/*! 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 1 An adiabatic compressor receives... [FREE SOLUTION] | 91Ó°ÊÓ

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Chapter 8: Problem 1

An adiabatic compressor receives \(20 \mathrm{~m}^{3} / \mathrm{min}\) of air from the atmosphere at \(20^{\circ} \mathrm{C}\) and compresses it to \(10 \mathrm{MPa}\). Calculate the minimum power requirement. SOLUTION An isentropic compression requires the minimum power input for an adiabatic compressor. The outlet temperature for such a process is $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(293)\left(\frac{10000}{100}\right)^{0.2857}=1092 \mathrm{~K} $$ To find the mass flux, we must know the density. It is \(\rho=P / R T=100 /(0.287)(293)=1.189 \mathrm{~kg} / \mathrm{m}^{3}\). The mass flux is then (the flow rate is given) \(\dot{m}=\rho(A V)=(1.189)(20 / 60)=0.3963 \mathrm{~kg} / \mathrm{s}\). The minimum power requirement is now calculated to be $$ \dot{W}_{\text {comp }}=\dot{m}\left(h_{2}-h_{1}\right)=\dot{m} C_{p}\left(T_{2}-T_{1}\right)=(0.3963)(1.00)(1092-293)=317 \mathrm{~kW} $$

Short Answer

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317 kW

Step by step solution

01

Determine Outlet Temperature

For an isentropic compression process in an adiabatic compressor, the outlet temperature can be found using:\[T_{2} = T_{1} \left( \frac{P_{2}}{P_{1}} \right)^{\frac{k-1}{k}} \]Given:\(T_1 = 293 \mathrm{~K}\), \(P_2 = 10,000 \mathrm{~kPa} = 10 \mathrm{~MPa}\), \(P_1 = 100 \mathrm{~kPa}\), and \(k = 1.4\)Using the equation:\[ T_{2} = 293 \left( \frac{10000}{100} \right)^{0.2857} = 1092 \mathrm{~K} \]
02

Calculate the Density of Air

The density \(\rho\) can be calculated using the ideal gas law:\[ \rho = \frac{P}{R T} \]Given:\(P = 100 \mathrm{~kPa}\), \(R = 0.287 \mathrm{~kJ/(kg·K)}\), and \(T = 293 \mathrm{~K}\)Calculating the density:\[ \rho = \frac{100}{0.287 \times 293} = 1.189 \mathrm{~kg/m^3} \]
03

Determine Mass Flow Rate

The mass flow rate \(\dot{m}\) can be calculated using the volumetric flow rate and the density:\[ \dot{m} = \rho \cdot \dot{V} \]Given:\(\rho = 1.189 \mathrm{~kg/m^3}\) and volumetric flow rate \(\dot{V} = 20 \mathrm{~m^3/min}\)Converting to \(\mathrm{m^3/s}\):\[ \dot{V} = \frac{20}{60} = 0.333 \mathrm{~m^3/s} \]Calculating the mass flow rate:\[ \dot{m} = 1.189 \times 0.333 = 0.3963 \mathrm{~kg/s} \]
04

Calculate Minimum Power Requirement

The minimum power requirement for the compressor can be found using the enthalpy change:\[ \dot{W}_{\text{comp}} = \dot{m} C_p (T_2 - T_1) \]Given:\(\dot{m} = 0.3963 \mathrm{~kg/s}\), \(C_p = 1.00 \mathrm{~kJ/(kg·K)}\), \(T_2 = 1092 \mathrm{~K}\), and \(T_1 = 293 \mathrm{~K}\)Calculating the power:\[ \dot{W}_{\text{comp}} = 0.3963 \cdot 1.00 \cdot (1092 - 293) = 317 \mathrm{~kW} \]

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Key Concepts

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

Isentropic Process
In thermodynamics, an isentropic process is one where the entropy of the system remains constant. This type of process is purely adiabatic, meaning no heat is transferred into or out of the system. It is highly idealized and represents the most efficient process for compressing or expanding gases.
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Most popular questions from this chapter

Air enters the compression process of a diesel cycle at \(120 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\). The pressure after compression is \(8 \mathrm{MPa}\) and \(1500 \mathrm{~kJ} / \mathrm{kg}\) is added during combustion. What are \((a)\) the cutoff ratio, \((b)\) the thermal efficiency, and ( \(c\) ) the MEP?

An air-standard cycle operates in a piston-cylinder arrangement with the following four processes: \(1 \rightarrow 2\) - isentropic compression from \(100 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\) to \(2 \mathrm{MPa} ; 2 \rightarrow 3\)-constant-pressure heat addition to \(1200^{\circ} \mathrm{C} ; 3 \rightarrow 4\)-isentropic expansion; and \(4 \rightarrow 1\)-constant-volume heat rejection. (a) Show the cycle on \(P-v\) and \(T-s\) diagrams, \((b)\) calculate the heat addition and \((c)\) calculate the cycle efficiency.

A compressor receives \(4 \mathrm{~kg} / \mathrm{s}\) of \(20^{\circ} \mathrm{C}\) air from the atmosphere and delivers it at a pressure of \(18 \mathrm{MPa}\). If the compression process can be approximated by a polytropic process with \(n=1.3\), calculate the power requirement and the rate of heat transfer. The power requirement is [see (8.6)] $$ \dot{W}_{\text {comp }}=\dot{m} \frac{n R}{n-1} T_{1}\left[\left(\frac{P_{2}}{P_{1}}\right)^{(n-1) / n}-1\right]=(4) \frac{(1.3)(0.287)}{1.3-1}(293)\left[\left(\frac{18000}{100}\right)^{0.3 / 1.3}-1\right]=3374 \mathrm{~kW} $$ The first law for the control volume [see (4.66)] surrounding the compressor provides us with $$ \begin{aligned} \dot{Q} &=\dot{m} \Delta h+\dot{W}_{\text {comp }}=\dot{m} C_{p}\left(T_{2}-T_{1}\right)+\dot{W}_{\text {comp }}=\dot{m} C_{p} T_{1}\left[\left(\frac{P_{2}}{P_{1}}\right)^{(n-1) / n}-1\right]+\dot{W}_{\text {comp }} \\ &=(4)(1.00)(293)\left[\left(\frac{18000}{100}\right)^{0.3 / 1.3}-1\right]-3374=-661 \mathrm{~kW} \end{aligned} $$ In the above, we have used the compressor power as negative since it is a power input. The expression of (8.6) is the magnitude of the power with the minus sign suppressed, but when the first law is used we must be careful with the signs. The negative sign on the heat transfer means that heat is leaving the control volume.

An inventor proposes a reciprocating engine with a compression ratio of 10 , operating on \(1.6 \mathrm{~kg} / \mathrm{s}\) of atmospheric air at \(20^{\circ} \mathrm{C}\), that produces \(50 \mathrm{hp}\). After combustion, the temperature is \(400^{\circ} \mathrm{C}\). Is the proposed engine feasible? We will consider a Carnot engine operating between the same pressure and temperature limits; this will establish the ideal situation without reference to the details of the proposed engine. The specific volume at state 1 (see Fig. 6.1) is $$ v_{1}=\frac{R T_{1}}{P_{1}}=\frac{(0.287)(293)}{100}=0.8409 \mathrm{~m}^{3} / \mathrm{kg} $$ For a compression ratio of 10 , the minimum specific volume must be \(v_{3}=v_{1} / 10=0.8409 / 10=0.08409\). The specific volume at state 2 is now found by considering the isothermal process from 1 to 2 and the isentropic process from 2 to 3 : $$ \begin{gathered} P_{2} v_{2}=P_{1} v_{1}=100 \times 0.8409=84.09 \quad P_{2} v_{2}^{k}=\frac{0.287(673)}{0.08409}(0.08409)^{1.4}=71.75 \\ \therefore v_{2}=0.6725 \mathrm{~m}^{3} / \mathrm{kg} \end{gathered} $$ The change in entropy is $$ \Delta s=R \ln \frac{v_{2}}{v_{1}}=0.287 \ln \frac{0.6725}{0.8409}=-0.0641 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} $$ The work output is then \(w_{\text {net }}=\Delta T|\Delta s|=(400-20)(0.0641)=24.4 \mathrm{~kJ} / \mathrm{kg}\). The power output is $$ \dot{W}=\dot{m} w_{\text {net }}=(1.6)(24.4)=39.0 \mathrm{~kW} \quad \text { or } \quad 52.2 \mathrm{hp} $$ The maximum possible power output is \(52.2 \mathrm{hp}\); the inventor's claims of \(50 \mathrm{hp}\) is highly unlikely, though not impossible.

A turbojet aircraft flies at a speed of \(300 \mathrm{~m} / \mathrm{s}\) at an elevation of \(10000 \mathrm{~m}\). If the compression ratio is 10 , the turbine inlet temperature is \(1000^{\circ} \mathrm{C}\), and the mass flux of air is \(30 \mathrm{~kg} / \mathrm{s}\), calculate the maximum thrust possible from this engine. Also, calculate the rate of fuel consumption if the heating value of the fuel is \(8400 \mathrm{~kJ} / \mathrm{kg}\). The inlet temperature and pressure are found from Table B.1 to be (see Fig. 8.20) $$ T_{1}=223.3 \mathrm{~K} \quad P_{1}=0.2615 \quad P_{0}=26.15 \mathrm{kPa} $$ The temperature exiting the compressor is $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(223.3)(10)^{0.2857}=431.1 \mathrm{~K} $$ Since the turbine drives the compressor, the two works are equal so that $$ C_{p}\left(T_{2}-T_{1}\right)=C_{p}\left(T_{3}-T_{4}\right) \quad \therefore T_{3}-T_{4}=T_{2}-T_{1} $$ Since \(T_{3}=1273\), we can find \(T_{4}\) as \(T_{4}=T_{3}+T_{1}-T_{2}=1273+223.3-431.1=1065.2 \mathrm{~K}\). We can now calculate the pressure at the turbine exit to be, using \(P_{3}=P_{2}=261.5 \mathrm{kPa}\), $$ P_{4}=P_{3}\left(\frac{T_{4}}{T_{3}}\right)^{k /(k-1)}=(261.5)\left(\frac{1065.2}{1273}\right)^{3.5}=140.1 \mathrm{kPa} $$ The temperature at the nozzle exit, assuming an isentropic expansion, is $$ T_{5}=T_{4}\left(\frac{P_{5}}{P_{4}}\right)^{(k-1) / k}=(1065.2)\left(\frac{26.15}{140.1}\right)^{0.2857}=659.4 \mathrm{~K} $$ The energy equation provides us with the exit velocity \(V_{5}=\left[2 C_{p}\left(T_{4}-T_{5}\right)\right]^{1 / 2}=[(2)(1000)(1065.2-659.4)]^{1 / 2}=\) \(901 \mathrm{~m} / \mathrm{s}\), where \(C_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) must be used in the expression. The thrust can now be calculated as $$ \text { thrust }=\dot{m}\left(V_{5}-V_{1}\right)=(30)(901-300)=18030 \mathrm{~N} $$ This represents a maximum since a cycle composed of ideal processes was used. The heat transfer rate in the burner is \(\dot{Q}=\dot{m} C_{p}\left(T_{3}-T_{2}\right)=(30)(1.00)(1273-431.1)=25.26\) MW. This requires that the mass flux of fuel \(\dot{m}_{f}\) be $$ 8400 \dot{m}_{f}=25260 \quad \therefore \dot{m}_{f}=3.01 \mathrm{~kg} / \mathrm{s} $$
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