Try the fastest way to create flashcards

Related questions with answers

Question

Air is compressed in an axial-flow compressor operating at steady state from 27C,1 bar27^{\circ} \mathrm{C}, 1\ \text{bar} to a pressure of 2.1 bar2.1\ \text{bar}. The work required is 94.6 kJ94.6 \mathrm{~kJ} per kg\mathrm{kg} of air flowing. Heat transfer from the compressor occurs at an average surface temperature of 40C40^{\circ} \mathrm{C} at the rate of 14 kJ14 \mathrm{~kJ} per kg of air flowing. The effects of motion and gravity can be ignored. Let T0=20C,p0=1 barT_0=20^{\circ} \mathrm{C}, p_0=1\ \text{bar}. Assuming ideal gas behavior, determine (a) the temperature of the air at the exit, in C{ }^{\circ} \mathrm{C}, and (b) the rate of exergy destruction within the compressor, in kJ\mathrm{kJ} per kg\mathrm{kg} of air flowing,

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 2

 a)\text{\color{#4257b2} a)} In this part of the problem we need to calculate the final temperature T2T_2. We will use the energy balance equation with the given assumptions that we can neglect motion and gravity effects and the ideal gas model for air. We will also need the given initial temperature T1=300KT_1=300\,\text{K}, work done w=94.6kJ/kgw=94.6\,\text{kJ/kg} and heat transfer q=14kJ/kgq=14\,\text{kJ/kg}.

0=qw+(h2h1)0=qw+cp(T2T1)0=14kJkg94.6kJkg+1.01kJkg K(T2300K)T2=380K\begin{align*} 0&=q - w + (h_2 - h_1) \\ 0&=q - w + c_p \cdot (T_2 - T_1) \\ 0&=14\,\frac{\text{kJ}}{\text{kg}} - 94.6\,\frac{\text{kJ}}{\text{kg}} + 1.01\,\frac{\text{kJ}}{\text{kg K}} \cdot (T_2 - 300\,\text{K}) \\ T_2&=\boxed{\color{#c34632}380\,\text{K}} \\ \end{align*}

 b)\text{\color{#4257b2} b)} In this part of the problem we will calculate the exergy destruction e˙d\dot{e}_d using the exergy rate balance equation. For the calculation we will also need the given surface temperature Tb=313KT_b=313\,\text{K} and ambient temperature T0=293KT_0=293\,\text{K} and pressures p1=1barp_1=1\,\text{bar} and p2=2.1barp_2=2.1\,\text{bar}.

e˙d=(1T0Tb)qw+(h2h1)T0(s2s1)e˙d=(1T0Tb)qw+cp(T2T1)T0(cplnT2T1Rlnp2p1)e˙d=(1293K313K)14kJkg94.6kJkg+1.01kJkg K(380K300K)293K(1.01kJkg Kln380K300K0.287kJkg Kln2.1bar1bar)e˙d=20.47kJkg\begin{align*} \dot{e}_d&=\left(1-\frac{T_0}{T_b}\right) \cdot q - w + (h_2 - h_1) - T_0 \cdot \left(s_2 - s_1\right) \\ \dot{e}_d&=\left(1-\frac{T_0}{T_b}\right) \cdot q - w + c_p \cdot (T_2 - T_1) - T_0 \cdot \left(c_p \ln \frac{T_2}{T_1} - R \ln \frac{p_2}{p_1}\right) \\ \dot{e}_d&=\left(1-\frac{293\,\text{K}}{313\,\text{K}}\right) \cdot 14\,\frac{\text{kJ}}{\text{kg}} - 94.6\,\frac{\text{kJ}}{\text{kg}} + 1.01\,\frac{\text{kJ}}{\text{kg K}} \cdot (380\,\text{K}- 300\,\text{K}) - \\ -& 293\,\text{K}\cdot \left(1.01\,\frac{\text{kJ}}{\text{kg K}} \ln \frac{380\,\text{K}}{300\,\text{K}} - 0.287\,\frac{\text{kJ}}{\text{kg K}} \ln \frac{2.1\,\text{bar}}{1\,\text{bar}}\right) \\ \dot{e}_d&=\boxed{\color{#c34632}-20.47\,\frac{\text{kJ}}{\text{kg}}} \\ \end{align*}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Fundamentals of Electric Circuits 6th Edition by Charles Alexander, Matthew Sadiku

Fundamentals of Electric Circuits

6th EditionISBN: 9780078028229 (2 more)Charles Alexander, Matthew Sadiku
2,120 solutions
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 4th Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (8 more)Randall D. Knight
3,508 solutions
Advanced Engineering Mathematics 10th Edition by Erwin Kreyszig

Advanced Engineering Mathematics

10th EditionISBN: 9780470458365 (5 more)Erwin Kreyszig
4,134 solutions
Fundamentals of Engineering Thermodynamics 7th Edition by Daisie D. Boettner, Howard N. Shapiro, Margaret B. Bailey, Michael J. Moran

Fundamentals of Engineering Thermodynamics

7th EditionISBN: 9780470917688 (2 more)Daisie D. Boettner, Howard N. Shapiro, Margaret B. Bailey, Michael J. Moran
1,848 solutions

More related questions

1/4

1/7