The volume flow rate through a water fountain on a college...

The volume flow rate through a water fountain on a college campus is $0.075 \mathrm{m}^{3} / \mathrm{s}$. Each water stream can rise to a height of $10 \mathrm{m}$. Estimate the daily cost to operate the fountain. Assume that the pump motor efficiency is 85 percent, the pump efficiency is 85 percent, and the cost of clectricity is $14 \& /(\mathrm{kW} \cdot \mathrm{hr})$

Key Concepts

Power and Energy Consumption Calculation

This involves determining the rate at which work is done or energy is consumed, often calculated using the energy required to lift the water and the time over which the work is performed. It helps in converting the physical parameters of the system into power, which is then used to assess total energy consumption over time.

Gravitational Potential Energy

This concept involves the energy required to raise a mass to a certain height in a gravitational field. In the context of pumping water, it is calculated as the product of the water's mass, gravitational acceleration, and the height it is raised. This energy represents the fundamental work that the pump must perform against gravity.

Volume Flow Rate and Mass Flow Conversion

Volume flow rate measures the amount of fluid passing a point per unit time. For energy calculations, it is often necessary to convert this to a mass flow rate using the density of the fluid. This conversion is crucial for determining the total mass of water being lifted, which directly affects the energy required.

Efficiency in Energy Systems

Efficiency accounts for the losses in energy conversion systems. In a pump or motor system, not all the electrical energy is converted into mechanical work due to various inefficiencies. Understanding efficiency is essential to accurately estimate the actual energy consumption required to perform a given task, as the input energy must be higher than the useful output energy.

Cost Estimation from Energy Consumption

After determining the total energy usage, the next step is to estimate the operating cost by converting the energy consumption into a monetary value. This is done using a cost per energy unit, such as cost per kilowatt-hour. This concept links the physical energy measurements to economic analysis.

Transcript

00:01 Here we will assume the flow to be steady and incompressible.

00:04 Let's calculate the pressure drop from the energy balance equation.

00:07 The pressure drop del p will be equals to rho g del h.

00:13 Del h.

00:14 So let's put the value of row as 998 multiplied by g is 9 .81 and delta h is given as 10.

00:24 So del p will be equals to 9, 98 ,000.

00:31 0 .9 newton per meter square or we can write it as 98 98 kilo newton per meter square per meter square now we will calculate we will calculate the power power so power of pump p can be given as q multiplied by del p so q is given as 0 .075 meter cube per second multiplied by del p is 98 kilo newton per meter square so it will come out to be 7 .35 kilo -neuton meter per second or we can write it as 7 .35 kilowatt now calculate the power required by using the equation we will calculate the required using the equation so p required p required is equal to p divided by efficiency of pump multiplied by mechanical efficiency so from here we get p is 7 .35 divided by n p is 85 which is equal to 0 .85 multiplied by nm is 85 which is also equal to 0 .85 so from here we get p required is equal to 10 .17 kilowatt 10 .17 kilowatt now we will calculate the daily cost to operate in the form 10 by using so daily cost daily cost calculation c will be equals to $14 dollar 14 dollar 14 dollar per kilo kilowatt divided by 100 multiplied by kilowatt hour multiplied by 10 .17 kilowatt.

02:44 So it will come out to be $1 .4 .2 .1 .42 dollar per hour.

02:55 We'll be $1 .42 dollars.

02:57 Let's convert it into per day...

Please give Ace some feedback