Question 1 |
An engine running on an air standard Otto cycle
has a displacement volume 250 cm^3 and a clearance volume 35.7 cm^3. The pressure and temperature at
the beginning of the compression process are 100
kPa and 300 K, respectively. Heat transfer during
constant-volume heat addition process is 800 kJ/kg. The specific heat at constant volume is 0.718 kJ/kg.K and the ratio of specific heats at constant
pressure and constant volume is 1.4. Assume the
specific heats to remain constant during the cycle. The maximum pressure in the cycle is ______ kPa
(round off to the nearest integer).
4811 | |
1254 | |
2589 | |
2547 |
Question 1 Explanation:
\begin{aligned} V_S&=250cm^3\\ V_C&=35.7 cm^3\\ T_1&=300K\\ P_1&=100kPa\\ Q_S&=800kJ/kg\\ C_v&=0.718kJ/kgK\\ \gamma &=1.4\\ P_3&=\_\_\_kPa\\ \frac{T_2}{T_1}&=\left ( \frac{P_2}{P_1} \right )^{\frac{\gamma -1}{\gamma }}=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}=\left ( \frac{V_S+V_C}{V_C} \right )^{\gamma -1}\\ \frac{T_2}{300}&=\left ( \frac{P_2}{100} \right )^{\frac{1.4-1}{1.4 }}=\left ( \frac{250+35.7}{35.7} \right )^{1.4 -1}\\ T_2&=689.31K\\ P_2&=1838.82kPa\\ Q_S&=c_v \times (T_3-T_2)\\ 800&=0.718(T_3-689.31)\\ T_3&=1803.516K\\ &\text{For Process 2-3(Volume is constant)}\\ \frac{P_3}{P_2}&=\frac{T_3}{T_2}\\ P_3&=\frac{1803.516}{689.31} \times 1838.82\\ P_3&=4811kPa \end{aligned}
Question 2 |
For an air-standard Diesel cycle,
heat addition is at constant volume and heat rejection is at constant pressure | |
heat addition is at constant pressure and heat rejection is at constant pressure | |
heat addition is at constant pressure and heat rejection is at constant volume | |
heat addition is at constant volume and heat rejection is at constant volume |
Question 2 Explanation:
Heat addition is at constant pressure and heat
rejection is at constant volume.
Question 3 |
The indicated power developed by an engine with compression ratio of 8, is calculated
using an air-standard Otto cycle {constant properties). The rate of heat addition is
10 kW. The ratio of specific heats at constant pressure and constant volume is 1.4.
The mechanical efficiency of the engine is 80 percent.
The brake power output of the engine is ________ kW (round off to one decimal place).
The brake power output of the engine is ________ kW (round off to one decimal place).
4.5 | |
6.2 | |
3.8 | |
5.6 |
Question 3 Explanation:
\begin{aligned} \eta &=1-\frac{1}{r_{c}^{\gamma-1}}=1-\frac{1}{8^{0.4}}=0.5647 \\ \frac{W}{Q_{1}} &=0.5647 \\ W &=\frac{10 \times 5647}{1000}=5.647 \mathrm{kW} \\ \mathrm{BP} &=\eta_{m} \times W=0.8 \times 5.647 \\ &=4.5175 \mathrm{kW} \end{aligned}
Question 4 |
An air standard Otto cycle has thermal efficiency of 0.5 and the mean effective pressure of the cycle is 1000 kPa. For air, assume specific heat ratio \gamma =1.4 and specific gas constant R=0.287 kJ/kg.K. If the pressure and temperature at the beginning of the compression stroke are 100 kPa and 300 K, respectively, then the specific net work output of the cycle is______ kJ/kg (round off to two decimal places).
105.48 | |
325.65 | |
635.25 | |
708.77 |
Question 4 Explanation:
Given:
\eta_{0}=0.5
\mathrm{p}_{\mathrm{m}}=100 \mathrm{kPa}
\gamma=1.4
\mathrm{R}=0.287 \mathrm{kJ} / \mathrm{kg} \mathrm{K}
\mathrm{P}_{1}=100 \mathrm{kPa}
\mathrm{T}_{1}=300 \mathrm{K}
\eta_{0}=0.5=1-\frac{1}{(\mathrm{r})^{\gamma-1}}
0.5=1-\frac{1}{(\mathrm{r})^{1.4-1}}
\mathrm{r}=5.656
At state(1)
\mathrm{P}_{1} \mathrm{v}_{1}=\mathrm{RT}_{1}
100 v_{1}=0.287 \times 300
\mathrm{v}_{1}=0.861 \mathrm{m}^{3} / \mathrm{kg}
\mathrm{r}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}
\mathrm{v}_{2}=0.1522 \mathrm{m}^{3} / \mathrm{kg}
P_{m}=\frac{W_{\text {net }}}{v_{1}-v_{2}}
1000=\frac{\mathrm{W}_{\text {net }}}{0.861-0.1522}
\mathrm{W}_{\mathrm{net}}=708.77 \mathrm{kJ} / \mathrm{kg}
Question 5 |
The figure shows a heat engine (HE) working between two reservoirs. The amount of heat (Q_2) rejected by the heat engine is drawn by a heat pump (HP). The heat pump receives the entire work output (W) of the heat engine. If temperatures, T_1 \gt T_3\gt T_2, then the relation between the efficiency (\eta) of the heat engine and the coefficient of performance (COP) of the heat pump is
COP=\eta | |
COP=1+ \eta | |
COP=\eta^{-1} | |
COP=\eta^{-1}-1 |
Question 5 Explanation:
\begin{array}{l} \eta=1-\frac{Q_{2}}{Q_{1}} \\ \frac{Q_{2}}{Q_{1}}=1-\eta \ldots \ldots(I) \\ \Rightarrow W=Q_{1}-Q_{2}=Q_{3}-Q_{2} \ldots \ldots (I I) \\ \begin{aligned} C O P&=\frac{R E}{W}=\frac{Q_{3}}{Q_{1}-Q_{2}} \\ &=\frac{Q_{3} / Q_{1}}{1-\frac{Q_{2}}{Q_{1}}}=\frac{1}{\eta} \end{aligned} \end{array}
There are 5 questions to complete.
does this practise set have all 30 years of gate questions???
This practice set have all PYQ from GATE 2001. Questions in the Previous paper before 2001 are too old to consider as the way of asking the questions are changed.
Some other topic questions are posted in this topic, Once look at those questions and arrange accordingly. Other than this everything is alright. 😊👌