Answer:
la escuela,en casa y listo...............
Answer:
The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Explanation:
Length of curve is given as
![L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft](https://tex.z-dn.net/?f=L%3D2%28PVT-PVI%29%5C%5CL%3D2%28242%2B30-240%2B00%29%5C%5CL%3D2%28230%29%5C%5CL%3D460%20ft)
is given as
![G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%](https://tex.z-dn.net/?f=G_2%3D%5Cfrac%7BE_%7BPVT%7D-E_%7BPVI%7D%7D%7B0.5L%7D%5C%5CG_2%3D%5Cfrac%7B127.5-122%7D%7B0.5%2A460%7D%5C%5CG_2%3D0.025%3D2.5%20%5C%25)
The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as
![A=\frac{L}{K}\\A=\frac{460}{115}\\A=4](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BL%7D%7BK%7D%5C%5CA%3D%5Cfrac%7B460%7D%7B115%7D%5C%5CA%3D4)
A is given as
![-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%](https://tex.z-dn.net/?f=-G_1%3DA-G_2%5C%5C-G_1%3D4.0-2.5%5C%5C-G_1%3D1.5%5C%5CG_1%3D-1.5%5C%25)
With initial grade, the elevation of PVC is
![E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\](https://tex.z-dn.net/?f=E_%7BPVC%7D%3DE_%7BPVI%7D%2BG_1%28L%2F2%29%5C%5CE_%7BPVC%7D%3D122%2B1.5%25%28460%2F2%29%5C%5CE_%7BPVC%7D%3D125.45%20ft%5C%5C)
The station is given as
![St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\](https://tex.z-dn.net/?f=St_%7BPVC%7D%3DSt_%7BPVI%7D-%28L%2F2%29%5C%5CSt_%7BPVC%7D%3D24000-%28230%29%5C%5CSt_%7BPVC%7D%3D237%2B70%5C%5C)
Low point is given as
![x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft](https://tex.z-dn.net/?f=x%3DK%20%5Ctimes%20%7CG_1%7C%5C%5Cx%3D115%20%5Ctimes%201.5%5C%5Cx%3D172.5%20ft)
The station of low point is given as
![St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\](https://tex.z-dn.net/?f=St_%7Blow%7D%3DSt_%7BPVC%7D-%28x%29%5C%5CSt_%7Blow%7D%3D23770%2B%28172.5%29%5C%5CSt_%7Blow%7D%3D239%2B42.5%20ft%5C%5C)
The elevation is given as
![E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft](https://tex.z-dn.net/?f=E_%7Blow%7D%3D%5Cfrac%7BG_2-G_1%7D%7B2L%7D%20x%5E2%2BG_1x%2BE_%7BPVC%7D%5C%5CE_%7Blow%7D%3D%5Cfrac%7B2.5-%28-1.5%29%7D%7B2%2A460%7D%20%281.72%29%5E2%2B%28-1.5%29%2A%281.72%29%2B125.45%5C%5CE_%7Blow%7D%3D124.16%20ft)
So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.
Answer:
There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.
Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.
Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.
Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.
Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
![1\ gallon = 0.13ft^3](https://tex.z-dn.net/?f=1%5C%20gallon%20%3D%200.13ft%5E3)
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.
Density can be defined as
![\rho = \frac{m}{V}](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D)
Where
m = Mass
V = Volume
For state one we know that
![\rho_1 = \frac{m_1}{V}](https://tex.z-dn.net/?f=%5Crho_1%20%3D%20%5Cfrac%7Bm_1%7D%7BV%7D)
![m_1 = \rho_1 V](https://tex.z-dn.net/?f=m_1%20%3D%20%5Crho_1%20V)
![m_1 = 1.18*1](https://tex.z-dn.net/?f=m_1%20%3D%201.18%2A1)
![m_1 = 1.18Kg](https://tex.z-dn.net/?f=m_1%20%3D%201.18Kg)
For state two we have to
![\rho_2 = \frac{m_2}{V}](https://tex.z-dn.net/?f=%5Crho_2%20%3D%20%5Cfrac%7Bm_2%7D%7BV%7D)
![m_2 = \rho_2 V](https://tex.z-dn.net/?f=m_2%20%3D%20%5Crho_2%20V)
![m_1 = 7.2*1](https://tex.z-dn.net/?f=m_1%20%3D%207.2%2A1)
![m_1 = 7.2Kg](https://tex.z-dn.net/?f=m_1%20%3D%207.2Kg)
Therefore the total change of mass would be
![\Delta m = m_2-m_1](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20m_2-m_1)
![\Delta m = 7.2-1.18](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%207.2-1.18)
![\Delta m = 6.02Kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%206.02Kg)
Therefore the mass of air that has entered to the tank is 6.02Kg