Answer:
(a) 21.44 ft/s
(b) 0 ft/s
(c) 19.51 ft/s
Explanation:
2 in = 2/12 ft = 0.167 ft
For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:
![v(r) = v_c\left[1 - \frac{r^2}{R^2}\right]](https://tex.z-dn.net/?f=v%28r%29%20%3D%20v_c%5Cleft%5B1%20-%20%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%5Cright%5D)
where R = 0.167 ft is the pipe radius and
is the constant fluid velocity at the center of the pipe.
We can integrate this over the cross-section area of the in order to find the volume flow
![\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%20%3D%20%5Cint%5Climits%20%7Bv%28r%29%7D%20%5C%2C%20dA%20%5C%5C%3D%20%5Cint%5Climits%5ER_0%20%7Bv_c%5Cleft%5B1%20-%20%5Cfrac%7Br%5E2%7D%7BR%5E2%7D%5Cright%5D2%5Cpi%20r%7D%20%5C%2C%20dr%5C%5C%20%20%3D%202%5Cpi%20v_c%5Cint%5Climits%5ER_0%20%7Br%20-%20%5Cfrac%7Br%5E3%7D%7BR%5E2%7D%7D%20%5C%2C%20dr%5C%5C%20%3D%202%5Cpi%20v_c%20%5Cleft%5B%5Cfrac%7Br%5E2%7D%7B2%7D%20-%20%5Cfrac%7Br%5E4%7D%7B4R%5E2%7D%5Cright%5D%5ER_0%5C%5C%3D%202%5Cpi%20v_c%20%5Cleft%28%5Cfrac%7BR%5E2%7D%7B2%7D%20-%20%5Cfrac%7BR%5E4%7D%7B4R%5E2%7D%5Cright%29%5C%5C%3D%202%5Cpi%20v_c%20%5Cleft%28%5Cfrac%7BR%5E2%7D%7B2%7D%20-%20%5Cfrac%7BR%5E2%7D%7B4%7D%5Cright%29%5C%5C%3D%202%5Cpi%20v_c%20R%5E2%2F4%5C%5C%3D%5Cpi%20v_c%20R%5E2%2F2%5C%5CA%20%3D%20%5Cpi%20R%5E2%5C%5C%5Cdot%7BV%7D%20%3D%20Av_c%2F2%5C%5C)
So the average velocity
![v = \dot{V} / A = v_c/2 = 10.72](https://tex.z-dn.net/?f=v%20%3D%20%5Cdot%7BV%7D%20%2F%20A%20%3D%20v_c%2F2%20%3D%2010.72)
![v_c = 10.72*2 = 21.44 ft/s](https://tex.z-dn.net/?f=v_c%20%3D%2010.72%2A2%20%3D%2021.44%20ft%2Fs)
b) At the wall of the pipe, r = R so ![v(R) = v_c(1 - 1) = 0 ft/s](https://tex.z-dn.net/?f=v%28R%29%20%3D%20v_c%281%20-%201%29%20%3D%200%20ft%2Fs)
c) At a distance of 0.6 in = 0.6/12 = 0.05 ft
![v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s](https://tex.z-dn.net/?f=v%280.05%29%20%3D%20v_c%281%20-%200.05%5E2%2F0.167%5E2%29%20%3D%200.91v_c%20%3D%200.91%2A21.44%20%3D%2019.51%20ft%2Fs)
Answer:
t = 25s
Explanation:
The required temperature change is 50°C. (80 –30). The temperature of the bar increased 2°C in the first second. Assuming uniformity in the properties of the bar, in 2s the temperature would now be 2+2 = 4°C, in 3s it will be 2+2+2 =6°C.
So for gain a temperature change of 50°C, it will take 50/2 = 25s
ΔT /t = 2
t = ΔT/2 = 50/2 = 25seconds
Answer:
past temperature, composition of gases in the atmosphere
Explanation:
Cylindrical cones can be drilled out from the core of glaciers to calculate past temperatures. The air bubbles in the cones can also be used to determine the composition of air in the past yars. This information provides us a means to analyze the trend of global warming and change in concentratinos of green house gases over the years.