# IC Engine

 Question 1
For an air-standard Diesel cycle,
 A heat addition is at constant volume and heat rejection is at constant pressure B heat addition is at constant pressure and heat rejection is at constant pressure C heat addition is at constant pressure and heat rejection is at constant volume D heat addition is at constant volume and heat rejection is at constant volume
GATE ME 2020 SET-2   Thermodynamics
Question 1 Explanation:
Heat addition is at constant pressure and heat rejection is at constant volume.
 Question 2
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).
 A 4.5 B 6.2 C 3.8 D 5.6
GATE ME 2020 SET-1   Thermodynamics
Question 2 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 3
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).
 A 105.48 B 325.65 C 635.25 D 708.77
GATE ME 2019 SET-2   Thermodynamics
Question 3 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 4
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 A $COP=\eta$ B $COP=1+ \eta$ C $COP=\eta^{-1}$ D $COP=\eta^{-1}-1$
GATE ME 2019 SET-2   Thermodynamics
Question 4 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} Question 5
A vehicle powered by a spark ignition engine follows air standard Otto cycle ($\gamma$ =1.4). The engine generates 70 kW while consuming 10.3 kg/hr of fuel. The calorific value of fuel is 44,000 kJ/kg. The compression ratio is _______ (correct to two decimal places).
 A 2.65 B 5.92 C 7.61 D 6.38
GATE ME 2018 SET-2   Thermodynamics
Question 5 Explanation:
\begin{aligned} \gamma &=1.4 \\ \mathrm{B.P.} &=70 \mathrm{kN} \\ \dot{m}_{f} &=10.3 \mathrm{kg} / \mathrm{hr} \\ C V &=44000 \mathrm{kJ} / \mathrm{kg} \\ \eta &=\frac{B P}{\dot{m}_{f} \times C V}=\frac{70}{\frac{10.3}{3600} \times 44000} \times 100 \% \\ \eta &=0.556 \\ \eta_{\mathrm{otto}} &=1-\frac{1}{(n)^{\gamma-1}}=0.556 \\ 1-\frac{1}{(n)^{1.4-1}} &=0.556 \\ \frac{1}{(n)^{0.4}} &=0.4439 \\ n &=7.61 \end{aligned}
 Question 6
Air is held inside a non-insulated cylinder using a piston (mass M=25 kg and area A=100 $cm^{2}$ and stoppers (of negligible area), as shown in the figure. The initial pressure $P_{i}$ and temperature $T_{i}$ of air inside the cylinder are 200 kPa and $400^{\circ}$C, respectively. The ambient pressure $P_{\infty }$ and temperature $T_{\infty }$ are 100 kPa and $27^{\circ}C$, respectively. The temperature of the air inside the cylinder $(^{\circ}C)$ at which the piston will begin to move is _________ (correct to two decimal places). A 100.05 B 185.12 C 146.02 D 200.25
GATE ME 2018 SET-2   Thermodynamics
Question 6 Explanation: \begin{aligned} P_{\infty} &=100 \mathrm{kPa} \\ P_{\text {piston }} &=\frac{M g}{A}=\frac{25 \times 9.81 \times 10^{-3} \mathrm{kN}}{0.01 \mathrm{m}^{2}} \\ P_{\text {piston }} &=24.525 \mathrm{kPa} \\ \text { Total external pressure } &=P_{\infty}+P_{\text {piston }}=124.525 \mathrm{kPa} \end{aligned}
Initial pressure of air, $P_{i}=200 \mathrm{kPa}$
Piston will be about to come down if system pressure is equal to total pressure i.e. 124.525kPa
$\Rightarrow$ Final pressure of air at which is about to move is $p_{f}=124.525 \mathrm{kPa}$
Applying mass conservation at two states $i \xi f$
\begin{aligned} M_{i} &=M_{f} \\ \frac{P_{i} V_{i}}{R T_{i}} &=\frac{P_{f} V_{f}}{R T_{f}}=V_{i}=V_{f} \\ \frac{200 \mathrm{kPa}}{673 \mathrm{K}} &=\frac{124.525 \mathrm{kPa}}{T_{f}} \\ T_{f} &=419.026 \mathrm{K} \\ T_{f} &=146.026^{\circ} \mathrm{C} \end{aligned}
 Question 7
A frictionless circular piston of area $10^{-2}\: m^{2}$ and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 1000 kg/m^{3} as shown in the figure.
The container has a hole of area $10^{-3}\:m^{2}$ at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be in compressible, the magnitude of the piston velocity (in m/s) at the instant shown is _____ (correct to three decimal places). A 0.522 B 1.456 C 2.257 D 3.952
GATE ME 2018 SET-2   Thermodynamics
Question 7 Explanation:
\begin{aligned} A_{1} V_{1} &=A_{2} V_{2} \\ V_{2} &=\left(\frac{A_{1}}{A_{2}}\right) V_{1} \\ \frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+Z_{1} &=\frac{P_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+Z_{2} \\ \left.\frac{P_{a t m}+\frac{100 \times 10}{10^{-2}}}{\rho \times g}\right] \times \frac{V_{1}^{2}}{2 g}+0.5 &=\frac{P_{a t m}}{\rho g}+\frac{A_{1}^{2}}{A_{2}^{2}} \times \frac{V_{1}^{2}}{2 g} \\ \frac{1000}{10^{-2} \times 100 \times 10}+0.5 &=\left(\frac{A_{1}^{2}}{A_{2}^{2}}-1\right) \frac{V_{1}^{2}}{2 g} \\ 10+0.5&=(10^{2}-1)\frac{V_{1}^{2}}{2g}\\ V_{1}^{2} &=\frac{10.5 \times 2 \times 10}{99}=2.12 \\ V_{1} &=1.456 \mathrm{m} / \mathrm{s} \end{aligned}
 Question 8
An engine working on air standard Otto cycle is supplied with air at 0.1 MPa and $35^{\circ}C$. The compression ratio is 8. The heat supplied is 500 kJ/kg. Property data for air: $c_{p}$ = 1.005 kJ/kg K, $c_{v}$ = 0.718 kJ/kg K, R = 0.287 kJ/kg K. The maximum temperature (in K) of the cycle is _________ (correct to one decimal place).
 A 1204.32 B 1620.5 C 1403.97 D 1113.32
GATE ME 2018 SET-1   Thermodynamics
Question 8 Explanation: $\begin{array}{c} P_{1}=0.1 \mathrm{MPa}, T_{1}=35^{\circ} \mathrm{C}=308 \mathrm{K} \\ \frac{V_{1}}{V_{2}}=n=8 \\ Q_{s}=500 \mathrm{kJ} / \mathrm{kg} \\ c_{p}=1.005 \mathrm{kJ} / \mathrm{kgK} \\ c_{v}=0.718 \mathrm{kJ} / \mathrm{kgK} \\ R=0.287 \mathrm{kJ} / \mathrm{kgK} \\ \gamma=\frac{c_{p}}{c_{v}}=1.399 \simeq 1.40 \\ T_{3}=T_{\max }=? \end{array}$
For process 1-2
\begin{aligned} P_{1} V_{1}^{\gamma} &=P_{2} V_{2}^{\gamma} \\ P_{2} &=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=0.1 \times(8)^{1.4} \\ P_{2} &=1.8379 \mathrm{MPa}\\ \text{and}\qquad \frac{P_{1} V_{1}}{T_{1}} &=\frac{P_{2} V_{2}}{T_{2}} \Rightarrow T_{2} \frac{P_{2} V_{2}}{P_{1} V_{1}} T_{1} \\ T_{2} &=\frac{1.8379 \times 1}{0.1 \times 8} \times 308 \\ T_{2} &=707.6 \mathrm{K}\\ \text{For process } 2 \rightarrow 3 \\ Q_{s}&=C_{V}\left(T_{3}-T_{2}\right)=500 \mathrm{kJ} / \mathrm{kg}\\ 0.718\left(T_{3}-707.6\right) &=500 \\ T_{3} &=1403.97 \mathrm{K} \end{aligned}
 Question 9
Propane (C3H8) is burned in an oxygen atmosphere with 10% deficit oxygen with respect to the stoichiometric requirement. Assuming no hydrocarbons in the products, the volume percentage of CO in the products is __________
 A 12.20% B 10.13% C 14.28% D 9.67%
GATE ME 2016 SET-1   Thermodynamics
Question 9 Explanation:
For stoichiometric burning (i.e., chemically correct),
number of moles of $\mathrm{O}_{2}$ required =5.
As it si burnt with 10 % deficient \mathrm{O}_{2} , it will generate CO.
That is, the new equation is
$\mathrm{C}_{3} \mathrm{H}_{8}+0.9 \times 5 \mathrm{O}_{2} \rightarrow \alpha \mathrm{CO}_{2}+\beta \mathrm{CO}+\gamma \mathrm{H}_{2} \mathrm{O}$
Balancing carbon atoms, we get
$3=\alpha+\beta \quad \cdots(i)$
Balancing hydrogen atoms, we get
\begin{aligned} 8&=2 \gamma\\ \text{or }\quad \gamma&=4\\ \end{aligned}
Balancing oxygen atoms, we get
\begin{aligned} 0.9 \times 10 &=2 \alpha+\beta+\gamma \\ 9 &=2 \alpha+\beta+4 \\ 5 &=2 \alpha+\beta\quad \cdots(ii) \end{aligned}
Eq. (ii) - Eq. (i), we get
\begin{aligned} 2&=\alpha\\ \text{or }\quad \alpha&=2 \end{aligned}
$\mathrm{C}_{3} \mathrm{H}_{8}+\mathrm{aO}_{2} \rightarrow \mathrm{bCO}_{2}+\mathrm{cH}_{2} \mathrm{O}$
Balancing carbon atoms, we get
\begin{aligned} 3&=b\\ \text{or } \quad b&=3\\ \end{aligned}
Balancing hydrogen atoms, we get
\begin{aligned} 8&=2 c\\ \text{or}\quad c&=4\\ \end{aligned}
Balancing oxygen atoms, we get
\begin{aligned} 2 a &=2 b+c \\ 2 a &=2 \times 3+4=10 \\ a &=5 \end{aligned}
Substituting $\alpha=2$ in Eq. (ii), we get
$\begin{array}{l} \quad 5=2 \times 2 \times \beta \\ \text { or } \quad \beta=1 \\ \therefore \mathrm{C}_{3} \mathrm{H}_{8}+4.5 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2}+\mathrm{CO}+4 \mathrm{H}_{2} \mathrm{O} \end{array}$
The volume percentage of CO in the product
$\begin{array}{l} =\left(\frac{\beta}{\alpha+\beta+\gamma}\right) \times 100 \\ =\left(\frac{1}{2+1+4}\right) \times 100 \\ =14.28 \% \end{array}$
 Question 10
An air-standard Diesel cycle consists of the following processes:
1-2: Air is compressed isentropically.
2-3: Heat is added at constant pressure.
3-4: Air expands isentropically to the original volume.
4-1: Heat is rejected at constant volume.
If $\gamma$ and T denote the specific heat ratio and temperature, respectively, the efficiency of the cycle is
 A $1-\frac{T_{4}-T_{1}}{T_{3}-T_{2}}$ B $1-\frac{T_{4}-T_{1}}{\gamma (T_{3}-T_{2})}$ C $1-\frac{\gamma (T_{4}-T_{1})}{T_{3}-T_{2}}$ D $1-\frac{T_{4}-T_{1}}{(\gamma -1)(T_{3}-T_{2})}$
GATE ME 2015 SET-3   Thermodynamics
Question 10 Explanation: Heat added to the system
$Q_H=c_p(T_3-T_2)$
Heat rejected to the system
$Q_L=c_v(T_4-T_1)$
Therefore, Efficinecy is given by
\begin{aligned} \eta&=1-\frac{Q_L}{Q_H}\\ \eta&=1-\frac{c_v}{c_p}\frac{T_4-T_1}{T_3-T_2}\\ \eta&=1-\frac{T_4-T_1}{\gamma (T_3-T_2)} \end{aligned}
There are 10 questions to complete.

### 3 thoughts on “IC Engine”

1. does this practise set have all 30 years of gate questions???

• 2. 