/*! 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 65 Saturated propane vapor at \(2.0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 65

Saturated propane vapor at \(2.00 \times 10^{2}\) psia is fed to a well- insulated heat exchanger at a rate of \(3.00 \times 10^{3} \mathrm{SCFH}\) (standard cubic feet per hour). The propane leaves the exchanger as a saturated liquid (i.e., a liquid at its boiling point) at the same pressure. Cooling water enters the exchanger at \(70^{\circ} \mathrm{F},\) flowing cocurrently (in the same direction) with the propane. The temperature difference between the outlet streams (liquid propane and water) is \(15^{\circ} \mathrm{F}\). (a) What is the outlet temperature of the water stream? (Use the Antoine equation.) Is the outlet water temperature less than or greater than the outlet propane temperature? Briefly explain. (b) Estimate the rate (Btu/h) at which heat must be transferred from the propane to the water in the heat exchanger and the required flow rate \(\left(1 \mathrm{b}_{\mathrm{m}} / \mathrm{h}\right)\) of the water. (You will need to write two separate energy balances.) Assume the heat capacity of liquid water is constant at \(1.00 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and neglect heat losses to the outside and the effects of pressure on the heat of vaporization of propane.

Short Answer

Expert verified
The outlet temperature of the water stream is \(85^{\circ} F\), which is greater than the outlet propane temperature indicating heat transfer from propane to water. The heat transfer rate and water flow rate need to be calculated step by step as per the solution.

Step by step solution

01

Determine the boiling point of propane

Using the Antoine equation and the given pressure, the boiling point of propane is obtained. The Antoine equation relates the vapour pressure and temperature for a pure substance. The constants for propane (found in thermodynamics book) can be used to solve the equation. This temperature is the outlet temperature for the propane as it's a saturated liquid.
02

Determine the outlet temperature of the water stream

The outlet temperature of the water stream is determined by adding the temperature difference between the outlet streams to the temperature at which the water enters the exchanger. Therefore, outlet water temperature = inlet water temperature + temperature difference = \(70^{\circ} F + 15^{\circ} F = 85^{\circ} F\).
03

Compare the outlet water and propane temperatures

The outlet water temperature is compared to the outlet propane temperature. If the water temperature is higher it means that heat transfer has occurred from the propane to the water.
04

Determine the heat transfer rate

The heat transfer rate is determined by the equation: Q = m * Cp * (T_out - T_in), where m is the mass flow rate of propane, Cp is the heat capacity of propane, T_out is the outlet temperature, and T_in is the inlet temperature. The mass flow rate of propane is given in SCFH and should be converted to lbm/h using conversion factors. Assume the Cp of propane to be near that of saturated liquid and can be found in thermodynamics book. The outlet and inlet temperatures are in \(^{\circ} F\).
05

Estimate the required water flow rate

A separate energy balance is written for the cooling water. Assuming its heat capacity is constant at 1 Btu / (lbm * \(^{\circ} F\)), the flow rate is the unknown. The formula used is: m_water = Q / (Cp_water * (T_out - T_in)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one substance to another. In this exercise, heat is transferred from propane to water in a heat exchanger.
Propane enters as saturated vapor and leaves as saturated liquid, implying it has released its latent heat to the water.
  • Heat transfer can occur through conduction, convection, or radiation.
  • In our context, convection is most common as propane and water flow along the heat exchanger.
To calculate heat transfer, you'll often use equations involving the mass flow rate, specific heat capacities, and temperature differences. A well-understood concept here ensures accurate control and optimization in systems like HVAC or energy generation.
Energy Balance
An energy balance involves setting up equations to account for energy entering and exiting a system, ensuring conservation of energy according to the first law of thermodynamics.
For the heat exchanger in this problem, you need to consider separate energy balances for propane and water.
  • For propane, you'll calculate the energy it releases when condensing from vapor to liquid.
  • For the water, you'll calculate the energy it gains as it absorbs the heat from propane.
The energy lost by propane must equal the energy gained by water, considering the heat exchanger is well-insulated and there are no losses to the surroundings.
Phase Change
Phase change refers to the transition of a substance from one phase to another, like from gas to liquid, which happens to propane in this scenario.
When propane undergoes phase change, it goes from vapor to liquid as it cools at constant pressure.
  • During this process, propane releases latent heat, which is absorbed by the cooling water.
  • This phase change occurs at a specific temperature known as the boiling point, which stays constant for a pure substance at a given pressure.
Understanding phase change is crucial, as it involves large amounts of energy transfer, indicating systems like heat exchangers are optimized for energy efficiency.
Antoine Equation
The Antoine equation is crucial in determining the relationship between vapor pressure and temperature for substances like propane.
This equation helps determine the boiling point at a given pressure, a key part of solving the problem.
  • The Antoine equation is expressed as: \( \log_{10}(P) = A - \frac{B}{C+T} \), where \( P \) is the vapor pressure, \( T \) the temperature, and \( A, B, C \) are substance-specific constants.
  • For propane, these constants can be found in thermodynamics tables.
Using this equation helps you accurately calculate temperatures for phase changes, ensuring precise energy management in thermal systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A stream of air at \(500^{\circ} \mathrm{C}\) and 835 torr with a dew point of \(30^{\circ} \mathrm{C}\) flowing at a rate of \(1515 \mathrm{L} / \mathrm{s}\) is to be cooled in a spray cooler. A fine mist of liquid water at \(15^{\circ} \mathrm{C}\) is sprayed into the hot air at a rate of \(110.0 \mathrm{g} / \mathrm{s}\) and evaporates completely. The cooled air emerges at \(1 \mathrm{atm}\) (a) Calculate the final temperature of the emerging air stream, assuming that the process is adiabatic. (Suggestion: Derive expressions for the enthalpies of dry air and water at the outlet air temperature, substitute them into the energy balance, and use a spreadsheet to solve the resulting fourth-order polynomial equation.) (b) At what rate (kW) is heat transferred from the hot air feed stream in the spray cooler? What becomes of this heat? (c) In a few sentences, explain how this process works in terms that a high school senior could understand. Incorporate the results of Parts (a) and (b) in your explanation.

A fuel gas containing 95 mole\% methane and the balance ethane is burned completely with 25\% excess air. The stack gas leaves the furnace at \(900^{\circ} \mathrm{C}\) and is cooled to \(450^{\circ} \mathrm{C}\) in a waste- heat boiler, a heat exchanger in which heat lost by cooling gases is used to produce steam from liquid water for heating, power generation, or process applications. (a) Taking as a basis of calculation 100 mol of the fuel gas fed to the fumace, calculate the amount of heat (kJ) that must be transferred from the gas in the waste heat boiler to accomplish the indicated cooling. (b) How much saturated steam at 50 bar can be produced from boiler feedwater at \(40^{\circ} \mathrm{C}\) for the same basis of calculation? (Assume all the heat transferred from the gas goes into the steam production.) (c) At what rate ( \(k\) mol/s) must fuel gas be burned to produce 1280 kg steam per hour (an amount required elsewhere in the plant) in the waste heat boiler? What is the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{s}\right)\) of the gas leaving the boiler? (d) Briefly explain how the waste-heat boiler contributes to the plant profitability. (Think about what would be required in its absence.)

A sheet of cellulose acetate film containing 5.00 wt\% liquid acetone enters an adiabatic dryer where \(90 \%\) of the acetone evaporates into a stream of dry air flowing over the film. The film enters the dryer at \(T_{\mathrm{f} 1}=35^{\circ} \mathrm{C}\) and leaves at \(T_{\mathrm{f} 2}\left(^{\circ} \mathrm{C}\right) .\) The air enters the dryer at \(T_{\mathrm{al}}\left(^{\circ} \mathrm{C}\right)\) and 1.01 atm and exits the dryer at \(T_{\mathrm{a} 2}=49^{\circ} \mathrm{C}\) and 1 atm with a relative saturation of \(40 \% . C_{p}\) may be taken to be \(1.33 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) for dry film and \(0.129 \mathrm{kJ} /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\) for liquid acetone. Make a reasonable assumption regarding the heat capacity of dry air. The heat of vaporization of acetone may be considered independent of temperature. Take a basis of \(100 \mathrm{kg}\) film fed to the dryer for the requested calculations. (a) Estimate the feed ratio [liters dry air (STP)/kg dry film]. (b) Derive an expression for \(T_{\mathrm{al}}\) in terms of the film temperature change, \(\left(T_{\mathrm{f} 2}-35\right),\) and use it to answer Parts (c) and (d). (c) Calculate the film temperature change if the inlet air temperature is \(120^{\circ} \mathrm{C}\). (d) Calculate the required value of \(T_{\mathrm{al}}\) if the film temperature falls to \(34^{\circ} \mathrm{C},\) and the value if it rises to \(36^{\circ} \mathrm{C}.\) (e) If you solved Parts (c) and (d) correctly, you found that even though the air temperature is consistently higher than the film temperature in the dryer, so that heat is always transferred from the air to the film, the film temperature can drop from the inlet to the outlet. How is this possible?

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton's rule and Chen's rule and compare the results with a tabulated value of this quantity. Calculate the percentage error that results from using each estimation. Then estimate \(\Delta \hat{H}_{\mathrm{v}}\) at \(100^{\circ} \mathrm{C}\) using Watson's correlation.

The heat capacity at constant pressure of hydrogen cyanide is given by the expression $$ C_{p}\left[J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=35.3+0.0291 T\left(^{\circ} \mathrm{C}\right) $$ (a) Write an expression for the heat capacity at constant volume for HCN, assuming ideal-gas behavior. (b) Calculate \(\Delta \hat{H}(\mathrm{J} / \mathrm{mol})\) for the constant- pressure process $$ \mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) $$(c) Calculate \(\Delta \hat{U}(\mathrm{J} / \mathrm{mol})\) for the constant- volume process $$\mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right)$$ (d) If the process of Part (b) were carried out in such a way that the initial and final pressures were each 0.80 atm but the pressure varied during the heating, the value of \(\Delta \hat{H}\) would still be what you calculated assuming a constant pressure. Why is this so?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.