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Gala2k [10]
3 years ago
8

Consider a section of muscle tissue of a cylindrical shape with a radius of 1.5 cm. During highly rigorous exercise, metabolic p

rocesses generate 15 kW/m3 of bulk tissue. The outer surface of the tissue is maintained at 37oC. The thermal conductivity of the muscle tissue is k =0.419 W/m*KDevelop a mathematical model to predict the steady-state temperature distribution, T(r), in the tissue along the radial position r. What is the temperature of the tissue at a distance of 0.75 cm from the surface?

Engineering
1 answer:
Tems11 [23]3 years ago
5 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a)The mathematical model to predict the steady temperature distribution is

       =  T + \frac{q^{*}r^{2}}{4k} = 312.014

b)The mathematical model to predict the maximum temperature inside the muscle tissue is

      = \frac{\delta T}{\delta r}  = 0 \implies  r = 0

c) The heat flux leaving the muscle is

      = 1.06 W

d) The new maximum temperature within the muscle is

     = 38.59°C

Explanation:

The explanation is shown on the second, third , fourth and fifth image

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The four strokes in a four stroke cycle engine in proper order.
Morgarella [4.7K]

Answer:

A four-stroke cycle engine is an internal combustion engine that utilizes four distinct piston strokes (intake, compression, power, and exhaust) to complete one operating cycle. The piston make two complete passes in the cylinder to complete one operating cycle.

Explanation:

6 0
2 years ago
A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c
Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice C_p=2.03\ KJ/kg.K

Latent heat for ice H=336 KJ/kg

Specific heat for ice C_p=4.187\ KJ/kg.K

We know that sensible heat given as

Q=mC_p\Delta T

Heat for -15F to 32 F:

Q=mC_p\Delta T

Q=9\times 2.03(32+15) KJ

Q=858.69 KJ

Heat for 32 Fto 200 F:

Q=mC_p\Delta T

Q=9\times 4.187(200-32) KJ

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So   7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0
3 years ago
Disconnecting a circuit while in operation can create a voltage blank
zlopas [31]

Answer:

what is the question

Explanation:

confused

5 0
2 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
A construction manager is looking for workers to build a series of walkways within a large garden apartment. What type of workme
Lesechka [4]

Answer:

<u>construction workers</u>

Explanation:

A construction worker is usually someone with the technical skills and abilities needed to manually construct physical infrastructures.

Since a walkway is an infrastructure, hiring skilled construction workers should perform the project of building the series of walkways within the large garden apartment.

3 0
3 years ago
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