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juin [17]
4 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]4 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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A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
Select the correct answer.
boyakko [2]
I think the answer is b
6 0
3 years ago
Read 2 more answers
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14,  pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

    = \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s

obtained Darcy friction factor  

calculate Reynold number (Re) ,

Re = \frac{\rho V D}{\mu}

where,\rho = density of water

\mu = Dynamic viscosity of water at 15 degree  C = 0.001 Ns/m2

so reynold number is

Re = \frac{1000\times 2.015\times 0.318}{0.001}

            = 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is  0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}

HL = 1.64 m

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Two units for measuring magnetic flux density are:
romanna [79]
The correct answer is A
Faraday and Weber
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Answer:

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8 0
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