A large tank, at 500 K and 200 kPa, supplies isentropic air flow to a nozzle. At section 1, the pressure is only 120 kPa. What is the temperature at section 1

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:

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:

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:

Explanation:

Pie charts generally should have no more than eight segments.

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°