Question 1 |

An adiabatic vortex tube, shown in the figure given below is supplied with 5 kg/s of air (inlet 1) at 500 kPa and 300 K. Two separate streams of air are leaving the device from outlets 2 and 3. Hot air leaves the device at a rate of 3 kg/s from outlet 2 at 100 kPa and 340 K, and 2 kg/s of cold air stream is leaving the device from outlet 3 at 100 kPa and 240 K.

Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).

Assume constant specific heat of air is 1005 J/kg.K and gas constant is 287 J/kg.K. There is no work transfer across the boundary of this device. The rate of entropy generation is ________kW/K (round off to one decimal place).

1.2 | |

4.3 | |

2.2 | |

3.8 |

Question 1 Explanation:

\begin{aligned} \left(\frac{d S}{d t}\right)_{C . V} &=\dot{S}_{i}+\dot{S}_{g e n}-\dot{S}_{\theta} \\ \dot{S}_{\text {gen }} &=\dot{S}_{e}-\dot{S}_{i} \\ &=\dot{m}_{2} s_{2}+\dot{m}_{3} s_{3}-\dot{m}_{1} s_{1} \\ &=3\left(s_{2}-s_{1}\right)+2\left(s_{3}-s_{1}\right) \\ &=3 \times 0.587+2(0.237) \\ &=2.235 \mathrm{~kW} / \mathrm{K} \simeq 2.2 \mathrm{~kW} / \mathrm{K} \end{aligned}

Question 2 |

The fundamental thermodynamic relation for a rubber band is given by dU=TdS+\tau dL, where T is the absolute temperature, S is the entropy, \tau is the tension in the rubber band, and L is the length of the rubber band. Which one of the following relations is CORRECT:

\tau =\left ( \frac{\partial U}{\partial S}\right )_L | |

\left ( \frac{\partial T}{\partial L}\right )_S =\left ( \frac{\partial \tau}{\partial S}\right )_L | |

\left ( \frac{\partial T}{\partial S}\right )_L =\left ( \frac{\partial \tau}{\partial L}\right )_S | |

T=\left ( \frac{\partial U}{\partial S}\right )_ \tau |

Question 2 Explanation:

\begin{aligned} d U&=T d s+\tau d L \\ \left(\frac{\partial T}{\partial L}\right)_{S}&=\left(\frac{\partial \tau}{\partial S}\right)_{L}\\ \text { Comparing with } M=\left(\frac{\partial z}{\partial x}\right)_{y}& \\ T&=\left(\frac{\partial U}{\partial S}\right)_{L}\\ \text { Comparing with } N=\left(\frac{\partial z}{\partial y}\right)_{x} \\ \tau&=\left(\frac{\partial U}{\partial L}\right)_{S} \\ M&=\left(\frac{\partial z}{\partial x}\right)_{y} \\ N&=\left(\frac{\partial z}{\partial y}\right)_{x} \\ d z&=M d x+N d y \end{aligned}

If z is exact different

\left(\frac{\partial M}{\partial y}\right)_{x}=\left(\frac{\partial N}{\partial x}\right)_{y}

If z is exact different

\left(\frac{\partial M}{\partial y}\right)_{x}=\left(\frac{\partial N}{\partial x}\right)_{y}

Question 3 |

In which of the following pairs of cycles, both cycles have at least one isothermal process?

Diesel cycle and Otto cycle | |

Carnot cycle and Stirling cycle | |

Brayton cycle and Rankine cycle | |

Bell-Coleman cycle and Vapour compression refrigeration cycle |

Question 3 Explanation:

1.Brayton

\begin{array}{ll} 1 \text { to } 2: & S=C \\ 2 \text { to } 3: & P=C \\ 3 \text { to } 4: & S=C \\ 4 \text { to } 1: & P=C \end{array}

2. VCRS

1 to 2 : Isentropic compression

2 to 3: constant pressure (P=C)

3 to 4: Isenthalpic expansion \left(h_{3}=h_{4}\right)

4 to 1: Constant pressure (P=C)

3. Carnot

1 to 2 : lsentropic compression

2 to 3 : Isothermal heat addition

3 to 4 : lsentropic expansion

4 to 1 : Isothermal heat rejection

4. Bell Coleman

1 to 2 : lsentropic compression

2 to 3 : Constant pressure

3 to 4 : lsentropic expansion

4 to 1 : Constant pressure

5. Rankine

1 to 2 : lsentropic compression

2 to 3 : Constant pressure

3 to 4 : lsentropic expansion

4 to 1 : Constant pressure

6. Diesel

1 to 2 : lsentropic compression

2 to 3 : Constant pressure (P = C)

3 to 4 : lsentropic expansion

4 to 1 : Constant volume

7. Otto

1 to 2 : lsentropic compression

2 to 3 : Constant pressure (V = C)

3 to 4 : lsentropic expansion

4 to 1 : Constant volume (V = C)

\begin{array}{ll} 1 \text { to } 2: & S=C \\ 2 \text { to } 3: & P=C \\ 3 \text { to } 4: & S=C \\ 4 \text { to } 1: & P=C \end{array}

2. VCRS

1 to 2 : Isentropic compression

2 to 3: constant pressure (P=C)

3 to 4: Isenthalpic expansion \left(h_{3}=h_{4}\right)

4 to 1: Constant pressure (P=C)

3. Carnot

1 to 2 : lsentropic compression

2 to 3 : Isothermal heat addition

3 to 4 : lsentropic expansion

4 to 1 : Isothermal heat rejection

4. Bell Coleman

1 to 2 : lsentropic compression

2 to 3 : Constant pressure

3 to 4 : lsentropic expansion

4 to 1 : Constant pressure

5. Rankine

1 to 2 : lsentropic compression

2 to 3 : Constant pressure

3 to 4 : lsentropic expansion

4 to 1 : Constant pressure

6. Diesel

1 to 2 : lsentropic compression

2 to 3 : Constant pressure (P = C)

3 to 4 : lsentropic expansion

4 to 1 : Constant volume

7. Otto

1 to 2 : lsentropic compression

2 to 3 : Constant pressure (V = C)

3 to 4 : lsentropic expansion

4 to 1 : Constant volume (V = C)

Question 4 |

One kg of air in a closed system undergoes an irreversible process from an initial state
of p_1 = 1 bar (absolute) and T_1 = 27^{\circ}C, to a final state of p_2 = 3 bar (absolute) and T_2= 127^{\circ}C. If the gas constant of air is 287 J/kg.K and the ratio of the specific heats
\gamma = 1.4, then the change in the specific entropy (in J/kg.K) of the air in the process is

-26.3 | |

28.4 | |

172 | |

indeterminate, as the process is irreversible |

Question 4 Explanation:

\begin{array}{c} P_{1}=1 \text { bar, } P_{2}=1 \text { bar, } T_{1}=300 \mathrm{K}, T_{2}=400 \mathrm{K} \\ c_{p}=\frac{\gamma R}{\gamma-1}=1005 \mathrm{J} / \mathrm{kgK} \\ \Delta S=c_{p} \ln \frac{T_{2}}{T_{1}}-R \ln \frac{P_{2}}{P_{1}}=1005 \ln \frac{400}{300}-287 \mathrm{ln} \frac{3}{1}=-26.32 \mathrm{J} / \mathrm{kgK} \end{array}

Question 5 |

If a reversed Carnot cycle operates between the temperature limits of 27^{\circ}C and -3^{\circ}C,
then the ratio of the COP of a refrigerator to that of a heat pump (COP of refrigerator/
COP of heat pump) based on the cycle is __________ (round off to 2 decimal places).

0.2 | |

0.8 | |

0.9 | |

1.5 |

Question 5 Explanation:

\begin{aligned} T_{H}&=27^{\circ} \mathrm{C}=300 \mathrm{K} \\ T_{L}&=-3^{\circ} \mathrm{C}=270 \mathrm{K} \\ \frac{\mathrm{COP}_{\mathrm{ref}}}{\mathrm{COP}_{\mathrm{HP}}}&=\frac{\frac{T_{L}}{T_{H}-T_{L}}}{\frac{T_{H}}{T_{H}-T_{L}}}=\frac{T_{L}}{T_{H}}=\frac{270}{300}=0.9 \end{aligned}

There are 5 questions to complete.

Please check answer for question 27. It should be (A) but answer is given (D).

please check the 19 question for u r doubt