Pressure and Head

The figure does not give every number legibly, so the reading used is stated below. Pressures are in gauge kPa, $g=9.81\ \text{m/s}^2$, and moving down a fluid adds $\gamma h$, moving up subtracts $\gamma h$.

Specific weights: oil $\gamma_o=0.85\times9.81=8.34$; mercury $\gamma_{Hg}=13.6\times9.81=133.4$; brine (assumed S.G. 1.1) $\gamma_b=10.79$; air $\gamma_a=0.00981$; water $\gamma_w=9.81$ kN/m³.

Assumed reading (heights from the figure):

• pipe 1 (oil) $\to$ down 20 cm in oil $\to$ up 60 cm in mercury $\to$ down 50 cm in air $\to$ up 10 cm in brine to pipe 2 (net rise = 0, since both pipe centres are at the same level)
• pipe 2 $\to$ down 50 cm in brine $\to$ up 10 cm in mercury $\to$ up 30 cm in mercury and 30 cm in water inside the tank, to the air at point 3

 (1) oil pipe          (2) brine pipe
   |                      |
   oil 20cm  Hg 60cm  air 50cm  brine 10cm

Pressure at point 2

$$\begin{aligned} p_2&=p_1+\gamma_o(0.20)-\gamma_{Hg}(0.60)+\gamma_a(0.50)-\gamma_b(0.10)\\ &=200+1.668-80.05+0.005-1.079\\ &=120.54\ \text{kPa} \end{aligned}$$

Pressure at point 3

$$\begin{aligned} p_3&=p_2+\gamma_b(0.50)-\gamma_{Hg}(0.10)-\gamma_{Hg}(0.30)-\gamma_w(0.30)\\ &=120.54+5.395-13.34-40.02-2.943\\ &=69.63\ \text{kPa} \end{aligned}$$

Answer: $p_2\approx120.5$ kPa and $p_3\approx69.6$ kPa (gauge). For any other reading of the figure, use the same step-by-step rule with the correct heights and brine S.G.

Answer: the pressure difference becomes zero only when the whole water body has a downward acceleration equal to $g$ (free fall), i.e. $a_z=-g$. For any upward acceleration it can never be zero; it only increases.

Take a fluid element in a body moving with vertical acceleration $a_z$ (upward positive). Newton's second law for a column of height $h$ and unit area:

$$p_2A-p_1A-\rho g h A=\rho h A\,a_z$$

where point 1 is the upper point and point 2 the lower point, a distance $h$ apart. So

$$\Delta p=p_2-p_1=\rho h\,(g+a_z)$$

Setting $\Delta p=0$:

$$g+a_z=0\ \Rightarrow\ a_z=-g$$

For an upward acceleration $a_z>0$, $\Delta p=\rho h(g+a_z)>\rho g h$, so the pressure difference is greater than the static value. At free fall (weightless condition) the fluid has no apparent weight and the pressure is uniform throughout.

The figure dimensions are not fully legible, so this reading is used and stated. Specific weight of water $\gamma_w=9.81$ kN/m³, $g=9.81$.

Assumed reading: the pipe centre lies 0.5 m below the free surface in the tank, so the pipe is under a static head of 0.5 m of water. Left leg of the inverted U (water) rises 0.25 m from the pipe to the lighter fluid ($S=0.83$); the manometer reading $R=0.15$ m (left interface is higher than the right one); the right leg water goes down 0.20 m to a mercury U-tube whose other limb is open. We find the differential mercury level $h$.

        ______ oil S=0.83 ______
       |                        |
 water |                  R     | water 0.20
 0.25  |                        |
 pipe  P                        Hg  |--h--| open

Pressure at pipe (gauge):

$$p_P=\gamma_w(0.5)=4.905\ \text{kPa}$$

At the left oil-water interface (rise of 0.25 m in water):

$$p_{L}=4.905-9.81(0.25)=2.453\ \text{kPa}$$

Across the lighter fluid to the right interface (it is $R=0.15$ m lower):

$$p_R=2.453+0.83(9.81)(0.15)=3.674\ \text{kPa}$$

Down 0.20 m of water to the mercury surface:

$$p_{Hg}=3.674+9.81(0.20)=5.636\ \text{kPa}$$

This is balanced by a mercury column $h$ in the open limb:

$$h=\frac{p_{Hg}}{\gamma_{Hg}}=\frac{5.636}{13.6\times9.81}=0.0422\ \text{m}$$

Answer: differential mercury level $h\approx42$ mm (for this reading). The method is the same for other dimensions: add $\gamma h$ going down, subtract going up, and equate to $\gamma_{Hg}h$.

Inverted U-tube manometer

Used to measure the small pressure difference between two points in a liquid pipeline carrying a light-density liquid such as water. The space above the two liquid columns is filled with a lighter fluid (air or oil), so a small pressure difference gives a large, easily read difference in column heights:

$$p_A-p_B=(\gamma_w-\gamma_m)\,h$$

where $\gamma_m$ is the specific weight of the light fluid. It is chosen when the fluids are liquids of low density and the pressure difference is very small.

Single-column inclined manometer

Used to measure very small gauge pressures, mostly of gases (low draught pressures), where ordinary manometer readings would be too small to read accurately. One limb is a large reservoir and the other a narrow tube inclined at angle $\theta$ to the horizontal. The liquid moves a length $L$ along the incline for a vertical rise $L\sin\theta$, so the scale is magnified by $1/\sin\theta$:

$$p=\gamma\,L\sin\theta\ \left(1+\frac{a}{A}\right)\approx\gamma L\sin\theta$$

where $a/A$ is the ratio of the tube to reservoir areas. It is therefore preferred where high sensitivity is needed.

Reading of the figure: gauge B is at the bottom of the main tank (under 55 mm of water). Gauge C is at the bottom of the small cylinder (under a depth $h$ of cleaning fluid, $S=0.8$). The air above both liquids is at gauge pressure $p_A$.

Gauge pressure $p_A$. From B:

$$p_B=p_A+\gamma_w(0.055)\ \Rightarrow\ p_A=13.4-9.81(0.055)=12.86\ \text{kPa}$$

Height $h$. From C:

$$p_C=p_A+\gamma_f h,\qquad \gamma_f=0.8\times9.81=7.848\ \text{kN/m}^3$$ $$h=\frac{p_C-p_A}{\gamma_f}=\frac{13.42-12.860}{7.848}=0.0713\ \text{m}$$

Answer: $p_A\approx12.86$ kPa (gauge) and $h\approx71.3$ mm.

Data: U-tube with base length $L$ rotating about the left vertical arm at $\omega=10$ rad/s. The relative density 0.85 of the liquid does not affect the result.

The liquid in the base is in rigid-body rotation. Measured from the axis, the pressure along the horizontal base increases with radius as

$$p=p_0+\tfrac{1}{2}\rho\omega^{2}r^{2}$$

At the outer arm ($r=L$) this pressure is balanced by a column of liquid rising a height $h$ above the base level (the surface in the inner arm lies at the axis, $r=0$, at the base level). Hence

$$\rho g h=\tfrac{1}{2}\rho\omega^{2}L^{2}$$ $$h=\frac{\omega^{2}L^{2}}{2g}=\frac{10^{2}L^{2}}{2\times9.81}=5.10\,L^{2}\ \text{(metres, with }L\text{ in metres)}$$

For example, $L=0.1$ m gives $h=0.051$ m. This is the standard result: the free surfaces of the two arms differ in height by $\omega^{2}L^{2}/2g$, independent of the liquid density. (It assumes the inner-arm level remains at the base; if the liquid withdraws appreciably from the axis, the rise is slightly smaller.)

Answer: rise in the outer arm $=\dfrac{\omega^{2}L^{2}}{2g}\approx5.1\,L^{2}$ m.

Data: tube $d=5$ mm, enlarged ends $D$ (equal), oil $S=0.97$, interface moves $x=25$ mm for a pressure difference of 1 mm of water.

Set-up. Water is in one limb and oil in the other, the interface lying in the tube. When a pressure difference $\Delta p$ is applied, the interface moves a distance $x$ in the tube. The displaced volume $\dfrac{\pi}{4}d^{2}x$ makes both free surfaces in the enlarged ends move by

$$y=x\left(\frac{d}{D}\right)^{2}$$

(the water surface falls $y$, the oil surface rises $y$). Before the pressure change the pressures balance at the initial interface level, so only the changes count. Balancing at the new interface level, with $s=0.97$:

$$\frac{\Delta p}{\gamma_w}=y(1+s)+x(1-s)$$ $$\frac{\Delta p}{\gamma_w}=x\left[(1+s)\left(\frac{d}{D}\right)^{2}+(1-s)\right]$$

Substitute $\dfrac{\Delta p}{\gamma_w}=1$ mm, $x=25$ mm:

$$1=25\left[1.97\left(\frac{d}{D}\right)^{2}+0.03\right]\ \Rightarrow\ \left(\frac{d}{D}\right)^{2}=\frac{0.04-0.03}{1.97}=0.005076$$ $$\frac{d}{D}=0.07125\ \Rightarrow\ D=\frac{5}{0.07125}=70.2\ \text{mm}$$

Answer: diameter of each enlarged end $\approx70$ mm.

Data: arms $b=60$ mm apart (fore-and-aft), alcohol density $\rho=0.79\times1000=790\ \text{kg/m}^3$, slope $\theta=20^\circ$ down, acceleration $a=2\ \text{m/s}^2$ forward (down the slope), level difference along the arms $=73$ mm, front limb higher.

Use axes fixed to the vehicle: $x$ along the slope (forward), $y$ perpendicular to the slope (along the arms). In this frame the fluid feels an effective body force per unit mass $\vec g-\vec a$:

• $x$ component: $g\sin\theta-a=9.81\sin20^\circ-2=1.355\ \text{m/s}^2$ (forward)
• $y$ component: $-g\cos\theta=-9.81\cos20^\circ=-9.218\ \text{m/s}^2$

Pressure varies as $dp=\rho(b_x\,dx+b_y\,dy)$ within the liquid. Go from the surface in the rear limb to the surface in the front limb: $\Delta x=+0.060$ m forward and $\Delta y=+0.073$ m (the front level is higher):

$$p_{front}-p_{rear}=790\left[1.355(0.060)-9.218(0.073)\right]=790(0.0813-0.6729)=-467.4\ \text{Pa}$$

The air pressure acting on each liquid surface is therefore lower in the front limb:

$$p_{rear}-p_{front}=467\ \text{Pa}$$

Answer: the air pressure difference is about 467 Pa (0.467 kPa), the rear limb being at the higher pressure.

Data: tube $d=7$ mm, enlarged ends $D=44$ mm, oil $S=0.83$ (right limb), water in the left limb and base, interface moves $x=1$ cm $=0.01$ m.

   p1 (water side)       p2 (oil side)
   [D]                   [D]
    |  water      oil     |
    |                     |
    |__________ interface moves x

Raise the pressure $p_1$ on the water side. The interface moves up the right limb by $x$. The volume displaced in the narrow tube moves the free surfaces in the enlarged ends by

$$y=x\left(\frac{d}{D}\right)^{2}=0.01\left(\frac{7}{44}\right)^{2}=2.53\times10^{-4}\ \text{m}$$

(the water surface falls $y$, the oil surface rises $y$). Initially the pressures balance at the interface level, so the change in the pressure difference is balanced by the change in column heights:

$$p_1-p_2=\gamma_w y+\gamma_o y+(\gamma_w-\gamma_o)x$$ $$p_1-p_2=\gamma_w\,x\left[(1+s)\left(\frac{d}{D}\right)^{2}+(1-s)\right]$$

With $s=0.83$:

$$p_1-p_2=9810(0.01)\left[1.83(0.02531)+0.17\right]=98.1\times0.2163$$ $$p_1-p_2=21.2\ \text{Pa}$$

Answer: a pressure difference of about 21.2 Pa (2.16 mm of water) moves the interface by 1 cm. The small value shows how sensitive the enlarged-end manometer is.

Reading of the figure: $B$ is on the axis, $C$ is on the axis 250 mm below $B$, and the quadrant $CD$ (radius $R=0.25$ m, centre $B$) curves from $C$ up to $D$, which is level with $B$ and 0.25 m from the axis. The water surface in $A$ is 0.3 m above $C$ (on the axis), and $A$ is open.

   A  o   <- water surface, 0.3 m above C
      |
      |
   B  +------- D (r = 0.25 m)
      |     /
      |   /  quadrant, centre B
   C  +--

For rigid-body rotation, the pressure head at a point at radius $r$ and height $z$ above $C$ is

$$h_p=h_C+\frac{\omega^{2}r^{2}}{2g}-z,\qquad h_C=0.3\ \text{m}$$

(a) Speed for $h_D=h_C$

At $D$: $r=0.25$ m and $z=0.25$ m.

$$\frac{\omega^{2}(0.25)^{2}}{2g}=0.25\ \Rightarrow\ \omega^{2}=\frac{2(9.81)}{0.25}=78.48$$ $$\omega=8.859\ \text{rad/s},\qquad N=\frac{60\omega}{2\pi}=84.6\ \text{rpm}$$

(b) Maximum head in the curved part

Let a point on the quadrant make angle $\theta$ at $B$ measured from $BC$. Then $r=R\sin\theta$, $z=R(1-\cos\theta)$. With $\dfrac{\omega^{2}R^{2}}{2g}=R$:

$$h_p=0.3+R\sin^{2}\theta-R(1-\cos\theta)=0.3+R(\cos\theta-\cos^{2}\theta)$$ $$\frac{dh_p}{d\theta}=0\ \Rightarrow\ \sin\theta(2\cos\theta-1)=0\ \Rightarrow\ \cos\theta=0.5,\ \theta=60^\circ$$ $$h_{p,max}=0.3+0.25(0.5-0.25)=0.3625\ \text{m of water}$$

Position: at $60^\circ$ from $C$ along the arc, i.e. 0.2165 m from the axis and 0.125 m above $C$.

Answer: (a) 8.86 rad/s (84.6 rpm); (b) maximum head 0.3625 m of water, at $60^\circ$ from C on the arc.

Data: tube $d=4$ mm, enlarged end $D=20$ mm (right limb, oil above the interface), left limb water open to atmosphere. $s_w=1.0$, $s_o=0.9$. The interface falls $x=25$ mm. The gas pressure $p$ acts on the oil surface.

Movement of the free surfaces

• Left water surface rises $x=25$ mm (water pushed into the left limb of equal bore).
• Oil surface falls: $y=x\left(\dfrac{d}{D}\right)^{2}=25\left(\dfrac{4}{20}\right)^{2}=1$ mm.

Before the gas is connected both sides are open to the atmosphere and balanced at the initial interface level, so $\gamma_wH_w=\gamma_oH_o$. After the change, balance pressures at the new interface level ($x$ below the old):

$$p+\gamma_o\,(H_o-y+x)=\gamma_w\,(H_w+x+x)$$

Subtracting the initial balance:

$$p=\gamma_w(2x)-\gamma_o(x-y)$$ $$p=9810(0.050)-0.9(9810)(0.025-0.001)$$ $$p=490.5-211.9=278.6\ \text{Pa}$$

Answer: gauge pressure in the cylinder $\approx278.6$ Pa (about 0.28 kPa, or 28 mm of water).

Data: drum radius $R=0.25$ m, pipe diameter 25 mm so $r_0=0.0125$ m, water level in the pipe $h=0.6$ m above the drum top, $N=600$ rpm so $\omega=\dfrac{2\pi\times600}{60}=62.83$ rad/s, $\rho=1000\ \text{kg/m}^3$.

        |  | <- water, 0.6 m above top
        |  |
    ____|  |____   drum top
   |             |
   |   water     |   0.5 m dia
   |_____________|

The drum is full and the free surface lies in the pipe on the axis, so on the underside of the drum top (gauge pressure) at radius $r$:

$$p=\rho g h+\tfrac{1}{2}\rho\omega^{2}r^{2}$$

The upward force on the top (annulus from $r_0$ to $R$):

$$F=\int_{r_0}^{R}p\,2\pi r\,dr=\rho g h\,\pi\left(R^{2}-r_0^{2}\right)+\frac{\pi\rho\omega^{2}}{4}\left(R^{4}-r_0^{4}\right)$$

Static part:

$$F_1=1000(9.81)(0.6)\,\pi\left(0.25^{2}-0.0125^{2}\right)=1152.8\ \text{N}$$

Rotational part:

$$F_2=\frac{\pi(1000)(62.83)^{2}}{4}\left(0.25^{4}-0.0125^{4}\right)=12\,111.8\ \text{N}$$ $$F=F_1+F_2=13\,264.6\ \text{N}$$

Answer: upward force on the drum top $\approx13.3$ kN (13.27 kN if the small pipe area is ignored). Rotation contributes about 91 % of it.

Fluid properties at 20 °C (White's table): gasoline $\rho=680\ \text{kg/m}^3$, $\gamma_g=680\times9.81=6671\ \text{N/m}^3$; glycerin $\rho=1260\ \text{kg/m}^3$, $\gamma_{gl}=12\,361\ \text{N/m}^3$. Air weight is neglected.

Reading of the figure: $z=0$ at the tank bottom; glycerin 1 m deep ($z=0$ to 1), gasoline 1.5 m deep ($z=1$ to 2.5), air above; $p_A=1.5$ kPa gauge acts on the gasoline surface at $z=2.5$ m. Piezometer B joins the gasoline layer, C the glycerin layer.

   A (1.5 kPa)  air
  ------------------ z = 2.5 m
   gasoline  <- tube B
  ------------------ z = 1.0 m
   glycerin  <- tube C
  ------------------ z = 0

Piezometer B (gasoline)

The pressure at the top of the gasoline is 1500 Pa. The gasoline rises in tube B to a height where the pressure is zero:

$$Z_B=2.5+\frac{1500}{6671}=2.5+0.225=2.725\ \text{m}$$

Piezometer C (glycerin)

Pressure at the gasoline-glycerin interface ($z=1$ m):

$$p=1500+6671(1.5)=11\,507\ \text{Pa}$$

The glycerin in tube C rises above $z=1$ m until the pressure is zero:

$$Z_C=1+\frac{11\,507}{12\,361}=1+0.931=1.931\ \text{m}$$

Answer: $Z_B\approx2.72$ m and $Z_C\approx1.93$ m.

Assumptions from the figure: the left limb (water side) is vertical, the right limb is inclined at $15^\circ$, the bore is uniform, and the pressure at B (air, large bulb) stays at 100 kPa. Initially the mercury is in balance at $p_A=p_B=100$ kPa. $\gamma_{Hg}=13.6\times9.81=133.4$ kN/m³, $\gamma_w=9.81$ kN/m³.

  A (water)                B (air)
     |                      /
     |  drop dh            /  dL along 15 deg
     |_____ mercury ______/

Geometry. The tube has uniform bore, so the mercury length moved in each limb is equal: the vertical drop on the water side is $\Delta h$, so the movement along the incline is $\Delta L=\Delta h$. The vertical rise on the inclined side is $\Delta L\sin15^\circ$.

Pressure balance. At the water-mercury interface the water column above has lengthened by $\Delta h$:

$$p_A+\gamma_w\Delta h=p_B+\gamma_{Hg}\left(\Delta h+\Delta L\sin15^\circ\right)$$ $$130-100=\Delta h\left[133.4(1+\sin15^\circ)-9.81\right]$$ $$30=\Delta h\,(167.95-9.81)=158.14\,\Delta h$$ $$\Delta h=0.1897\ \text{m}$$ $$\Delta L=\Delta h=0.1897\ \text{m},\qquad \text{vertical rise on right}=0.1897\sin15^\circ=0.0491\ \text{m}$$

Answer: mercury on the water side drops about 19.0 cm; on the inclined side it moves about 19.0 cm along the tube (a rise of 4.9 cm).

Assumption: the tank is completely full and closed, and the pressure at the centre of the top cover (on the axis) remains atmospheric (0 gauge). Then the parabolic pressure variation starts from zero at the axis on the top.

Data: $R=0.5$ m, height 2 m, $N=100$ rpm so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s, $\rho=1000$ kg/m³.

Pressure at a point at radius $r$ and depth $z$ below the top:

$$p=\tfrac{1}{2}\rho\omega^{2}r^{2}+\rho g z$$

Along AC (top cover, $z=0$)

$$p=\tfrac{1}{2}(1000)(10.472)^{2}r^{2}=54\,831\,r^{2}\ \text{Pa}$$

The diagram is a parabola, zero at the centre and 13.71 kPa at A and at C (symmetrical).

Along AB (vertical wall, $r=0.5$ m)

$$p=13.71+9.81\,z\ \text{kPa}$$

The diagram is a straight line (trapezium) from 13.71 kPa at A to 33.33 kPa at B.

  A |->13.71 kPa          C
    |  \     parabola    /
    |   \_______________/  (0 at centre)
    |->23.52 (1 m)
    |->33.33 kPa at B

Answer: at A and C, 13.71 kPa; at B and D, 33.33 kPa; AC is parabolic (0 at the centre) and AB linear.

Definitions

• Absolute pressure is measured from absolute zero (complete vacuum): $p_{abs}=p_{gauge}+p_{atm}$.
• Gauge pressure is measured relative to the local atmospheric pressure: $p_{gauge}=p_{abs}-p_{atm}$. It is negative for vacuum.

Numerical

Reading of the figure: gauge $G$ and the liquid-mercury interface of the U-tube are at the same level, 1 m below the liquid surface; the mercury stands $h$ above that interface in the open limb. $\rho_l=800\ \text{kg/m}^3$, $p_v=120$ kPa (abs), $p_{atm}=101$ kPa.

(i) Gauge reading. Absolute pressure at G:

$$p_G=p_v+\rho_l g(1)=120+\frac{800\times9.81\times1}{1000}=127.85\ \text{kPa abs}$$ $$p_{G,gauge}=127.85-101=26.85\ \text{kPa}$$

(ii) Mercury height $h$. The same absolute pressure is balanced by atmosphere plus the mercury column:

$$p_G=p_{atm}+\gamma_{Hg}h\ \Rightarrow\ h=\frac{26.85}{13.6\times9.81}=0.2012\ \text{m}$$

Answer: (i) gauge pressure $=26.85$ kPa; (ii) $h\approx201$ mm of mercury.

Reading of the figure: the U-tube is filled with water; limb AB (sealed at the top A, height 3 m) is 1 m from the axis Y-Y; limb CD (open at the top D) is 2 m on the other side. Take B and C at the bottom, joined by a horizontal tube through the axis, and D at the same level as A (3 m above C). $N=60$ rpm so $\omega=2\pi=6.283$ rad/s. Pressures are gauge.

 A(sealed)            D(open)
  |                     |
  | 3 m                 | 3 m
  |                     |
 B+----------Y----------+C
  1 m                2 m

At rest the open end D is at atmospheric pressure. During rotation, pressure follows $p=\tfrac12\rho\omega^{2}r^{2}-\rho g z+\text{constant}$. At D ($r=2$, $z=3$ above the base), $p=0$.

At C (bottom of the open limb, $r=2$ m)

Hydrostatic in the vertical limb (no change with $r$ along a vertical line):

$$p_C=\rho g(3)=1000(9.81)(3)=29.43\ \text{kPa}$$

At B (bottom of the sealed limb, $r=1$ m, same level as C)

$$p_B=p_C+\tfrac{1}{2}\rho\omega^{2}\left(r_B^{2}-r_C^{2}\right)$$ $$p_B=29.43+\tfrac12(1000)(6.283)^{2}(1-4)=29.43-59.22=-29.79\ \text{kPa}$$

At A (top of the sealed limb)

$$p_A=p_B-\rho g(3)=-29.79-29.43=-59.22\ \text{kPa}$$

Answer: $p_A\approx-59.2$ kPa, $p_B\approx-29.8$ kPa, $p_C\approx+29.4$ kPa (gauge). The absolute pressure at A is still about 42 kPa, above the vapour pressure of water, so the column does not break.

Data: brass cylinder $D=37.5$ mm, $H=76.2$ mm, $S_{brass}=8.5$, mercury $S=13.6$, larger tube $d_1=39.5$ mm. The second tube diameter $d_2$ is not legible in the figure; the effect of the second tube is shown below.

(a) Pressure on the bottom of the cylinder

The floating cylinder's weight acts over its base area $A_c$, so the pressure on the bottom is weight divided by area:

$$p=\frac{W}{A_c}=S_{brass}\,\gamma_w\,H=8.5(9810)(0.0762)=6354\ \text{Pa}$$

(about 6.35 kPa gauge.) This equals the hydrostatic pressure of mercury at the immersion depth $d$.

(b) New equilibrium level

Depth of immersion (weight = buoyancy):

$$S_{brass}H=S_{Hg}\,d\ \Rightarrow\ d=\frac{8.5}{13.6}(76.2)=47.6\ \text{mm}$$

Volume of mercury displaced:

$$V=\frac{\pi}{4}D^{2}d=\frac{\pi}{4}(37.5)^{2}(47.6)=52\,570\ \text{mm}^3$$

This volume raises the mercury level by $h$ over the free mercury area. Around the cylinder the free area in the larger tube is the annulus:

$$A_{ann}=\frac{\pi}{4}\left(d_1^{2}-D^{2}\right)=\frac{\pi}{4}\left(39.5^{2}-37.5^{2}\right)=120.9\ \text{mm}^2$$

If the other tube is also connected, its area adds to the free surface:

$$h=\frac{V}{A_{ann}+\frac{\pi}{4}d_2^{2}}$$

With the narrow second tube neglected, $h=\dfrac{52\,570}{120.9}\approx435$ mm. Using the actual $d_2$ from the figure in the formula above gives the exact value.

Answer: (a) $p\approx6.35$ kPa; (b) immersion $47.6$ mm and rise $h=\dfrac{52\,570}{120.9+0.785\,d_2^{2}}$ mm (about 435 mm if $d_2$ is small).

Reading of the figure: axis through limb AB (A sealed at the top, on the axis); base BC has length $L=2$ cm; limb CD is vertical with the open end D at the same level as A, $H=4$ cm above the base. The tube is full of water. The clause "1600 rpm" is not needed; the question is for $N=300$ rpm, so $\omega=\dfrac{2\pi\times300}{60}=31.42$ rad/s.

 A (sealed)          D (open)
 |                    |
 | H = 4 cm           |
 B+------- L = 2 cm ---+C

Pressure at A

Pressure at D is zero. The pressure varies with radius and height as $p=\tfrac12\rho\omega^{2}r^{2}-\rho g z+\text{const}$. Between D ($r=L$, $z=H$) and A ($r=0$, $z=H$), the heights are equal, so only the rotational term acts:

$$p_A=p_D-\tfrac{1}{2}\rho\omega^{2}L^{2}$$ $$p_A=-\tfrac12(1000)(31.42)^{2}(0.02)^{2}=-197.4\ \text{Pa}$$

Pressure at A $\approx-197$ Pa (gauge), i.e. a small suction.

Water lost if A leaks

When A is opened to the atmosphere ($p_A=0$), the water surface at A (on the axis) and the surface at D (at $r=L$) must differ in level by

$$\Delta z=\frac{\omega^{2}L^{2}}{2g}=\frac{(31.42)^{2}(0.02)^{2}}{2(9.81)}=0.0201\ \text{m}=20.1\ \text{mm}$$

The surface at D cannot rise above the rim, so the water surface in AB falls by 20.1 mm and the same volume spills out at D.

Answer: $p_A\approx-197$ Pa; water lost at D equals a length of 20.1 mm of tube, i.e. volume $=\dfrac{\pi}{4}d^{2}\times0.0201\ \text{m}^3$ for tube bore $d$.