Fluid Mechanics Test 4

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Fluid Mechanics Test 4

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Question 1
1 / -0

In an orifice meter, coefficient of discharge is 0.58 and coefficient of vena contracta is 0.61, what is coefficient of velocity?

A
0.94-0.96

B
0.944-0.961

C
0.947-0.966

D
0.949-0.962

Solution

Concept:
C_d = C_c × C_v
Where, C_d = Coefficient of discharge, C_c = Coefficient of vena contracta, C_v = Coefficient of velocity
Calculation:
C_d = C_c × C_v
0.58 = 0.61 × C_v
∴ C_v = 0.95
Question 2
1 / -0

Water is flowing through a pipe with a velocity of 2 m/s at section (1) which is 3 m above the ground and at section (2) with a velocity of 5 m/s which is 0.5 m above the ground. If the pressure at section (1) is 25 kPa & there is energy loss of 0.5 m, what is the pressure at section (2)?

A
34.1 kPa

B
17.06 kPa

C
38.6 kPa

D
19.3 kPa

Solution

Explanation:
Applying Bernoulli’s equation between (1) & (2)
\(\frac{{{p_1}}}{{\rho g}} + \frac{{v_1^2}}{{2g}} + {z_1} = \frac{{{p_2}}}{{\rho g}} + \frac{{v_2^2}}{{2g}} + {z_2} + {h_{loss}}\)
\(\frac{{25\; \times \;{{10}^3}}}{{{{10}^3}\left( {9.81} \right)}} + \frac{{{2^2}}}{{2\left( {9.81} \right)}} + 3 = \frac{{{p_2}}}{{{{10}^3}\left( {9.81} \right)}} + \frac{{{5^2}}}{{2\left( {9.81} \right)}} + 0.5 + 0.5\)
p₂ = 34.12 × 10³
∴ p₂ = 34.12 kPa
Question 3
1 / -0

A horizontal venturimeter with inlet diameter 250 mm and throat diameter 100 mm is used to measure the flow of an oil of specific gravity 0.85. The oil-mercury differential manometer shows a reading of 20 cm. The rate of flow of oil (in litres/sec) through the venturimeter is ____ (Take C_d = 0.98)

A
58-62

B
584-621

C
587-626

D
589-622

Solution

Concept:
For venturimeter, discharge
\(Q = {C_d} \cdot \frac{{{A_1}{A_2}\sqrt {2gh} }}{{\sqrt {A_1^2 - A_2^2} }}\)
Calculation:
Given:
S₀ = 0.85, x = 20 cm = 0.2 m, D₁ = 0.25 m, D₂ = 0.1 m, C_d = 0.98
Now,
\({A_1} = \frac{\pi }{4}D_1^2 = \frac{\pi }{4}\left( {{{0.25}^2}} \right) = 0.0490\)
\({A_2} = \frac{\pi }{4}\left( {{{0.1}^2}} \right) = 0.00785\)
Now,
\({\rm{Pressure\;head\;difference\;}}\left( h \right) = x \cdot \left( {\frac{{{S_m}}}{{{S_0}}} - 1} \right)\)
\(h = 0.2\left( {\frac{{13.6}}{{0.85}} - 1} \right)\)
∴ h = 3m
Note:
The given manometer reading (x = 20 cm) is not pressure head. We have to calculate pressure head using above formula.
\({\rm{Rate\;of\;flow\;}}\left( Q \right) = \frac{{{C_d} \cdot {A_1}{A_2}\sqrt {2gh} }}{{\sqrt {A_1^2 - A_2^2} }}\)
\(Q = \frac{{\left( {0.98} \right)\left( {0.0490} \right)\left( {0.00785} \right)\sqrt {2\: \times \:9.81\: \times \:3} }}{{\sqrt {{{0.0490}^2}\: - \:{{0.00785}^2}} }}\)
Q = 0.05979 m³/s
∴ Q = 59.79 litres/sec
Question 4
1 / -0

In a hydro project, a turbine (η = 100) has a head of 50 m and discharge in the feeding penstock is 3 m³/s. If head loss of 5 m take place in penstock, and a power of 1000 kW is extracted, the residual head loss of turbine is______

A
7m

B
11m

C
15m

D
19m

Solution

Concept:
Power (P) = ρQgH
Calculation:
P = ρQgH × η_T
∴ 1000 × 10³ = 1000 × 3 × 9.81 × H
∴ H = 33.98 m
Applying Bernoulli’s equation,
\(\frac{{{P_1}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_1} = \frac{{{P_2}}}{{\rho g}} + \frac{{V_1^2}}{{2g}} + {Z_2} + {H_{loss}}\)
∴ 50 = 5 + H_loss + 33.98
∴ H_loss = 50 – 5 – 33.98
∴ H_Loss = H_Residual = 11.02 m