From 8a92779a679f3d40eb8ab973730ea8a39d7470c1 Mon Sep 17 00:00:00 2001 From: Alvie Rahman Date: Thu, 21 Oct 2021 10:20:31 +0100 Subject: [PATCH] Update headers --- mechanical/mmme1026_maths_for_engineering.md | 8 +- mechanical/mmme1048_fluid_mechanics.md | 89 ++++++++++++++++++++ 2 files changed, 92 insertions(+), 5 deletions(-) diff --git a/mechanical/mmme1026_maths_for_engineering.md b/mechanical/mmme1026_maths_for_engineering.md index ee26426..df80864 100755 --- a/mechanical/mmme1026_maths_for_engineering.md +++ b/mechanical/mmme1026_maths_for_engineering.md @@ -5,9 +5,9 @@ title: MMME1026 // Mathematics for Engineering tags: [ uni, nottingham, mmme1026, maths, complex_numbers ] --- -# Lecture 1 // Complex Numbers (2021-10-04) +# Complex Numbers -## Complex Numbers +## What is a Complex Number? - $i$ is the unit imaginary number, which is defined by: @@ -26,7 +26,7 @@ tags: [ uni, nottingham, mmme1026, maths, complex_numbers ] e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$ -### Complex Conjugate +### The Complex Conjugate Given complex number $z$: @@ -118,8 +118,6 @@ always hold true as there are many solutions. 2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent. If it puts you in the wrong quadrant, add or subtract $\pi$. -# Lecture 2 // Complex Numbers (2021-10-12) - ## Exponential Functions - The exponential function $f(x) = \exp x$ may be wirtten as an infinite series: diff --git a/mechanical/mmme1048_fluid_mechanics.md b/mechanical/mmme1048_fluid_mechanics.md index d136710..6254ffa 100755 --- a/mechanical/mmme1048_fluid_mechanics.md +++ b/mechanical/mmme1048_fluid_mechanics.md @@ -244,6 +244,95 @@ p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\ $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$ +
+ + +## Exercise Sheet 1 + + + +1. If 4 m$^3$ of oil weighs 35 kN calculate its density and relative density. + Relative density is a term used to define the density of a fluid relative + + > $$ \frac{35000}{9.81\times4} = 890 \text{ kgm}^{-3} $$ + > + > $$1000 - 891.9... = 108 \text{ kgm}^{-3}$$ + +2. Find the pressure relative to atmospheric experienced by a diver + working on the sea bed at a depth of 35 m. + Take the density of sea water to be 1030 kgm$^{-3}$. + + > $$ + > \rho gh = 1030\times 9.81 \times 35 = 3.5\times10^5 + > $$ + +3. An open glass is sitting on a table, it has a diameter of 10 cm. + If water up to a height of 20 cm is now added calculate the force exerted onto the table by + the addition of the water. + + > $$V_{cylinder} = \pi r^2h$$ + > $$m_{cylinder} = \rho\pi r^2h$$ + > $$W_{cylinder} = \rho g\pi r^2h = 1000\times9.8\pi\times0.05^2\times0.2 = 15.4 \text{ N} $$ + +4. A closed rectangular tank with internal dimensions 6m x 3m x 1.5m + high has a vertical riser pipe of cross-sectional area 0.001 m2 in + the upper surface (figure 1.4). The tank and riser are filled with + water such that the water level in the riser pipe is 3.5 m above the + + Calulate: + + i. The gauge pressure at the base of the tank. + + > $$\rho gh = 1000\times9.81\times(1.5+3.5) = 49 \text{ kPa}$$ + + ii. The gauge pressure at the top of the tank. + + > $$\rho gh = 1000\times9.81\times3.5 = 34 \text{ kPa}$$ + + iii. The force exercted on the base of the tank due to gauge water pressure. + + > $$F = p\times A = 49\times10^3\times6\times3 = 8.8\times10^5 \text{ N}$$ + + iv. The weight of the water in the tank and riser. + + > $$V = 6\times3\times1.5 + 0.001\times3.5 = 27.0035$$ + > $$W = \rho gV = 1000\times9.8\times27.0035 = 2.6\times10^5\text{ N}$$ + + v. Explain the difference between (iii) and (iv). + + *(It may be helpful to think about the forces on the top of the tank)* + + > The pressure at the top of the tank is higher than atmospheric pressure because of the + > riser. + > This means there is an upwards force on the top of tank. + > The difference between the force acting up and down due to pressure is equal to the + > weight of the water. + +6. A double U-tube manometer is connected to a pipe as shown below. + + Taking the dimensions and fluids as indicated; calculate + the absolute pressure at point A (centre of the pipe). + + Take atmospheric pressure as 1.01 bar and the density of mercury as 13600 kgm$^{-3}$. + + ![](./images/vimscrot-2021-10-13T10:42:52,999793176+01:00.png) + + > \begin{align*} + P_B &= P_A + 0.4\rho_wg &\text{(1)}\\ + P_C = P_{C'} &= P_B - 0.2\rho_mg &\text{(2)}\\ + P_D = P_{at} &= P_A + 0.4\rho_wg &\text{(3)}\\ + \\ + \text{(2) into (3)} &\rightarrow P_B -0.2\rho_mg-0.1\rho_wg = P_{at} &\text{(4)}\\ + \text{(1) into (4)} &\rightarrow P_A + 0.4\rho_wg - 0.2\rho_mg - 0.1\rho_wg = P_{at} \\ + \\ + P_A &= P_{at} + 0.1\rho_wg + 0.2\rho_mg - 0.4\rho_wg \\ + &= P_{at} + g(0.2\rho_m - 0.3\rho_w) \\ + &= 1.01\times10^5 + 9.81(0.2\times13600 - 0.3\times1000)\\ + &= 124.7\text{ kPa} + > \end{align*} + +
+ # Lecture 3 // Submerged Surfaces ## Prepatory Maths