____ are intended to help provide the correct amount of ventilation and also help reduce cooling costs by allowing outside air to be introduced to the conditioned space when conditions are correct

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, [tex]$h_0=2.2$[/tex] inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

[tex]$\Delta h = H_0-h_f=\mu^2R$[/tex]

     [tex]$=0.2^2 \times 14 = 0.56$[/tex] inches

[tex]$h_f = 2.2 - 0.56$[/tex]

     = 1.64 inches

Roll strip contact length (L) = [tex]$\sqrt{R(h_0-h_f)}$[/tex]

                                             [tex]$=\sqrt{14 \times 0.56}$[/tex]

                                             = 2.8 inches

Absolute value of true strain, [tex]$\epsilon_T$[/tex]

[tex]$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$[/tex]

Average true stress, [tex]$\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$[/tex] Psi

Roll force, [tex]$L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$[/tex]

                                 = 788,900 lb

For SI units,

Power = [tex]$\frac{2 \pi FLN}{60}$[/tex]  

           [tex]$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$[/tex]

           = 10399.81168 W

Horse power = 13.9357

Answer:

ICC

Explanation:

The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.

Answer:

W18 * 106

Explanation:

Section modulus of wide flange = 200 m^3

Determine the value of the lightest beam possible

The lightest beam possible that will satisfy the given condition will  have a section modulus ≥ 200m^3 ( note: it will also be the nearest to 200 in^3 )

From Beam Table ; The Lightest beam with its section modulus( 204 in^3) > 200in^3  is   W18 * 106

Answer:

- the undrained friction angles for the soil is 25.02°

- the drained friction angles for the soil is 18.3°

Explanation:

Given the data in the question;

First we determine the major principle stress using the express;

σ₁ = σ₃ + (Δσ[tex]_d[/tex] )[tex]_f[/tex]

where σ₃ is the total minor principle stress at failure ( 15 lb/in² )

(Δσ[tex]_d[/tex] )[tex]_f[/tex] is the deviator stress ( -9 lb/in² )

so

σ₁ = 15 lb/in² + 22 lb/in²

σ₁ = 37 lb/in²

Now, we calculate the consolidated-undrained friction angle as follows;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ ) ]

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 ) ]

∅ = sin⁻¹[ 22 / 52  ]

∅ = sin⁻¹[ 0.423 ]

∅ = 25.02°

Therefore, the undrained friction angles for the soil is 25.02°

-  The drained friction angles for the soil;

∅ = sin⁻¹[ (σ₁ - σ₃ ) / ( σ₁ + σ₃ - 2(Δσ[tex]_d[/tex] )[tex]_f[/tex] ) ]

so we substitute

∅ = sin⁻¹[ (37 - 15 ) / ( 37 + 15 - 2( -9 ) ]

∅ = sin⁻¹[ 22 / ( 37 + 15 + 18 ) ]  

∅ = sin⁻¹[ 22 / 70 ]

∅ = sin⁻¹[ 0.314 ]

∅ = 18.3°

Therefore, drained friction angles for the soil is 18.3°