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2 years ago

11

Concrete ___ support and anchor the bottom of steel columns and wood post, which support beams that are pare of framing system o

f building.
A.anchor
B.piers
C.footing
D.post

1 answer:

dolphi86 [110]2 years ago

3 0

I wanna say it’s D post support

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2 years ago

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

3 years ago

Suggest appropriate materials for a cylinder head and give four reason for your choice

ycow [4]

Answer:

head and give four reason for your choice

7 0

3 years ago

Which of the following are true about the American Wire Gauge?

harina [27]

Answer:

A. smallest wire is No. 12

7 0

2 years ago

A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg

Eva8 [605]

Answer:

$\dot{w}$ = -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

$\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}$

Given that tank is insulated so $\dot{Q}=0$ and no mass is leaving so

$\dot{m_e}=0$

$\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt$

$m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t$

Mass conservation $m_2-m_1=m_e-m_i$

$m_1,m_2$ is the initial and final mass in the system respectively.

Initially tank is evacuated so $m_1=0$

We know that for air $u=C_vT ,h=C_p T$ , $P_2v_2=m_2RT_2$

$m_2=0.42 kg$

So now putting values

$0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300$

$\dot{w}$ = -0.303 KW

3 0

3 years ago