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juin [17]
3 years ago
9

A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem

perature is 37.0°C, and it has a surface area of 2.23 m2. What is the average thickness of its blubber? The conductivity of fatty tissue without blood is 0.20 (J/s · m · °C).
Engineering
1 answer:
Llana [10]3 years ago
7 0

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

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In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).

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3 years ago
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =
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Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

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3 years ago
A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a
OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS        A                B                    C

Benzene 1             4.72583     1660.652        -1.461

Toluene  2            4.07827     1343.943         -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5%    = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis     for all components

So for Benzene ; p × y1=x1 × p1s   ------let this be equation 1

for Toluene ; p × y2=x2 × p2s   ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0
2 years ago
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