A non-viscous liquid of constant density 1000 kg/m^3 flows in streamline motion along a tube of vari

Q: Why is the work done by pressure equal to the pressure difference $p_1 - p_2$?

Work done is Force times displacement ($W = F \cdot d$). For a fluid, force is pressure times area ($F = p \cdot A$). Thus, work done is $W = p \cdot A \cdot d = p \cdot V$, where $V$ is the volume. Dividing by volume, the work done per unit volume is simply the pressure $p$. For a flow from 1 to 2, the net work done by pressure is $p_1 - p_2$.

Q: Why is the work done by gravity negative?

Gravity acts downwards, while the fluid is moving upwards (from $2\text{ m}$ to $5\text{ m}$). Since the force of gravity and the displacement are in opposite directions, the work done by gravity is negative.

Q: Does the shape of the tube between P and Q affect the work done?

No, because both gravity and pressure forces (in a non-viscous, streamline flow) are conservative or state-dependent. The work done depends only on the initial and final states (heights, velocities, and pressures) and not on the path taken.

Q: What would change if the fluid were viscous?

If the fluid were viscous, some mechanical energy would be dissipated as heat due to internal friction. This would require a larger pressure difference ($p_1 - p_2$) to maintain the same flow, meaning the work done by pressure would be higher.

Q: Can we use $g = 10\text{ m/s}^2$ instead of $9.8\text{ m/s}^2$?

Unless specified otherwise in the question, it is safer to use $g = 9.8\text{ m/s}^2$ for standard JEE subjective questions to get the exact textbook answer. Using $10\text{ m/s}^2$ gives $W_{\text{pressure}} = 29625\text{ J/m}^3$ and $W_{\text{gravity}} = -30000\text{ J/m}^3$.

Question

A non-viscous liquid of constant density $1000	ext{ kg/m}^3$ flows in streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points P and Q at heights of $2	ext{ m}$ and $5	ext{ m}$ are respectively $4 	imes 10^{-3}	ext{ m}^2$ and $8 	imes 10^{-3}	ext{ m}^2$. The velocity of the liquid at point P is $1	ext{ m/s}$. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q.

Sigma Percentile · Accepted Answer

### Analyzing the Flow

Imagine a fluid traveling through an inclined, tapering pipe. As it moves from a lower point $P$ to a higher point $Q$, several physical principles govern its motion.

First, we must recognize that the fluid is **non-viscous** and **incompressible**, which allows us to apply the ideal fluid dynamics equations without worrying about energy losses due to friction.

Let's write down the given parameters at both points:

*   **Density of the liquid ($ho$):** $1000	ext{ kg/m}^3$
*   **At point P:** 
    *   Height $h_1 = 2	ext{ m}$
    *   Area of cross-section $A_1 = 4 	imes 10^{-3}	ext{ m}^2$
    *   Velocity $v_1 = 1	ext{ m/s}$
*   **At point Q:**
    *   Height $h_2 = 5	ext{ m}$
    *   Area of cross-section $A_2 = 8 	imes 10^{-3}	ext{ m}^2$

### The Continuity Connection

Before we can calculate the work done, we need to know how fast the fluid is moving when it reaches point $Q$. Since the fluid is incompressible, the mass entering point $P$ per unit time must equal the mass exiting point $Q$ per unit time. This is mathematically expressed by the **Equation of Continuity**:

$$A_1 v_1 = A_2 v_2$$

Let's substitute the known values into this equation to find $v_2$:

$$(4 	imes 10^{-3}	ext{ m}^2) 	imes (1	ext{ m/s}) = (8 	imes 10^{-3}	ext{ m}^2) 	imes v_2$$

$$v_2 = \frac{4 	imes 10^{-3}}{8 	imes 10^{-3}} = 0.5	ext{ m/s}$$

As the cross-sectional area doubles from $P$ to $Q$, the velocity is halved to maintain a constant volume flow rate. This makes intuitive sense!

### Applying Bernoulli's Principle

Now, we turn to **Bernoulli's Equation**, which is essentially the conservation of energy statement for a flowing fluid:

$$p_1 + \frac{1}{2}ho v_1^2 + ho g h_1 = p_2 + \frac{1}{2}ho v_2^2 + ho g h_2$$

We want to find the work done per unit volume by the pressure forces. The work done by pressure forces per unit volume is equal to the pressure difference between the two points:

$$W_{	ext{pressure}} = p_1 - p_2$$

Let's rearrange Bernoulli's equation to solve for $p_1 - p_2$:

$$p_1 - p_2 = ho g (h_2 - h_1) + \frac{1}{2}ho (v_2^2 - v_1^2)$$

Now, substitute the numerical values into this expression:

$$W_{	ext{pressure}} = (1000	ext{ kg/m}^3) 	imes (9.8	ext{ m/s}^2) 	imes (5	ext{ m} - 2	ext{ m}) + \frac{1}{2} 	imes (1000	ext{ kg/m}^3) 	imes ((0.5	ext{ m/s})^2 - (1	ext{ m/s})^2)$$

$$W_{	ext{pressure}} = 1000 	imes 9.8 	imes 3 + 500 	imes (0.25 - 1)$$

$$W_{	ext{pressure}} = 29400 - 375 = 29025	ext{ J/m}^3$$

Thus, the work done per unit volume by the pressure forces is **$29025	ext{ J/m}^3$**.

### Work Done by Gravity

Gravity is a conservative force, so the work done by gravity per unit volume is equal to the negative change in gravitational potential energy per unit volume:

$$W_{	ext{gravity}} = -\Delta U = ho g (h_1 - h_2)$$

Substituting the values:

$$W_{	ext{gravity}} = 1000 	imes 9.8 	imes (2 - 5)$$

$$W_{	ext{gravity}} = 1000 	imes 9.8 	imes (-3) = -29400	ext{ J/m}^3$$

The negative sign indicates that gravity is opposing the motion of the fluid as it climbs upwards from $2	ext{ m}$ to $5	ext{ m}$.

Therefore, the work done per unit volume by gravity is **$-29400	ext{ J/m}^3$**.

Visualized Solution

Identify Given Parameters

Apply Equation of Continuity

Substitute Values in Continuity Equation

Calculate Velocity at Point Q

State Bernoulli's Equation

Relate Work Done by Pressure to Pressure Difference

Rearrange Bernoulli's Equation for Pressure Difference

Calculate Work Done by Pressure

Define Work Done by Gravity

Calculate Work Done by Gravity

Final Conclusion

The Sigma Insight: Flow of Fluid

Analyzing the Flow

The Continuity Connection

Applying Bernoulli's Principle

Work Done by Gravity

Similar Questions

Water flows in a horizontal tube (see figure). The pressure of water changes by $700 Nm^{- 2}$ between $A$ and $B$ , where the area of cross-section are $40 cm^{2}$ and $20 cm^{2}$ , respectively. Find the rate of flow of water through the tube. (Take, density of water = $1000 kgm^{- 3}$ )

Water from a tap emerges vertically downwards with an initial speed of $1.0 m/s$ . The cross-sectional area of tap is $1 0^{- 4} m^{2}$ . Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross-sectional area of stream $0.15 m$ below the tap is

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4 cm$ and $4.8 cm$ , respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is

A cylinder of height $20 m$ is completely filled with water. The velocity of efflux of water (in $ms^{- 1}$ ) through a small hole on the side wall of the cylinder near its bottom, is

A liquid of density $ρ$ is coming out of a hose pipe of radius $a$ with horizontal speed $v$ and hits a mesh. 50\% of the liquid passes through the mesh unaffected 25\% losses all of its momentum and, 25\% comes back with the same speed. The resultant pressure on the mesh will be

Water is flowing continuously from a tap having an internal diameter $8 \times 1 0^{- 3} m$ . The water velocity as it leaves the tap is $0.4 ms^{- 1}$ . The diameter of the water stream at a distance $2 \times 1 0^{- 1} m$ below the tap is close to