Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Using constant specific heats at room temperature, determine the thermal efficiency of this cycle and the rate of heat input if the cycle is to produce \(90 \mathrm{kW}\) of power. An ideal Otto cycle has a compression ratio of \(8 .\) At the beginning of the compression process, air is at \(95 \mathrm{kPa}\) and \(27^{\circ} \mathrm{C},\) and \(750 \mathrm{kJ} / \mathrm{kg}\) of heat is transferred to air during the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine \((a)\) the pressure and temperature at the end of the heat-addition process, \((b)\) the net work output, \((c)\) the thermal efficiency, and \((d)\) the mean effective pressure for the cycle.

Short Answer

Expert verified
661\, \mathrm{K})}{0.113\, \mathrm{m^3/kg}} = 4,191\, \mathrm{kPa}\) Step 5: Calculate the Temperature after Heat Addition #tag_title# Calculate the Temperature after Heat Addition #tag_content# We know that heat is added at constant volume, so we can calculate the temperature after the heat addition process using the relation: \(q_{in} = C_v (T_3 - T_2)\) Where \(C_v\) is the specific heat at constant volume for air, which is \(0.718 \, \mathrm{kJ/kg \cdot K}\). Now, we can calculate the temperature after heat addition: \(T_3 = T_2 + \frac{q_{in}}{C_v} = 1661\, \mathrm{K} + \frac{750\, \mathrm{kJ/kg}}{0.718\, \mathrm{kJ/kg \cdot K}} = 2,711\, \mathrm{K}\) Step 6: Calculate the Pressure after Heat Addition #tag_title# Calculate the Pressure after Heat Addition #tag_content# We can calculate the pressure after heat addition using the ideal gas law: \(p_3 = \frac{RT_3}{v_2}\) \(p_3 = \frac{(0.287\, \mathrm{kJ/kg \cdot K})(2,711\, \mathrm{K})}{0.113\, \mathrm{m^3/kg}} = 6,866\, \mathrm{kPa}\) Step 7: Calculate the Net Work Output #tag_title# Calculate the Net Work Output #tag_content# The net work output for an ideal Otto cycle can be calculated using the relationship: \(w_{net} = q_{in} - C_v (T_4 - T_1)\) We need to find temperature after the expansion process \(T_4\): \(\frac{T_4}{T_3} = \left(\frac{v_4}{v_3}\right)^{k-1} = \left(\frac{v_1}{v_2}\right)^{k-1} = \frac{T_1}{T_2}\) \(T_4 = T_1 \times \frac{T_3}{T_2} = 300\, \mathrm{K} \times \frac{2,711\, \mathrm{K}}{1,661\, \mathrm{K}} = 489\, \mathrm{K}\) Now we can find the net work output: \(w_{net} = q_{in} - C_v (T_4 - T_1) = 750\, \mathrm{kJ/kg} - (0.718\, \mathrm{kJ/kg \cdot K})(489\, \mathrm{K} - 300\, \mathrm{K}) = 493\, \mathrm{kJ/kg}\) Step 8: Calculate the Thermal Efficiency #tag_title# Calculate the Thermal Efficiency #tag_content# The thermal efficiency of the ideal Otto cycle can be calculated using the relationship: \(\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{493\, \mathrm{kJ/kg}}{750\, \mathrm{kJ/kg}} = 0.657 = 65.7\%\) Step 9: Calculate the Mean Effective Pressure #tag_title# Calculate the Mean Effective Pressure #tag_content# The mean effective pressure (MEP) is a measure of the average pressure exerted during the power stroke of an engine. It can be calculated using the relationship: \(MEP = \frac{w_{net}}{v_1 - v_2} = \frac{493\, \mathrm{kJ/kg}}{0.905\, \mathrm{m^3/kg} - 0.113\, \mathrm{m^3/kg}} = 638\, \mathrm{kPa}\) Now we have calculated all the required parameters for this ideal Otto cycle: - Pressure after heat addition: \(p_3 = 6,866\, \mathrm{kPa}\) - Temperature after heat addition: \(T_3 = 2,711\, \mathrm{K}\) - Net work output: \(w_{net} = 493\, \mathrm{kJ/kg}\) - Thermal efficiency: \(\eta_{th} = 65.7\%\) - Mean effective pressure: \(MEP = 638\, \mathrm{kPa}\)

Step by step solution

01

Calculate Initial Volume and Specific Volume

We know that \(pV = mRT\), and the problem wants us to calculate the properties per unit mass, so we can rewrite the ideal gas equation as: \(p_1v_1 = RT_1\) \(v_1 = \frac{RT_1}{p_1}\) where \(R\) is the specific gas constant for air, which is \(0.287 \, \mathrm{kJ/kg\cdot K}\). Now, we can calculate the initial specific volume: \(v_1 = \frac{(0.287 \,\mathrm{kJ/kg \cdot K})(300\, \mathrm{K})}{95 \, \mathrm{kPa}} = 0.905 \, \mathrm{m^3/kg}\) Step 2: Calculate the Specific Volume after Compression
02

Calculate the Specific Volume after Compression

As we know the compression ratio is 8, we can calculate the specific volume after the compression process: \(v_2 = \frac{v_1}{r_c} = \frac{0.905\, \mathrm{m^3/kg}}{8} = 0.113 \, \mathrm{m^3/kg}\) Step 3: Calculate the Temperature after Compression
03

Calculate the Temperature after Compression

The temperature after the compression can be found using the relationship for the isentropic compression process: \(\frac{T_2}{T_1} = \left(\frac{v_1}{v_2}\right)^{k-1}\) where \(k\) is the specific heat ratio (\(k = 1.4\) for air). Now, we calculate the temperature after compression: \(T_2 = T_1 \left(\frac{v_1}{v_2}\right)^{k-1} = 300\, \mathrm{K} \times \left(\frac{0.905\, \mathrm{m^3/kg}}{0.113\, \mathrm{m^3/kg}}\right)^{0.4} = 1661\, \mathrm{K}\) Step 4: Calculate Pressure after Compression
04

Calculate Pressure after Compression

We can calculate the pressure after compression using the ideal gas law: \(p_2 = \frac{RT_2}{v_2}\) \(p_2 = \frac{(0.287\, \mathrm{kJ/kg \cdot K})(1\)

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 Vaia!

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

Develop an expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100 percent. Use constant specific heats at room temperature.

An ideal Stirling engine using helium as the working fluid operates between temperature limits of 300 and 2000 K and pressure limits of \(150 \mathrm{kPa}\) and 3 MPa. Assuming the mass of the helium used in the cycle is \(0.12 \mathrm{kg}\), determine \((a)\) the thermal efficiency of the cycle, \((b)\) the amount of heat transfer in the regenerator, and \((c)\) the work output per cycle.

A four-cylinder spark-ignition engine has a compression ratio of \(10.5,\) and each cylinder has a maximum volume of 0.4 L. At the beginning of the compression process, the air is at \(98 \mathrm{kPa}\) and \(37^{\circ} \mathrm{C}\), and the maximum temperature in the cycle is 2100 K. Assuming the engine to operate on the ideal Otto cycle, determine \((a)\) the amount of heat supplied per cylinder, ( \(b\) ) the thermal efficiency, and \((c)\) the number of revolutions per minute required for a net power output of \(45 \mathrm{kW}\). Assume variable specific heats for air

When we double the compression ratio of an ideal Otto cycle, what happens to the maximum gas temperature and pressure when the state of the air at the beginning of the compression and the amount of heat addition remain the same? Use constant specific heats at room temperature.

An ideal Brayton cycle has a net work output of \(150 \mathrm{kJ} / \mathrm{kg}\) and a back work ratio of \(0.4 .\) If both the turbine and the compressor had an isentropic efficiency of 85 percent, the net work output of the cycle would be \((a) 74 \mathrm{kJ} / \mathrm{kg}\) \((b) 95 \mathrm{kJ} / \mathrm{kg}\) \((c) 109 \mathrm{kJ} / \mathrm{kg}\) \((d) 128 \mathrm{kJ} / \mathrm{kg}\) \((e) 177 \mathrm{kJ} / \mathrm{kg}\)

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free