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Chapter 7: Problem 5

Liquid ethanol is pumped from a large storage tank through a 1 -inch ID pipe at a rate of 3.00 gal/min. (a) At what rate in (i) \(\mathrm{ft} \cdot \mathrm{lb}_{\mathrm{f}} / \mathrm{s}\) and (ii) hp is kinetic energy being transported by the ethanol in the pipe? (b) The electrical power input to the pump transporting the ethanol must be greater than the amount you calculated in Part (a). What would you guess becomes of the additional energy? (There are several possible answers.)

Short Answer

Expert verified
The ethanol transports the kinetic energy through the pipe at a rate of \(x ft路lb_f/s\) and \(y hp\). Extra electrical input power that exceeds the kinetic energy being transported by the ethanol could become thermal energy due to inefficiencies and lost due to friction which causes heating, vibration, and sound. It might also store potential energy if the ethanol is pumped upwards or used to overcome the pressure drop in the pipe.

Step by step solution

01

Convert Flow Rate

First, convert the flow rate from gallons per minute to cubic feet per second. The formula to convert gallons per minute (gpm) to cubic feet per second (ft鲁/s) is \[Q= (flow rate in gal/min) 脳 (0.004329004329 ft鲁/gal)/ (60 s/min)\]
02

Find Velocity

Next, find the velocity of the liquid using the formula \[ V = Q/A\] where A is the cross-sectional area of the pipe. The cross-sectional area of a 1-inch internal diameter (ID) pipe can be found from the formula for the area of a circle, \(A=\pi r^2\), where r is the radius of the pipe = diameter/2.
03

Calculate Kinetic Energy

Next, calculate the kinetic energy being transported by ethanol using the formula for kinetic energy \(KE=\frac{1}{2}\rho V^2\), where \(\rho\) is the density of ethanol (approximately \(62.37 lb_m/ft^3\) and \(V\) is the velocity of ethanol. Since the exercise requires the kinetic energy in foot-pounds force per second, multiply the resulting kinetic energy by \(g = 32.174 ft/s^2\)
04

Convert to Horsepower

Now, convert the kinetic energy from foot-pounds force per second to horsepower using the formula \(1 hp = 550 ft路lbf/s\). This gives the kinetic energy being transported by the ethanol in the pipe in horsepower.
05

Guess What Happens to Extra Power Input

Finally, if the electrical power input to the pump is greater than the amount of kinetic energy being transported by the ethanol in the pipe, the additional input power might be converted to thermal energy due to the inefficiencies of the pump and may also be lost due to the friction between the moving fluid and the pipe which causes heating, vibration, and sound. It may also be storing potential energy if the ethanol is being pumped upwards or is used to overcome the pressure drop in the pipe.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flow Rate Conversion
In chemical engineering, accurately converting flow rates is essential for understanding fluid dynamics in systems. One common conversion is from gallons per minute (gpm) to cubic feet per second (ft鲁/s). This exercise involves pumping liquid ethanol through a pipe. To convert the flow rate from gpm to ft鲁/s, use the formula:
  • Multiply the flow rate by 0.004329004329 to convert gallons to cubic feet.
  • Divide by 60 to change the time unit from minutes to seconds.
Given the flow rate of 3.00 gpm, the calculation becomes: \[ Q = 3.00 \times 0.004329004329 \div 60 \approx 0.0002165 \text{ ft}^3/ ext{s} \]Now, you have the flow rate in a more useful unit for kinetic energy calculations. Understanding and performing this conversion accurately ensures that subsequent calculations for properties like velocity and energy will be correct.
Kinetic Energy Calculation
Kinetic energy in fluid systems is an important parameter because it reflects the energy that is being transported by the fluid. To calculate the kinetic energy of ethanol being transported in the pipe, we first need to determine the velocity. This is done by dividing the flow rate by the cross-sectional area of the pipe: \[ V = \frac{Q}{A} \]For a 1-inch diameter pipe, convert the diameter to feet (1 inch = 1/12 feet). The radius is half of the diameter, so:
  • Calculate the radius as 0.5/12 feet.
  • Calculate the area using the formula for area of a circle: \( A = \pi r^2 \).
After finding the area, use it to calculate the velocity. Once velocity, \( V \), and density, \( \rho \) (approximately 62.37 lbm/ft鲁 for ethanol), are known, plug them into the formula for kinetic energy: \[ KE = \frac{1}{2} \rho V^2 \times g \]Here, \( g \) is the acceleration due to gravity (32.174 ft/s虏), ensuring that kinetic energy is measured in ft路lbf/s. Useful note: Converting kinetic energy to horsepower (hp) involves dividing by 550 since 1 hp equals 550 ft路lbf/s.
Energy Efficiency in Pumps
Understanding the energy efficiency of pumps is crucial in chemical engineering, as pumps consume a significant amount of power in process systems. While calculating the kinetic energy of ethanol, we noticed that the pump's electrical power input exceeds the energy being transported. This additional power could transform into several other forms, such as:
  • Thermal energy due to electrical inefficiencies.
  • Heat generation from friction in the pipes.
  • Vibration and noise resulting from operational dynamics.
  • Potential energy if the fluid is being pumped to a higher elevation.
Recognizing these transformations helps in identifying losses and optimizing pump performance. By improving efficiency, we can achieve significant energy and cost savings in fluid transport systems.

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Most popular questions from this chapter

You recently purchased a large plot of land in the Amazon jungle at an extremely low cost. You are quite pleased with yourself until you arrive there and find that the nearest source of electricity is 1500 miles away, a fact that your brother-in-law, the real estate agent, somehow forgot to mention. since the local hardware store does not carry 1500 -mile-long extension cords, you decide to build a small hydroelectric generator under a 75-m high waterfall located nearby. The flow rate of the waterfall is 10 \(^{5} \mathrm{m}^{3} / \mathrm{h}\), and you anticipate needing \(750 \mathrm{kW} \cdot \mathrm{h} / \mathrm{wk}\) to run your lights, air conditioner, and television. Calculate the maximum power theoretically available from the waterfall and see if it is sufficient to meet your needs.

Jets of high-speed steam are used in spray cleaning. Steam at 15.0 bar with \(150^{\circ} \mathrm{C}\) of superheat is fed to a well-insulated valve at a rate of \(1.00 \mathrm{kg} / \mathrm{s}\). As the steam passes through the valve, its pressure drops to 1.0 bar. The outlet stream may be totally vapor or a mixture of vapor and liquid. Kinetic and potential energy changes may be neglected. (a) Draw and label a flowchart, assuming that both liquid and vapor emerge from the valve. (b) Write an energy balance and use it to determine the total rate of flow of enthalpy in the outlet stream \(\left(\dot{H}_{\text {out }}=\dot{m}_{1} \hat{H}_{1}+\dot{m}_{v} \hat{H}_{v}\right) .\) Then determine whether the outlet stream is in fact a mixture of liquid and vapor or whether it is pure vapor. Explain your reasoning. (c) What is the temperature of the outlet stream? (d) Assuming that your answers to Parts (b) and (c) are correct and that the pipes at the inlet and outlet of the valve have the same inner diameter, would \(\Delta E_{\mathrm{k}}\) across the valve be positive, negative, or explain.

Liquid water at \(30.0^{\circ} \mathrm{C}\) and liquid water at \(90.0^{\circ} \mathrm{C}\) are combined in a ratio \((1 \mathrm{kg} \text { cold water/2 } \mathrm{kg}\) hot water). (a) Use a simple calculation to estimate the final water temperature. For this part, pretend you never heard of energy balances. (b) Now assume a basis of calculation and write a closed-system energy balance for the process, assuming that the mixing is adiabatic. Use the balance to calculate the specific internal energy and hence (from the steam tables) the final temperature of the mixture. What is the percentage difference between your answer and that of Part (a)?

Liquid water is fed to a boiler at \(24^{\circ} \mathrm{C}\) and 10 bar and is converted at constant pressure to saturated steam. (a) Use the steam tables to calculate \(\Delta \hat{H}(\mathrm{kJ} / \mathrm{kg})\) for this process, and then determine the heat input required to produce \(15,800 \mathrm{m}^{3} / \mathrm{h}\) of steam at the exit conditions. Assume the kinetic energy of the entering liquid is negligible and that steam is discharged through a 15 -cm ID pipe. (b) How would the calculated value of the heat input change if you did not neglect the kinetic energy of the inlet water and if the inner diameter of the steam discharge pipe were \(13 \mathrm{cm}\) (increase, decrease, stay the same, or no way to tell without more information)?

Your friend has asked you to help move a 60 inch \(\times 78\) inch mattress with a mass of 75 Ib \(_{\mathrm{m}}\). The two of you position it horizontally in an open flat-bed trailer that you hitch to your car. There is nothing available to tie the mattress to the trailer, but you know there is a risk of the mattress being lifted from the trailer by the air flowing over it and perform the following calculations: (a) Although the conditions do not exactly match those for which the Bemoulli equation is applicable, use the equation to get a rough estimate of how fast you can drive (miles/h) before the mattress is lifted. Assume the velocity of air above the mattress equals the velocity of the car, the pressure difference between the top and bottom of the mattress equals the weight of the mattress divided by the mattress cross-sectional area, and air has a constant density of \(0.075 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}\). What is your result? (b) You see that your friend also has several boxes of books. Since you would like to drive at 60 miles per hour, what weight of books ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) do you need to put on the mattress to hold it in place?
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