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Ann [662]
3 years ago
9

Consider the problem of oxygen transfer from the interior lung cavity, across the lung tissue, to the network of blood vessels o

n the opposite side. The lung tissue (species B) may be approximated as a plane wall of thickness L. The inhalation process may be assumed to maintain a constant molar concentration CA(0) of oxygen (species A) in the tissue at its inner surface (x = 0), and assimilation of oxygen by the blood may be assumed to maintain a constant molar concentration CA(L) of oxygen in the tissue at its outer surface (x = L). There is oxygen consumption in the tissue due to metabolic processes, and the reaction is zero order, with N_A=-k_0.
Obtain expressions for the distribution of the oxygen concentration in the tissue and for the rate of assimilation of oxygen by the blood per unit tissue surface area.

Engineering
1 answer:
aalyn [17]3 years ago
8 0

Answer:

See attached images

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5 0
2 years ago
Read 2 more answers
Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.
zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

T_{sat}=179.88\°c

H_{fg} = 2014.6kJ/kg

c_p=4.18 kJkg^{-1}{K^{-1}

\nu_g = 0.19436m^3/kg

Replacing,

\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})

\Delta h = 2014.6+4.18(179.88-24)

\Delta h=2666.17kJ/kg

With the specific volume we know can calculate the mass flow, that is

\dot{m}=\frac{\frac{15000}{3600}}{0.19436}

\dot{m} = 21.4378kg/s

Then the heat required in input is,

Q=\dot{m}\Delta h

Q=21.4378*2666.17

Q=57157.036kW

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

V= \frac{\dotV}{A}

V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}

V=235.79m/s

Finally we can apply the steady flow energy equation, that is

\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2

Re-arrange for Q,

Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})

Q=\dot{m}(\Delta h-\frac{V^2}{2000})

Q= (21.4378)(2666.17-\frac{235.79^2}{2000})

Q= 56560.88kW

We can note that consider the Kinetic Energy will decrease the heat input.

4 0
3 years ago
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%

Where:

V_{gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{fe} - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%

Where:

m_{fe}, m_{gr} - Masses of the ferrite and graphite phases, measured in grams.

\rho_{fe}, \rho_{gr} - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%

\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%

If \rho_{gr} = 2.3\,\frac{g}{cm^{3}}, \rho_{fe} = 7.9\,\frac{g}{cm^{3}}, m_{gr} = 3.2\,g and m_{fe} = 96.8\,g, the volume percentage of graphite is:

\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%

\%V_{gr} = 10.197\,\%V

The volume percentage of graphite is 10.197 per cent.

5 0
3 years ago
If engineering is easy then why don't most people join?
Mazyrski [523]
Engineering requires a lot of school
5 0
3 years ago
Read 2 more answers
5. Assume that you and your best friend ench have $1000 to invest. You invest your money
Bezzdna [24]

Correct question reads;

Assume that you and your best friend each have $1000 to invest. You invest your money in a fund that pays 10% per year compound interest. Your friend invests her money at a bank that pays 10% per year simple interest. At the end of 1 year, the difference in the total amount for each of you is:

(a) You have $10 more than she does

(b) You have $100 more than she does

(c) You both have the same amount of money

(d) She has $10 more than you do

<u>Answer:</u>

<u>(d) She has $10 more than you do</u>

<u>Explanation</u>:

Using the compound interest formula

A= P [ (1-i)^n-1

Where P = Principal/invested amount, i = annual interest rate in percentage, and n = number of compounding periods.

<u>My compound interest is:</u>

= 1000 [ (1-0.1)^1-1

= $1000

$1,000 + $1,000 invested= $2,000 total amount received.

<u>My friend's simple interest is;</u>

To determine the total amount accrued we use the formula:

P(1 + rt) Where:

P = Invested Amount (1000)

I = Interest Amount (10,000)

r = Rate of Interest per year (10% or 0.2)

t = Time Period (1 )

= 1000 (1 + rt)

= 1000 (1 + 0.1x1)

= $1100 + $1000 invested = $2100 total amount received.

Therefore, we observe that she (my friend) has $100 more than I do.

5 0
3 years ago
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