to express molar volumes or reciprocal molar volumes. (One amagat is the molar volume Vm of a real gas at p = 101 325 Pa and T = 273.15 K and is approximately equal to 22.4 × 10-3 m3/mol. The name 'amagat' is also given to 1/Vm of a real gas at p = 101 325 Pa and T = 273.15 K and in this case is approximately equal to 44.6 mol/m3.) solvent substance S can itself be a mixture.
8.6.4 Molar mass
Quantity symbol:M. SI unit: kilogram per mole (kg/mol).
Definition: mass of a substance divided by its amount of substance: M = m/n.
Notes:
1. For a mixture, this term is often called 'mean molar mass.'
2. The molar mass of a substance B of definite chemical composition is given by M(B) = Mr(B) × 10-3 kg/mol = Mr(B) kg/kmol = Mr g/mol, where Mr(B) is the relative molecular mass of B (see Sec. 8.4). The molar mass of an atom or nuclide X is M(X) = Ar(X) × 10-3 kg/mol = Ar(X) kg/kmol = Ar(X) g/mol, where Ar(X) is the relative atomic mass of X (see Sec. 8.4).
8.6.5 Concentration of B; amount-of-substance concentration of B
Quantity symbol:cB. SI unit: mole per cubic meter (mol/m3).
Definition: amount of substance of B divided by the volume of the mixture: cB = nB/V.
Notes:
1. This Guide prefers the name 'amount-of-substance concentration of B' for this quantity because it is unambiguous. However, in practice, it is often shortened to amount concentration of B, or even simply to concentration of B. Unfortunately, this last form can cause confusion because there are several different 'concentrations,' for example, mass concentration of B, ρB = mB/V; and molecular concentration of B, CB = NB/V, where NB is the number of molecules of B.
2. The term normality and the symbol N should no longer be used because they are obsolete. One should avoid writing, for example, 'a 0.5 N solution of H2SO4' and write instead 'a solution having an amount-of-substance concentration of c [(1/2)H2SO4]) = 0.5 mol/dm3' (or 0.5 kmol/m3 or 0.5 mol/L since 1 mol/dm3 = 1 kmol/m3 = 1 mol/L).
3. The term molarity and the symbol M should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of B and such units as mol/dm3, kmol/m3, or mol/L. (A solution of, for example, 0.1 mol/dm3 was often called a 0.1 molar solution, denoted 0.1 M solution. The molarity of the solution was said to be 0.1 M.)
8.6.6 Volume fraction of B
Quantity symbol: φB. SI unit: one (1) (volume fraction is a quantity of dimension one).
Definition: for a mixture of substances A, B, C, . . . ,
$$varphi_{rm B} = x_{rm B} V_{rm m,B}^* /sum x_{rm A} V_{rm m,A}^*$$
where xA, xB, xC, . . . are the amount-of-substance fractions of A, B, C, . . ., V*m,A , V*m,B , V*m,C , . . . are the molar volumes of the pure substances A, B, C, . . . at the same temperature and pressure, and where the summation is over all the substances A, B, C, . . . so that ΣxA = 1.
8.6.7 Mass density; density
Quantity symbol: ρ. SI unit: kilogram per cubic meter (kg/m3). World clock deluxe 4 10 1.
Definition: mass of a substance divided by its volume: ρ = m / V.
Notes:
1. This Guide prefers the name 'mass density' for this quantity because there are several different 'densities,' for example, number density of particles, n = N / V; and charge density, ρ = Q / V.
2. Mass density is the reciprocal of specific volume (see Sec. 8.6.9): ρ = 1 / ν.
8.6.8 Molality of solute B
Quantity symbol:bB (also mB). SI unit: mole per kilogram (mol/kg).
Definition: amount of substance of solute B in a solution divided by the mass of the solvent: bB = nB / mA.
Note: The term molal and the symbol m should no longer be used because they are obsolete. One should use instead the term molality of solute B and the unit mol/kg or an appropriate decimal multiple or submultiple of this unit. (A solution having, for example, a molality of 1 mol/kg was often called a 1 molal solution, written 1 m solution.)
8.6.9 Specific volume
Quantity symbol:ν. SI unit: cubic meter per kilogram (m3/kg).
Definition: volume of a substance divided by its mass: ν = V / m.
Note: Specific volume is the reciprocal of mass density (see Sec. 8.6.7): ν = 1 / ρ.
8.6.10 Mass fraction of B
Quantity symbol:wB. SI unit: one (1) (mass fraction is a quantity of dimension one).
Definition: mass of substance B divided by the mass of the mixture: wBB = mB / m.
8.7 Logarithmic quantities and units: level, neper, bel
This section briefly introduces logarithmic quantities and units. It is based on Ref. [5: IEC 60027-3], which should be consulted for further details. Two of the most common logarithmic quantities are level-of a-field-quantity, symbol LF, and level-of-a-power-quantity, symbol LP; and two of the most common logarithmic units are the units in which the values of these quantities are expressed: the neper, symbol Np, or the bel, symbol B, and decimal multiples and submultiples of the neper and bel formed by attaching SI prefixes to them, such as the millineper, symbol mNp (1 mNp = 0.001 Np), and the decibel, symbol dB (1 dB = 0.1 B).
Level-of-a-field-quantity is defined by the relation LF = ln(F/F0), where F/F0 is the ratio of two amplitudes of the same kind, F0 being a reference amplitude. Level-of-a-power-quantity is defined by the relation LP = (1/2) ln(P/P0), where P/P0 is the ratio of two powers, P0 being a reference power. (Note that if P/P0 = (F/F0)2, then LP = LF.) Similar names, symbols, and definitions apply to levels based on other quantities which are linear or quadratic functions of the amplitudes, respectively. In practice, the name of the field quantity forms the name of LF and the symbol F is replaced by the symbol of the field quantity. For example, if the field quantity in question is electric field strength, symbol E, the name of the quantity is 'level-of-electric-field-strength' and it is defined by the relation LE = ln(E/E0).
The difference between two levels-of-a-field-quantity (called 'field-level difference') having the same reference amplitude F0 is ΔLF = LF1 - LF2 = ln(F1/F0) - ln(F2/F0) = ln(F1/F2), and is independent of F0. This is also the case for the difference between two levels-of-a-power-quantity (called 'power-level difference') having the same reference power P0: ΔLP1 = LP2 = ln(P1/P0) - ln(P2/P0) = ln(P1/P2).
It is clear from their definitions that both LF and LP are quantities of dimension one and thus have as their units the unit one, symbol 1. However, in this case, which recalls the case of plane angle and the radian (and solid angle and the steradian), it is convenient to give the unit one the special name 'neper' or 'bel' and to define these so-called dimensionless units as follows:
One neper (1 Np) is the level-of-a-field-quantity when F/F0 = e, that is, when ln(F/F0) = 1. Equivalently, 1 Np is the level-of-a-power-quantity when P/P0 = e2, that is, when (1/2) ln(P/P0) = 1. These definitions imply that the numerical value of LF when LF is expressed in the unit neper is {LF}Np = ln(F/F0), and that the numerical value of LP when LP is expressed in the unit neper is {LP}Np = (1/2) ln(P/P0); that is
LF = ln(F/F0) Np
LP = (1/2) ln(P/P0) Np.
LP = (1/2) ln(P/P0) Np.
One bel (1 B) is the level-of-a-field-quantity when $$F/F_0 = sqrt{10}$$ that is, when 2 lg(F/F0) = 1 (note that lg x = log10x – see Sec. 10.1.2). Equivalently, 1 B is the level- of-a-power-quantity when P/P0 = 10, that is, when lg(P/P0) = 1. These definitions imply that the numerical value of LF when LF is expressed in the unit bel is {LF}B = 2 lg(F/F0) and that the numerical value of LP when LP is expressed in the unit bel is {LP}B = lg(P/P0); that is
LF = 2 lg(F/F0) B = 20 lg(F/F0) dB LP = lg(P/P0) B = 10 lg(P/P0) dB.
Since the value of LF (or LP) is independent of the unit used to express that value, one may equate LF in the above expressions to obtain ln(F/F0) Np = 2 lg(F/F0) B, which implies
$$begin{eqnarray*} 1~{rm B}&=&frac{ln 10}{2} ~ {rm Np~exactly} & approx&1.151 , 293 ~ {rm Np} 1~{rm dB} &approx& 0.115 , 129 , 3 ~ {rm Np} ~ . end{eqnarray*}$$
When reporting values of LF and LP, one must always give the reference level. According to Ref. 5:IEC 60027-3, this may be done in one of two ways: Lx (re xref) or L x / xref where x is the quantity symbol for the quantity whose level is being reported, for example, electric field strength E or sound pressure p, and xref is the value of the reference quantity, for example, 1 μV/m for E0, and 20 μPa for p0. Thus
$$begin{eqnarray*} 1~{rm B}&=&frac{ln 10}{2} ~ {rm Np~exactly} & approx&1.151 , 293 ~ {rm Np} 1~{rm dB} &approx& 0.115 , 129 , 3 ~ {rm Np} ~ . end{eqnarray*}$$
When reporting values of LF and LP, one must always give the reference level. According to Ref. 5:IEC 60027-3, this may be done in one of two ways: Lx (re xref) or L x / xref where x is the quantity symbol for the quantity whose level is being reported, for example, electric field strength E or sound pressure p, and xref is the value of the reference quantity, for example, 1 μV/m for E0, and 20 μPa for p0. Thus
LE (re 1 μV/m) = - 0.58 Np or LE/(1 μV/m) = - 0.58 Np
means that the level of a certain electric field strength is 0.58 Np below the reference electric field strength E0 = 1 μV/m. Similarly
Lp (re 20 μPa) = 25 dB or Lp/(20 μPa) = 25 dB
means that the level of a certain sound pressure is 25 dB above the reference pressure p0 = 20 μPa.
Notes:
1. When such data are presented in a table or in a figure, the following condensed notation may be used instead: - 0.58 Np (1 μV/m); 25 dB (20 μPa).
2. When the same reference level applies repeatedly in a given context, it may be omitted if its value is clearly stated initially and if its planned omission is pointed out.
3. The rules of Ref. [5: IEC 60027-3] preclude, for example, the use of the symbol dBm to indicate a reference level of power of 1 mW. This restriction is based on the rule of Sec. 7.4, which does not permit attachments to unit symbols.
0.1 Fraction Form
8.8 Viscosity
The proper SI units for expressing values of viscosity η (also called dynamic viscosity) and values of kinematic viscosity ν are, respectively, the pascal second (Pa·s) and the meter squared per second (m2/s) (and their decimal multiples and submultiples as appropriate). The CGS units commonly used to express values of these quantities, the poise (P) and the stoke (St), respectively [and their decimal submultiples the centipoise (cP) and the centistoke (cSt)], are not to be used; see Sec. 5.3.1 and Table 10, which gives the relations 1 P = 0.1 Pa·s and 1 St = 10-4 m2/s.
8.9 Massic, volumic, areic, lineic
Reference [4: ISO 31-0] has introduced the new adjectives 'massic,' 'volumic,' 'areic,' and 'lineic' into the English language based on their French counterparts: 'massique,' 'volumique,' 'surfacique,' and 'linéique.' They are convenient and NIST authors may wish to use them. They are equivalent, respectively, to 'specific,' 'density,' 'surface . . . density,' and 'linear . . . density,' as explained below.
(a) The adjective massic, or the adjective specific, is used to modify the name of a quantity to indicate the quotient of that quantity and its associated mass.
Examples:
massic volume or specific volume: ν = V / m
massic entropy or specific entropy: s = S / m
(b) The adjective volumic is used to modify the name of a quantity, or the term density is added to it, to indicate the quotient of that quantity and its associated volume.
Examples:
volumic mass or (mass) density: ρ = m / V
volumic number or number density: n = N / V
Note: Parentheses around a word means that the word is often omitted.
(c) The adjective areic is used to modify the name of a quantity, or the terms surface . . . density are added to it, to indicate the quotient of that quantity (a scalar) and its associated surface area.
Examples:
areic mass or surface (mass) density: ρA = m / A
areic charge or surface charge density: σ = Q / A
(d) The adjective lineic is used to modify the name of a quantity, or the terms linear . . . density are added to it, to indicate the quotient of that quantity and its associated length.
Examples:
lineic mass or linear (mass) density: ρl = m / l
lineic electric current or linear electric current density: A = I / b
Introduction
If your spacecraft design needs more delta-V a glance at the Tsiolkovsky rocket equation tells you you have to increase the exhaust velocity or increase the mass ratio. Or both. Since the exhaust velocity depends upon the propulsion system, often the only option is to somehow raise the mass ratio.
Staging is a desperate attempt to increase the rocket's mass ratio in order to increase the delta V to a point where the rocket can perform the desired mission. For a single stage rocket, it is very difficult to get the mass ratio above 15, and it is probably impossible to get it above 20. After all, there is only so much structural mass you can shave off before the force of acceleration will make the struts snap like toothpicks and the tanks pop like balloons.
Without staging, there is also a tendency to make huge rockets, due to the constraints of 'minimum gauge'.
Pournelle goes on to mention that multi stage rockets make rocket designers subconsciously think of spacecraft as 'ammunition' instead of as 'aircraft', that is, disposable as opposed to reusable.
By using staging, the mass ratio of the rocket as a whole can be above 15, even though the mass ratio of each individual stage is below 15. When the current lowest stage has expended all its reaction mass, that stage is discarded. This eliminates the dead weight of that stage's engines, tanks, and structural members. In essence, in a three stage rocket, stage one is a rocket who's payload is another rocket composed of stages two and three.
Back in the old days instead of the term 'staging' the common term was 'step rockets.'
You generally only see staging in extreme circumstances, such as trying to boost a worth-while payload into Terra's orbit with anemic chemical rockets (though in Joan Vinge's THE OUTCASTS OF HEAVEN BELT there was the High Delta V Defense Force, which was an orbit to orbit flotilla of staged vehicles). You do not need staging if you are boosting payloads with atomic rockets. Nor do you need them in deep space, where you have the luxury of using those propulsion systems with high exhaust velocities (and high specific impulses) which regrettably have a thrust to weight ratio below one and thus cannot be used for lift-off.
Which is good because no self respecting Rocketeer wants to ride a disintegrating totem pole if they can possible avoid it. Not to mention the cost if the discarded stages cannot be recovered. Staging also decreases reliability, as you are multiplying the number of propulsion units that can malfunction.
In theory, one could jettison individual tanks and engines as the tanks went dry, but in practice it is more efficient to jettison an entire engine cluster. Having said that, this is the method used by the Space Shuttle with the strap on solid rocket boosters, and by the Otrag. This is called 'Parallel Staging'.
With staged chemical rockets, the exhaust velocity for the engines of each stage is often different. This is due to complicated optimizations that I will only lightly touch on (since I do not understand it myself). However, I get the impression that most of these optimizations only apply to lift-off, not to deep space missions.
With staging, to figure the effective mass ratio of the entire rocket, you take the individual mass ratios of each stage and multiply them. Now you know why staging is so popular with rocket scientists, despite the many draw backs.
When figuring the delta V contribution of a given stage, you use the standard delta V equation, but when figuring the mass ratio you use the total mass of the rocket after discarding the previous stage to plug into the equation.
The total delta V of the multi-stage rocket is obviously the sum of the delta V contributions of each of the stages. Since delta V is Ve * ln[R], a three stage rocket's total delta V would be:
Δv = (Ve1 * ln[R1]) + (Ve2 * ln[R2]) + (Ve3 * ln[R3])
where
- Δv = ship's total deltaV capability (m/s)
- Ve1 = exhaust velocity of propulsion system in stage 1 (m/s)
- Ve2 = exhaust velocity of propulsion system in stage 2 (m/s)
- Ve3 = exhaust velocity of propulsion system in stage 3 (m/s)
- R1 = mass ratio of propellant in stage 1 to (dry mass of stage 1 plus total mass of stage 2 and 3)
- R2 = mass ratio of propellant in stage 2 to (dry mass of stage 2 plus total mass of 3)
- R3 = mass ratio of propellant in stage 3 to dry mass of stage 3
Your next question is how does one determine the optimum mass ratios of each stage. I was afraid you'd ask that. It's a bit complicated. You are given the Δv requirements for the mission Δvm, and the mass of the actual mission payload (the weather satellite, the ICBM, the Apollo command and lunar module, the interplanetary warship, the orbital fighter, whatever).
To optimize a multi-stage rocket, go through the following steps, taken from the SMAD. You'll quickly discover why they call it 'rocket science.'
- Choose the number of stages
- Choose specific impulse for each stage
- Choose the inert-mass fraction for each stage
- Allocate a fraction of Δv to each stage
- Size the stages and the vehicle
- Minimize the vehicle mass by optimizing the Δv fraction alloted to each stage
This will be called nstage. Choose the minimum number of stages that is practical. Choose different values for nstage and compare the marginal differences.
The primary choice is the highest specific impulse, and if this is for lift-off you'll need a thrust to weight ratio above one.
But sometimes it is advantageous to use a propellant that has a higher density (which generally means a lower specific impulse). Sometimes if a higher density propellant is used for the lower stages, it will give you a better inert-mass fraction, which gives you a lighter first stage. But only sometimes. In other cases it may give you a lighter first stage but a heavier total vehicle mass. You will have to try and see.
A stage's 'inert mass' is the total mass of the stage minus the mass of the propellant and the mass of the payload. Inert mass is used because while we know the payload mass, we generally have no way of knowing the structural mass.
To find the 'inert mass fraction', divide the inert mass by the total mass of the stage.
In existing chemical rocket designs, the inert-mass fraction of a given stage tends to be between 0.08 and 0.7.
The goal is to divide up the Δv contributions in such a way as to minimize the vehicle mass. But the only way to discover this is by trying different combinations (see step 6 below).
In the special case where each stage has the same inert-mass fractions and the same specific impulse, they will also have the same Δv fraction. The fraction will be 1 / nstage. For example, if there are four stages, each stage will contribute 1/4 = 0.25 = 25% of the total Δv.
Start with the final (uppermost) stage and work backwards to the first stage. The payload of the uppermost stage is the actual mission payload. The payload for each lower stage includes all the previous upper stages plus the actual mission payload.
So, starting with the uppermost final stage, the payload mass is equal to the actual mission payload. The specific impulse (Isp) for this stage comes from step 2. The inert-mass fraction (finert) for this stage comes from step 3. And the Δv fraction (fΔv) for this stage comes from step 4.
Using those values, we can calculate the propellant mass, the inert mass, and the initial mass of that stage.
To calculate the propellant mass: specific Impulse gives us the Exhaust Velocity:
Ve = g0 * Isp
Δv fraction and mission Δv gives us the stage Δv:
Δvs = fΔv * Δvm
Δvs and Exhaust Velocity gives us the Mass Ratio:
R = e(Δvs/Ve)
Payload mass, mass ratio, and inert mass fraction gives us the Mass of the Propellant:
Mpt = (Mpl * (R-1) * (1 - finert)) / (1 - (finert * R))where:
- Ve = Exhaust velocity (m/s)
- Δvm = total mission Δv
- fΔv = Δv fraction for this stage
- Δvs = Δv for this stage
- g0 = Acceleration due to gravity at Earth's surface = 9.81 (m/s2)
- Isp = Specific impulse (s)
- ex = the inverse of the Ln() function, called 'exp(x)' in spreadsheets
- finert = inert-mass fraction
- Mpl = mass of 'payload' (kg)
- Mpt = mass of propellant (kg)
To calculate the inert mass:
Minert = (finert / (1 - finert)) * Mpt
Finally, to calculate the initial mass of the stage:
M = Mpl + Mpt + Minert
For the next lower, and all subsequently lower stages, use the value of the previous stage's initial mass M in the place of the payload mass Mpl, use the new stage's values forΔv fraction, specific impulse, and inert-mass fractions. Re-calculate all the equationswith the new values.
Keep doing this until you have finished all the stages. The initial mass M of the last stage is the vehicle's initial mass. The goal is to get the vehicle's initial mass to be as low as possible.
Minimize the vehicle mass by optimizing the delta-V fraction alloted to each stage
![Atomic 1 0 4 Fraction Atomic 1 0 4 Fraction](https://media.cheggcdn.com/media/f6f/f6fdd1cd-8cc8-4f4a-a906-e5a36bbfd4b7/phpmr870i.png)
The Δv fractions of each of the stages must be varied in order to determine the combination that minimizes the vehicle's initial mass. This generally is done with an 'iterative process,' i.e., a computer program, a grad student, or unlucky you has to repeatedly run various combinations of Δv fractions through steps 4 and 5 until the minimum initial mass is discovered.
Fractions Equal To 1 4
With a two stage rocket, it is, well, I won't say 'easy' but it is a lot easier than a three or more stage rocket. Obviously with a two stage rocket, once you've decided on a value for the Δv fraction of stage one, you have automatically determined stage two's Δv fraction (i.e., it is 1.0 - fΔv1).
So you set a range of values for the Δv fraction of stage one (say, from 0.2 to 0.8) and step through it at convenient intervals (say, increment by 0.1 each step). Subtract from 1.0 to figure the fraction for stage two. Send these through the battery of equations above and plot them on a graph like this:
By inspecting the graph, you can identify the minimum value for initial vehicle mass, and the associated Δv fractions. Your optimization is complete.
Performing this for three or more stages is a major pain in the gluteus maximus.
We need to vary two Δvs over some range to find a minimum. For a three stage system, we repeat the above graphing technique for a range of fΔv2 values, choosing the minimum fΔv1 value for each fΔv2(remember fΔv3 = 1.0 - fΔv1 - fΔv2). We can then plot the initial vehicle mass (each point being minimized for fΔv1)and choose the fΔv2 with the minimum initial mass value.
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