Question

An electrically-driven pump operating at steady state draws water from a pond at a pressure of 1 bar and a rate of \(40 \mathrm{~kg} / \mathrm{s}\) and delivers the water at a pressure of 4 bar. There is no significant heat transfer with the surroundings, and changes in kinetic and potential energy can be neglected. The isentropic pump efficiency is \(80 \%\). Evaluating electricity at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), estimate the hourly cost of running the pump.

Short answer

The hourly cost of running the pump is $1.2 per hour.

Step-by-step solution

- Understand Given Data

Identify all the given quantities: - Inlet pressure, \(P_{in} = 1 \text{ bar} = 100 \text{ kPa}\)- Outlet pressure, \(P_{out} = 4 \text{ bar} = 400 \text{ kPa}\)- Water flow rate, \(\text{mass flow rate} = 40 \text{ kg/s}\)- Isentropic efficiency, \( \text{efficiency} = 80\text{ \%} = 0.8 \)- Electricity cost, \(8 \text{ cents/kW} \times \text{ h}\)

- Use Pump Work Formula for Isentropic Process

For an isentropic process for an incompressible fluid like water: \[ W_s = \frac{(P_{out} - P_{in}) \times \text{mass flow rate}}{\text{density}} \] Given density of water, \( \rho \text{ is } 1000 \text{ kg/m}^3 \)\[ W_s = \frac{(400 \text{ kPa} - 100 \text{ kPa}) \times 40 \text{ kg/s}}{1000 \text{ kg/m}^3} \]Calculating, we get:\[ W_s = \frac{(300 \text{ kPa} \times 40 \text{ kg/s})}{1000 \text{ kg/m}^3} = 12 \text{ kW} \]

- Real Pump Work

The actual work done by the pump considering efficiency: \[ W_{actual} = \frac{W_s}{\text{efficiency}} \]\[ W_{actual} = \frac{12 \text{ kW}}{0.8} = 15 \text{ kW} \]

- Calculate Cost

The cost of running the pump is calculated by multiplying the power consumption with the operational time and unit cost:Electricity rate is \(8 \text{ cents/kW} \times\text{ h}\)For one hour, \[ \text{Cost} = 15 \text{ kW} \times 8 \text{ cents/kW} \times 1 \text{ h} = 120 \text{ cents} = 1.2 \text{ dollars/hour} \]

Key concepts

Pump Work Formula

Calculating the work required by a pump is crucial in understanding its efficiency and operational cost. The pump work formula for an isentropic process helps us determine the theoretical work done when there is no heat transfer, kinetic, or potential energy changes. For an incompressible fluid like water, we can use the formula:
\[ W_s = \frac{(P_{out} - P_{in}) \times \text{mass flow rate}}{\text{density}} \]
This formula estimates the amount of work needed to move the fluid from the inlet pressure \(P_{in}\) to the outlet pressure \(P_{out}\). Given the density of water (\(\rho\)), and the mass flow rate, we can plug in these values to find the required isentropic work (\(W_s\)). For this problem, substituting the values:
- Inlet pressure, \(P_{in}\) = 100 kPa (1 bar)
- Outlet pressure, \(P_{out}\) = 400 kPa (4 bar)
- Mass flow rate = 40 kg/s
- Density of water, \(\rho\) = 1000 kg/m³
We get:
\[ W_s = \frac{(400 \text{ kPa} - 100 \text{ kPa}) \times 40 \text{ kg/s}}{1000 \text{ kg/m}^3} = 12 \text{ kW} \]

Electrically-Driven Pump

An electrically-driven pump converts electrical energy into mechanical work to move fluids. It's essential to understand that not all the electrical energy consumed by the pump translates into useful mechanical work because there's always some loss. Pump efficiency helps us quantify these losses and the actual work achieved.

For our pump, the isentropic work (\(W_s\)) calculated is 12 kW. But due to efficiency losses (80% efficiency in our case), the real work (\(W_{actual}\)) done by the pump will be higher. We can find the actual work using the formula:

\[ W_{actual} = \frac{W_s}{\text{efficiency}} \]

By substituting the values, we get:
\[ W_{actual} = \frac{12 \text{ kW}}{0.8} = 15 \text{ kW} \]

This shows that the electrically-driven pump needs to consume 15 kW of electricity to achieve the 12 kW of mechanical work due to its 80% efficiency.

Isentropic Process

An isentropic process is a thermodynamic process in which entropy remains constant. This process is ideal because no energy is lost due to friction, heat transfer, or other inefficiencies. For an incompressible fluid like water, the isentropic process helps in computing the minimum work required for pumping the fluid.

To calculate the isentropic work (\(W_s\)), we assume the fluid undergoes no changes in kinetic and potential energy and there is no significant heat transfer with the surroundings. The formula used is:

\[ W_s = \frac{(P_{out} - P_{in}) \times \text{mass flow rate}}{\text{density}} \]

This process assumes an ideal scenario, helping us estimate the best-case scenario for the pump's work requirements without any real-world inefficiencies.

Hourly Cost Calculation

Calculating the hourly cost of running an electrically-driven pump involves understanding both the power consumption and the cost of electricity. We already know the actual work the pump does is 15 kW. The electricity cost is given as 8 cents per kWh.

To calculate the hourly cost, use the formula:

\[ \text{Cost} = W_{actual} \times \text{electricity rate} \times \text{operational time} \]

Given the electricity rate is 8 cents/kWh and we are interested in the cost for one hour of operation, we get:
\[ \text{Cost} = 15 \text{ kW} \times 8 \text{ cents/kW} \times 1 \text{ h} = 120 \text{ cents} = 1.2 \text{ dollars/hour} \]

Thus, the hourly cost of running the pump is $1.2, considering both the efficiency of the pump and the cost of electricity.