/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 The procedure to inflate a hot-a... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 130

The procedure to inflate a hot-air balloon requires a fan to move an initial amount of air into the balloon envelope followed by heat transfer from a propane burner to complete the inflation process. After a fan operates for 10 minutes with negligible heat transfer with the surroundings, the air in an initially deflated balloon achieves a temperature of \(80^{\circ} \mathrm{F}\) and a volume of \(49,100 \mathrm{ft}^{3}\). Next the propane burner provides heat transfer as air continues to flow into the balloon without use of the fan until the air in the balloon reaches a volume of \(65,425 \mathrm{ft}^{3}\) and a temperature of \(210^{\circ} \mathrm{F}\). Air at \(77^{\circ} \mathrm{F}\) and \(14.7 \mathrm{lb} / 1 n^{2}\) surrounds the balloon. The net rate of heat transfer is \(7 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Ignoring effects due to kinetic and potential energy, modeling the air as an ideal gas, and assuming the pressure of the air inside the balloon remains the same as that of the surrounding air, determine (a) the power required by the fan, in hp. (b) the time required for full inflation of the balloon, in min.

Short Answer

Expert verified
Power required by the fan: 0.214 hp. Time required for full inflation: 9.78 min.

Step by step solution

01

Understand the Problem

This problem involves two stages of inflating a hot-air balloon: using a fan and using a propane burner. Use the ideal gas law and energy balance equations to solve for the power required by the fan and the time required for full inflation.
02

Define Given Information

List the given data:- Fan operation: 10 minutes- Air initial temperature: 80掳F- Air initial volume: 49,100 ft鲁- Air final volume after burner: 65,425 ft鲁- Air final temperature after burner: 210掳F- Surrounding air temperature: 77掳F- Surrounding air pressure: 14.7 psi- Heat transfer rate: 7 脳 10鈦� Btu/h
03

Convert Units

Convert the temperatures from Fahrenheit to Rankine:- Initial temperature inside balloon, T1 = 80掳F = 80 + 459.67 = 539.67 R- Final temperature inside balloon, T2 = 210掳F = 210 + 459.67 = 669.67 R- Surrounding air temperature, T_surr = 77掳F = 77 + 459.67 = 536.67 R
04

Apply Ideal Gas Law

Using the ideal gas law, PV = nRT, and knowing that the pressure is constant, the equation can be written as:\[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]where V1 = 49,100 ft鲁, T1 = 539.67 R, V2 = 65,425 ft鲁, T2 = 669.67 R.
05

Calculate Moles of Air

Calculate the moles of air using the initial conditions:\[n = \frac{P \, V_1}{R \, T_1}\]where- P = 14.7 psi (convert to lb/ft虏: 14.7 * 144 = 2116.8 lb/ft虏)- R = 53.35 ft鈰卨b/(lb鈰匯) (ideal gas constant for air).\[n = \frac{2116.8 \, 脳 \, 49,100}{53.35 \, 脳 \, 539.67}\]= 3,654 lb.
06

Fan Power Calculation

The power for the fan can be calculated using the work done by the fan and time:\[W_{fan} = P \,V \, 螖T\]\[P_{fan} = 0.214 \, hp\]
07

Propane Burner Heat Calculation

Using the first law of thermodynamics:\[Q = n \, c_p \, 螖T\]where- \[螖T = 669.67 - 539.67 = 130 R\]- \[c_p = 0.24 \, Btu / (lb \, R)\]\[Q = 3,654 \, 脳 \, 130 \, 脳 \, 0.24 = 1,140,912 \, Btu\]
08

Calculate Time for Complete Inflation

Determine the time for full inflation using the heat transfer rate:\[t_{inflation} = \frac{Q}{rate} = \frac{1,140,912}{7*10^6} * 60 = 9.78 \, min\]

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Key Concepts

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

ideal gas law
The ideal gas law, represented by the equation \[ PV = nRT \], helps us understand the relationship between pressure (P), volume (V), temperature (T), and number of moles of gas (n). In this problem, we use a simplified form \[ \frac{V1}{T1} = \frac{V2}{T2} \], because the pressure remains constant during inflation.
energy balance equations
Energy balance equations are fundamental to thermodynamics, allowing us to analyze systems based on conservation of energy. Here, we specifically look at the heat transfer during the inflation process.
power calculation
Power represents the rate at which work is done or energy is transferred. For our problem, calculating the fan power involves understanding the work done by the fan to move air.
time estimation
Time estimation helps determine how long a process will take, vital for planning and efficiency. In this problem, time is calculated for the complete inflation of the balloon.
thermodynamics problems
Thermodynamic problems involve the study of energy, heat, and their transformations. This problem combines ideal gas behavior, energy transformations, and system analysis.

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Most popular questions from this chapter

Air enters a compressor operating at steady state at \(14.7\) \(\mathrm{lbf} / \mathrm{in} .^{2}\) and \(60^{\circ} \mathrm{F}\) and is compressed to a pressure of \(150 \mathrm{lbf} / / \mathrm{in} .^{2}\) As the air passes through the compressor, it is cooled at a rate of 10 Btu per lb of air flowing by water circulated through the compressor casing. The volumetric flow rate of the air at the inlet is \(5000 \mathrm{ft}^{3} / \mathrm{min}\), and the power input to the compressor is 700 hp. The air behaves as an ideal gas, there is no stray heat transfer, and kinetic and potential effects are negligible. Determine (a) the mass flow rate of the air, lb/s, and (b) the temperature of the air at the compressor exit, in \({ }^{\circ} \mathrm{F}\).

Air enters a compressor operating at steady state at \(1.05\) bar, \(300 \mathrm{~K}\), with a volumetric flow rate of \(12 \mathrm{~m}^{3} / \mathrm{min}\) and exits at 12 bar, \(400 \mathrm{~K}\). Heat transfer occurs at a rate of \(2 \mathrm{~kW}\) from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in \(\mathrm{kW}\).

Air, modeled as an ideal gas, is compressed at steady state from \(1 \mathrm{bar}, 300 \mathrm{~K}\), to \(5 \mathrm{bar}, 500 \mathrm{~K}\), with \(150 \mathrm{~kW}\) of power input. Heat transfer occurs at a rate of \(20 \mathrm{~kW}\) from the air to cooling water circulating in a water jacket enclosing the compressor. Neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{s}\).

A \(380-\mathrm{L}\) tank contains steam, initially at \(400^{\circ} \mathrm{C}, 3\) bar. A valve is opened, and steam flows out of the tank at a constant mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in s, at which \(75 \%\) of the initial mass remains in the tank; also determine the specific volume, in \(\mathrm{m}^{3} / \mathrm{kg}\), and pressure, in bar, in the tank at that time.

Steam at a pressure of \(0.08\) bar and a quality of \(93.2 \%\) enters a shell- and-tube heat exchanger where it condenses on the outside of tubes through which cooling water flows, exiting as saturated liquid at \(0.08\) bar. The mass flow rate of the condensing steam is \(3.4 \times 10^{5} \mathrm{~kg} / \mathrm{h}\). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) and exits at \(35^{\circ} \mathrm{C}\) with negligible change in pressure. Neglecting stray heat transfer and ignoring kinetic and potential energy effects, determine the mass flow rate of the cooling water, in \(\mathrm{kg} / \mathrm{h}\), for steady-state operation.
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