Answer:
The maximum theoretical height that the pump can be placed above liquid level is ![\Delta h=9.975\,m](https://tex.z-dn.net/?f=%5CDelta%20h%3D9.975%5C%2Cm)
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:
![\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2} \Delta v^2 =0](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20P%7D%7B%5Crho%7D%2Bg%5C%2C%20%5CDelta%20h%20%2B%5Cfrac%7B1%7D%7B2%7D%20%20%5CDelta%20v%5E2%20%3D0)
(
stands here for density,
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:
![\frac{\Delta P}{\rho}+g\, \Delta h =0](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20P%7D%7B%5Crho%7D%2Bg%5C%2C%20%5CDelta%20h%20%20%3D0)
![\Delta P= -g\, \rho\, \Delta h](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20-g%5C%2C%20%5Crho%5C%2C%20%5CDelta%20h)
This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:
![\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%202.34%5C%2CkPa-%20100%20%5C%2CkPa%20%3D%20-97.66%20%5C%2C%20kPa%5C%5C)
We insert that into our last equation and get:
![\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5CDelta%20P%7D%7B%20-g%5C%2C%20%5Crho%5C%2C%7D%3D%20%5CDelta%20h%5C%5C%5CDelta%20h%3D%5Cfrac%7B97.66%20%5C%2C%20kPa%7D%7B998%20kg%2Fm%5E3%20%5C%2C%20%5C%2C%209.81%20m%2Fs%5E2%7D%20%5C%5C%5CDelta%20h%3D9.975%5C%2Cm)
And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
I think the answer would be A if its wrong I’m sorry
120 volt divided by 22 ampere
= 5.4545454545455 ohm (Ω)
P = V × I
= 120 volt × 22 ampere
= 2640 watt (W)
Answer:
Heat gain of 142 kJ
Explanation:
We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.
Δ
⇒
Δ![U -W](https://tex.z-dn.net/?f=U%20-W)
![Q = 79 - (-63) = 142 kJ](https://tex.z-dn.net/?f=Q%20%3D%2079%20-%20%28-63%29%20%3D%20142%20kJ)
Therefore, the gas gained heat by an amount of 142 kJ.
Complete Question
The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h
Answer:
![v=23.6m/s](https://tex.z-dn.net/?f=v%3D23.6m%2Fs)
Explanation:
Velocity ![v_c=18.0km/h](https://tex.z-dn.net/?f=v_c%3D18.0km%2Fh)
Radius ![r=21m](https://tex.z-dn.net/?f=r%3D21m)
initial velocity u![u=19=>5.27778](https://tex.z-dn.net/?f=u%3D19%3D%3E5.27778)
Generally the equation for Angle is mathematically given by
![\theta=\frac{v_c}{2r}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7Bv_c%7D%7B2r%7D)
![\theta=\frac{18}{2*21}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cfrac%7B18%7D%7B2%2A21%7D)
![\theta=0.45](https://tex.z-dn.net/?f=%5Ctheta%3D0.45)
![\theta=25.7831 \textdegree](https://tex.z-dn.net/?f=%5Ctheta%3D25.7831%20%5Ctextdegree)
Generally
Height of mass
![h=\frac{rsin\theta}{\theta}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7Brsin%5Ctheta%7D%7B%5Ctheta%7D)
![h=\frac{21sin25.78}{0.45}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B21sin25.78%7D%7B0.45%7D)
![h=20.3m](https://tex.z-dn.net/?f=h%3D20.3m)
Generally the equation for Work Energy is mathematically given by
![0.5mv_0^2+mgh=0.5mv^2](https://tex.z-dn.net/?f=0.5mv_0%5E2%2Bmgh%3D0.5mv%5E2)
Therefore
![v=\sqrt{u^2+2gh}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2gh%7D)
![v=\sqrt{=5.27778^2+2*9.81*20.3}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%3D5.27778%5E2%2B2%2A9.81%2A20.3%7D)
![v=23.6m/s](https://tex.z-dn.net/?f=v%3D23.6m%2Fs)