Update headers

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:

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 \\
 
   $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
 
+<details>
+<summary>
+
+## Exercise Sheet 1
+
+</summary>
+
+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*}
+
+</details>
+
 # Lecture 3 // Submerged Surfaces
 
 ## Prepatory Maths