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aleksklad [387]

3 years ago

12

Where are revolved sections placed in a print? A) in between break lines B) cutting planes are used to identify their locations

C) in between section lines D) stand alone

1 answer:

Svetach [21]3 years ago

6 0

Answer:

B. Cutting planes are used to identify their locations.

Explanation:

Revolved view is a cross section view of revolved 90 degrees around a cutting plane projections. The revolved view of print will differ from a cross sectional view. It includes a line nothing the axis of revolution for the view. The correct answer is B. The revolved section in the prints has cutting planes that will be used to identify their location.

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For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, $A=2\pi rL\;\cdots(ii)$

Variation of temperature w.r.t the radius of the annalus is

$\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}$

$\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)$

Putting the values from the equations (ii) and (iii) in the equation (i), we have

$q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}$

$\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}$

$\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)}$ [as $q=2.4\pi L$ , and $T_2-T_1=10 ^{\circ}C$ ]

$\Rightarrow k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0

3 years ago

What’s Statistics<br> What are the 2 Source of error in data collection

FromTheMoon [43]

Answer:

<em>The main sources of error in the collection of data are as follows : Due to direct personal interview. Due to indirect oral interviews. Information from correspondents may be misleading.</em>

4 0

3 years ago

Read 2 more answers

A sample of municipal sewage is diluted to 1% by volume prior to running a BOD5 analysis. After 5 days the oxygen consumption is

liberstina [14]

Answer:

BOD5 = 200 mg/L

Explanation:

given data

diluted = 1% = 0.01

time = 5 day

oxygen consumption = 2.00 mg · L−1

solution

we get here BOD5 that is BOD after 5 day

and here total volume is 100% = 1

so dilution factor is $\frac{100}{1}$ = 100

so BOD5 is

BOD5 = oxygen consumption × dilution factor

BOD5 = 2 × 100

BOD5 = 200 mg/L

4 0

3 years ago

6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

7 0

3 years ago

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

$Q=\dot{m}C_p\Delta T$

$404.5=\dot{m}\times 4.187(13.9-6.1)$

$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

$\dot{m}=\rho\times volume\ flow\ rate$

12.38=1000 x volume flow rate

$volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}$

So

volume flow rate = 12.38 L/s

So the option B is correct.

8 0

3 years ago