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Air enters the compressor of an ideal cold air-standard Brayton cycle at \(100 \mathrm{kPa}, 300 \mathrm{~K}\), with a mass flow rate of \(6 \mathrm{~kg} / \mathrm{s}\). The compressor pressure ratio is 10 , and the turbine inlet temperature is \(1400 \mathrm{~K}\). For \(k=1.4\), calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in \(\mathrm{kW}\).

Short Answer

Expert verified
Thermal Efficiency: 78.6%, Back Work Ratio: 0.565, Net Power Developed: 2278.44 kW

Step by step solution

01

- Determine Compressor Exit Conditions

First, use the compressor pressure ratio to find the exit temperature of the compressor using the relation: \[ T_2 = T_1 \times \text{PR}^{(k-1)/k} \] Given: \( T_1 = 300 \text{ K}, \text{PR} = 10, k = 1.4 \) Plugging in the values, \[ T_2 = 300 \text{ K} \times 10^{(\frac{1.4-1}{1.4})} \] \[ T_2 \approx 300 \text{ K} \times 2.639 \approx 791.7 \text{ K} \]
02

- Turbine Exit Temperature

Use the turbine inlet temperature to determine the exit temperature of the turbine. Given: \( T_3 = 1400 \text{ K} \) We use the same pressure ratio reversed for the turbine: \[ T_4 = T_3 \times \frac{1}{\text{PR}^{(k-1)/k}} \] Plugging in the values, \[ T_4 = 1400 \text{ K} \times 10^{-(\frac{1.4-1}{1.4})} \] \[ T_4 \approx 1400 \text{ K} \times \frac{1}{2.639} \approx 530.6 \text{ K} \]
03

- Thermal Efficiency Calculation

The thermal efficiency is given by: \[ \text{Efficiency} (\text{η}) = 1 - \frac{T_1}{T_2} \] Plugging in the values: \[ \text{η} = 1 - \frac{300 \text{ K}}{1400 \text{ K}} = 1 - 0.214 \approx 0.786 \text{ or } 78.6\text{%} \]
04

- Back Work Ratio

The back work ratio (BWR) is the ratio of compressor work to turbine work: \[ \text{BWR} = \frac{W_c}{W_t} \] The work done per kg of air can be found using: \[ W_c = C_p (T_2 - T_1) \] \[ W_t = C_p (T_3 - T_4) \] Given: \( C_p = 1.005 \text{ kJ/kg K} \) Plugging in the values: \[ W_c = 1.005 \times (791.7 - 300) \approx 493.96 \text{ kJ/kg} \] \[ W_t = 1.005 \times (1400 - 530.6) \approx 873.7 \text{ kJ/kg} \] Thus, \[ \text{BWR} = \frac{493.96}{873.7} \approx 0.565 \]
05

- Net Power Developed

The net power developed (\text{P}_{net}) can be calculated using the mass flow rate: \[ \text{P}_{net} = \text{mass flow rate} \times (W_t - W_c) \] Given: \( \text{mass flow rate} = 6 \text{ kg/s} \) Plugging in the values: \[ \text{P}_{net} = 6 \times (873.7 - 493.96) \approx 6 \times 379.74 \approx 2278.44 \text{ kW} \]

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Key Concepts

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

Compressor Pressure Ratio
The compressor pressure ratio is a key parameter in the Brayton cycle. It represents the ratio of the pressure after compression to the pressure before compression. In mathematical terms, it is denoted as \text{PR} and defined as \text{PR}=\frac{P_2}{P_1}\. This ratio is crucial because it determines the temperature rise in the compressor according to the relation: \ T_2 = T_1 \times \text{PR}^{(\frac{k-1}{k})}\, where \(k=1.4\). The higher the pressure ratio, the higher the temperature rise, and consequently, the higher the efficiency of the cycle. Given the compressor inlet temperature (\(T_1=300\) K) and pressure ratio (\(\text{PR}=10\)), the compressor exit temperature is determined using the formula above, resulting in \(T_2=791.7\) K. This high exit temperature forms the basis for subsequent calculations in the Brayton cycle.
Turbine Inlet Temperature
The turbine inlet temperature (TIT) is the temperature at which air enters the turbine. It is a critical parameter for the Brayton cycle as it directly affects the power output of the turbine and the overall efficiency of the cycle. For our exercise, the turbine inlet temperature is given as \(T_3=1400\, K\). To find the turbine exit temperature (\(T_4\)), we use the pressure ratio in reverse and the formula: \ T_4 = T_3 \times \frac{1}{\text{PR}^{(\frac{k-1}{k})}}\. This gives us an exit temperature of \(T_4\approx 530.6\, K\). This temperature is lower than the inlet temperature, showing that the turbine produces work by expanding high-temperature air to a lower temperature.
Thermal Efficiency Calculation
The thermal efficiency of the Brayton cycle indicates how well the cycle converts heat into work. It is calculated using the formula: \ \text{Efficiency} (\eta) = 1 - \frac{T_1}{T_3}\, where \(T_1\) is the compressor inlet temperature and \(T_3\) is the turbine inlet temperature. From the given values (\(T_1=300\, K\) and \(T_3=1400\, K\)), the efficiency is: \ \eta = 1 - \frac{300}{1400} = 1 - 0.214 \approx 0.786\ or \ 78.6\text{%}\. This relatively high efficiency highlights the effectiveness of the Brayton cycle under the given conditions.
Back Work Ratio
The back work ratio (BWR) is the ratio of the work consumed by the compressor to the work produced by the turbine. It is a measure of how much of the turbine's work is used to drive the compressor. The BWR is given by: \ \text{BWR} = \frac{W_c}{W_t}\, where \(W_c\) is the compressor work and \(W_t\) is the turbine work. In the exercise, we calculate these as follows: \(W_c = C_p (T_2 - T_1)\ and\ W_t = C_p (T_3 - T_4)\). Given \(C_p=1.005\, kJ/kg \cdot K\), we find: \(W_c \approx 493.96\, kJ/kg\) and \(W_t \approx 873.7\, kJ/kg\). Thus, the BWR is \( \text{BWR} \approx 0.565\). This ratio tells us that about 56.5% of the turbine work is used to drive the compressor.
Net Power Developed
The net power developed (\text{P}_net) is the useful power output of the Brayton cycle after accounting for the power used by the compressor. It is calculated by the formula: \ \text{P}_{net} = \text{mass flow rate} \times (W_t - W_c)\. Given the mass flow rate of \(6\, kg/s\), \(W_t=873.7\, kJ/kg\), and \(W_c=493.96\, kJ/kg\), the net power developed is: \ \text{P}_{net} = 6 \times (873.7 - 493.96) \approx 2278.44\, kW\. This is the effective power that can be used for doing useful work, such as driving generators or propulsion systems.

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Most popular questions from this chapter

An ideal air-standard Brayton cycle operating at steady state produces \(10 \mathrm{MW}\) of power. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.9. Sketch the \(T-5\) diagram for the cycle and determine (a) the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\) (b) the rate of heat transfer, in \(\mathrm{kW}\), to the working fluid passing through the heat exchanger. (c) the thermal efficiency. $$ \begin{array}{crcr} \text { State } & p(\mathrm{kPa}) & T(\mathrm{~K}) & h(\mathrm{k}] / \mathrm{kg}) \\ \hline 1 & 100 & 300 & 300.19 \\ 2 & 1200 & 603.5 & 610.65 \\ 3 & 1200 & 1450 & 1575.57 \\ 4 & 100 & 780.7 & 800.78 \end{array} $$

Consider an air-standard Otto cycle. Operating data at principal states in the cycle are given in the table below. The states are numbered as in Fig. 9.3. The mass of air is \(0.002 \mathrm{~kg}\). Determine (a) the heat addition and the heat rejection, each in kJ. (b) the net work, in kJ. (c) the thermal efficiency. (d) the mean effective pressure, in \(\mathrm{kPa}\). $$ \begin{array}{cccc} \text { State } & T(\mathrm{~K}) & p(\mathrm{kPa}) & u(\mathrm{k}) / \mathrm{kg}) \\ \hline 1 & 305 & 85 & 217.67 \\ 2 & 367.4 & 767.9 & 486.77 \\ 3 & 960 & 2006 & 725.02 \\ 4 & 458.7 & 127.8 & 329.01 \end{array} $$

Air is the working fluid in an Ericsson cycle. Expansion through the turbine takes place at a temperature of \(2000^{\circ} \mathrm{R}\). Heat transfer from the compressor occurs at \(520^{\circ} \mathrm{R}\). The compressor pressure ratio is 10 . Assuming the ideal gas model and ignoring kinetic and potential energy effects, determine (a) the net work, in Btu per lb of air flowing. (b) the thermal efficiency.

Air enters the compressor of a cold air-standard Brayton cycle with regeneration, intercooling, and reheat at \(100 \mathrm{kPa}\), \(300 \mathrm{~K}\), with a mass flow rate of \(6 \mathrm{~kg} / \mathrm{s}\). The compressor pressure ratio is 10 , and the pressure ratios are the same across each compressor stage. The intercooler and reheater both operate at the same pressure. The temperature at the inlet to the second compressor stage is \(300 \mathrm{~K}\), and the inlet temperature for each turbine stage is \(1400 \mathrm{~K}\). The compressor and turbine stages each have isentropic efficiencies of \(80 \%\) and the regenerator effectiveness is \(80 \%\). For \(k=1.4\), calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in \(\mathrm{kW}\). (d) the rates of exergy destruction in the compressor and turbine stages as well as the regenerator, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\).

Air enters a two-stage compressor operating at steady state at 1 bar, \(290 \mathrm{~K}\). The overall pressure ratio across the stages is 16 and each stage operates isentropically. Intercooling occurs at the pressure that minimizes total compressor work, as determined in Example 9.10. Air exits the intercooler at \(290 \mathrm{~K}\). Assuming ideal gas behavior with \(k=1.4\), determine (a) the intercooler pressure, in bar, and the heat transfer, in kJ per kg of air flowing. (b) the work required for each compressor stage, in kJ per \(\mathrm{kg}\) of air flowing.

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