Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech II CHEMICAL ENGINEERING FLUID MECHANICS Previous Question Paper (July/August - 2021 - R15)

B.Tech | II | II | R15 | July/August - 2021

Answer PaperPaper Code: 124CMMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Get an expression for pressure drop in U-tube manometer.

7 marks

Answer

Consider a U-tube manometer containing a manometric fluid of density ρm, connected to a pipe carrying a fluid of density ρf. Let the difference in the levels of the manometric fluid in the two limbs be R. Applying the principle of hydrostatic equilibrium, the pressure at the same horizontal level in a continuous static fluid is equal. If the pressure at the two tapping points are P1 and P2, and the distance from the center of the pipe to the manometric fluid level in the limb connected to P1 is h1, and the distance to the level in the limb connected to P2 is h2, the pressure balance is:

P1 + ρfg(h1) = P2 + ρfg(h2 - R) + ρmgR

Rearranging for the pressure drop ΔP = P1 - P2:

ΔP = R g (ρm - ρf)

where g is the acceleration due to gravity.

b

A simple U-tube manometer is installed across an orifice meter. The manometer is filled with mercury (specific gravity 13.6) and liquid above the mercury is carbon tetrachloride (specific gravity 1.6). The manometer reads 200 mm. What is the pressure difference over the manometer in newtons per square meter?

8 marks

Answer

Given data:

  • Specific gravity of mercury (ρm/ρw) = 13.6
  • Specific gravity of carbon tetrachloride (ρf/ρw) = 1.6
  • Manometer reading (R) = 200 mm = 0.2 m
  • Acceleration due to gravity (g) = 9.81 m/s2
  • Density of water (ρw) = 1000 kg/m3

The pressure difference is calculated using the formula:

ΔP = R g (ρm - ρf)

Substituting the values:

ΔP = 0.2 × 9.81 × (13600 - 1600)
ΔP = 0.2 × 9.81 × 12000
ΔP = 23544 N/m2

The pressure difference over the manometer is 23,544 N/m2.

2

Derive the Bernoulli’s equation without friction.

15 marks

Answer

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3

Water at 200C is pumped at a constant rate of 9 m3/hr from a large reservoir resting on the floor to the open top of an experimental absorption tower. The point of discharge is 5 m above the floor and friction losses in the 50 mm pipe from the reservoir to the tower amount to 2.5 J/kg. At what height in the reservoir must the water level to be kept if the pump can develop only 0.1 KW?

15 marks

Answer

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415 marks
a

Write short notes on adiabatic frictional flow.

5 marks

Answer

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b

Methane gas is being pumped through a 305 m length of 52.5 mm diameter steel pipe at a rate of 41 kg/m2-sec. The inlet pressure is 345 K.Pa abs. Assume isothermal flow at 288.8 K. (μ = 1.045 × 10-5 Pa.sec, ε = 0.0014). Calculate the pressure at the end of the pipe.

10 marks

Answer

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5

Derive Ergun equation for the pressure drop in a packed bed and mention the limitations of the equation.

15 marks

Answer

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6

Starting from fundamentals, derive an equation for the terminal settling velocity of a solid particle falling in a liquid column for various ranges of settling.

15 marks

Answer

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7

Explain the principle and working of a centrifugal pump. Discuss about the characteristic curves in detail.

15 marks

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815 marks
a

Water flows through an orifice of diameter 1 inch installed in a pipe of 75 mm diameter at a rate of 1.0 kg/s. What will be the difference in the level of manometer with CCl4 as manometric liquid? The specific gravity of CCl4 is 1.6 and that of water is 1.0. The coefficient of orifice meter is 0.61.

10 marks

Answer

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b

What are merits and demerits of venturi meter?

5 marks

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This page contains the solved question paper with answers of CHEMICAL ENGINEERING FLUID MECHANICS from Jawaharlal Nehru Technological University Hyderabad for B.Tech PTM II II, July/August - 2021, R15.

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