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Transport Phenomena in Biological Systems Views 94MB Size Report. DOWNLOAD PDF Rotary Kilns Transport Phenomena and Transport Processes. DOWNLOAD PDF. Instructor's Solutions Manual for Transport Phenomena in Biological Systems Second Edition George A. Truskey, Fan Yuan, and David F. PDF | On Jan 1, , G A Truskey and others published Transport Phenomena in Biological Systems.

Reaction 1 has high sensitivity in two regions where the denominator in equation S For diffusion-limited dissociation of a ligand from a receptor, Equation 6. Mass enters through region 1 and exits through region 3. For the case of conduction only, energy transport through the liquid is unchanged. If the receptor can be labeled, then the disappearance from the surface can be studied. The AUC during the first 6 hours is listed in the table. Short contact time release in a rectangular channel.

First, determine the oxygen concentrations in arteries and veins. The concentration in blood is: In terms of the total air inspired in each breadth, it is 1. DV corresponds to the stroke volume. Although the pressure drops from mm Hg to mm Hg, the partial pressures are unchanged.

The inspired air at 3, m is For a 30 mm Hg drop, the alveolar air is at Glucose is at a lower concentration in urine than plasma, suggesting that its transport across the glomerulus is restricted. Assuming that inulin is not reabsorbed by the kidneys and returned to the blood, then the mass flow rate of inulin across the glomerulus must equal the mass flow rate in urine. Since the velocity does not vary with angular position, substitution for vz and integration in the angular direction yields: Integrating in the radial direction yields: Evaluating equation 2.

Rearranging equation 2. From Equation 2. The pressure difference needed to draw the smaller cell into the micropipette is smaller than the pressure difference needed to draw in a larger cell as in part a. Each side is equal to a constant C1. Thus the velocity is: The volumetric flow rate for each fiber can be written as: Reynolds number is calculated from the equation: From equation 2.

This value is much larger than the length of the hollow fiber unit. For a power law fluid, the average velocity and shear rate equal equations 2. A schematic of a volume element is shown below. This is because the viscosity of the gas is much less than the viscosity of the liquid. For fluid 1, since the resulting velocity is linearly related to the original velocity, the fluid is Newtonian.

For fluid 2, there is no linear or power dependence between the velocities suggesting that the fluid is a Bingham plastic. Applying Equation 2. For fluid 3, the resulting velocity is proportional to the square of the original velocity. The fluid is a power law fluid. For a power law fluid, equation 2. For a Bingham plastic the momentum balance is unchanged from equation 2.

The inner cylinder is not moving. Integrating we have: For a Bingham plastic the result is: For a power law fluid the shear stress is related to the shear rate as: The shear stress is: Since the apparent viscosity depends upon the shear rate, the fluid is not Newtonian. Taking the logarithm of each side yields.

The value of m is A plot of the shear stress versus shear rate revealed that while a straight line gives a good fit, there is some curvature to the data suggesting a shear thinning fluid. So the fluid does exhibit some shear thinning behavior. The enzyme functions by clipping the hyaluronic acid chains decreasing their length. As a result the hyaluronic acid offers much less resistance to flow.

The error is 0. Thus, the torque for a power law fluid is: This velocity is a function of r only. There is no angular variation in velocity or pressure.

Applying a momentum balance, Equation 2. C1 r2 We need the velocity profile in order to determine the constant. Thus, the velocity field is: Note that the velocity is constant and the unit normal is in the - r direction. One is to neglect the fluid in the gaps.

The velocity profiles, shear rate and apparent viscosity are dependent on the constitutive equation. The shear rate is 7. The relation between flow and pressure drop is: Assuming that the enzyme decreased the inner radius of the blood vessel by removing the glycocalyx completely, the change is resistance is due solely to a change in the effective radius of the blood vessel.

Assuming that the radius after enzyme treat is Section 3. For flow in a cylindrical tube, the friction factor is defined as: The coordinate system originates at the centerline as shown in Figure 3. The boundary conditions are stated below. Substituting for a Newtonian fluid: To find the volumetric flow rate, compute the average velocity in the fluid between r and R and compare to the value obtained in the absence of the catheter. The average velocity is: The term on the right hand side is less than or equal to one.

Integrating Equation 3. The shear stress gradients are very different. The unsteady velocity profile for flow in a cylindrical tube of radius R is: The shorter time to reach steady state in the cylindrical geometry arises as a result of the deceasing cross-sectional area which improves momentum transport. The force balance is the same that presented in equation 3.

Thus, the velocity vz is obtained from equation 3. The solution becomes: Thus, according to the Pi theorem there are two dimensionless groups.

The second dimensionless group can be found, using the same approach, to be RcRP The force F acting on the leukocyte is a function of the following variables: So there are 7 dimensional variables and three characteristic dimensions, mass, length and time. Thus, in addition, there are four other dimensionless groups. Thus, the dimensionless relation is: The resulting shape is this result yields the following: This latter condition ensures that the shear stress is continuous.

The resulting velocity profile is: Using the same approach on equation 3. From equation S3. As a result, equation 3. Equation 3.

The final result is: The resulting drag force is 4. This problem is the inverse of the example discussed in Section 3. That is, the fluid far from the sphere has a zero velocity and the sphere is moving at a constant speed U0. To find the pressure field, substitute equations 3. These derivatives are: The second term is then integrated by parts. Thus, the component of the drag due to the shear stress is: The minus sign indicates that the drag force is in the direction opposite the motion of the sphere.

The velocity can be obtained from the definition of the wall shear stress and equation 2. The Reynolds number is 0. Thus the assumption of low Reynolds number flow is valid. Same as in unmodified solution. Solve the force balances to determine the distance traveled by a cell which is initially at the centerline.

To determine the distance traveled, we need to compute the change in x and y position. The cell velocity in the x direction is just the change in x position with time. Likewise, the cell velocity in the y direction is just the change in y position with time.

The fluid velocity is given by equation 2. Thus, the force balances are rewritten as: Notice that the displacement of the cell in the x direction depends upon the vertical position of the cell.

Thus, the y location of the cell is now known as a function of time. Note that the terminal velocity is negative so time is still a positive quantity! This result means that if you measured the number of adherent or rolling cells within 0. Thus, the criterion is met.

For steady, fully developed laminar flow in a cylindrical tube using rectangular coordinates, flow is in the z direction and normal to the x-y plane. Since p z and vz x,y , each side of Equation S3. This solution does satisfy the boundary conditions.

Using this in Equation S3. The solution for an elliptical tube can be obtained by a variation on the approach used in Problem 3. Equation S. The tube surface is given by: Begin with equation 4. Since integration is over the entire surface, the second and third terms on the right hand side of equation S4. Oxford University Press. Assume that the flow is laminar, the centerline is horizontal so that gravity does not influence flow and flow is steady.

Equation 4. With these simplifications, the analysis uses the same development presented in Example 4. The control volume can be divided into three regions, 1, 2 and 3. Mass enters through region 1 and exits through region 3. There is no flow across the walls of the vessel, region 2. The average velocity in the inlet and outlet can be related by the conservation of mass.

The z component of the left hand side of equation S4. Because the detailed velocity field through the taper is not known, the shear stress cannot be computed. The force depends strongly on the taper ratio E. The net pressure force acting on the fluid is obtained by rearranging equation S4.

For steady, laminar flow through the entrance region of a cylindrical tube that is horizontally so that gravity is negligible, the integral form of the conservation of linear momentum, eEquation 4. The integral on the left hand side is nonzero at the entrance and downstream where the flow is fully developed.

The integral of the shear stress represents the net drag force acting on the fluid. Thus, the drag force equals: From equations 4. The results are presented in Figure S4. The net force is smallest at an angle of zero degrees. The net force at zero degrees is not zero because the 51 cross-sectional areas in sections 4 and 5 are greater than 1, resulting in a decrease in velocity through the tubes.

As the branch angle increases, the net force increases, reaching a maximum at degrees. Figure S4. The reasons for this discrepancy are that the centerline velocity increases as the fluid decelerates near the surfaces and the boundary layer thickness is not small.

This coefficient is much closer to the expected value. By comparison of equations 4. Substituting equation S4.

This corresponds to The viscous losses are much smaller than the pressure drops induced by conductive acceleration. Note that this analysis is quite simplified. A more thorough analysis would include the effect of viscous losses are due to the jet as discussed in [11]. From equation 4. This problem is similar to the previous except that the atrial area must be calculated from the diameter.

First calculate the Reynolds number treating blood as a Newtonian fluid. Thus, from Table 2. We do this assuming that the turbulent profile is uniform. If the flow is laminar, we need to recalculate assuming that the centerline is two times the average.

The conclusion still holds. Choosing a streamline along the centerline, gravity can be neglected. Flow upstream is laminar. So, the centerline velocity is 2. Flow in the constriction could be laminar, transitional or turbulent. Assume that the flow is laminar.

The radius decrease by 5. Consequently, the Reynolds number increases by 2. Assuming laminar flow through is acceptable but there may be bursts of turbulence. If the flow were turbulent, would equal the centerline velocity. At 1 and 2 the result is: The centerline is horizontal so that gravity does not influence flow and flow is steady. Mass enters through region 1 and exits through region 2.

There is no flow across the walls of the vessel, region 3. The z component of the left hand side of Equation S4. Neglect any frictional resistance acting on the plunger. Making these substitutions: The expression for the pressure drop is from Equation 4.

To convert to mm Hg, divide by The acceleration term is then 4. The wall shear stress is reported as a positive quantity. Thus, blood flow in the human does not represent a dynamically scaled model. From the other two conditions: The flow is steady and two-dimensional. Gravity forces can be neglected.

The characteristic length in the x-direction is L and the characteristic length in the y-direction is h. The characteristic velocity in the x-direction is U. The characteristic velocity in the y-direction can be determined from the dimensionless form of the conservation of mass.

Define the following dimensionless variables as follows. As a result, the y-component of the Navier-Stokes equation reduces to: It should be 0. To support a net force of Clearly, by increasing the fluid viscosity, the sliding velocity can be significantly reduced. Beginning with Equation 4. Te pressure gradient can be rewritten as: We can rewrite equation S.

Thus, there are, at most, four dimensionless groups. These can be identified using the procedures outlined in Section 3. Alternatively, the groups can be identified from the known dimensionless groups. The dimensionless pressure drop depends on two dimensionless groups. There are two possible groupings. At high values of curvature, there is a steep boundary layer near the outer wall. Figure S5. For vigorous breathing, the length of the branch exceeds the entrance length for branch generation 15 and greater.

Case 1 has positive pressures throughout each cycle. The velocity varies about the mean value and shapes are similar at all time. For case 2 there is a significant negative pressure during the early part of the cycle period causing fluid deceleration. Fluid accelerates during the pressure upswing and the velocity reaches a maximum by 0. The change in fluid velocity with time lags the pressure change due to inertia.

The pressure gradient as a function of time is shown in Figure S5. The pressure pulse wave can be well represented by six to eight harmonics. The pressure is unusual in that it becomes negative during the early part of the cycle. This is due to vessel compression arising from contraction of the left ventricle. The flow tracks the pressure, although there is a slight lag. For other results see the paper by He et al.

BME Pressure during the period T. Solid line are original data. Squares represent the predicted values from the inverse FFT. The dotted line represents the pressure obtained from the first 8 harmonics. Six to eight harmonics are sufficient to generate reasonably accurate reconstructions of the pulse although it was not possible to duplicate the pulse even with more harmonics. Including higher harmonics in the Fourier series did not improve results.

Going above the Nyquist frequency of 11 harmonics actually resulted in some degradation of the signal. Note that rc is defined in Equation 2. Shown in Figure S5. The velocity profiles for the Casson fluid exhibits a slight amount of flattening. The Newtonian velocities are smaller than the corresponding values for the Casson fluid. The flow rate for the Casson fluid is: As expected from the velocity profiles, the flow rate for the Casson fluid is larger than the value for the Newtonian fluid.

The shear stress was calculated using the velocity profiles for the Newtonian fluid equation 5. At early times, the shear stress for the Casson fluid is greater than the value for the Newtonian fluid. This difference declines as the magnitude of the velocity declines. When flow is negative, the shear stress for the Casson fluid again becomes slightly greater in magnitude than the value for the Newtonian fluid. Flow A and shear stress B for Casson and Newtonian fluids.

For the right coronary artery, data in Table 5. The average fluid velocity is 7. Assume a Newtonian fluid of viscosity of 0.

For these data, equations 5. The effect of the curvature on the shear stress in the z direction is modest. To solve, use equation 5. The net force on the cells is: For an ellipsoid shape, this represents one-half the surface area of an ellipsoid. Integration of S5. The quasi-steady state velocity is Equation 5. Equation 5. For the quasi-steady result, the phase lag is 0 radians. From equation 5. The wall shear stress for laminar flow in a cylindrical tube is obtained from Equation 2.

Q is a monotonically increasing function of radius to the third power. Therefore, when F is minimized, Q reaches its maximum value. For oxygen, the partial molar volume is Application of the Wilkie-Chang equation equation 6.

Comparing with the measured diffusion coefficient in Table 6. An estimate for the radius is needed to apply the Stokes-Einstein equation. If the measurements for both molecules are assumed to occur at the same temperature and both can be treated as spheres, then the diffusion coefficients are related by: Shown below are four step simulations using the second relation. Forty two simulations were performed and the root mean square displacement was Figure S6. The scatter represents the variability arising from the random number generation.

The cylinder is a slightly more accurate model than the prolate ellipsoid but both shapes do not accurately predict the diffusion coefficient. A mass balance on the volume element shown in Figure 6. Because the area changes with x, the flux decreases with increasing cross-sectional area. Using this relation and Equation 6. Since the mole fraction of NO is 0.

The term 1xi in the expression for the flux is initially 0. Thus, the flux can be approximated as: For the values given the time is 1. For the problem given this is the time that it would take for the gas to leave the alveolus. This is much shorter than the time between breathes, 5 sec. Since the same solvent is on both sides of the membrane, K should be the same on both sides.

Further the flux through the membrane is in the opposite direction as the flux in the fluid phase. Submit Search. Successfully reported this slideshow. We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime. Transport phenomena in biological systems 2nd edition [pdf] download.

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