JNTU
JNTU B.Tech II CHEMICAL ENGINEERING FLUID MECHANICS Previous Question Paper (March - 2022 - R13)
Answer any Five Questions. All Questions Carry Equal Marks.
Main
Water at 20°C 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 6 m above the floor, and friction losses in the 50mm pipe from the reservoir to tower amount to 2.5 J/kg. At what height in the reservoir must the water level be kept if the pump can deliver only 0.1 kW.
Answer
To determine the required water level height in the reservoir, we apply the mechanical energy balance equation between the surface of the reservoir (point 1) and the discharge point at the top of the tower (point 2):
Given values:
- Flow rate (Q) = 9 m3/hr = 0.0025 m3/s
- Density of water (ρ) = 998 kg/m3
- Mass flow rate (ṁ) = ρ × Q = 998 × 0.0025 = 2.495 kg/s
- Pump power (P) = 0.1 kW = 100 W
- Pump work (Wp) = Power / ṁ = 100 / 2.495 = 40.08 J/kg
- Discharge height (z2) = 6 m
- Friction loss (ΣF) = 2.5 J/kg
- Pressure at both points (P1 = P2) = Atmospheric pressure
- Velocity at reservoir surface (v1) ≈ 0
First, calculate the velocity at the discharge point (v2) using the pipe diameter (d = 0.05 m):
Substituting these values into the energy balance:
The water level in the reservoir must be maintained at a height of 2.252 meters above the floor.
Prove that, for static fluid, pressure turns out to be independent of the orientation of any internal surface on which the pressure is assumed to act, (P = Px = Py = Pz).
Answer
To prove that pressure at a point in a static fluid is independent of orientation, consider a small wedge-shaped fluid element of unit thickness (perpendicular to the page) at rest. Let the dimensions of the wedge be Δx, Δy, and Δs, where Δs is the length of the inclined face making an angle θ with the vertical.
Let Px, Py, and Ps be the pressures acting on the faces perpendicular to the x-axis, y-axis, and the inclined face, respectively. The forces acting on the element are due to pressure and gravity.
The force balance in the x-direction is:
Since Δs sinθ = Δy, this simplifies to:
The force balance in the y-direction, including the weight of the element (W = ρg × volume), is:
Since Δs cosθ = Δx, this becomes:
As the element size shrinks to a point (Δx, Δy → 0), the weight term (involving the product of Δx and Δy) vanishes faster than the pressure terms. Thus, in the limit:
Combining these results, we obtain:
Because the angle θ is arbitrary, the pressure at any point in a static fluid is the same in all directions, confirming that pressure is a scalar quantity.
For a fluid flowing in a circular pipe, the pressure drop per unit length is found to depend on the diameter of the pipe, the average velocity, the density and viscosity of the fluid. Using the method of dimensional analysis, determine the functional relationship between these variables.
Answer
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Ammonia at atmosphere pressure and 300K with a bulk stream velocity of 10 m/s flows through a pipe of ID 25 cm. Calculate the pressure drop per 100m length of the pipe and the power consumed. Friction factor
Answer
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Derive the equations for adiabatic frictional flow.
Answer
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Derive an expression for velocity in nozzle for an isentropic flow.
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Derive the equations for one-dimensional motion of a particle through a fluid by writing the force balance on the particle.
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An orifice meter consisting of 10 cm diameter orifice in a 25 cm diameter pipe has coefficient of 0.65. The pipe delivers oil of 0.8 specific gravity. The pressure difference on the two sides of the orifice plate is measured by a mercury oil differential manometer. If the differential gauge reads 80 cm of mercury, calculate the rate of flow in lit/sec.
Answer
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