/*! 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 15 The air in the cylinder of an ai... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 15

The air in the cylinder of an air compressor is compressed from \(100 \mathrm{kPa}\) to \(10 \mathrm{MPa}\). Estimate the final temperature and the work required if the air is initially at \(100^{\circ} \mathrm{C}\). Assume constant specific heats. SOLUTION Since the process occurs quite fast, we assume an adiabatic quasi-equilibrium process. Then $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(373)\left(\frac{10000}{100}\right)^{(1.4-1) / 1.4}=1390 \mathrm{~K} \quad \text { or } \quad 1117^{\circ} \mathrm{C} $$ The work is found by using the first law with \(Q=0\) : $$ w=-\Delta u=-C_{v}\left(T_{2}-T_{1}\right)=-(0.717)(1390-373)=-729 \mathrm{~kJ} / \mathrm{kg} $$ The work per unit mass is calculated since the mass (or volume) was not specified.

Short Answer

Expert verified
Final temperature is approx. \( 1117^{\circ} C \) and work done is approx. \( -729 \mathrm{kJ/kg} \).

Step by step solution

01

Identify Initial and Final Conditions

Identify the initial and final pressures and temperatures. Initial pressure, \( P_1 = 100 \ \mathrm{kPa} \), final pressure, \( P_2 = 10 \ \mathrm{MPa} \), and initial temperature, \( T_1 = 100^{\circ}C = 373 \ \mathrm{K}. \)
02

Assume Adiabatic Process and Use Corresponding Equation

Since this is an adiabatic process based on the assumption given, use the equation for adiabatic processes: \[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] where \( k \) (specific heat ratio) for air is 1.4.
03

Calculate Final Temperature

Substitute the values into the adiabatic equation: \[ T_2 = 373 \left( \frac{10000}{100} \right)^{\frac{1.4-1}{1.4}} \] Simplify: \[ T_2 = 373 \times 100^{0.2857} \] Use a calculator to find \( T_2 \): \( T_2 鈮� 1390 \ \mathrm{K} = 1117^{\circ}C \).
04

Find Work Done Using First Law of Thermodynamics

Apply the first law of thermodynamics for an adiabatic process with \( Q = 0 \): \[ w = -\Delta u = -C_v (T_2 - T_1) \] where \( C_v = 0.717 \ \mathrm{kJ/(kg \cdot K)} \). Substitute the values: \[ w = -(0.717)(1390 - 373) \] Calculate the work: \( w 鈮� -729 \ \mathrm{kJ/kg} \).

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Key Concepts

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

temperature calculation
To find the final temperature in an adiabatic process, we use the relationship between temperature and pressure as given for ideal gases. For the initial temperature \( T_1 = 373 K \), and initial pressure \( P_1 = 100 \ kPa \), and final pressure \( P_2 = 10,000 \ kPa \), the formula becomes: \[ T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] . Substituting the values, we have: \[ T_2 = 373 \( \frac{10000}{100} \)^{0.2857} \] Simplifying further, \[ T_2 = 373 \times 100^{0.2857} \] Using a calculator, we find \( T_2 鈮� 1390 K \), which is also \( 1117^{\text{掳}C \). This step-by-step process shows the direct impact of pressure changes on temperature in an adiabatic process.

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Most popular questions from this chapter

Air travels through the \(4 \times 2 \mathrm{~m}\) test section of a wind tunnel at \(20 \mathrm{~m} / \mathrm{s}\). The gage pressure in the test section is measured to be \(-20 \mathrm{kPa}\) and the temperature \(20^{\circ} \mathrm{C}\). After the test section, a diffuser leads to a 6-m-diameter exit pipe. Estimate the velocity and temperature in the exit pipe. SOLUTION The energy equation (4.72) for air takes the form $$ V_{2}^{2}=V_{1}^{2}+2 C_{p}\left(T_{1}-T_{2}\right)=20^{2}+(2)(1.00)\left(293-T_{2}\right) $$ The continuity equation, \(\rho_{1} A_{1} V_{1}=\rho_{2} A_{2} V_{2}\), yields $$ \frac{P_{1}}{R T_{1}} A_{1} V_{1}=\rho_{2} A_{2} V_{2} \quad \therefore \rho_{2} V_{2}=\left[\frac{80}{(0.287)(293)}\right]\left[\frac{8}{\pi(3)^{2}}\right](20)=5.384 \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s} $$ The best approximation to the actual process is the adiabatic quasi- equilibrium process. Using (4.49), letting \(\rho=1 / v\), we have $$ \frac{T_{2}}{T_{1}}=\left(\frac{\rho_{2}}{\rho_{1}}\right)^{k-1} \quad \text { or } \quad \frac{T_{2}}{\rho_{2}^{0.4}}=\frac{293}{[80 /(0.287)(293)]^{0.4}}=298.9 $$ The above three equations include the three unknowns \(T_{2}, V_{2}\), and \(\rho_{2}\). Substitute for \(T_{2}\) and \(V_{2}\) back into the energy equation and find $$ \frac{5.384^{2}}{\rho_{2}^{2}}=20^{2}+(2)(1.00)\left[293-(298.9)\left(\rho_{2}^{0.4}\right)\right] $$ This can be solved by trial and error to yield \(\rho_{2}=3.475 \mathrm{~kg} / \mathrm{m}^{3}\). The velocity and temperature are then $$ V_{2}=\frac{5.384}{\rho_{2}}=\frac{5.384}{3.475}=1.55 \mathrm{~m} / \mathrm{s} \quad T_{2}=(298.9)\left(\rho_{2}^{0.4}\right)=(298.9)(3.475)^{0.4}=492 \mathrm{~K} \quad \text { or } \quad 219^{\circ} \mathrm{C} $$

Steam with a mass flux of \(600 \mathrm{lbm} / \mathrm{min}\) exits a turbine as saturated steam at 2 psia and passes through a condenser (a heat exchanger). What mass flux of cooling water is needed if the steam is to exit the condenser as saturated liquid and the cooling water is allowed a \(15^{\circ} \mathrm{F}\) temperature rise? SOLUTION The energy equations (4.75) are applicable to this situation. The heat transfer rate for the steam is, assuming no pressure drop through the condenser, $$ \dot{Q}_{s}=\dot{m}_{s}\left(h_{s 2}-h_{s 1}\right)=(600)(94.02-1116.1)=-613.200 \mathrm{Btu} / \mathrm{min} $$ This energy is gained by the water. Hence, $$ \dot{Q}_{w}=\dot{m}_{w}\left(h_{w 2}-h_{w 1}\right)=\dot{m}_{w} C_{p}\left(T_{w 2}-T_{w 1}\right) \quad \frac{613,200}{60}=\dot{m}_{w}(1.00)(15) \quad \dot{m}_{w}=681 \mathrm{lbm} / \mathrm{s} $$

The pressure of \(200 \mathrm{~kg} / \mathrm{s}\) of water is to be increased by \(4 \mathrm{MPa}\). The water enters through a 20 -cm-diameter pipe and exits through a \(12-\mathrm{cm}\)-diameter pipe. Calculate the minimum horsepower required to operate the pump. SOLUTION The energy equation (4.68) provides us with $$ -\dot{W}_{p}=\dot{m}\left(\frac{\Delta P}{\rho}+\frac{V_{2}^{2}-V_{1}^{2}}{2}\right) $$ The inlet and exit velocities are calculated as follows: $$ V_{1}=\frac{\dot{m}}{\rho A_{1}}=\frac{200}{(1000)(\pi)(0.1)^{2}}=6.366 \mathrm{~m} / \mathrm{s} \quad V_{2}=\frac{\dot{m}}{\rho A_{2}}=\frac{200}{(1000)(\pi)(0.06)^{2}}=17.68 \mathrm{~m} / \mathrm{s} $$ The energy equation then gives $$ \dot{W}_{P}=-200\left[\frac{4000000}{1000}+\frac{(17.68)^{2}-(6.366)^{2}}{2}\right]=-827200 \mathrm{~W} \quad \text { or } \quad 1109 \mathrm{hp} $$ Note: The above power calculation provides a minimum since we have neglected any internal energy increase. Also, the kinetic energy change represents only a 3 percent effect on \(\dot{W}_{P}\) and could be neglected.

Water enters a radiator through a 4-cm-diameter hose at \(0.02 \mathrm{~kg} / \mathrm{s}\). It travels down through all the rectangular passageways on its way to the water pump. The passageways are each \(10 \times 1 \mathrm{~mm}\) and there are 800 of them in a cross section. How long does it take water to traverse from the top the bottom of the 60 -cm-high radiator? SOLUTION The average velocity through the passageways is found from the continuity equation, using \(\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^{3}\) : $$ \dot{m}=\rho_{1} V_{1} A_{1}=\rho_{2} V_{2} A_{2} \quad \therefore V_{2}=\frac{\dot{m}}{\rho_{2} A_{2}}=\frac{0.02}{(1000)[(800)(0.01)(0.001)]}=0.0025 \mathrm{~m} / \mathrm{s} $$ The time to travel \(60 \mathrm{~cm}\) at this constant velocity is $$ t=\frac{L}{V}=\frac{0.60}{0.0025}=240 \mathrm{~s} \quad \text { or } \quad 4 \mathrm{~min} $$

Estimate the constant-pressure specific heat and the constant-volume specific heat for R134a at 30 psia and \(100^{\circ} \mathrm{F}\). SOLUTION We write the derivatives in finite-difference form and, using values on either side of \(100^{\circ} \mathrm{F}\) for greatest accuracy, we find $$ \begin{aligned} &C_{p} \equiv \frac{\Delta h}{\Delta T}=\frac{126.39-117.63}{120-80}=0.219 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \\ &C_{v} \equiv \frac{\Delta u}{\Delta T}=\frac{115.47-107.59}{120-80}=0.197 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \end{aligned} $$
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