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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

A Thomas flowmeter is a device in which heat is transferred at a measured rate from an electric coil to a flowing fluid, and the flow rate of the stream is calculated from the measured increase of the fluid temperature. Suppose a device of this sort is inserted in a stream of nitrogen, the current through the heating coil is adjusted until the wattmeter reads \(1.25 \mathrm{kW},\) and the stream temperature goes from \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) before the heater to \(34^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) after the heater. (a) If the specific enthalpy of nitrogen is given by the formula \(\hat{H}(\mathrm{kJ} / \mathrm{kg})=1.04\left[T\left(^{\circ} \mathrm{C}\right)-25\right]\) what is the volumetric flow rate of the gas (L/s) upstream of the heater (i.e., at \(30^{\circ} \mathrm{C}\) and \(110 \mathrm{kPa}\) )? (b) List several assumptions made in the calculation of Part (a) that could lead to errors in the calculated flow rate.

Water is to be pumped from a lake to a ranger station on the side of a mountain (see figure). The length of pipe immersed in the lake is negligible compared to the length from the lake surface to the discharge point. The flow rate is to be \(95 \mathrm{gal} / \mathrm{min}\), and the flow channel is a standard 1-inch. Schedule 40 steel pipe (ID \(=1.049\) inch). A pump capable of delivering \(8 \mathrm{hp}\left(=\dot{W}_{\mathrm{s}}\right)\) is available. The friction loss \(\tilde{F}\left(\mathrm{ft} \cdot \mathrm{lb}_{\mathrm{f}} / \mathrm{lb}_{\mathrm{m}}\right)\) equals \(0.041 L,\) where \(L(\mathrm{ft})\) is the length of the pipe. (a) Calculate the maximum elevation, \(z\), of the ranger station above the lake if the pipe rises at an angle of \(30^{\circ}\) (b) Suppose the pipe inlet is immersed to a significantly greater depth below the surface of the lake, but it discharges at the elevation calculated in Part (a). The pressure at the pipe inlet would be greater than it was at the original immersion depth, which means that \(\Delta P\) from inlet to outlet would be greater, which in turn suggests that a smaller pump would be sufficient to move the water to the same elevation. In fact, however, a larger pump would be needed. Explain (i) why the pressure at the inlet would be greater than in Part (a), and (ii) why a larger pump would be needed.

A fuel oil is burned with air in a boiler furnace. The combustion produces \(813 \mathrm{kW}\) of thermal energy, of which \(65 \%\) is transferred as heat to boiler tubes that pass through the furnace. The combustion products pass from the furnace to a stack at \(550^{\circ} \mathrm{C}\). Water enters the boiler tubes as a liquid at \(30^{\circ} \mathrm{C}\) and leaves the tubes as saturated steam at 20 bar absolute. (a) Calculate the rate ( \(\mathrm{kg} / \mathrm{h}\) ) at which steam is produced. (b) Use the steam tables to estimate the volumetric flow rate of the steam produced. (c) Repeat the calculation of Part (b), only assume ideal-gas behavior instead of using the steam tables. Would you have more confidence in the estimate of Part (b) or Part (c)? Explain. (d) What happened to the \(35 \%\) of the thermal energy released by the combustion that did not go to produce the steam?

Liquid water at 60 bar and \(250^{\circ} \mathrm{C}\) passes through an adiabatic expansion valve, emerging at a pressure \(P_{\mathrm{f}}\) and temperature \(T_{\mathrm{f}} .\) If \(P_{\mathrm{f}}\) is low enough, some of the liquid evaporates. (a) If \(P_{\mathrm{f}}=1.0\) bar, determine the temperature of the final mixture \(\left(T_{\mathrm{f}}\right)\) and the fraction of the liquid feed that evaporates \(\left(y_{\mathrm{v}}\right)\) by writing an energy balance about the valve and neglecting \(\Delta \dot{E}_{\mathrm{k}}\) (b) If you took \(\Delta \dot{E}_{\mathrm{k}}\) into account in Part (a), how would the calculated outlet temperature compare with the value you determined? What about the calculated value of \(y_{\mathrm{v}} ?\) Explain. (c) What is the value of \(P_{\mathrm{f}}\) above which no evaporation would occur? (d) Sketch the shapes of plots of \(T_{\mathrm{f}}\) versus \(P_{\mathrm{f}}\) and \(y_{\mathrm{v}}\) versus \(P_{\mathrm{f}}\) for 1 bar \(\leq P_{\mathrm{f}} \leq 60\) bar. Briefly explain your reasoning.

Write and simplify the closed-system energy balance (Equation \(7.3-4\) ) for each of the following processes, and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. The solution of Part (a) is given as an illustration. (a) The contents of a closed flask are heated from \(25^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). (b) A tray filled with water at \(20^{\circ} \mathrm{C}\) is put into a freezer. The water tums into ice at \(-5^{\circ} \mathrm{C}\). (Note: When a substance expandsit does work on its surroundings and when it contracts the surroundings do work on it.) (c) A chemical reaction takes place in a closed adiabatic (perfectly insulated) rigid container. (d) Repeat Part (c), only suppose that the reactor is isothermal rather than adiabatic and that when the reaction was carried out adiabatically, the temperature in the reactor increased.
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