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Pavlova-9 [17]
3 years ago
14

Consider a drug-eluting balloon catheter deployed into a blood vessel. The balloon is inflated to perfectly adhere to the vessel

walls. Drug is released from the outer surface of the balloon and diffuses radially through the endothelial tissue only in the r-direction. Let the outside radius of the balloon be r = R_0, where the drug has a constant concentration c_0. The drug diffuses into the tissue as far as r = R, where the concentration of the drug becomes zero.
Assume cylindrical geometry, steady state conditions and no chemical reaction involving the drug. The drug diffusion coefficient in tissue is D.
(a) Find an expression for the profile c(r) inside the tissue (i.e. from R_0 to R).
(b) Determine the diffusive flux at the outer surface of the balloon.
Engineering
1 answer:
GaryK [48]3 years ago
6 0

Answer:

a)  Cr = Co - Fx / D

b)   ΔC / Δx = ( CR - Cr )  / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X  = D ( Co - Cr ) / X   ------ 1

where : F = flux , D = drug diffusion coefficient

            X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface =  ΔC / Δx = CR - Cr / xR - xRo

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Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

Hence, length of each side of a square,

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4 0
3 years ago
The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
leva [86]

Answer:

The distance measure from the wall = 36ft

Explanation:

Given Data:

w = 10

g =32.2ft/s²

x = 2

Using the principle of work and energy,

T₁ +∑U₁-₂ = T₂

0 + 1/2kx² -wh = 1/2 w/g V²

Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

170 = 0.15528V²

V² = 170/0.15528

V²     = 1094.796

V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

   = 36.87°

From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

19.85t - 16.1t² + 3 = 0

16.1t² - 19.85t - 3 = 0

Solving it quadratically, we obtain t = 1.36s

The distance measure from the wall is given by the formula

d = VCos∅*t

Substituting, we have

d = 33.09 * cos 36. 87 * 1.36

d = 36ft

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Answer:

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