True/False

In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface

ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0

3 years ago

What are the reasons why fine grained of alkali igneous rocks can not be used in cement

Luden [163]

Fine grained of alkali igneous rocks cannot be used in cement because of the volume expansion caused by the Alkali-silica reaction, fine-grained igneous rocks cannot be used as aggregates in cement.

<h3>What does fine grained mean in an igneous rock?</h3>

Extrusive igneous rocks have a fine-grained or aphanitic texture, with grains that are too small to see without a magnifying glass. The fine-grained texture suggests that the rapidly cooling lava did not have enough time to form large crystals. A petrographic microscope can be used to examine these tiny crystals.

The texture of an igneous rock (fine-grained vs coarse-grained) is determined by the rate at which the melt cools: slow cooling produces large crystals, while fast cooling produces small crystals.

The chemical reaction that occurs in both alkali cations and hydroxyl ions in the pore solution of hydrated cement paste and certain reactive silica phases present in concrete aggregates is known as the alkali-silica reaction (ASR).

Learn more about cement on:

brainly.com/question/14323034

#SPJ1

8 0

2 years ago

What is 220 C in degrees Fahrenheit (F)?

8_murik_8 [283]

Answer:

428°F

Explanation:

The equation to convert degrees Celsius to degrees fahrenheit is

°F (degrees fahrenheit) = (9/5 * °C (degrees celsius) ) + 32

°F = (9/5 * 220 °C (degrees celsius)) +32 = 428 °F (degrees fahrenheit)

4 0

3 years ago

Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a

defon

Answer:

Explanation:

Wow

5 0

3 years ago

A soil has the following Green-Ampt parameters Effective porosity 0.400 Initial volumetric moisture content-15% Hydraulic Conduc

Rudik [331]

Answer:

The graphs are attached below

Explanation:

Ans) We know,

F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]

where K = hydraulic conductivity =0.1 cm/hr

z = capillary suction =20cm

Δ∅ = ∅ (1 - Se)

∅ = effective porosity , Se = initial moisture content

Δ∅ = 0.40(1 - 0.15)

Δ∅ = 0.34

Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]

F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]

Also, infiltration rate (f),

f = K [(zΔ∅ + F)/F]

f = 0.40 [6.8 + F]/F

Condition of ponding,

Ponding time tp = K zΔ∅ i(i-k)

where, i = rainfall rate (1cm/hr)

tp = 0.40(6.8) / 0.60

tp= 4.53 hours

Now, cumulative infiltraton at ponding Fp = i tp

Fp = 1 x 4.53 or 4.53 cm

For infiltration at time less then ponding time , infiltration rate = rainfall rate

For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm

For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm

For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm

For t = 1 hr , f = 1cm/hr ; F = 1cm/hr

For t > tp ,

Equivalent time origin(to),

to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)

to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)

to = 4.53 - 2.5 ( 4.53 - 3.44)

to = 1.82 hr

Hence,

F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]

Solving above equation for t by assuming F, and further solving equation for infiltration rate

Ans (a) Following is required curve :

O.S 1-5 2 time Chr)ホー Lt 2 time Chr)

Ans (c) Following is required table :

Ans (d) For time step t = 0.25 hrs

At t < tp ; i = f = 1 cm/hr

F = i x t

F = 1 x 0.25

F = 0.25 cm

5 0

4 years ago