34  The Kidney: Clearance and Urea Balance

A 60 kg woman whose kidneys have failed is connected to a dialysis machine. How long must the session run to remove enough urea? The clinical aim is to remove about 70% of the urea in one session, a dose written \(Kt/V = 1.2\) and defined below.

The body produces nitrogen-containing waste daily from the breakdown of protein. The dominant waste product is urea, a small, water-soluble molecule that becomes toxic when it accumulates. In a healthy person the kidneys filter urea and many drugs out of the blood and excrete them in urine, holding urea concentration steady. When the kidneys fail, urea and other wastes build up (a condition called uremia), and without intervention the concentration keeps rising. Hemodialysis replaces the missing function and lowers the concentration to a safe level within a few hours.

The body is modeled as a single well-mixed compartment with a urea mass balance, as in one-compartment drug models.

Urea, the kidney, and dialysis

Urea is the end product of protein metabolism. Protein the body does not need is broken down, and the nitrogen it releases is converted in the liver to urea, which is released into the blood and forms the largest circulating pool of waste nitrogen in the body (Weiner et al. 2015). On a typical protein intake an adult generates on the order of 20–30 g of urea per day; the rate is lower on the protein-restricted diets often prescribed in kidney disease (Weiner et al. 2015). This generation rate is denoted \(G\) and treated as constant, because it changes slowly compared with the processes below.

The kidneys are the principal organ of excretion for urea and for most water-soluble drugs (Grogan and Preuss 2023). Removal is described by the clearance \(CL\), the volume of blood cleared of a solute per unit time (mL/min or L/h). The rate of removal is proportional to the concentration:

\[\text{rate out} = CL \cdot C .\]

When the kidneys cannot remove enough urea, external filtration can. This is dialysis: blood flows past a semipermeable membrane, and urea diffuses across the membrane, down its concentration gradient, into a cleaning fluid called the dialysate. The dialyzer, the filter that holds the membrane, therefore adds an artificial clearance to whatever the kidneys still provide (Figure 34.1).

Figure 34.1: The kidney (left) and a dialyzer (right). In the kidney, filtered urea leaves with the urine. In the dialyzer, blood flows past a semipermeable membrane and urea diffuses across it into the dialysate.

A mass balance on the body

A boundary is drawn around the body and everything that crosses it is tracked (Figure 34.2).

Figure 34.2: One-compartment model of urea in the blood. Urea enters at generation rate \(G\) and leaves through two parallel first-order routes: renal removal (\(k_{\text{renal}}\)) and, during treatment, dialysis (\(k_{\text{dialysis}}\)). The compartment has volume of distribution \(V_d\) and holds urea at concentration \(C\) (initially \(C_0\)).

1. The body is a single, well-mixed compartment: urea has the same concentration \(C\) everywhere in it. This is a 1-compartment model.
2. The volume \(V_d\) is constant. (Dialysis also removes some water; this is neglected.)
3. The liver is drawn outside the boundary, so the urea it makes enters as an input. No reaction makes or destroys urea inside the boundary.

The balance in words.

\[\frac{\Delta\,\text{urea}}{\Delta t} = \underbrace{\dot m_{\text{in}}}_{\text{generation by liver}} - \underbrace{\dot m_{\text{out}}}_{\text{dialysis + kidney}} + \underbrace{\text{reaction}}_{\text{none}}\]

The removal rates are first order. The amount of urea in the body is \(CV_d\) (concentration times volume). As for a drug, removal is first order: the amount removed per unit time is proportional to the amount present, so twice as much urea is removed twice as fast. The dialyzer and the kidney are two such routes acting at the same time:

NoteKey result: urea mass balance

\[\frac{d\,(C V_d)}{dt} = G - k_r\,C\,V_d - k_d\,C\,V_d\]

symbol	meaning	units
\(C\)	concentration of urea in the blood	mg/L
\(V_d\)	volume of the compartment (volume of distribution of urea)	L
\(CV_d\)	amount (mass) of urea in the body	mg
\(G\)	rate at which the liver releases urea (generation)	mg/h
\(k_r\)	first-order rate constant for removal by the kidney (renal)	1/h
\(k_d\)	first-order rate constant for removal by dialysis	1/h

Check the units. Every term must be in mg/h: \(G\) is already in mg/h, and \(k\,CV_d\) has units \(\frac{1}{\text{h}}\cdot\frac{\text{mg}}{\text{L}}\cdot\text{L} = \frac{\text{mg}}{\text{h}}\).

Two exits add. Both removal terms are proportional to \(CV_d\), so it can be factored out:

\[\frac{d(CV_d)}{dt} = G - (k_r + k_d)\,CV_d = G - k\,CV_d, \qquad k \equiv k_r + k_d \quad\text{(combined rate constant, 1/h).}\]

To keep the algebra short, define the total amount of urea \(U \equiv CV_d\) (units \(\frac{\text{mg}}{\text{L}}\cdot\text{L} = \text{mg}\)):

\[\frac{dU}{dt} = G - kU .\]

Solving the mass balance

Separate variables, then integrate. Every \(U\) moves to the left and every \(t\) to the right:

\[\frac{dU}{G - kU} = dt \qquad\Longrightarrow\qquad \int \frac{dU}{G - kU} = \int dt .\]

NoteDerivation: the integral of \(\frac{1}{G - kU}\) by substitution

The integral is in standard tables, so it does not need to be memorized. To derive it, let \(u = G - kU\). Then \(du = -k\,dU\), so \(dU = -du/k\), and

\[\int \frac{dU}{G - kU} = -\frac{1}{k}\int \frac{du}{u} = -\frac{1}{k}\ln|u| = -\frac{1}{k}\ln|G - kU| \;(+\text{ a constant}).\]

The factor \(-1/k\) comes from the chain rule: \(u\) changes by \(-k\) for each unit change in \(U\). The absolute value is there because the logarithm is only defined for positive numbers, and \(G - kU\) can have either sign.

Check by differentiating.

\[\frac{d}{dU}\left[-\frac{1}{k}\ln|G - kU|\right] = -\frac{1}{k}\cdot\frac{-k}{G - kU} = \frac{1}{G - kU},\]

which is the integrand.

Integrating leaves one constant of integration, \(c\):

\[-\frac{1}{k}\ln|G - kU| = t + c .\]

Multiplying by \(-k\) and exponentiating gives \(|G - kU| = e^{-kc}\,e^{-kt}\). The absolute value allows either sign, so the result can be written with a single constant \(B\) that may be positive or negative:

\[G - kU = B\,e^{-kt} .\]

Initial condition. At \(t = 0\), the amount of urea is \(U(0) = U_0\). Since \(e^{0} = 1\):

\[B = G - kU_0 .\]

Substituting \(B\) back in and solving for \(U\):

\[G - kU = (G - kU_0)\,e^{-kt} \quad\Longrightarrow\quad U(t) = \frac{G}{k} - \left(\frac{G}{k} - U_0\right)e^{-kt} = \frac{G}{k} + \left(U_0 - \frac{G}{k}\right)e^{-kt} .\]

Since \(U(t) = C(t)\,V_d\) (mass of urea = concentration of urea \(\times\) volume), dividing by \(V_d\) with \(C_0 = U_0/V_d\) gives the concentration:

NoteKey result: urea concentration

\[C(t) = \frac{G}{kV_d} + \left(C_0 - \frac{G}{kV_d}\right)e^{-kt}\]

symbol	meaning	units
\(C(t)\)	urea concentration at time \(t\)	mg/L
\(C_0\)	urea concentration at \(t = 0\)	mg/L
\(G/(kV_d)\)	a concentration: \(\dfrac{\text{mg/h}}{(1/\text{h})\cdot\text{L}} = \dfrac{\text{mg}}{\text{L}}\)	mg/L
\(k\)	combined rate constant, \(k_r + k_d\)	1/h

Nephrology calls this one-compartment model the single-pool model and uses it to judge whether a session removes enough urea (Gotch 2001; Locatelli et al. 2005).

The shape of the curve

• At \(t = 0\): \(e^{0} = 1\), so \(C(0) = C_0\).
• As \(t \to \infty\): \(e^{-kt} \to 0\), so \(C(\infty) = \dfrac{G}{kV_d}\).

In between, the gap between \(C\) and \(G/(kV_d)\) decays as \(e^{-kt}\). Urea approaches the steady state \(G/(kV_d)\) exponentially (Figure 34.3).

Figure 34.3: The shape of the urea curve for \(C_0 > G/(kV_d)\). The concentration decays exponentially toward the steady state \(G/(kV_d)\); the time constant is \(1/k\).

The steady state, directly. The plateau can be found without solving anything. At steady state the concentration stops changing, \(d(CV_d)/dt = 0\), so \(0 = G - kC_{\text{ss}}V_d\) and

\[C_{\text{ss}} = \frac{G}{kV_d} = \frac{G}{CL_{\text{tot}}},\]

where \(CL_{\text{tot}} = V_d(k_r + k_d)\) is the total clearance: the removal rates add, so the rate constants add, and clearances in parallel add. At steady state, removal (\(kCV_d\)) exactly balances generation (\(G\)). This is the steady-state relation for drug infusions in one-compartment models, with urea generation in place of the infusion rate. The steady state is inversely proportional to clearance: doubling the clearance halves it.

How fast? The exponent \(kt\) must be unitless, so \(1/k\) has units of time. It is called the time constant, \(\tau = 1/k\). After one time constant, \(e^{-1} = 0.37\), so the concentration has covered 63% of the distance from \(C_0\) to the steady state. After \(3\tau\) it has covered 95%. The half-life of the one-compartment model, \(t_{1/2} = \ln 2/k\), is also set by \(k\).

ImportantKnowledge check

A patient’s kidneys have failed completely and she is not on dialysis, so \(k_r = k_d = 0\). The closed form above divides by \(k\), so return to the mass balance: set \(k = 0\) and integrate. What does it predict for her urea?

NoteAnswer

With \(k = 0\) the balance is \(d(CV_d)/dt = G\), so \(C(t) = C_0 + (G/V_d)\,t\). Urea rises in a straight line and never levels off: there is no steady state.

The volume of distribution of urea

Urea is small, uncharged, and very water soluble, so it spreads through the water in the body, inside cells as well as outside. Its volume of distribution \(V_d\) is therefore the total body water (TBW): the volume that holds the urea at the common concentration \(C\). It is taken as constant (Gotch 2001). On average,

\[\text{TBW} \approx 0.60~\text{L/kg (men)}, \qquad \text{TBW} \approx 0.50~\text{L/kg (women)} .\]

For a 70 kg man, \(V_d = 0.60 \times 70 = 42\) L. The model also assumes that equilibrium is instantaneous: as dialysis removes urea from the blood, urea from the rest of the body water moves into the blood immediately.

A clearance and a rate constant are linked through this volume: \(k = CL/V_d\), so \(k_r = CL_r/V_d\) for the kidney.

Steady state in numbers

With a generation rate of \(G = 1000\) mg/h (24 g/day), healthy kidneys with a renal clearance \(CL_r\) (the subscript marks renal) of about 3.6 L/h (60 mL/min) hold the blood at

\[C_{\text{ss}} = \frac{1000}{3.6} \approx 278\ \text{mg/L},\]

a normal, non-toxic level. If disease drops the renal clearance to \(CL_r \approx 0.6\) L/h (10 mL/min), the same generation drives the steady state to

\[C_{\text{ss}} = \frac{1000}{0.6} \approx 1670\ \text{mg/L},\]

roughly six times higher, and uremic. The steady state is lowered by reducing \(G\) (a protein-restricted diet) or raising the clearance (dialysis).

Urea removal, clearance, and \(Kt/V\)

For a patient, the goal is a target fraction remaining: the fraction of the starting urea that is still there at the end of the session,

\[R = \frac{C_{\text{post}}}{C_{\text{pre}}},\]

where \(C_{\text{pre}}\) is measured before the session and \(C_{\text{post}}\) after it. Clinics often report the urea reduction ratio (URR), the fraction removed: \(\text{URR} = 1 - R\).

Neglecting generation during the session. During dialysis, generation is small compared with removal, \(G \ll kC_0V_d\), so \(G \approx 0\). The first term of the solution then vanishes, leaving

\[C(t) = C_0\,e^{-kt} .\]

Residual kidney function is neglected during the session, so all removal is by the dialyzer and \(k = k_d\). This is the standard single-pool description of a short session (Dębowska et al. 2011). Section 34.10 quantifies the error: for the 4 h session in Table 34.1, setting \(G = 0\) at \(G = 1000\) mg/h makes the computed end-of-session concentration about 12% too low.

Clearance: removal measured as a volume per time. The dialyzer’s performance is usually not given as a rate constant \(k\) (units 1/h) but as a clearance \(K\) (units mL/min or L/h) (Locatelli et al. 2005). Clearance is the volume of blood that the dialyzer completely clears of urea per unit time. If the dialyzer clears \(K\) liters per hour of blood at concentration \(C\), it removes urea at the rate

\[\text{removal rate} = K\,C \qquad \left[\tfrac{\text{L}}{\text{h}}\cdot\tfrac{\text{mg}}{\text{L}} = \tfrac{\text{mg}}{\text{h}}\right].\]

In the model the same removal rate is \(k\,CV_d\). Setting the two equal:

NoteKey result: clearance and the rate constant

\[K = k\,V_d \qquad\Longleftrightarrow\qquad k = \frac{K}{V_d}\]

symbol	meaning	units
\(K\)	clearance: volume of blood cleared of urea per unit time	mL/min or L/h
\(k\)	rate constant: fraction of the body’s urea removed per unit time	1/h
\(V_d\)	volume of the compartment	L

The rate constant is the clearance divided by the volume the urea is spread through: a dialyzer clears a fixed volume of blood each minute, and the larger the body’s pool, the smaller the fraction of it that volume represents. Clearance is written \(CL\) in general, \(CL_r\) for the kidney, and \(K\) for the dialyzer. Lowercase \(k\) is a rate constant, and the total clearance is \(CL_{\text{tot}} = CL_r + K = kV_d\).

Substituting \(k = K/V_d\):

\[C(t) = C_0\,e^{-(K/V_d)\,t} \qquad\Longrightarrow\qquad R = \frac{C(t)}{C_0} = e^{-Kt/V_d} \qquad\Longrightarrow\qquad \ln R = -\frac{Kt}{V_d}\]

NoteKey result: the dialysis dose

\[\frac{Kt}{V} = -\ln R\]

Here \(V\) is the urea distribution volume \(V_d\), and \(Kt/V\) is the dialysis dose. It is unitless: \(\dfrac{(\text{L/h})\cdot\text{h}}{\text{L}}\).

The quantity \(Kt/V\) was introduced in the analysis of the National Cooperative Dialysis Study (Gotch and Sargent 1985).

The clinical target. Dialysis prescriptions are written as a target value of \(Kt/V\). For patients dialyzed three times a week, the current U.S. guideline sets \(Kt/V = 1.2\) per session as the minimum delivered dose and recommends aiming somewhat higher (National Kidney Foundation 2015). At \(Kt/V = 1.2\) the fraction of urea remaining is \(R = e^{-1.2} \approx 0.30\), so the session removes about 70%, within the 65–70% reduction that guidelines associate with adequate dialysis (Locatelli et al. 2005). The target is chosen from clinical outcomes, not derived from the model. Clinics calculate \(Kt/V\) with a more complete version of this model that includes urea generation and the water removed during the session (Daugirdas 1993); the model here omits both.

Measuring clearance from two blood samples. Neglecting generation, \(\ln C = \ln C_0 - kt\), so two urea measurements, \(C_1\) and then \(C_2\), taken \(\Delta t\) apart give the rate constant, \(k = \ln(C_1/C_2)/\Delta t\). The delivered clearance then follows from \(K = kV_d\).

ImportantKnowledge check

Two patients have the same dialyzer clearance \(K\) and the same session length \(t\). One weighs twice as much as the other. Which one ends the session with a larger fraction of urea remaining?

NoteAnswer

The heavier patient. \(V\) doubles, so \(Kt/V\) halves and \(R = e^{-Kt/V}\) is larger. The same volume of blood cleared is a smaller fraction of a larger pool.

Example: how long must the session run?

NoteExample: session length

Given values. A 60 kg woman; TBW \(= 0.5\) L/kg; dialyzer clearance \(K = 200\) mL/min; target \(Kt/V = 1.2\). How long must the session run?

Solve for \(t\).

\[\frac{Kt}{V} = 1.2 \qquad\Longrightarrow\qquad t = \frac{1.2\,V}{K}\]

Volume. \(V = (0.5~\text{L/kg})(60~\text{kg}) = 30\) L.

Watch the units. \(V\) is in liters and \(K\) is in milliliters per minute. Plugging in the bare numbers gives \(1.2 \times 30/200 = 0.18\), wrong by a factor of 1000. Convert \(V\) to mL before dividing by \(K\) in mL/min: \(V = 30~\text{L} = 30{,}000\) mL.

\[t = \frac{1.2\,V}{K} = \frac{1.2 \times 30{,}000~\text{mL}}{200~\text{mL/min}} = 180~\text{min} = 3~\text{h}\]

The milliliters cancel and the answer is in minutes.

Result. The session must run about 3 hours to bring urea down to 30% of its starting value (Figure 34.4).

Figure 34.4: Urea during the session, \(C/C_{\text{pre}} = e^{-Kt/V}\) with \(K = 200\) mL/min and \(V = 30\) L. At \(Kt/V = 1.2\), reached at 180 min, 30% of the urea remains.

Back to the question

For the 60 kg woman with a 200 mL/min dialyzer, reaching the target \(Kt/V = 1.2\) takes 180 min, about 3 hours, and leaves 30% of the urea.

ImportantKnowledge check

A patient’s residual renal clearance is \(0.6~\text{L/h}\), and a dialyzer adds \(12~\text{L/h}\). Suppose the dialyzer ran continuously. At a fixed generation rate, by what factor would it lower the steady-state urea concentration?

NoteAnswer

\(C_{\text{ss}}\) is inversely proportional to total clearance, so the factor is the ratio of clearances, \((0.6 + 12)/0.6 = 21\). Dialysis lowers the steady-state urea 21-fold.

Optional: how good is \(G \approx 0\)?

The liver keeps producing urea during the session, so setting \(G = 0\) makes the computed end-of-session concentration too low. Table 34.1 compares the solution with generation and the \(G = 0\) solution for a 4 h session with \(C_0 = 1400\) mg/L, dialyzer clearance \(K = kV_d = 12.5\) L/h (residual kidney function neglected), and \(V_d = 42\) L, a separate case from the example above. The solution with generation is \(C(t) = G/(kV_d) + (C_0 - G/(kV_d))\,e^{-kt}\), and the initial removal rate is \(kV_dC_0 = 17{,}500\) mg/h.

Table 34.1: End-of-session urea concentration with and without generation. The column \(G/(kV_dC_0)\) is the generation rate as a fraction of the initial removal rate. The last column is how far the \(G = 0\) value falls below the value with \(G\), as a percentage of the value with \(G\).

\(G\) (mg/h)	\(G/(kV_dC_0)\)	\(C_{\text{post}}\) with \(G\) (mg/L)	\(C_{\text{post}}\) with \(G = 0\) (mg/L)	underestimate from \(G = 0\)
500	2.9%	454	426	6.1%
1000	5.7%	481	426	11.6%
1500	8.6%	509	426	16.4%
2000	11.4%	537	426	20.7%

In Figure 34.5, the slider sets \(G\) for the session in Table 34.1. The solid curve includes generation and the dashed curve sets \(G = 0\). The sentence below the plot reports the end-of-session concentration and urea reduction ratio both ways, and the gap between them; at \(G = 1000\) mg/h it reproduces the second row of the table.

viewof genG = Inputs.range([0, 2000], {value: 1000, step: 50, label: "urea generation rate G (mg/h)"})

dial = ({V: 42, CL: 12.5, C0: 1400, T: 4})

Plot.plot({
  width: 624, marginTop: 30, style: {fontSize: "16px"},
  height: 340,
  marginLeft: 82,
  marginBottom: 65,
  x: {domain: [0, dial.T], label: "time in session (h)", grid: true},
  y: {domain: [0, 1500], label: "blood urea concentration (mg/L)", grid: true},
  marks: [
    Plot.line(d3.range(0, dial.T + 0.001, 0.02).map(t => {
      const k = dial.CL / dial.V, css = genG / dial.CL;
      return {t, c: css + (dial.C0 - css) * Math.exp(-k * t)};
    }), {x: "t", y: "c", stroke: "#3B7EA1", strokeWidth: 3}),
    Plot.line(d3.range(0, dial.T + 0.001, 0.02).map(t => ({t, c: dial.C0 * Math.exp(-(dial.CL / dial.V) * t)})),
      {x: "t", y: "c", stroke: "#FDB515", strokeWidth: 3, strokeDasharray: "6,4"})
  ]
})

Figure 34.5: Urea concentration during a 4 h session for the chosen \(G\). The solid curve solves the mass balance with generation. The dashed curve sets \(G = 0\).

{
  const k = dial.CL / dial.V, css = genG / dial.CL, e = Math.exp(-k * dial.T);
  const exact = css + (dial.C0 - css) * e, approx = dial.C0 * e;
  const urrE = (1 - exact / dial.C0) * 100, urrA = (1 - approx / dial.C0) * 100;
  return html`<p><b>With G = ${genG} mg/h</b>, the urea concentration at the end of the 4 h session is <b>${exact.toFixed(0)} mg/L</b>, a ${urrE.toFixed(1)}% reduction from the starting 1400 mg/L (solid curve). If generation is ignored (G = 0, dashed curve), the calculation gives <b>${approx.toFixed(0)} mg/L</b>, a ${urrA.toFixed(1)}% reduction. Ignoring generation therefore underestimates the end-of-session urea by ${(exact - approx).toFixed(0)} mg/L (${((exact - approx) / exact * 100).toFixed(1)}% of the true value) and overstates the reduction by ${(urrA - urrE).toFixed(1)} percentage points.</p>`;
}

The underestimate is zero at \(G = 0\) and grows with \(G\) and with time in the session, because the removal rate \(kCV_d\) falls as \(C\) falls while \(G\) stays fixed. At \(G = 1000\) mg/h it is 1.9% after 1 h, 4.4% after 2 h, and 11.6% after 4 h. At the end of the session, the urea reduction ratio is 69.6% with \(G = 0\) and 65.6% with \(G = 1000\) mg/h.

NoteOptional: urea rebound, a second compartment in dialysis

In real dialysis patients, the single well-mixed compartment is an approximation. Dialysis removes urea from the blood faster than urea can move out of cells into the blood, so equilibrium is not instantaneous. When the session ends, urea keeps moving into the blood and the concentration rises again: urea rebound. The single-compartment \(Kt/V\) therefore overestimates the dose delivered (Dębowska et al. 2011). Modeling rebound needs two compartments, with flow in both directions.

Exercises

Problem 1 uses the dialysis example above. The remaining problems take the urea distribution volume as \(V_d = 42~\text{L}\) (total body water of a \(70~\text{kg}\) adult) unless stated otherwise, and express concentrations in mg/L. Recall \(C_{\text{ss}} = G/(V_d k)\), \(C(t) = C_0 e^{-kt}\) for the no-generation case, \(t_{1/2} = \ln 2 / k\), and \(CL = k V_d\). Evaluate \(e^{x}\) and \(\ln x\) with Python (np.exp, np.log) or a scientific calculator; see the Python and calculus appendices.

TipProblem 1

For the patient in the example (\(V = 30\) L, \(K = 200\) mL/min), convert \(K\) to L/h and find the rate constant \(k\), the time constant \(\tau\), and the half-life of urea during dialysis.

NoteAnswer 1

\(K = 0.2~\text{L/min} \times 60~\text{min/h} = 12\) L/h. \(k = K/V = 12/30 = 0.40\) h\(^{-1}\). \(\tau = 1/k = 2.5\) h. \(t_{1/2} = \ln 2/k = 1.73\) h.

TipProblem 2

A clinic sets the goal as a urea reduction ratio of at least 65% (given value). What \(Kt/V\) does that correspond to in the model?

NoteAnswer 2

URR \(= 0.65\) means \(R = 0.35\), so \(Kt/V = -\ln 0.35 = 1.05\).

TipProblem 3

During a 4-hour session, neglect generation (\(G \approx 0\)). A patient starts at \(C_0 = 1400~\text{mg/L}\) with a dialyzer clearance of \(K = 12.5~\text{L/h}\) (neglect residual kidney function).

1. Find \(k\) and the half-life.
2. Find the urea concentration at the end of 4 hours and the urea reduction ratio (URR).
3. How long would be needed to reach a URR of 65%?

NoteAnswer 3

1. \(k = K/V_d = 12.5/42 = 0.298~\text{h}^{-1}\); \(t_{1/2} = \ln 2 / k = 2.33~\text{h}\).
2. \(C(4) = 1400\,e^{-0.298\times4} = 1400\,e^{-1.190} = 426~\text{mg/L}\); \(\text{URR} = 1 - 426/1400 = 0.70\), about 70%.
3. \(\text{URR} = 0.65\) means \(C/C_0 = 0.35\), so \(t = -\ln(0.35)/k = 1.050/0.298 = 3.53~\text{h}\).

TipProblem 4

Clinical dialysis dose is the dimensionless group \(Kt/V\): clearance \(K\) times session time \(t\) divided by volume \(V\). Take \(K = 12.5~\text{L/h}\), \(V = 42~\text{L}\).

1. Compute \(Kt/V\) for a 4-hour session.
2. The single-pool model gives \(\text{URR} = 1 - e^{-Kt/V}\). Compute the URR.
3. What session time achieves the guideline minimum \(Kt/V = 1.2\)?

NoteAnswer 4

1. \(Kt/V = (12.5)(4)/42 = 1.19\).
2. \(\text{URR} = 1 - e^{-1.19} = 0.70\), about 70% (consistent with Problem 3, since \(Kt/V = kt\)).
3. \(t = (Kt/V)\,V/K = (1.2)(42)/12.5 = 4.03~\text{h}\).

TipProblem 5

Immediately after dialysis a patient’s urea is \(430~\text{mg/L}\). Dialysis stops; the residual renal clearance is \(CL_r = 0.6~\text{L/h}\) and generation resumes at \(G = 1000~\text{mg/h}\).

1. What steady state would the urea now approach?
2. Write \(C(t)\) using the general solution.
3. What is the urea concentration 24 hours later?
4. Compare this with the starting value. What does it imply about repeating dialysis?

NoteAnswer 5

1. \(C_{\text{ss}} = G/CL_r = 1000/0.6 = 1670~\text{mg/L}\).
2. \(k = 0.6/42 = 0.0143~\text{h}^{-1}\), so \(C(t) = 1670 + (430 - 1670)\,e^{-0.0143 t} = 1670 - 1240\,e^{-0.0143 t}\).
3. \(C(24) = 1670 - 1240\,e^{-0.343} = 1670 - 880 = 790~\text{mg/L}\).
4. Urea nearly doubles within a day (430 to 790 mg/L) and keeps climbing toward \(1670~\text{mg/L}\), so dialysis must be repeated to bring it down again.

TipProblem 6

Clearance can be measured from the decay itself. During a session (neglect generation and residual kidney function), a patient’s urea falls from \(C_0 = 1500~\text{mg/L}\) to \(450~\text{mg/L}\) in \(4~\text{hours}\), with \(V_d = 42~\text{L}\). Infer the delivered clearance \(K\).

NoteAnswer 6

From \(C(t) = C_0 e^{-kt}\), the rate constant is \(k = \ln(C_0/C)/t = \ln(1500/450)/4 = 1.204/4 = 0.301~\text{h}^{-1}\). Then \(K = k\,V_d = 0.301 \times 42 = 12.6~\text{L/h}\) (about \(210~\text{mL/min}\)).

NoteFurther reading

• Weiner, Mitch & Sands, “Urea and Ammonia Metabolism and the Control of Renal Nitrogen Excretion,” Clin. J. Am. Soc. Nephrol. 10(8), 1444–1458 (2015): where urea comes from and how the kidney handles it (Weiner et al. 2015).
• Gotch, “Evolution of the single-pool urea kinetic model,” Semin. Dial. 14(4), 252–256 (2001): the one-compartment model as used in nephrology (Gotch 2001).
• Locatelli et al., “Dialysis dose and frequency,” Nephrol. Dial. Transplant. 20(2), 285–296 (2005): \(Kt/V\), the urea reduction ratio, and how dialysis adequacy is defined (Locatelli et al. 2005).

References

Daugirdas, John T. 1993. “Second Generation Logarithmic Estimates of Single-Pool Variable Volume Kt/V: An Analysis of Error.” Journal of the American Society of Nephrology 4 (5): 1205–13. https://doi.org/10.1681/ASN.V451205.

Dębowska, Małgorzata, Bengt Lindholm, and Jacek Waniewski. 2011. “Kinetic Modeling and Adequacy of Dialysis.” In Progress in Hemodialysis – from Emergent Biotechnology to Clinical Practice, edited by Angelo Carpi, Carlo Donadio, and Gianfranco Tramonti. IntechOpen. https://doi.org/10.5772/24771.

Gotch, Frank A. 2001. “Evolution of the Single-Pool Urea Kinetic Model.” Seminars in Dialysis 14 (4): 252–56. https://doi.org/10.1046/j.1525-139X.2001.00064.x.

Gotch, Frank A., and John A. Sargent. 1985. “A Mechanistic Analysis of the National Cooperative Dialysis Study (NCDS).” Kidney International 28 (3): 526–34. https://doi.org/10.1038/ki.1985.160.

Grogan, Sean, and Charles V. Preuss. 2023. “Pharmacokinetics.” In StatPearls. StatPearls Publishing. https://www.ncbi.nlm.nih.gov/books/NBK557744/.

Locatelli, Francesco, Umberto Buoncristiani, Bernard Canaud, et al. 2005. “Dialysis Dose and Frequency.” Nephrology Dialysis Transplantation 20 (2): 285–96. https://doi.org/10.1093/ndt/gfh550.

National Kidney Foundation. 2015. “KDOQI Clinical Practice Guideline for Hemodialysis Adequacy: 2015 Update.” American Journal of Kidney Diseases 66 (5): 884–930. https://doi.org/10.1053/j.ajkd.2015.07.015.

Weiner, I. David, William E. Mitch, and Jeff M. Sands. 2015. “Urea and Ammonia Metabolism and the Control of Renal Nitrogen Excretion.” Clinical Journal of the American Society of Nephrology 10 (8): 1444–58. https://doi.org/10.2215/CJN.10311013.