Fluid Mechanics Test 6

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Fluid Mechanics Test 6

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Weekly Quiz Competition

Question 1
1 / -0

Two pipes of lengths 600 m and 300 m and of diameters 400 mm and 200 mm are connected in series. These pipe are to be replaced by a single pipe of length 900 m, then the diameter of new pipe is _______mm.

A
244-248

B
2444-2481

C
2447-2486

D
2449-2482

Solution

Explanation:
\({h_f} = {h_{{f_1}}} + {h_{{f_2}}}\)
\(\frac{{fL{Q^2}}}{{12.1{d^5}}} = \frac{{f{L_1}{Q^2}}}{{12.1d_1^5}} + \frac{{f{L_2}{Q^2}}}{{12.1\;d_2^5}}\)
\(\frac{L}{{{d^5}}} = \frac{{{L_1}}}{{d_1^5}} + \frac{{{L_2}}}{{d_2^5}}\)
\(\frac{{900}}{{{d^5}}} = \frac{{600}}{{{{0.4}^5}}} + \frac{{300}}{{{{0.2}^5}}}\)
d = 0.2461 m
∴ d = 246.1 mm
Question 2
1 / -0

Water at 5° (ρ = 1000 kg/m³ and μ = 1.519 × 10^-3 kg/m.s) is flowing steadily through a 0.3 cm diameter, 9-m long horizontal pipe at an average velocity of 0.9 m/s. Find the head loss?

A
1.23 m

B
1.7 m

C
4.45 m

D
1.12 m

Solution

Concept:
\(Head\;loss\;{h_f} = \frac{{fL{V^2}}}{{2gD}}\)
Where, V is average velocity, f = Friction factor
\(For\;laminar\;flow,f = \frac{{64}}{{Re}}\)
Calculation:
Given:
ρ = 1000 kg/m³, μ = 1.519 × 10^-3 kg/m, D = 0.3 m, L = 9 m, V_avg = 0.9 m/s
Now, initially checking if the flow is laminar or turbulent
\(Rd = \frac{{\rho VD}}{\mu } = \frac{{1000 \times 0.9 \times 0.3 \times {{10}^{ - 2}}}}{{1.519 \times {{10}^{ - 3}}}} = 1777.48\)
As Reynold’s number Re < 2000, so flows laminar
So \(f = \frac{{64}}{{Re}} = \frac{{64}}{{1777}} = 0.0360\)
\({h_f} = \frac{{fL{V^2}}}{{2gD}} = \frac{{0.0360 \times 9 \times {{\left( {0.9} \right)}^2}}}{{2 \times 9.81 \times 0.3 \times {{10}^{ - 2}}}}\)
∴ h_f= 4.45 m
Question 3
1 / -0

It is required to carry out model studies on a boat having a characteristic length of 3.6 m and travelling at a speed of 3 m/s. Assume the acceleration due to gravity as 10 m/s² and neglect the effects due to viscous and surface tension forces. The value of appropriate non-dimensional number is ____.

A
0.5-0.5

B
0.54-0.51

C
0.57-0.56

D
0.59-0.52

Solution

Concept:
Note that model studies on a boat involve the gravity force as the predominant force and the governing non-dimensional number in this case is Froude No. \(Fr = \frac{V}{{\sqrt {gL} }}\)
Calculation:
Given, L = 3.6 m/s, V = 3 m/s, g = 10 m/s²
\(Fr = \frac{V}{{\sqrt {gL} }} = \frac{3}{{\sqrt {10 \times 3.6} }} = \frac{1}{2} = 0.5\)
Thus, Fr = 0.5
Question 4
1 / -0

A pump is lifting water from pond to overhead reservoir situated at 15 m above pond. If head loss through pump is 5 m and flow rate is 170 m³/hr with efficiency of 60 %. Find power (kW) required to drive pump?

A
15-16

B
154-161

C
157-166

D
159-162

Solution

Concept:
\(P = \frac{{\rho QgH}}{\eta }\)
Calculation:
Total head = h + hf
Total head = 15 m + 5m
Given: H = 20 m, ρ = 1000 kg/m³, Q = 170 m³/hr = 0.04722 m³/s, g = 9.81 m/s², η = 60%
Now,
\(P = \frac{{\rho QgH}}{\eta }\)
\(\therefore P = \frac{{1000\; \times \;0.04722\; \times \;9.81\; \times\; 20}}{{0.6}}\)
∴ P = 15.44 kW
Question 5
1 / -0

A 1:16 scale model of a submarine moving far below the surface of water is tested in a water tunnel. If the speed of prototype is 8 m/s, the corresponding velocity of water in tunnel is____
[Fluids are same in testing as in actual model]

A
128 m/s

B
64 m/s

C
16 m/s

D
32 m/s

Solution

Concept:
As submarine is completely submerged in water, thus Reynold’s number need to be same.
∴ (R_e)_model = (R_e)_prototype
\({\left( {\frac{{VL}}{v}} \right)_m} = {\left( {\frac{{VL}}{v}} \right)_p}\)
V_mL_m = V_pL_p
\(\frac{{{L_m}}}{{{L_p}}} = \frac{1}{{16}}\)
\(\therefore \frac{{{V_m}}}{{{V_p}}} = \frac{{{L_p}}}{{{L_m}}}\)
\(\therefore {V_m} = {V_p}\left( {\frac{{{L_p}}}{{{L_m}}}} \right)\)
V_m = 8 (16)
∴ V_m = 128 m/s
Point to remember:
1. When body submerged ⇒ Reynold’s numbers are equal
2. When body is floating ⇒ Froude’s numbers are equal.
Question 6
1 / -0

Water flows through a pipe of length 3 km at the rate of 250 litres/s and head lost due to friction is 5 m. Find the diameter of this pipe (in cm) ________ (Take chezy’s constant, c = 50)

A
60-65

B
604-651

C
607-656

D
609-652

Solution

Concept:
\({\rm{Chezy's\;formula\;}}\left( v \right) = c\sqrt {mi} \)
\({\rm{m\;}} = {\rm{\;Hydraulic\;mean\;depth}} = \frac{{Area}}{{Wetted\;perimeter}}\)
\(\therefore m = \frac{{\frac{\pi }{4}{d^2}}}{{\pi d}} = \frac{d}{4}\)
Now,
\(i = \frac{{{h_f}}}{L}\)
Calculation:
Given:
L = 3 km = 3000 m
Q = 250 litres/s = 0.25 m³/s
h_f = 5m, c = 50
\(v = \frac{Q}{A} = \frac{{0.25}}{{\frac{\pi }{4}{d^2}}} = \frac{1}{{\pi {d^2}}}\)
\(i = \frac{{{h_f}}}{L} = \frac{5}{{3000}} = \frac{1}{{600}}\)
Using, \(v = c\sqrt {mi} \)
\(\frac{1}{{\pi {d^2}}} = 50\sqrt {\frac{d}{4} \times \frac{1}{{600}}} \)
\(\frac{1}{{{\pi ^2}{d^4}}} = \frac{{\left( {2500} \right)d}}{{2400}}\)
\(d = {\left( {\frac{{2400}}{{2500\;{\pi ^2}}}} \right)^{1/5}}\)
∴ d = 0.627 m = 62.7 cm
Question 7
1 / -0

A pipe of diameter 1.8 m is required to transport an oil of relative density 0.8 and viscosity 0.04 poise at the rate of 4 m3/s. Tests were conducted on a 20 cm diameter pipe using water. The discharge (litres/s) through the model is _____. (Correct up to 2 decimals)
Take viscosity of water = 0.01 poise, density = 1000 kg/m3

A
88-89

B
884-891

C
887-896

D
889-892

Solution

Concept:
Specific Gravity or Relative Density: The ratio of the density of the fluid to the density of water—usually 1000 kg/m³ at a standard condition—is defined as Specific Gravity or Relative Density of fluids
∴ The density of fluid = Relative Density × Density of water
Poise = 0.1 Ns/m²= 0.1 kg/ms
Calculation:
Given data,
RD = 0.8 ⇒ ρ = 0.8 × 1000 = 800 kg/m3
Prototype
Model
D_P = 1.8 m
D_m = 0.2 m
ρ_P = 800 kg/m³
ρ_m = 1000 kg/m³
μ_P = 0.004 kg/ms
μ_m = 0.001 kg/ms
Q_P = 4 m³/s
Q_m = ?

(Re)_P = (Re)_m
\({{\left( \frac{ρ VD}{\mu } \right)}_{P}}={{\left( \frac{ρ VD}{\mu } \right)}_{m}}\) ...(1)
\(\frac{{{\rho _m}}}{{{\rho _P}}}.\frac{{{V_m}}}{{{V_P}}}.\frac{{{D_m}}}{{{D_P}}}.\frac{{{\mu _P}}}{{{\mu _m}}} = 1\)
\(\frac{{1000}}{{800}} × \frac{{{V_m}}}{{{V_P}}} × \frac{{0.2}}{{1.8}} × \frac{{0.004}}{{0.001}} = 1\)
\(\frac{{{V_m}}}{{{V_P}}} = 1.8\)
\(\frac{{{Q_m}}}{{{Q_P}}} = \frac{{{A_m}{V_m}}}{{{A_P}{V_P}}} = {\left( {\frac{{{D_m}}}{{{D_P}}}} \right)^2} × \frac{{{V_m}}}{{{V_P}}}\)
\(\frac{{{Q_m}}}{{{Q_P}}} = {\left( {\frac{{0.2}}{{1.8}}} \right)^2} × 1.8 = 0.0222\)
Q_m = 0.0222 × Q_P= 0.0222 × 4 = 0.08888 m3/s
1 litre = 1000 cm³ = 10^-3 m³
∴ 1 m3/s = 1000 litre/s
Q_m = 0.08889 m³/s
∴ Qm = 88.89 litres/s

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Re-Attempt Weekly Quiz Competition

Prototype	Model
D_P = 1.8 m	D_m = 0.2 m
ρ_P = 800 kg/m³	ρ_m = 1000 kg/m³
μ_P = 0.004 kg/ms	μ_m = 0.001 kg/ms
Q_P = 4 m³/s	Q_m = ?

Fluid Mechanics Test 6

Result

Fluid Mechanics Test 6

Score

-

Rank

-

SHARING IS CARING

Question 1

244-248

2444-2481

2447-2486

2449-2482

Solution

Question 2

1.23 m

1.7 m

4.45 m

1.12 m

Solution

Question 3

0.5-0.5

0.54-0.51

0.57-0.56

0.59-0.52

Solution

Question 4

15-16

154-161

157-166

159-162

Solution

Question 5

128 m/s

64 m/s

16 m/s

32 m/s

Solution

Question 6

60-65

604-651

607-656

609-652

Solution

Question 7

88-89

884-891

887-896

889-892

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

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