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:
4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Explanation:
we have given 4140 steel contains 0.4% C
we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element
and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel
while in 1045 steel contains 0.45 % c is plain carbon steel
and it do not contain any alloying element
so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Overcurrent protective devices, or OCPDs
Answer:
Following are the proving to this question:
Explanation:
using the energy equation for entry and exit value
:
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%20%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D)
where
![= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\](https://tex.z-dn.net/?f=%3D%20%28%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%20%29%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%29%5E4%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C%5C%5C)
![= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7B%282f%20%28%5Cfrac%7Bl%7D%7BD%7D%29%20%20%29%7D%20%5C%20%20V%5E%7B2%7D_%7B1%7D%5C%5C)
![\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7Bp_o%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7Bo%7D%7D%7B2g%7D%2BZ_0%20%20%3D%5Cfrac%7Bp_1%7D%7By%7D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7BV%5E%7B2%7D%7D%7B2g%7D%20%5C%5C%5C%5C)
![\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})](https://tex.z-dn.net/?f=%5Cto%200%2B0%2BZ_0%20%3D%200%20%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%20%7D%7B2g%7D%20%2BZ_1%2B%20f%20%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7B%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%29%7D%5C%20V%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%20%20%5C%5C%5C%5C%5Cto%20Z_0%20-Z_1%20%3D%20%2B%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%5C%20%281%2Bf%5Cfrac%7Bl%7D%7BD%7D%5Cfrac%7B1%7D%7B%282f%28%5Cfrac%7Bl%7D%7BD%7D%29%20%29%7D%20%29%20%20%5C%5C%5C%5C%5Cto%20H%3D%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%28%5Cfrac%7B3%7D%7B2%7D%29%20%5C%5C%5C%5C%5Cto%20%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2g%7D%20%3D%20H%28%5Cfrac%7B3%7D%7B2%7D%29)
L.H.S = R.H.S
All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing
Answer: parallel