Answer:
Explanation:
Given
Pressure drop ![\Delta P=650\ kPa](https://tex.z-dn.net/?f=%5CDelta%20P%3D650%5C%20kPa)
inlet diameter ![d_1=31\ cm](https://tex.z-dn.net/?f=d_1%3D31%5C%20cm)
Outlet diameter ![d_2=19\ cm](https://tex.z-dn.net/?f=d_2%3D19%5C%20cm)
density of water ![\rho=10^3\ kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D10%5E3%5C%20kg%2Fm%5E3)
Suppose
and
be the inlet and outlet velocity
According to continuity equation
![A_1v_1=A_2v_2](https://tex.z-dn.net/?f=A_1v_1%3DA_2v_2)
where A=cross-section of Pipe
![A=\frac{\pi d^2}{4}](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7D)
thus ![d_1^2v_1=d_2^2v_2](https://tex.z-dn.net/?f=d_1%5E2v_1%3Dd_2%5E2v_2)
![v_2=v_1\times (\frac{d_1}{d_2})^2](https://tex.z-dn.net/?f=v_2%3Dv_1%5Ctimes%20%28%5Cfrac%7Bd_1%7D%7Bd_2%7D%29%5E2)
![v_2=v_1\times (\frac{31}{19})^2](https://tex.z-dn.net/?f=v_2%3Dv_1%5Ctimes%20%28%5Cfrac%7B31%7D%7B19%7D%29%5E2)
Also from Bernoulli's Equation
![\Delta P=\frac{1}{2}\rho (v_2^2-v_1^2)](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cfrac%7B1%7D%7B2%7D%5Crho%20%28v_2%5E2-v_1%5E2%29)
![650\times 10^3=\frac{1}{2}\times 10^3\times (v_1^2(\frac{31}{19})^4-v_1^2)](https://tex.z-dn.net/?f=650%5Ctimes%2010%5E3%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%2010%5E3%5Ctimes%20%28v_1%5E2%28%5Cfrac%7B31%7D%7B19%7D%29%5E4-v_1%5E2%29)
![v_1=\sqrt{106.79}](https://tex.z-dn.net/?f=v_1%3D%5Csqrt%7B106.79%7D)
![v_1=10.33\ m/s](https://tex.z-dn.net/?f=v_1%3D10.33%5C%20m%2Fs)
Two explanations might include two sets of colorless solutions that might yield a gas. What comes to mind is the production of carbon dioxide which could come from a soluble carbonate, such as sodium carbonate solution, and an acid, such as vinegar or hydrochloric acid. A second possibility is a gas such as ammonia. This could be produced by the reaction of a soluble ammonium compound, such as ammonium chloride, and a base such as sodium hydroxide.
Answer:
The difference of power is
ΔP = 172.767 kPa
Explanation:
ρ = 1390 kg / m³
v = 9.63 m/s
d₁ = 10.1 cm , d₂ = 15.3 cm
Δz = 8.85 m
To find the difference ΔP between the fluid pressure at locations 2 and the fluid pressure at location 1
ΔP = ρ * g * Z + ¹/₂ * ρ * v² * ( 1 - (d₁ / d₂)⁴ )
ΔP = 1390 kg / m³ * 9.8 m/s² * 8.85 m + 0.5 * 1390 kg / m³ *(9.63 m /s)² * (1 - (0.101 m / 0.153 m )⁴ )
ΔP = 172.767 x 10 ³ Pa
ΔP = 172.767 kPa
Answer:
because its not going down a long hill instead its going on a leveled street
Answer:
Geometrical representation