Test: Pressure & Fluid Statics Level - 3 - Mechanical Engineering MCQ

By hydro-static law

dp / dh = ρg ⋯ ①

By the definition of Bulk Modulus

K = dp / dρ/ρ

⇒ dp = Kdρ / ρ ⋯②

Substituting ② in ①

Kdρ / ρdh = ρg

⇒Kdρ / ρ 2 = g dh

⇒−K / ρ = g h + C

Given at h = 0, ρ = ρ0

Given at h = 0, ρ = ρ0 ⇒

−K/ ρ0 = C

⇒−K / ρ = gh − K / ρ0

K / ρ =K / ρ0 − gh

ρ0K / K − ρ0gh = ρ ⋯ ③

Substituting the same in ①

dp / dh =ρ0Kg / K − ρ0gh

Given, K = 2070 MN/m²

ρ0g = 10050 N/m³

∴ at h = 1500

p = 2070 ln ( 2070 × 106 / 2070 × 106 − 10050 × 1500)

p = 15.13 MPa

dp / dy = −ρg ⋯ ②

But as the atmospheric air may be treated as an ideal gas ρ = p / RT

Substituting the same in ② dp / dy = −pg / RT

Using ①

dp / dy =−pg / R(T0 − βy )

∴dp / p = −g dy / R(T0 − βy )

Given, p0 = 101.3 kPa

T0 = 15°C i. e. 288 K

R = 287 J / kg‒ K

Also β can be predicted from the graph.

β = −56.5 − 15 / 11000

= 0.0065°C/m or 0.0065 K/m

∴ at y = 828 m

p = 101.3 ( 1 − 0.0065 × 828 / 288) 9.81/^{9.81/0.0065x287}

p = 101.3 × 0.9813^5.256

p = 91.73 kPa

T + FB = mg

FB = FB,Oil + FB,water

FB = (0.8 × 103) × g × Vsolid / 2 + 10³ × g × Vsolid / 2

FB = (400 + 500)g VSolid

FB = 900 g VSolid

T = mg − FB

T = (25.93 × 103 − 900 g)VSolid

T = 57.71 N

= ρghC(Area) + (pair Gauge)(Area)

= Area (10³ × g × (3.7 + 0.9) − 0.147 × 105)

= 65.72 kN

Force on gate due to oil

= ρoil × g × hC × Area

= 0.75 × 10³ × 9.81 × 1.8 / 2 × 1.2 × 1.8

= 14.30 kN

The center of pressure for water force can be calculated using pressure triangle

yP = h / 3×(2 × 21.6 + 39.25) / (21.6 + 39.25)

= 0.81 m from bottom

Similarly centre of pressure on oil side

= 0.6 m from the bottom i. e

For equilibrium about A Fw × (1.8 − 0.81)

= Foil (1.8 − 0.6) + FH (1.8)

⇒ 65.72 × 0.99

= 14.3(1.2) + FH × 1.8

FH = 26.61 kN

h cm of air +(23.5 − h) cm of gasoline +1.5 cm of water= 25 cm of gasoline

∴ 0.0188 × h + 6.65 × (23.5 − h) + 9.81 × 1.5 = 6.65 × 225

∴ h = 0.713 cm

pC + ρg × 0.2 = 0

pC = −0.2 × ρg

Equation for manometer D

pB = pC + ρgx

2.1 × 105 = −0.2 × ρg + ρgx

ρ = 13600 kg/m³

2.1 × 105 + 2.66 × 105 = 13.6 × 103 × g × x

x = 1.77 m

Using hydro-static law pA = pB

⇒ p + 10 cm of water = 10 cm of Hg ⋯ ① If p decreases by 5 kPa, level of Hg in left limb i.e. A will rise up to A′ and C will fall to C′ . Thus new configuration will be as follows.

Equating pressure at !′ and B′ pA′ = pB′ p′ + (10 − x)cm of water = (10 − 2x)cm of Hg p′ + 10 cm of water − x cm of water = 10 cm of Hg − 2x cm of Hg Also p′ = p − 5 kPa ∴ p − 5 kPa + 10 cm of water − x cm of water = 10 cm of Hg − 2x cm of Hg ⋯ ② Substituting ① in ②

−5 kPa − x cm of water = −2x cm of Hg

−5 kPa = x cm of water − 2x cm of Hg

∴ −5000 =x / 100 × 1000 × 9.81 − 2x / 100 × 13600 × 9.81

∴ 5000 = 2x / 100 × 13600 × 9.81 − x / 100 × 1000 × 9.81

5000 = x ( 2/100 × 13600 × 9.81 −1 / 100× 1000 × 9.81)

x = 5000 / 2570.22

x = 1.9453 cm

∴ New Manometric height

10 − 2x = 6.1 cm

For the surface the gate

hc = 2 m

∴ hp = hc + IG / Ahc sin2θ

hp = 2 + 5×(3√2) 3 / 12 / 5 × 3√2 × 2 × 1 / 2

hp = 2 + 9 × 2 / 12 × 2×1 / 2

hp = 2.375 m

Also Fp = ρgAhc

Fp = 1000 × 9.81 × (5 × 3√2) × 2

Fp = 416.2 kN

Fp × 1.875√2 + w × 1.5 = F × 1.5√2

F = 521 kN

After adding the oil, ρoil × g(18 + h1) = ρHg × g(h1 + h2)

2.72 × (18 + h1) = 13.6 (h1 + h2)

Also h1 × (2D)2 = D2 × h2

h1 × 4 = h2

2.72 × (18 + h1) = 13.6(5h1)

48.96 + 2.72 h1 = 68 h1

h1 = 0.75 cm

Quantity of oil added

=π / 4 × (0.04)² × (0.1875

= 2.35 × 102^-4m³

= 0.2356 lts.

Hydrostatic forces on the circular gate

= ρ × g × hC × Area

= 103 × g × 10 × π / 4× 42

= 1232.76 kN

yP = IxC / A × yC + yC = πD4 /64 /π 4D² × 11.54 + 11.54

= 11.62 m

(Note: this is ?? and not the depth from free surface)

Fp = 1232.76 kN acts 0.08 m Below the centroid, measured along the plane of the gate The moment required to turn the gate about the shaft = Fp × (yp − yc)

= 1232.76 × (11.62 − 11.54) × 103

= 98.62 kN-m

The pressure diagram on any side of the tank will be

Using the concept of pressure prism, the total force is equal to total volume of the pressure prism.

i.e. volume (a + b + c)

Fp = ρ1g × h /4 × ( h/ 2× 0.20) + ρ2g × h /4 (h/ 2 × 0.20) +ρ1g h /2× (h/ 2× 0.20)

Fp = (1.2 + 1.4 + 2.4)g = 49.05 N

To calculate the center of pressure Fp × yP = Moment due to a + Moment of b +Moment of c

Fp × yP = (1.2 × g × 2 /3×h/ 2) + (1.4 × g × [ 2/3×h/ 2+h/ 2]) + (2.4g × [ h /2+0/h 4])

All distances measured from free surface 49.05 × yP = 0.7848 + 2.289 + 3.5316 yP = 0.134 m = 13.46 cm from top.

The vertical component of pressure force is equal to the weight of liquid above the curved surface. i.e. the volume between the curved surface and y-axis

⇒ Fv = ρg ∫_y=0²⁰ x dy × 1

⇒ Fv = 4 × 9810 / 3× 20^3/2

⇒ Fv = 1169.9 × 10³ N Or 1169.9 kN

FHS = [π /4× d² × (h + 3) − 2 /3 πr³] × ρW × g

To lift the dome FHS = mg

So,

[28.27(h + 3) − 56.54]ρW × g = 50 × 10³ × g

28.27 (h + 3) = 106.54

h + 3 = 3.768

h = 0.77 m

W_A = W₁ + W₂

W_A = ρwV₁g + ρiV₂g WB = ρwV_bg

_{Consider A, as ice is floating, weight of ice is equal to weight of water displaced. ρiV2g = ρwV4g ∴} W_{A =} ρwV₁g ₊ ρwV₄g

_{WA = ρw(V1 + V4)g}

_{WA = ρwV3g}

W_{A =} W_B

_{Hence the two tumblers weigh the same.}

Also by buoyancy

ρiV₂g = ρwV₄g

∴ ρwV₅g = ρwV₄g

i.e. V₅ = V₄

After ice melts, total volume of water in A,

V_T = V₁ + V₅

V_T = V₁ + V₄

V_T = V₃

Hence volume of water is the same in both tumblers even if ice in A melts.

Considering a small fluid element

Resolving forces in z-direction, pdA − (p + ∂p / ∂z dz) dA − ρdAdzg = ρdAdz. a

⇒ − ∂p / ∂z dz = ρ(g + a)dz

Again resolving forces along r − (p + ∂p ∂r dr) dA + pdA + ρdAdrω2r = 0

[ρdAdrω2 r is the centrifugal force]

⇒∂p / ∂r dr = ρω2 rdr

Also we can write dp = ∂p / ∂r dr + ∂p / ∂z dz

∴ dp = ρω2rdr − ρ(a + g)dz

Which gives p = ρω²r² / 2− ρ(a + g)z + C

(a) The difference of pressure at the center (r = 0) of top & bottom surface

Δp = Pbottom − Ptop

⇒ Δp = [−ρ(a + g)0 + C] − [−ρ(a + g)⁽²⁾ + C]

⇒ Δp = ρ(a + g)(2)

⇒ Δp = 21.91 kPa

(b) The difference of pressure at the center and edge of bottom surface (Z = 0)

Δp = ρω²(0.5) ²/ 2

⇒ Δp = ρ × (N/60× 2π)²× (0.5)² / 2

⇒ Δp = 8.21 kPa

p1 = p2 = patm For forced vortex

p = ρω2r2 / 2 − ρgZ + C

Equating pressures at 1 & 2

ρω2(0)2 /2− ρg(0) = ρω2r2/2 /2 − ρgh

i. e ω2r2/2 / 2 = gh

(4.2)2(0.25)2 /2 = gh

h = 5.61 cm

For right arm, ρ = 103, r = 0.15, Z = 0.1 and pguage = 0

0 = 103 ( N/60²π)²× (0.15)² /2− 10³ × g(0.1) + C

0 = 111.03 − 981 + C ⟹ C = 870

For left arm, p = 0, Z = 0.1 cm, r = 0.05m

0 = ρ × 0.012 − ρ(0.981) + 870

ρ(0.969) = 870 ρ

= 897.83 kg/m³

Volume of parallelepiped

=1 / 2× πr²Z₂ = πr²Z₁

Z₂ = 2Z₁

_{Taking 0 as origin at gauge pressure}

_{p = ρω}²_r² _{/ 2 − ρg Z + C}

_{atO,p = 0,r = 0,Z = 0so,C = 0}

_{at A p = 0 = ρω}²_r² _{/ 2 − lgZ1 ⟹ Z2 = ω}²_r² _{/ 2g}

_{he rise of liquid above original level (Z1)}

₌ Z_{2 −} Z_{1 ⟹} Z_{2 −} Z_{2 / 2 =} Z_{2 / 2 = ω}²_r² _{/ 4g}

∴ h = 13.6 × 9810 × 0.2 / 9810

= 2.72 m

For case (b) when the vessel is full of water: Let the mercury level fall by a distance δ in the right limb with a corresponding rise δ in the left limb. Again writing the governing manometric equation for the points 1′ and 2′:

Patm + 13.6 × 9810 × (0.2 + 2δ)

= 9810 (3 + 2.72 + δ) + Patm

Solution gives: δ = 0.1145 m Hence the manometer reading would be

0.20 + 2 × 0.1145 = 0.429 m

= 42.90 cm