Book name: Construction Methods and Management Stephens W. Nunnally Eighth Edition

The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt

Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0

2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".