Question 1 |

Consider adiabatic flow of air through a duct. At a given point in the duct, velocity of air is 300 m/s, temperature is 330 K and pressure is 180 kPa. Assume that the air behaves as a perfect gas with constant c_p=1.005 kJ/kg.K. The stagnation temperature at this point is ______K (round off to two decimal places).

374.71 | |

352.24 | |

874.65 | |

458.32 |

Question 1 Explanation:

\begin{aligned} M_{1}&=\frac{V_{1}}{\sqrt{\gamma R T_{1}}}=\frac{300}{\sqrt{1.4 \times 287 \times 330}}=0.823 \\ \frac{T_{o1}}{T_{1}}&=1+\frac{\gamma-1}{2} M_{1}^{2}=1+\frac{1.4-1}{2}(0.823)^{2} \\ \frac{T_{o1}}{T_{1}}&=1.154 \\ T_{01}&=374.7037 \mathrm{~K} \end{aligned}

Question 2 |

A rigid insulated tank is initially evacuated. It is connected through a valve to a supply line that carries air at a constant pressure and temperature of 250 kPa and 400 K respectively. Now the valve is opened and air is allowed to flow into the tank until the pressure inside the tank reaches to 250 kPa at which point the valve is closed. Assume that the air behaves as a perfect gas with constant properties (c_p=1.005\; kJ/kg.K, c_v=0.718\; kJ/kg.K, R=0.287 kJ/kg.K). Final temperature of the air inside the tank is _______K (round off to one decimal place).

512 | |

248 | |

688 | |

560 |

Question 2 Explanation:

\begin{array}{l} T_{2}=\frac{C_{P}}{C_{V}} T_{1} \\ T_{2}=\left(\frac{1.005}{0.718}\right) \times 400 \\ T_{2}=559.888 \mathrm{~K} \approx 560 \mathrm{~K} \end{array}

Question 3 |

A gas is heated in a duct as it flows over a resistance heater. Consider a 101 kW electric heating system. The gas enters the heating section of the duct at 100 kPa and 27^{\circ}C with a volume flow rate of 15 m^3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 51 kW, the exit temperature of the gas is

(Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; C_p=1 kJ/kg.K; R=0.5 kJ/kg.K)

(Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; C_p=1 kJ/kg.K; R=0.5 kJ/kg.K)

32^{\circ}C | |

37^{\circ}C | |

53^{\circ}C | |

76^{\circ}C |

Question 3 Explanation:

\dot{\mathrm{m}}\left[\mathrm{h}_{1}+\frac{\mathrm{V}_{1}^{2}}{2000}\right]+\frac{\mathrm{d} \mathrm{Q}}{\mathrm{dt}}=\dot{\mathrm{m}}\left[\mathrm{h}_{2}+\frac{\mathrm{V}_{2}^{2}}{2000}\right]+\frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}

where, \dot{\mathrm{m}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{RT}_{1}}=\frac{100 \times 15}{0.5 \times 300}=10

Thus,

10\left[\mathrm{c}_{\mathrm{p}} \mathrm{T}_{1}\right]+101-51=10 \times\left[\mathrm{c}_{\mathrm{p}} \times \mathrm{T}_{2}\right] \qquad\left[\because \mathrm{v}_{1}=0, \mathrm{v}_{2}=0 \text { and } \frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}=0\right]

10 \times 1 \times 300+50=10 \times 1 \times \mathrm{T}_{2}

\mathrm{T}_{2}=305 \mathrm{K}=32^{\circ} \mathrm{C}

where, \dot{\mathrm{m}}=\frac{\mathrm{P}_{1} \mathrm{V}_{1}}{\mathrm{RT}_{1}}=\frac{100 \times 15}{0.5 \times 300}=10

Thus,

10\left[\mathrm{c}_{\mathrm{p}} \mathrm{T}_{1}\right]+101-51=10 \times\left[\mathrm{c}_{\mathrm{p}} \times \mathrm{T}_{2}\right] \qquad\left[\because \mathrm{v}_{1}=0, \mathrm{v}_{2}=0 \text { and } \frac{\mathrm{d} \mathrm{W}}{\mathrm{dt}}=0\right]

10 \times 1 \times 300+50=10 \times 1 \times \mathrm{T}_{2}

\mathrm{T}_{2}=305 \mathrm{K}=32^{\circ} \mathrm{C}

Question 4 |

During a non-flow thermodynamic process(1-2) executed by a perfect gas, the heat interaction is equal to the work interaction (Q_{1-2}=W_{1-2}) when the process is

Isentropic | |

Polytropic | |

Isothermal | |

Adiabatic |

Question 4 Explanation:

Given, \mathrm{Q}_{1-2}=\mathrm{W}_{1-2}

\therefore \Delta \mathrm{U}_{1-2}=0

\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0

\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}

So, the process is isothermal.

\therefore \Delta \mathrm{U}_{1-2}=0

\Rightarrow \mathrm{c}_{\mathrm{v}}\left[\mathrm{T}_{2}-\mathrm{T}_{1}\right]=0

\Rightarrow \mathrm{T}_{1}=\mathrm{T}_{2}

So, the process is isothermal.

Question 5 |

The volume and temperature of air (assumed to be an ideal gas) in a closed vessel is 2.87 m^{3} and 300K, respectively. The gauge pressure indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the gas constant of air is R=287\, J/kg\! \cdot \! K and the atmospheric pressure is 1 bar, the mass of air (in kg) in the vessel is

1.67 | |

3.33 | |

5.04 | |

6.66 |

Question 5 Explanation:

\begin{aligned} P_{g}&=5 \mathrm{bar}=50 \mathrm{kPa} \\ R&=287 \mathrm{J} / \mathrm{kgK}=0.287 \mathrm{kJ} / \mathrm{kg}^{\mathrm{K}} \end{aligned}

Ideas gas equation:

P V=m R T

Here P is absolute pressure:

\begin{aligned} P &=P_{a b s}=P_{\text {gauge }}+P_{\text {atm }} \\ &=50+101.325=151.325 \mathrm{kPa} \\ m &=\frac{P V}{R T}=\frac{151.325 \times 2.87}{0.287 \times 300}=5.04 \mathrm{kg} \end{aligned}

Ideas gas equation:

P V=m R T

Here P is absolute pressure:

\begin{aligned} P &=P_{a b s}=P_{\text {gauge }}+P_{\text {atm }} \\ &=50+101.325=151.325 \mathrm{kPa} \\ m &=\frac{P V}{R T}=\frac{151.325 \times 2.87}{0.287 \times 300}=5.04 \mathrm{kg} \end{aligned}

Question 6 |

Which of the following statements are TRUE with respect to heat and work?

(i) They are boundary phenomena

(ii) They are exact differentials

(iii) They are path functions

(i) They are boundary phenomena

(ii) They are exact differentials

(iii) They are path functions

both (i) and (ii) | |

both (i) and (iii) | |

both (ii) and (iii) | |

only (iii) |

Question 6 Explanation:

Heat and work are boundary phenomenon and
depend upon the path followed between initial
and final states. Both heat and work are inexact
differential since

\begin{array}{l} \int_{1}^{2} d W \neq W_{2}-W_{1} \\ \int_{1}^{2} d Q \neq Q_{2}-Q_{1} \end{array}

Hence statement (ii) alone is wrong.

Answer should be (B).

\begin{array}{l} \int_{1}^{2} d W \neq W_{2}-W_{1} \\ \int_{1}^{2} d Q \neq Q_{2}-Q_{1} \end{array}

Hence statement (ii) alone is wrong.

Answer should be (B).

Question 7 |

Two identical metal blocks L and M (specific heat = 0.4 kJ/kg.K), each having a mass of 5 kg, are initially at 313 K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293 K. The final temperature (in K) of block M is _______

334.36K | |

985.6K | |

452.6K | |

125.6K |

Question 7 Explanation:

Given data:

Specified heat,

\begin{aligned} c&=0.4 \mathrm{kJ} / \mathrm{kgK} \text{ each body}\\ \text{Mass:}m&=5 \mathrm{kg} \quad \text { each body } \end{aligned}

Initial temperature,

T_{1}=313 \mathrm{K} \quad \text { each body }

Final temprature of block L

T_{2}=293 \mathrm{K}

Let \quad T_{f}= Final temperature of block M,

Entropy change of block L

\Delta S_{L}=m c \log _{e} \frac{T_{2}}{T_{1}}

Entropy change of block M

\Delta S_{M}=m c \log _{e} \frac{T_{f}}{T_{1}}

Entropy change of universe,

\begin{aligned} \Delta S_{\text {univ }} &=\Delta S_{L}+\Delta S_{M} \\ &=mc \log _{e} \frac{T_{2}}{T_{1}}+mc \log _{e} \frac{T_{f}}{T_{1}} \end{aligned}

For a reversible refrigeration,

\begin{aligned} \Delta S_{\text {univ }}&=0 \\ m c \log _{e} \frac{T_{2}}{T_{1}}&+m_{c} \log _{e} \frac{T_{f}}{T_{1}}=0\\ mc \log _{e} \frac{T_{2} T_{f}}{T_{1}^{2}}&=0\\ \text{or}\quad \log _{e} \frac{T_{2} T_{f}}{T_{1}^{2}}&=0\\ \text{or}\quad \frac{T_{2} T_{f}}{T_{1}^{2}} &=e^{0}=1 \\ T_{f} &=\frac{T_{1}^{2}}{T_{2}}=\frac{(313)^{2}}{293}=334.36 \mathrm{K} \end{aligned}

Question 8 |

A certain amount of an ideal gas is initially at a pressure p_{1} and temperature T_{1} . First, it undergoes a constant pressure process 1-2 such that T_{2}=3T_{1}/4. Then, it undergoes a constant volume process 2-3 such that T_{3}=T_{1}/2. The ratio of the final volume to the initial volume of the ideal gas is

0.25 | |

0.75 | |

1 | |

1.5 |

Question 8 Explanation:

Process 1-2: Cooling at p=C

T=\frac{3 T_{1}}{4} \text { given condition }

\text{Or} \quad \frac{T_{2}}{T_{1}}=\frac{3}{4}

According to Charle's law

\frac{V_{2}}{V_{1}}=\frac{T_{2}}{T_{1}}=\frac{3}{4}

Process 2-3: Cooling at V=C

\begin{aligned} T_{3} &=\frac{T_{2}}{2} \text { given condition } \\ \frac{\text { Final volume }}{\text { Initial volume }} &=\frac{V_{3}}{V_{1}}=\frac{V_{2}}{V_{1}} \quad \because V_{3}=V_{2} \\ &=\frac{3}{4}=0.75 \end{aligned}

Question 9 |

Heat and work are

Intensive properties | |

Extensive properties | |

Point functions | |

Path functions |

Question 9 Explanation:

Heat and work are path function. These are not point function.

Question 10 |

If a closed system is undergoing an irreversible process, the entropy of the system

Must increase | |

Always remains constant | |

Must decrease | |

Can increase, decrease or remain constant |

Question 10 Explanation:

If a closed system is undergoing an irreversible
process, the entropy of the system can increase,
decrease or remain constant

There are 10 questions to complete.

great work,user friendly site, but one suggestion for some questions (like q1) ,please add some explanation like formula name or concept usedlike that 👍👏