Answer:
A renewable electricity generation technology harnesses a naturally existing energy. But they have other features that a few fringe customers value.
Explanation:
Answer:
Power output, ![P_{out} = 178.56 kW](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20178.56%20kW)
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, ![T = 350^{\circ}C](https://tex.z-dn.net/?f=T%20%3D%20350%5E%7B%5Ccirc%7DC)
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, ![\dot{m} = 0.1 kg.s^{- 1}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%200.1%20kg.s%5E%7B-%201%7D)
Diameter of exhaust pipe, ![d_{h} = 15 cm = 0.15 m](https://tex.z-dn.net/?f=d_%7Bh%7D%20%3D%2015%20cm%20%3D%200.15%20m)
Pressure at exhaust, P' = 50 kPa
temperature, T' = ![100^{\circ}C](https://tex.z-dn.net/?f=100%5E%7B%5Ccirc%7DC)
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:
![\dot{m} = \frac{Av_{i}}{V_{1}}](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%20%3D%20%5Cfrac%7BAv_%7Bi%7D%7D%7BV_%7B1%7D%7D)
![0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}](https://tex.z-dn.net/?f=0.1%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%20d%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7B0.2004%7D)
![0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}](https://tex.z-dn.net/?f=0.1%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%200.08%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7B0.2004%7D)
![v_{i} = 3.986 m/s](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D%203.986%20m%2Fs)
Now, eqn for compressible fluid:
![\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}](https://tex.z-dn.net/?f=%5Crho_%7B1%7Dv_%7Bi%7DA_%7B1%7D%20%3D%20%5Crho_%7B2%7Dv_%7Be%7DA_%7B2%7D)
Now,
![\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BA_%7B1%7Dv_%7Bi%7D%7D%7BV_%7B1%7D%7D%20%3D%20%5Cfrac%7BA_%7B2%7Dv_%7Be%7D%7D%7BV_%7B2%7D%7D)
![\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B%5Cpi%20d_%7Bi%7D%5E%7B2%7D%7D%7B4%7Dv_%7Bi%7D%7D%7BV_%7B1%7D%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%20d_%7Be%7D%5E%7B2%7D%7D%7B4%7Dv_%7Be%7D%7D%7BV_%7B2%7D%7D)
![\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B%5Cpi%20%5Ctimes%200.08%5E%7B2%7D%7D%7B4%7D%5Ctimes%203.986%7D%7B0.2004%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B%5Cpi%200.15%5E%7B2%7D%7D%7B4%7Dv_%7Be%7D%7D%7B3.418%7D)
![v_{e} = 19.33 m/s](https://tex.z-dn.net/?f=v_%7Be%7D%20%3D%2019.33%20m%2Fs)
Now, the power output can be calculated from the energy balance eqn:
![P_{out} = -\dot{m}W_{s}](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%5Cdot%7Bm%7DW_%7Bs%7D)
![P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%5Cdot%7Bm%7D%28H_%7B2%7D%20-%20H_%7B1%7D%29%20%2B%20%5Cfrac%7Bv_%7Be%7D%5E%7B2%7D%20-%20v_%7Bi%7D%5E%7B2%7D%7D%7B2%7D)
![P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW](https://tex.z-dn.net/?f=P_%7Bout%7D%20%3D%20-%200.1%283.4181%20-%200.2004%29%20%2B%20%5Cfrac%7B19.33%5E%7B2%7D%20-%203.986%5E%7B2%7D%7D%7B2%7D%20%3D%20178.56%20kW)
Answer:
Toeboards, debris nets, or canopies
Explanation:
Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Technical Drawings give a better understanding of what is needed and required in the project.
Explanation: