Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
![Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]](https://tex.z-dn.net/?f=Q%20-%20%5Cdot%7BW%20%7D%20%3D%20%5Cdot%7Bm%7D%5Cleft%20%5B%20%5Cleft%20%28h_%7B2%7D-h_%7B1%7D%20%20%5Cright%20%29%2B%5Cdfrac%7Bv_%7B2%7D%5E%7B2%7D-%20v_%7B1%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20g%28z_%7B2%7D-z_%7B1%7D%29%5Cright%20%5D)
Which gives;
![560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]](https://tex.z-dn.net/?f=560%20-%20%5Cdot%7BW%20%7D%20%3D%208%5Cleft%20%5B%20%5Cleft%20%282599.2418-3755.39%20%20%5Cright%20%29%2B%5Cdfrac%7B15%5E%7B2%7D-%2060%5E%7B2%7D%7D%7B2%7D%20%5Cright%20%5D)
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
= -22749.1856 - 560 = -23309.1856 kJ
= 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Answer:
I = 1205.69 Lx
Explanation:
The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀
I = I₀ sin θ
in this case with the initial data we can calculate the initial irradiance
I₀ = 
I₀ = 1600 /sin 53
I₀ = 2003.42 lx
for when the angle is θ = 37º
I = 2003.42 sin 37
I = 1205.69 Lx
Answer:
4 times around
Explanation:
The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.
If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...
(48 teeth)/(12 teeth/turn) = 4 turns
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
Answer:
A force must s applied to a wall or roof rafters to add strength and keep the building straight and plumb
