Calculating Isentropic Work Done by a Pump

This question asks us to determine the isentropic work done by a pump when it increases the pressure of a liquid from a starting pressure to a final pressure, given the liquid's density. We need to find the work done per unit mass, expressed in kJ/kg.

For a liquid pump operating ideally (isentropically), the work done per unit mass can be calculated using the formula related to the change in pressure and the specific volume of the liquid. Liquids are often treated as incompressible fluids, meaning their density (and thus specific volume) remains nearly constant during the compression process by a pump.

Formula for Isentropic Work Done

For an incompressible fluid, the isentropic work done per unit mass ($w$) is given by:

$w = \int_{P_1}^{P_2} v \, dP$

Where:

• $w$ is the work done per unit mass (in J/kg or kJ/kg)
• $v$ is the specific volume of the liquid (in m³/kg)
• $P_1$ is the initial pressure
• $P_2$ is the final pressure
• $dP$ is the differential change in pressure

Since the liquid is assumed incompressible, the specific volume ($v$) is constant. The specific volume is the reciprocal of the density ($\rho$).

$v = \frac{1}{\rho}$

So, the formula simplifies to:

$w = v (P_2 - P_1) = \frac{1}{\rho} (P_2 - P_1)$

Step-by-Step Calculation

Let's use the given values in the formula.

• Initial pressure, $P_1 = 1 \text{ bar}$
• Final pressure, $P_2 = 30 \text{ bar}$
• Density of liquid, $\rho = 990 \text{ kg/m}^3$

First, convert the pressures from bar to Pascal (Pa) or kN/m² for consistency in units.

• $1 \text{ bar} = 10^5 \text{ Pa} = 10^5 \text{ N/m}^2$
• $1 \text{ N/m}^2 = 1 \text{ Pa}$
• $1000 \text{ N} \cdot \text{m} = 1000 \text{ J} = 1 \text{ kJ}$
• So, $1 \text{ N/m}^2 \times 1 \text{ m}^3 = 1 \text{ N} \cdot \text{m} = 1 \text{ J}$
• Also, $1 \text{ kN/m}^2 \times 1 \text{ m}^3 = 1 \text{ kN} \cdot \text{m} = 1 \text{ kJ}$

Let's use kN/m² (which is kPa):

• $P_1 = 1 \text{ bar} = 1 \times 100 \text{ kN/m}^2 = 100 \text{ kN/m}^2$
• $P_2 = 30 \text{ bar} = 30 \times 100 \text{ kN/m}^2 = 3000 \text{ kN/m}^2$

The specific volume is:

$v = \frac{1}{\rho} = \frac{1}{990} \text{ m}^3\text{/kg}$

Now, calculate the work done:

$w = \frac{1}{990} \text{ m}^3\text{/kg} \times (3000 \text{ kN/m}^2 - 100 \text{ kN/m}^2)$

$w = \frac{1}{990} \text{ m}^3\text{/kg} \times (2900 \text{ kN/m}^2)$

$w = \frac{2900}{990} \text{ } \frac{\text{kN} \cdot \text{m}}{\text{kg}}$

Since $1 \text{ kN} \cdot \text{m} = 1 \text{ kJ}$, the units are kJ/kg.

$w = \frac{2900}{990} \text{ kJ/kg}$

Let's calculate the numerical value:

$w \approx 2.92929... \text{ kJ/kg}$

Comparing with Options

The calculated value is approximately 2.929 kJ/kg. We compare this with the given options:

The value 2.93 is the closest option to our calculated isentropic work done.