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Chapter 8: Problem 20

A stream of air at \(77^{\circ} \mathrm{F}\) and 1.2 atm absolute flowing at a rate of \(225 \mathrm{ft}^{3} / \mathrm{h}\) is blown through ducts that pass through the interior of a large industrial motor. The air emerges at \(500^{\circ} \mathrm{F}\). Calculate the rate at which the air is removing heat generated by the motor. What assumption have you made about the pressure dependence of the specific enthalpy of air?

Short Answer

Expert verified
The rate at which the air is removing heat generated by the motor is 1713 Btu/h. The assumption made about the pressure dependence of the specific enthalpy of air is that it is negligible at these pressures and temperatures.

Step by step solution

01

- Conversion of temperatures to Rankine

To make the computations, convert all the temperatures from Fahrenheit to Rankine. Utilize the conversion formula \( R = F + 459.67 \). Hence, the inlet temperature is \(77^{\circ}F + 459.67 = 536.67^{\circ}R\) and the outlet temperature is \(500^{\circ}F + 459.67 = 959.67^{\circ}R\).
02

- Calculation of enthalpy

To calculate the enthalpy, use the formula \( \Delta h = cp \Delta T \), where \(\Delta h\) is the change in enthalpy, \(cp\) is the specific heat capacity at constant pressure, and \(\Delta T\) is the difference in temperature. For standard air, \(cp\) is approximately 0.24 Btu/lb掳F. Therefore, \(\Delta h = 0.24 (959.67 - 536.67) = 101.52 Btu/lb\).
03

- Determining the flow rate

The properties of standard air at 60掳F and 14.696 psia can be found in standard tables. The density of air is \(0.075 lb/ft^3\), therefore the mass flow rate can be calculated as \(225 ft^3/h * 0.075 lb/ft^3 = 16.875 lb/h\).
04

- Calculation of heat removal rate

Calculate the heat removal rate using the formula \( Q = \dot{m} \Delta h \), where \( Q \) is the heat removal rate and \( \dot{m} \) is the mass flow rate. This gives you \( Q = 16.875 lb/h * 101.52 Btu/lb = 1713 Btu/h \).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property of a material that tells you how much heat energy is needed to change its temperature. For air, the specific heat capacity is the amount of heat required to raise the temperature of one pound of air by one degree Fahrenheit when pressure remains constant.
In our enthalpy calculations, we denote specific heat capacity with the symbol \( c_p \). This is because we're assuming pressure remains constant while air flows through the ducts. The value for \( c_p \) for air, at standard conditions, is roughly 0.24 Btu/lb掳F.
Understanding specific heat capacity is crucial because it is part of the formula used to calculate the change in enthalpy, \( \Delta h = c_p \Delta T \). This tells us how much heat is absorbed or released as the air's temperature changes as a result of the motor's heat.
  • Is material-specific
  • Helps compute energy changes
  • Involves in constant pressure conditions
Temperature Conversion
Temperature conversions are vital when working with thermodynamic calculations, especially in contexts like enthalpy calculations that require consistent units.
In the given exercise, we initially have temperatures in degrees Fahrenheit and we need to convert these to Rankine for our calculations. This is because Rankine is the absolute temperature scale used in this kind of thermodynamic problem involving British Thermal Units (BTU).
We use the conversion formula \( R = F + 459.67 \) to change Fahrenheit to Rankine. This transition standardizes the temperature values and allows for the direct application of enthalpy formulas.
  • Converts Fahrenheit (掳F) to Rankine (掳R).
  • Ensures compatibility with the BTU unit of energy.
  • Simplifies thermodynamic equations.
Heat Transfer
Heat transfer dictates how thermal energy moves from one place to another. In our scenario, the stream of air moves heat from the motor's interior through the ducts, cooling the motor.
The calculation for heat transfer relies on mass flow rate and specific enthalpy change. It follows the formula \( Q = \dot{m} \Delta h \). Here, \( Q \) is the rate of heat removal, \( \dot{m} \) represents the mass flow rate of the air, and \( \Delta h \) is the change in enthalpy.
This setup assumes the motor's heat is being transferred to the air efficiently, indicated by consistent pressure (hence we use specific heat at constant pressure). Through this assumption, enthalpy is considered only dependent on temperature change. It's also important to note that efficient heat transfer ensures the motor works optimally without overheating.
  • Involves moving heat energy efficiently.
  • Calculated using mass flow and enthalpy change.
  • Requires assumptions about pressure dependency.

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Most popular questions from this chapter

A mixture of \(n\) -hexane vapor and air leaves a solvent recovery unit and flows through a \(70-\mathrm{cm}\) diameter duct at a velocity of \(3.00 \mathrm{m} / \mathrm{s}\). At a sampling point in the duct the temperature is \(40^{\circ} \mathrm{C}\), the pressure is \(850 \mathrm{mm}\) Hg, and the dew point of the sampled gas is \(25^{\circ} \mathrm{C}\). The gas is fed to a condenser in which it is cooled at constant pressure, condensing \(70 \%\) of the hexane in the feed. (a) Perform a degree-of-freedom analysis to show that enough information is available to calculate the required condenser outlet temperature \(\left(^{\circ} \mathrm{C}\right)\) and cooling rate \((\mathrm{kW})\) (b) Perform the calculations. (c) If the feed duct diameter were \(35 \mathrm{cm}\) for the same molar flow rate of the feed gas, what would be the average gas velocity (volumetric flow rate divided by cross-sectional area)? (d) Suppose you wanted to increase the percentage condensation of hexane for the same feed stream. Which three condenser operating variables might you change, and in which direction?

The heat required to raise the temperature of \(m\) (kg) of a liquid from \(T_{1}\) to \(T_{2}\) at constant pressure is $$ Q=\Delta H=m \int_{T_{1}}^{T_{2}} C_{p}(T) d T $$ In high school and in first-year college physics courses, the formula is usually given as $$ Q=m C_{p} \Delta T=m C_{p}\left(T_{2}-T_{1}\right) $$ (a) What assumption about \(C_{p}\) is required to go from Equation 1 to Equation \(2 ?\) (b) The heat capacity \(\left(C_{p}\right)\) of liquid \(n\) -hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well- insulated vessel containing \(2.00 \mathrm{L}\) of liquid \(n-\mathrm{C}_{6} \mathrm{H}_{14}\) at \(T=300 \mathrm{K} .\) A combustion reaction known to release \(16.73 \mathrm{kJ}\) of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be \(3.10 \mathrm{K}\). In a separate experiment, it is found that \(6.14 \mathrm{kJ}\) of heat is required to raise the temperature of everything in the system except the hexane by \(3.10 \mathrm{K}\). Use these data to estimate \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) for liquid \(n\) -hexane at \(T \approx 300 \mathrm{K},\) assuming that the condition required for the validity of Equation 2 is satisfied. Compare your result with a tabulated value.

Propane gas enters a continuous adiabatic heat exchanger \(^{17}\) at \(40^{\circ} \mathrm{C}\) and \(250 \mathrm{kPa}\) and exits at \(240^{\circ} \mathrm{C}\). Superheated steam at \(300^{\circ} \mathrm{C}\) and 5.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure. (a) Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed \(\left(\mathrm{m}^{3}\right),\) the mass of steam fed \((\mathrm{kg}),\) and the volume of steam fed \(\left(\mathrm{m}^{3}\right)\) (b) Calculate values of the labeled specific enthalpies in the following inlet-outlet enthalpy table for this process. $$\begin{array}{|l|cc|cc|} \hline \text { Species } & n_{\text {in }} & \hat{H}_{\text {in }} & n_{\text {out }} & \hat{H}_{\text {out }} \\ \hline \mathrm{C}_{3} \mathrm{H}_{8} & 100 \mathrm{mol} & \hat{H}_{\mathrm{a}}(\mathrm{kJ} / \mathrm{mol}) & 100 \mathrm{mol} & \hat{H}_{\mathrm{c}}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{H}_{2} \mathrm{O} & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{b}}(\mathrm{kJ} / \mathrm{kg}) & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{d}}(\mathrm{kJ} / \mathrm{kg}) \\ \hline \end{array}$$ (c) Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams ( \(\mathrm{m}^{3}\) steam fed \(/ \mathrm{m}^{3}\) propane fed). Assume ideal-gas behavior for the propane but not the steam and recall that the exchanger is adiabatic. (d) Calculate the heat transferred from the water to the propane ( \(k J / m^{3}\) propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.) (e) Over a period of time, scale builds up on the heat-transfer surface, resulting in a lower rate of heat transfer between the propane and the steam. What changes in the outlet streams would you expect to see as a result of the decreased heat transfer?

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?

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.
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