Option 4 : 2.0

__Concept:__

\({\rm{COP\;of\;Carnot\;Heat\;pump}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}\)

__Calculation:__

**Given: **

T_{1} = 327° C = 600 K, T_{2} = 27° C = 300 K

\(\therefore COP = \frac{{{T_1}}}{{{T_1} - {T_2}}}\)

\(\therefore COP = \frac{{600}}{{600 - 300}} = \frac{{600}}{{300}} = 2\)

**∴ ****COP of Carnot heat pump = 2**

Option 3 : 6 kW

__Concept:__

\(CO{P_{HP}} = \frac{{Desired\;Output}}{{Required\;Input}} = \frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}}\)

**Calculation:**

**Given:**

T_{1} = 27°C = 300 K, T_{2} = -23°C = 250 K, W = 1 kW

Now

\(\frac{{{Q_1}}}{1} = \frac{{{300}}}{{{300} - {250}}} \Rightarrow Q = 6 \ kW\)

Option 2 : 4

__Concept__:

As there is no mention that it works on the Carnot cycle, therefore the **COP of Heat Pump is given by,**

**\(COP = \frac{{Heat\;rejection}}{{Heat\;rejection - Heat\;addition}}\)**

__Calculation__:

__Given__:

Q_{add} = 750 kW, Q_{rej }= 1000 kW, T_{H} = 30°C, T_{L} = 15°C

∴ \(COP~=~\frac{1000}{1000-750}\)

**COP = 4**

Option 2 : 6.4 lakh kCal / hour

__Concept:__

- Waste heat is the heat that is produced by a machine, or other process that uses energy, as a by product of doing work.
- All such processes gives off some waste heat as a fundamental result of the laws of thermodynamics.
- Waste heat occurs in almost all thermal and mechanical process.
- Sources of waste heat include hot combustion gases discharged to the atmosphere, heated water released into the environment, heated product existing industrial processes and heat transfer from hot equipment surfaces.
- The most significant amount of waste heat are being host in industrial and energy generation processes.
- Exact amount of waste heat is difficult to qualify, but around 20 to 50 percent of industrial energy consumption is ultimately discharged as waste heat and 18 to 30 percent of this waste heat could be utilized.

__Formula:__

Waste Heat Potential = Real Power × 8 × 0.25 × (temperature – 180)

__Calculation:__

Given that, Apparent power = 2200 kVA

Real Power = 1000 kW

Temperature = 500°C

Waste Heat Potential = 1000 × 8 × 0.25 × (500 - 180)

= 8 × 25 × 10 (320)

= 6,40,000 kcal / hr

**= 6.4 lakh kcal / hr**

Option 1 : 81.4 kg

**Concept:**

Formation of ice takes place as:

∴ Total heat, Q = m c ΔT + m (L.H.)

Refrigeration cycle is working between - 8˚C (265 K) and 20˚C (293 K).

COP \(=\frac{{{{\rm{T}}_{\rm{L}}}}}{{{{\rm{T}}_{{\rm{H}} - {{\rm{T}}_{\rm{L}}}}}}} = \frac{Q}{W}\)

**Calculation:**

Given:

m = 1 kg, Cpw = 4.18 kJ/kg.K, Latent heat (L.H.) = 335 KJ/kg

Q = m (20 × 4.18 + 335) = 418.6 m kJ

Power Input, W = 1 kWh = 3600 J

COP = \(\frac{{265}}{{293 - 265}}\) = 9.464

COP = ^{\(\frac{{{\rm{Q\;}}}}{{{\rm{W\;}}}} = \frac{{418.6\;m}}{{3600}}\)}

9.464 = \(\frac{{418.6\;m}}{{3600}}\)

A reversed Carnot cycle refrigerator maintains a temperature of -5°C. The ambient air temperature is 35°C. The heat gained by the refrigerator at a continuous rate is 2.5 kJ/s. The power (in watt) required to pump this heat out continuously is _______

__Concept:__

For a reversible Carnot cycle:

\(\oint \frac{{dQ}}{T} = 0\)

__Calculation:__

__Given:__

Q_{1} = -2.5 kJ/s, T_{1} = -5°c = 268 K, T_{2} = 35°c = 308 K

As the cycle is reversible

\(\oint \frac{{dQ}}{T} = 0\)

\(\frac{{{Q_1}}}{{{T_1}}} + \frac{{{Q_2}}}{{{T_2}}} = 0\)

\( - \frac{{2.5}}{{268}} + \frac{{{Q_2}}}{{308}} = 0\)

\( \Rightarrow {Q_2} = \frac{{2.5 \times 308}}{{268}} = 2.873\frac{{kJ}}{s}\)

W = Q_{1} – Q_{2}

W = 2.873 – 2.5 = 0.37313 kJ/s = 373.13 J/s = 373.13 W

Option 4 : 7.75

__Concept:__

COP of Heat Pump is given as, \(COP=\frac{T_H}{T_H-T_L}\)

__Calculation__:

Given:

T_{L} = -3°C = 270 K, T_{H} = 37°C = 310 K

\(COP = \frac{{{T_H}}}{{{T_H} - {T_L}}} = \frac{{310}}{{310-270}} = \frac{{310}}{{40}} = 7.75\)

Option 4 : 3.5

__Concept:__

If Carnot refrigerator and a Carnot heat pump are operating between the same two thermal reservoirs.

(C.O.P)_{heat pump} = 1 + (C.O.P)_{refrigerator}

__Calculation:__

__Given:__

Carnot refrigerator and a Carnot heat pump are operating between the same two thermal reservoirs.

(C.O.P)_{refrigerator} = 2.5

(C.O.P)_{heat pump} = 1 + (C.O.P)_{refrigerator}

(C.O.P)_{heat pump} = 2.5 + 1 = **3.5**

Option 1 : 5

__Concept:__

Heat Rejected into the condenser = Heat absorbed into the Evaporator + Work input to the compressor

Coefficient of Performance = \(\frac{Desired~Effect~\left( Heat~absorbed~in~the~Evaporator \right)}{Work~Input~to~the~Compressor}\)

__Calculation:__

Given that Heat Rejected in Condenser = 1500 kJ/Kg

Work input to the compressor = 250 kJ/kg

Heat absorbed by refrigerant in the evaporator = 1500 - 250 = 1250 kJ/kg

Now \(COP=\frac{1250}{250}=5\)A heat pump with refrigerant R22 is used for space heating between temperature limits of −20°C and 25°C. The heat required is 200 MJ/h. Assume specific heat of vapour at the time of discharge as 0.98 kJ/kg.K. Other relevant properties are given below. The enthalpy (in kJ/kg) of the refrigerant at isentropic compressor discharge is _______

Saturation temperature |
Pressure |
Specific enthalpy |
Specific entropy |
||

T |
P(MN/m |
h |
h |
s |
S |

-20 |
0.2448 |
177.21 |
397.53 |
0.9139 |
1.7841 |

25 |
1.048 |
230.07 |
413.02 |
1.1047 |
1.7183 |

__Given:__

T_{1 }= -20°C = 253 K, T3 = 25°C = 298 K

S_{1} = S_{2} = 1.7841 kJ/kg.K, S_{3g} = 1.7183 kJ/kg.K

h_{3g} = 413.02 kJ/kg

\(\begin{array}{l} {{\rm{S}}_1} = {{\rm{S}}_2} = {{\rm{S}}_{3{\rm{g}}}} + {{\rm{C}}_{\rm{p}}}\ln \frac{{{{\rm{T}}_2}}}{{{{\rm{T}}_3}}}\\ ⇒ 1.7841 = 1.7183 + 0.98\ln \frac{{{{\rm{T}}_2}}}{{298}} \end{array}\)

⇒ \(\ln \left( {\frac{{{T_2}}}{{298}}} \right) = 0.0671\)

⇒ \(\frac{{{T_2}}}{{298}} = {e^{0.067}} = 1.069\)

⇒ T_{2} = 1.069 × 298 = 318.65 K

h_{2} = h_{3g} + C_{p} (T_{2} – T_{3})

⇒ h_{2} = 413.02 + 0.98 (318.68 - 298)

⇒ h_{2} = 433.2864 kJ/kg

Option 3 : Heated

**Concept:**

As the room is closed, so no heat will go outside. Now, the refrigerator working is as shown.

Q_{1} = Q_{2} + W

Now, in our case when the door is open, then the Q_{2} is heat absorbed from the entire room instead of just the refrigerator compartment and Q_{1} is the heat rejected to the room.

As of Q_{1} > Q_{2}

Option 1 : 3

**Concept:**

A **heat pump** is a device used to warm places by transferring thermal energy from a cooler space to a warmer space using the refrigeration cycle.

The **coefficient of performance** or **COP** of a heat pump is a ratio of useful heating provided to work (energy) required.

\(\mathbf{COP_{HP}={Heat\: Rejected \over Work\: Provided}}\)\(\mathbf{= {Q_1\over W}}\)

**Calculation:**

**Given:**

Rate of heat rejected, Q_{1} = 360 kJ/min = \( {360 \over 60}\) kJ/sec = 6 kW, power supplied, W = 2 kW

coefficient of performance of a heat pump

COP = \({Q_1 \over W}\)

= \( {6\over 2}\)

= **3**

__Additional Information__

The coefficient of performance or COP of a refrigerator is a ratio of heat absorbed to work (energy) required.

\(\mathbf{COP_{RE}={Heat\: Absorbed \over Work\: Provided}}\)\(\mathbf{= {Q_2 \over W}}\)

∴ **COP _{HP} = COP_{RE} + 1**

Option 4 : 10

__ Concept__:

**Coefficient of Performance of Heat Pump**:

\({\left( {C.O.P} \right)_{H.P}} = \frac{{{T_H}}}{{{T_H} - {T_L}}}\)

Where TL is the lower temperature limit and TH is the high-temperature limit.

__Calculation:__

__ Given__:

TL = -3°C = 273 – 3 = 270 K, TH = 27°C = 273 + 27 = 300 K

COP of the Heat Pump:

\({\left( {C.O.P} \right)_{HP}} = \frac{{{T_H}}}{{{T_H} - {T_L}}} = \frac{{300}}{{300 - 270}} = 10\)

Option 3 : 70 kW

__Concept:__

The COP of refrigerator is given as,

Ideal \(COP = \frac{{{T_L}}}{{{T_H} - {T_L}}}\) and Actual \(COP = \frac{Q}{W_{input}}\)

__Calculation:__

**Given:**

Q̇ = 20 × 3.5 = 70 kW, TL = 200 K, TH = 300 K

Ideal \(COP = \frac{{{200}}}{{{300} - {200}}}\) = 2

Actual COP = \(\frac{1}{2}\) × Ideal COP = \(\frac{1}{2}\) × 2 = 1

Actual COP = 1

Also, Actual COP = \(\frac{{\dot Q}}{{\dot W}} = \frac{{70}}{{\dot W}}\)

∴ \(\dot {W}\) = **70**** **kW.

Option 3 : Decreases

Explanation:

- The latent heat is the amount of heat transfer required to cause a
**phase change per unit mass**of a substance at constant pressure and temperature. **With the increase in pressure**the boiling point of the liquid increases and a lesser amount of energy needed to overcome the intermolecular force thus the**latent heat**of steam required is**decreased.**- Conversely with the decrease in pressure the boiling point of the liquid decreases thus higher energy needed to overcome the intermolecular forces thus the latent heat of steam required is more to cause a phase change.
- If we keep increases the pressure a the point will come at which latent heat of steam becomes zero this point is called
**Critical point.**

****

Option 3 : 7.5

**Concept:**

\({\rm{CO}}{{\rm{P}}_{{\rm{heat\;pump}}}} = \frac{{{\rm{Refrigeration\;Effect}}}}{{{\rm{Work\;Input}}}} = \frac{{{{\rm{Q}}_{\rm{H}}}}}{{{{\rm{Q}}_{\rm{H}}} - {{\rm{Q}}_{\rm{L}}}}} = \frac{{{{\rm{T}}_{\rm{H}}}}}{{{{\rm{T}}_{\rm{H}}} - {{\rm{T}}_{\rm{L}}}}}\)

COP of Heat Pump is defined as the ratio of heat transfer to the hot reservoir to network transfer to the heat pump.

\(COP = \frac{{{Q_S}}}{W} = \frac{60}{8}=7.5\)

**Concept:**

Heat Pump:

- A heat pump is a device that works on a reversed Carnot cycle and transfers heat from a lower temperature body to a higher temperature body.
- \(CO{P_{HP}} = \frac{{{Q_1}}}{W} \)

**Calculation:**

**Given:**

Q_{1} (heat rejected from HP) = 10 + 15 = 25 kW, W = 15 kW

\(CO{P_{HP}} = \frac{{{Q_1}}}{W} = \frac{{10 + 15}}{{15}} = 1.67\)

__Important Points__

COP of heat pump = \(\frac{{{T_H}}}{{{T_H} - {T_L}}}\) is used only for reversible heat pump.

\(CO{P_{HP}} = \frac{{{Q_1}}}{W}\;\) Can be used for any type of heat pump.

Option 2 : Decreasing the higher temperature

__Explanation:__

Refrigerator:

A refrigerator is a device that works on a reverse Carnot cycle and extracts heat from a lower temperature body to keep the temperature of the body lower than the surrounding temperature and by taking work input it transfers heat to the higher temperature body or surrounding.

Refrigerating Effect R.E. = Q1

Work input = Q2 - Q1

\({\bf{Coefficient}}\;{\bf{of}}\;{\bf{Performance}}\;\left( {{\bf{COP}}} \right) = \frac{{{\bf{Refrigerating}}\;{\bf{Effect}}}}{{{\bf{Work}}\;{\bf{Input}}}} = \frac{{{{\bf{Q}}_{\bf{1}}}}}{{\bf{W}}}\)

For a reversible engine:

\(\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}\)

\({\bf{CO}}{{\bf{P}}_{{\bf{refrigerator}}}} = \frac{{{{\bf{Q}}_{\bf{1}}}}}{{{{\bf{Q}}_{\bf{2}}} - {{\bf{Q}}_{\bf{1}}}}} = \frac{{{{\bf{T}}_{\bf{1}}}}}{{{{\bf{T}}_{\bf{2}}} - {{\bf{T}}_{\bf{1}}}}}\)

If lower temperature ie T_{1 }is fixed then COP depends on higher temperature T_{2}.

\({\bf{CO}}{{\bf{P}}_{{\bf{refrigerator}}}} = \frac{{{{\bf{T}}_{\bf{1}}}}}{{{{\bf{T}}_{\bf{2}}} - {{\bf{T}}_{\bf{1}}}}}\)

We can see with the decrease in higher temperature denominator will decrease and COP will increase.

Option 1 : 9

__Concept:__

Refrigerator:

A refrigerator is a device that works on a reverse Carnot cycle and extracts heat from a lower temperature body to keep the temperature of the body lower than the surrounding temperature and by taking work input it transfers heat to the higher temperature body or surrounding.

Refrigerating Effect R.E. = Q_{1}

Work input = Q_{2} - Q_{1}

\({\bf{Coefficient}}\;{\bf{of}}\;{\bf{Performance}}\;\left( {{\bf{COP}}} \right) = \frac{{{\bf{Refrigerating}}\;{\bf{Effect}}}}{{{\bf{Work}}\;{\bf{Input}}}} = \frac{{{{\bf{Q}}_{\bf{1}}}}}{{\bf{W}}}\)

For a reversible engine:

\(\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}\)

\({\bf{CO}}{{\bf{P}}_{{\bf{refrigerator}}}} = \frac{{{{\bf{Q}}_{\bf{1}}}}}{{{{\bf{Q}}_{\bf{2}}} - {{\bf{Q}}_{\bf{1}}}}} = \frac{{{{\bf{T}}_{\bf{1}}}}}{{{{\bf{T}}_{\bf{2}}} - {{\bf{T}}_{\bf{1}}}}}\)

__Calculation:__

__Given:__

T_{1} = 270 K, T_{2} = 300 K

\({\bf{CO}}{{\bf{P}}_{{\bf{refrigerator}}}} = \frac{{{{\bf{T}}_{\bf{1}}}}}{{{{\bf{T}}_{\bf{2}}} - {{\bf{T}}_{\bf{1}}}}}\)

\({\bf{CO}}{{\bf{P}}_{{\bf{refrigerator}}}} = \frac{{{{\bf{270}}_{\bf{}}}}}{{{{\bf{300}}_{\bf{}}} - {{\bf{270}}_{\bf{}}}}}\)

COP = 9.

A Carnot refrigeration is used to maintain temperature at -30°C that requires 1.28 kW per ton of refrigeration. The heat rejected in kW per tonne refrigeration is:

Option 3 : 4.78

__Concept:__

Where, T_{1 }= Low-temperature reservoir, T_{2} = High-temperature reservoir, Q_{1} = Heat supplied to refrigerator, Q_{2} = Heat rejected by refrigerator, W_{R} = Work input given to refrigerator

Heat rejected to the higher temperature in Carnot refrigeration = Refrigeration effect + Work input

**Q _{2} = Q_{1 }+ W_{R}**

__Calculation:__

**Given:**

T_{L}= - 30°C , W_{input }= 1.28 kW , Q_{S} = 1 TR = 3.5167 kW

**Q _{R} = **Q