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Consider an original model of a cold section enclosing a throttle and collecting region, with the precooling heat exchanger outside its boundary. The inlet has mass flow rate $\dot{m}>0$ and specific enthalpy $h_{\mathrm{in}}$. The outlets are equilibrium liquid and vapor at one fixed pressure, with specific enthalpies $h_l$ and $h_g$. Let $y$ be the outlet liquid mass fraction, and let $q=\dot{Q}/\dot{m}\ge 0$ be incoming unwanted heat per unit incoming mass. Assume steady flow, no shaft work, and negligible bulk kinetic energy and gravitational potential energy changes. The steady-flow energy balance givesHere $L$ is the latent heat per unit mass at the chosen outlet pressure. The mass flow rates of the outlets are $y\dot{m}$ and $(1-y)\dot{m}$. The formula describes a flowing cold section; a filling vessel with appreciable energy or mass accumulation requires storage terms.

For a deliberately dimensionless example, choosePer unit incoming mass, one quarter leaves as liquid and three quarters as vapor. With zero incoming heat, the same inlet would give $y=0.30$. Increasing $q/L$ to $0.20$ reduces $y$ to $0.10$, and $q/L=0.30$ reaches the no-liquid boundary. These are chosen model values, not helium property data or a historical yield. In normalized enthalpy units with $h_l=0$ and $h_g=1$, conservation checks directly: $0.70+0.05=0.25\times 0+0.75\times 1$.

The admissible interval is $h_l\le h_{\mathrm{in}}+q\le h_g$, or $0\le y\le 1$. Outside it, the assumed equilibrium two-phase outlet fails: an enthalpy above $h_g$ requires a hotter single-vapor outlet under these assumptions, rather than a negative amount of liquid. Below $h_l$, a colder liquid or a different thermodynamic phase must be considered. The balance does not predict its own outlet pressure, temperature or helium property functions.

At fixed outlet pressure, hence fixed $L$, the derivatives are $\partial y/\partial h_{\mathrm{in}}=-1/L$ and $\partial y/\partial q=-1/L$. Lowering inlet specific enthalpy by regeneration and reducing incoming heat therefore buy the same increase in liquid mass fraction per unit energy per unit mass in this model. Adding any constant to all three specific enthalpies leaves $y$ unchanged. This independence of the arbitrary enthalpy reference makes the balance useful even when property tables choose different zero points.

Apply the steady-flow energy balance to an entire liquefaction apparatus, allowing a warm gas outlet and a separate cold liquid outlet. With incoming heat $q$ per unit feed mass, no shaft work, and negligible changes in bulk kinetic energy and gravitational potential energy, conservation of mass and conservation of energy give $h_{\mathrm{in}}+q=y{h}_l+(1-y)h_g$. Solving this equation gives the displayed liquid mass fraction.

The two outlet streams need not share a temperature. For example, a counterflow heat exchanger can warm returning gas before it crosses the overall control volume boundary, while collected liquid leaves at the cold end. When the outlet temperatures differ, $h_g-h_l$ includes sensible heating as well as phase change; it is not simply a latent heat. A colder two-phase separator nested inside the apparatus has its own boundary states and a different local enthalpy balance.

Mass concentration is the mass of a specified constituent divided by the final sample volume, ${\rho }_i=m_i/V$. Its SI unit is $\mathrm{kg}{\mathrm{m}}^{-3}$; laboratory data also commonly use mass per litre.^[ref]

For example, $10\mathrm{g}$ of a solute in a final solution volume of $1\mathrm{L}$ gives $10\mathrm{g}{\mathrm{L}}^{-1}$. The denominator is the solution volume after mixing, not necessarily the volume of solvent initially added.

If $w_i$ is mass fraction and $\rho$ the mixture mass density, then ${\rho }_i=w_i\rho$. Heating or compressing the sample can change the mass concentration through a change of volume while leaving the mass fraction unchanged.

The mass fraction of component $i$ in a mixture is $w_i=m_i/{\sum }_j{m}_j$. It is dimensionless, lies between zero and one, and the component mass fractions sum to one when all material is included.

For example, a mixture containing $5$ grams of solute and $95$ grams of solvent has solute mass fraction $0.05$, or five percent by mass. This differs from a mole fraction, which counts amounts of substance, and from a mass-per-volume concentration. A temperature-dependent volume change does not by itself change the mass fraction of a closed sample.

The mass fraction multiplied by 100%.

A concrete motivating experiment is the acetone–chloroform mixture in Truman State University's Liquid-Vapor Equilibrium of a Binary System, Procedure and Figure 1. A two-neck flask feeds a distillation head with a water-cooled laboratory condenser, condensate sampling arm and temperature probe. Starting with one pure liquid, the procedure adds the other component successively and waits for stable temperature with returning condensate. Small paired samples from the flask and condensate are read on an Abbé refractometer; a tabulated refractive index–composition relation gives the two compositions. Controls include three-point thermometer calibration, corrected atmospheric pressure readings, and agreement of the initial pure-liquid and condensate refractive indices with the reference value. Small withdrawals and little retained condensate reduce disturbance of composition. Sample evaporation can bias the readout. The inferred paired mole fractions, temperature and pressure describe vapor-liquid equilibrium, subject to the conversion from mass fraction to mole fraction and to representative sampling. This teaching protocol reports no measured results of a particular student run.^[ref]

How does such a measurement expose activity coefficients? Let $x_i$ and $y_i$ be the mole fractions in the liquid and gas thermodynamic phases, respectively, and let $p_i^{\ast }(T)$ be the pure-component vapor pressure. Assume the gas thermodynamic phase is an ideal gas mixture and neglect the change of the pure-liquid reference chemical potential between its saturation pressure and the experimental pressure. Equality of each component's chemical potential between the two thermodynamic phases then givesThis turns the paired composition readings into estimates of the activity coefficients. The approximation needs refinement when the gas departs appreciably from ideal gas behavior or the neglected liquid pressure effect matters.

The apparatus measures a boiling-composition series in which temperature changes as the mixture changes. Its readings therefore cannot be inserted directly into the fixed-temperature, fixed-pressure activity coefficient constraint. They motivate the more general task: test whether the measured mixture properties can arise from a common Gibbs free energy while accounting for changing state variables. A failure can reveal incompatible property fits or experimental bias; a pass supplies a consistency check, not proof that every reading is correct.

For a dry background inventory of column weight $P_d$, a mixed layer of fixed pressure thickness $\Delta p$, and saturated surface vapor partial pressure $E$, put $x=E/P$ and let $ϵ$ be the vapor-to-background molar mass ratio. The vapor mass fraction is $ϵx/[1+(ϵ-1)x]$, so hydrostatic balance and mass conservation give the displayed quadratic equation. Physical solutions require $x\le 1$ and a nonnegative upper-layer pressure. At $x=1$, $P=E=P_d+\Delta p$. A real positive algebraic root beyond that value need not be physically admissible.

The solution-released mass of an element divided by its initial glass mass fraction and the exposed glass area: $N{L}_e=m_e/(f_{e}S)$. It has units of glass-equivalent mass per area; it does not alone establish congruent dissolution.

Parts per million, written ppm, expresses a dimensionless ratio multiplied by $10^6$. A mass fraction of $3\times 10^{-6}$ is 3 ppm by mass; an amount fraction of the same numerical value is 3 ppm by amount of substance. The two compare different quantities and need not describe the same mixture composition.

A concentration written in milligrams per litre is a mass concentration, not automatically a parts-per-million ratio. Its approximate numerical equality to ppm by mass for a dilute aqueous sample uses an additional mass density approximation of $1kg/L$. Stating the basis prevents a unit conversion from silently becoming an assumption about the sample.

At equilibrium at room temperature the phases are pure Ag and pure Cu, with mass fractions $0.80$ and $0.20$. A representative solidification structure contains primary Ag-rich grains or dendrites surrounded by a finer two-phase eutectic aggregate. Subsequent equilibrium solute rejection modifies both constituents as they cool.

The $42.1\%$ primary-$\alpha$ and $57.9\%$ eutectic-constituent fractions calculated at the eutectic arrest describe how the liquid froze. They are not the final pure-Ag and pure-Cu phase fractions. For example, just below the arrest the total copper-rich phase fraction is $(20-9)/(92-9)=11/83=0.1325$; it rises to $0.20$ as solubility is lost on further equilibrium cooling. The distinction separates a microstructural constituent from a thermodynamic phase.

Faster solidification usually refines the dendrite and eutectic spacings. If solid-state diffusion cannot keep up, composition gradients, supersaturated solids and delayed precipitation can persist. Their exact fractions then depend on the thermal history; faster cooling does not change the alloy's overall $20\%$ copper inventory. Complete long-term equilibration would restore the same equilibrium phase compositions and mass fractions.

Sample B should show two FCC peak families, one from a copper-rich phase and one from a silver-rich phase. Their lattice parameters are set by the compositions of the coexisting phases, not by one lattice parameter at the overall alloy composition.

For corresponding Miller indices the silver-rich phase has a larger lattice parameter and hence a larger interplanar spacing, so its reflection occurs at smaller $2\theta$. Some reflections may overlap or be weak. Relative intensities depend on phase fractions, atomic scattering factors, absorption and texture; a raw peak-height ratio is not directly a mass fraction.

At room temperature the mutual solubilities in the supplied Cu–Ag diagram are small, so the two families would be near the elemental copper and silver positions. Slow cooling is important: a metastable supersaturated single phase would be a different interpretation. Quantitative phase or composition analysis would require calibrated diffraction and an appropriate refinement, not merely counting the visible peaks.

Convert atomic fractions to mass fractions before using the diagram's weight-percent scale:With $M_{\mathrm{Sn}}=118.71$ and $M_{\mathrm{Bi}}=208.98\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the overall $x=0.30$ is $19.58wt\%$ Sn. At $150{}^{\circ }\mathrm{C}$, the diagram gives approximately $x_B=0.028$ and $x_L=0.555$. These are about $1.61$ and $41.47wt\%$ Sn. The lever rule givesThe microstructure has primary Bi-rich crystals surrounded by remaining liquid. The phase compositions and fractions are approximate readings of a printed phase diagram, not exact thermodynamic values. Faster cooling generally gives finer crystals, solute segregation and departure from these equilibrium compositions.

For complete neutralization, the required masses of anhydrous potassium carbonate, anhydrous sodium carbonate and magnesium oxide are in the ratio $138:106:40=69:53:20$. Here “carbonate of potash” means potassium carbonate, “carbonate of soda” means sodium carbonate, and “caustic magnesia” means magnesium oxide, rather than magnesium hydroxide or a magnesium carbonate preparation. The hydration state of the sodium carbonate matters considerably.

Each formula unit of either carbonate consumes two protons; each formula unit of magnesium oxide also consumes two protons. Thus one mole of each base neutralizes one mole of sulfuric acid when the final salts are the normal sulfates:The carbon dioxide evolved in the first two chemical reactions escapes, whereas magnesium oxide produces no gas. The stoichiometric coefficients, rather than the visibly different behaviour of the bases, determine the required amounts.

Let $m$ be the mass of pure ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ in the given chemical solution. Using the rounded relative atomic masses $\mathrm{H}=1$, $\mathrm{C}=12$, $\mathrm{O}=16$, $\mathrm{Na}=23$, $\mathrm{Mg}=24$ and $\mathrm{K}=39$ givesThus 98 mass units of pure sulfuric acid require respectively 138, 106 and 40 mass units. Present atomic weights slightly change these rounded values, without changing the one-to-one mole ratios. If $m_s$ denotes the mass of the supplied acid and $w$ its mass fraction of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, substitute $m=w{m}_s$. A volume alone is insufficient without its concentration or an equivalent density and composition.

If “carbonate of soda” is instead the crystalline decahydrate, sodium carbonate decahydrate, its rounded molar mass is $286\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, so the corresponding factor is $286/98$, rather than $106/98$. Likewise an impure base requires correction for its assayed content. The unspecified given quantity can therefore be handled exactly by proportional reasoning; no missing numerical measurement is needed.

The 1864 preparations contain either crude opium or its aqueous Extractum Opii; those quantities must be distinguished from a quantity of isolated morphine. The following are the officinal opium preparations. Ratios for solids mean the mass fraction in the compounded preparation; liquid strengths specify mass per volume or the actual ingredients used.

The arithmetic uses the new avoirdupois ounce of 437.5 grains, not the old apothecaries' ounce of 480 grains. These are historical formulation proportions, not modern assays of morphine delivered after solvent extraction. Confectio Opii was omitted in 1864, and this edition's Vinum Opii contains no spices. The separate morphine hydrochloride preparations—Liquor Morphiae Hydrochloratis, Suppositoria Morphiae, Trochisci Morphiae, and Trochisci Morphiae et Ipecacuanhae—contain an isolated alkaloid salt rather than crude opium. ^[ref]

Confirm hydrogen cyanide by chemical tests on a suitable separated fraction, then estimate its amount with a defined quantitative assay. Here “hydrocyanic acid” means $\mathrm{HCN}$ in water, also historically called prussic acid. Smelling or tasting is not an analytical test: hydrogen cyanide is acutely poisonous, and its characteristic odour is not reliably perceived.^[ref]

A coloured or complex liquid can mask the tests or contain interfering ions. Historically, controlled distillation separated volatile hydrogen cyanide from many nonvolatile constituents; a receiving solution captures the analyte. In a modern analytical apparatus, the closed transfer system includes a condenser and an alkaline gas absorber. The receiver converts volatile hydrogen cyanide into less volatile cyanide:The measured fraction must be defined: free hydrogen cyanide in the original liquid is not automatically identical to “total cyanide” released from complexes by a more aggressive preparation. A blank, a known positive control and recovery checks help distinguish a negative sample from failed separation or reagent failure.^[ref]

For the historical iron test, make a separate test portion alkaline, introduce fresh iron(II) salt together with an iron(III) source, and finally acidify as specified by the historical procedure. Cyanide first forms ferrocyanide with iron(II); iron(III) then gives the characteristic Prussian blue precipitate. Idealized balances explaining the recognition chemistry areReal Prussian blue can be hydrated and nonstoichiometric; the second chemical equation is a useful idealization. The acid stage also removes obscuring iron hydroxide precipitates. These steps explain the old mixed iron-salt test, rather than assuming that cyanide simply reacts directly with any iron salt to give blue material.^[ref]

A second aliquot can be examined with silver nitrate. Silver cyanide is white, but whiteness alone is not specific because silver chloride and other precipitates can interfere:For a purified aliquot, precipitate the cyanide quantitatively with silver nitrate, separate the silver cyanide, wash it free of soluble salts, dry it to constant mass and weigh it. The precipitation conditions must avoid losses as volatile hydrogen cyanide and dissolution as dicyanoargentate; interfering precipitates must be excluded. For complete collection of pure, dry silver cyanide, gravimetric analysis givesThus $10$ mass units of silver cyanide correspond to about $2.02$ mass units of hydrogen cyanide. The 1867 pharmacopoeial check was $10$ grains of dry silver cyanide from $100$ grains of the diluted acid, corresponding approximately to its stated $2\%$ strength by mass.^[ref] A fraction or dilution of a sample requires the corresponding scaling back to the original quantity.

Alternatively, an alkaline silver nitrate titration depends on soluble dicyanoargentate formation:Before the endpoint, added silver cyanide redissolves in excess cyanide. In the classical version, the first persistent turbidity marks the end of this clear-solution stage. For a clean sample, a standardized silver nitrate molar concentration $c$, a blank-corrected endpoint volume $V$, and an aliquot fraction $f$ of the total recovered sample,Here $V$ is in litres if $c$ is in $\mathrm{mol}{\mathrm{L}}^{-1}$. Divide by the original sample mass for a mass fraction, or by its original volume for a mass concentration. No numerical sample amount is supplied, so no particular measured concentration can be inferred.

Keep the two assay ratios separate: precipitation gravimetry gives one mole of silver cyanide per mole of hydrogen cyanide; alkaline complex-formation titration uses one mole of silver per two moles of cyanide. A modern version of the latter uses a rhodanine indicator and a blank, as in EPA Method 9014; that indicator procedure is a modern comparison, not a detail assigned to 1867.^[ref]

Write $SA=S$ for the exposed glass area. If $C_e$ is the blank-corrected dissolved mass concentration of element $e$, $x_e$ its initial mass fraction in the glass and $V$ the leachate volume, thenThe numerator is the released element mass measured in solution. Dividing by $x_e$ converts this to an equivalent mass of the original glass, while dividing by $SA$ expresses that mass per exposed area. With $C_e$ in $\mathrm{g}{\mathrm{L}}^{-1}$ and $SA/V$ in ${\mathrm{m}}^2{\mathrm{L}}^{-1}$, $N{L}_e$ has units $\mathrm{g}{\mathrm{m}}^{-2}$ of glass-equivalent release. The fraction $x_e$ must be used as a fraction, not an unconverted percentage.

These corrections allow comparison of specimens of different size and composition without confusing a lower concentration with better durability. State how the area was measured; geometric area and a rough powder's accessible area need not coincide. For repeated sampling, include element mass removed in earlier aliquots and the volume remaining; for an initial blank in the leachant, subtract its corresponding inventory. Surface cracking and evolving porosity can make a rate based on initial area differ from a rate based on the true reactive area.

Under congruent glass dissolution, constituents enter solution in their bulk proportions, so their correctly normalised releases coincide. Preferential sodium release by ion exchange can instead give $N{L}_{\mathrm{Na}}>N{L}_{\mathrm{Si}}$, while silicon retention in a gel or secondary precipitate can depress its dissolved release. Noncoincident curves show nonstoichiometric aqueous release, but do not uniquely identify the mechanism. Conversely, solution coincidence does not by itself establish that no secondary reactions occurred. A complete inference compares solution inventories with altered solids.

Normalisation removes composition and first-order area/volume scaling; it does not make different pH, temperature, solution composition or contact histories equivalent. In a batch test, even the same initial water changes composition and pH as reaction proceeds. Reporting those conditions and time is essential for a transferable comparison rather than treating $N{L}_e$ as a condition-independent material constant.

Use the thermal power when applying nuclear fuel burnup. No capacity factor is supplied, so take a full-power year of $365$ days; if the operating fraction were $u$, the annual quantities in this part would be multiplied by $u$.The corresponding initial heavy-metal nuclear fuel requirement isAll nuclear fuel and concentrate masses below are uranium masses, excluding oxygen. Let $F,P,W$ denote uranium feed, product and enrichment tails, and $x_f,x_p,x_w$ the uranium-235 mass fractions. Conservation of total uranium and uranium-235 givessoWith $P=M$, the uranium oxide concentrate requirement is thereforeThis is not the gross mass of oxide concentrate. Both $F-P-W=0$ and $0.007F-0.050P-0.003W=0$ provide independent balance checks.

The recycle feed is the recovered uranium, not the whole initial nuclear fuel charge. After the stated loss it isUsing the same enrichment tails assay and the stipulated higher recycle-product assay, the isotopic mass balance givesTreat the $5.3\%$ recycled product as replacing the same uranium mass of ordinary $5.0\%$ nuclear fuel: the additional uranium-235 compensates for the supplied uranium-236 penalty. It displaces fresh natural-uranium feed in the ratio calculated in part (b), henceThe remaining natural feed is $313.36245\mathrm{t}\mathrm{U}\mathrm{y}{\mathrm{r}}^{-1}$, a 9.885% resource saving relative to the full-power once-through requirement. This credits one uranium recycle only; it adds no second-pass recycling or plutonium credit.

Separative work must be charged for producing the recycled nuclear fuel as well as credited for displaced fresh nuclear fuel. With uranium-235 mass fraction $x$, use the separative value functionMasses in kilograms give work in kg-SWU. For ordinary fresh product the work per product kilogram isThus the fresh work avoided isThe recycled product still requiresThe net separative-work saving is consequentlyAs a whole-cycle check, the fresh route requires $215668.0851$ kg-SWU per year, while the mixed fresh-plus-recycled supply requires $214075.3261$ kg-SWU. The net saving is only 0.7385% of the original work. The recovered stream has a higher uranium-235 assay than natural feed, but the higher required product assay uses most of the work credit. These equations are resource accounting and do not specify an uranium enrichment plant or its operation.

burnup refers to thermal energy per initial heavy-metal mass. Correct the electrical rating for thermal efficiency before using it, and apply the capacity factor once:Consequently the annual initial heavy-metal fuel requirement isLet $F,P,W$ be feed, enriched product and enrichment tails masses, all measured as uranium, and $x_f,x_p,x_w$ the corresponding uranium-235 mass fractions. The isotopic mass balance givesWith $P=M$, the uranium oxide concentrate requirement is thereforeThese are uranium or heavy-metal masses, not tonnes of uranium oxide. The check $327.394(0.007)=35.394(0.040)+292.000(0.003)$ closes the uranium-235 balance as well as the total mass balance.

Let $c_e$ be the measured mass concentration of element $e$, $c_{e,b}$ its blank concentration, $V$ the leachate volume, $f_e$ the element's mass fraction in the original glass, and $S$ the exposed glass surface area. For one closed test, the normalised elemental release isIf concentration is in $\mathrm{g}{\mathrm{L}}^{-1}$, volume in litres and area in square metres, $N{L}_e$ is in $\mathrm{g}{\mathrm{m}}^{-2}$ of glass-equivalent release. Dividing by $f_e$ removes the trivial difference caused by loading different amounts of the measured element; dividing by $S$ removes the first-order geometrical scale difference. Normalising only by initial element content or reporting raw solution concentration does not accomplish both corrections.

For repeated sampling with leachant replacement, use cumulative released mass: add the mass removed in all earlier aliquots to the mass present in the final remaining solution, with the corresponding blank correction. A flow-through experiment instead integrates $Q(t)[c_{e,\mathrm{out}}(t)-c_{e,\mathrm{in}}(t)]$ over time. State how surface area was determined and whether it is held at its initial value in the reported normalisation; cracks, fines and alteration can change the genuinely reactive area.

Compare $N{L}_e(t)$ for several constituents. In congruent glass dissolution, the different element-normalised curves coincide, provided the released elements remain in solution. Preferential early alkali release can indicate ion exchange, whereas lower dissolved silicon or actinide release can reflect an alteration layer, sorption or secondary-phase precipitation. boron is often useful as a relatively mobile dissolution tracer, but not if the leachant contains a large boron background or if it is itself retained. A low aqueous concentration of one constituent is therefore not automatically a low glass-dissolution rate.

A near-linear $N{L}_e(t)$ interval gives an approximately constant normalised elemental release rate. A square-root time dependence can be consistent with diffusion-limited transport, and decreasing slopes can indicate saturation or a protective alteration layer; no one curve shape proves a unique mechanism. Compare solution inventories with the altered solid by microscopy, composition and phase analysis. Differences in pH, temperature, leachant composition, duration and surface-area-to-volume ratio remain even after this mass-and-area normalisation. Careful background subtraction and control of solution conditions are experimentally consequential, as demonstrated in primary flow-through measurements.^[ref]