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:
![\dot W_{out} = 133.327\,kW](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20133.327%5C%2CkW)
Explanation:
The model for the turbine can be derived by means of the First Law of Thermodynamics:
![-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0](https://tex.z-dn.net/?f=-%5Cdot%20Q_%7Bout%7D-%5Cdot%20W_%7Bout%7D%20%2B%5Cdot%20m%20%5Ccdot%20%5Cleft%5B%28h_%7Bin%7D-h_%7Bout%7D%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%28v_%7Bin%7D%5E%7B2%7D-v_%7Bout%7D%5E%7B2%7D%29%20%2B%20g%5Ccdot%20%28z_%7Bin%7D-z_%7Bout%7D%29%5Cright%5D%20%3D0)
The work produced by the turbine is:
![\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%3D-%5Cdot%20Q_%7Bout%7D%20%2B%5Cdot%20m%20%5Ccdot%20%5Cleft%5B%28h_%7Bin%7D-h_%7Bout%7D%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%28v_%7Bin%7D%5E%7B2%7D-v_%7Bout%7D%5E%7B2%7D%29%20%2B%20g%5Ccdot%20%28z_%7Bin%7D-z_%7Bout%7D%29%5Cright%5D)
The mass flow and heat transfer rates are, respectively:
![\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%20%2810%5Cfrac%7Bkg%7D%7Bmin%7D%29%5Ccdot%20%28%5Cfrac%7B1%5C%2Cmin%7D%7B60%5C%2Cs%7D%20%29)
![\dot m = 0.167\,\frac{kg}{s}](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%200.167%5C%2C%5Cfrac%7Bkg%7D%7Bs%7D)
![\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )](https://tex.z-dn.net/?f=%5Cdot%20Q_%7Bout%7D%20%3D%20%280.167%5C%2C%5Cfrac%7Bkg%7D%7Bs%7D%20%29%5Ccdot%20%281.1%5Ctimes%2010%5E%7B3%7D%5C%2C%5Cfrac%7BJ%7D%7Bkg%7D%20%29)
![\dot Q_{out} = 183.7\,W](https://tex.z-dn.net/?f=%5Cdot%20Q_%7Bout%7D%20%3D%20183.7%5C%2CW)
Finally:
![\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20-183.7%5C%2CW%20%2B%20%280.167%5C%2C%5Cfrac%7Bkg%7D%7Bs%7D%20%29%5Ccdot%20%5Cleft%288%5Ctimes%2010%5E%7B5%7D%5C%2C%5Cfrac%7BJ%7D%7Bkg%7D%20-562%2C5%5C%2C%5Cfrac%7BJ%7D%7Bkg%7D%20%2B29.43%5C%2C%5Cfrac%7BJ%7D%7Bkg%7D%20%5Cright%29)
![\dot W_{out} = 133.327\,kW](https://tex.z-dn.net/?f=%5Cdot%20W_%7Bout%7D%20%3D%20133.327%5C%2CkW)
Answer:
676 ft
Explanation:
Minimum sight distance, d_min
d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time
v_max is given as 50 mi/h
t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds
Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft
For practical purposes, this distance is taken as 676 ft
Answer:
David should obtain a bachalor's degree of training
Explanation:
It says which of the following paths should he take to obtain a higher-level metrology position.
Based on the maximum capacity as a result of the pipe-laying operations, and the maximum capacity without obstruction to the four-lane dual carriageway, the mean speed of traffic in the bottleneck is -2.73km.
<h3>What is the mean speed of traffic?</h3>
This can be found as:
= (Maximum restricted capacity - freeflowing rate) / (Kb - Ka)
= (1,100 - 1,500) / (209 - 62.5)
= -2.73 km/h
The rate that the queue outside the bottleneck grows is:
= Free flowing rate - (mean speed of traffic x ka)
= 1,500 - (-2.73 x 62.5)
= 1,670 veh/hour
Find out more on bottleneck at brainly.com/question/9551615
#SPJ1