Gas and Vapor Power Cycles MCQ Quiz in मराठी - Objective Question with Answer for Gas and Vapor Power Cycles - मोफत PDF डाउनलोड करा

Explanation:

Steam trap:

• It is a device which operates automatically and used to removes condensate and non-condensable gases like air without allowing steam to escape.

Types of steam trap:

• Mechanical steam trap: In this type following are the steam traps;
  • Float steam trap
  • Bucket steam trap
  • Free float steam trap​
• Thermostatic steam trap:
  • Liquid expansion
  • Balance pressurized
  • Bi-metallic
  • Bellowed
• Thermodynamic steam trap:
  • Disc type
  • Steam jacket

Thus, option (2) is correct answer.

Concept:

Work ratio is defined as net work output and work done by turbine

Work ratio = \(\frac{{Net\;work\;output}}{{positive\;workdone\;\left( {Turbine\;work} \right)}}\)

\(WR = \frac{{{C_p}\left[ {{T_3} - {T_4}} \right] - {C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}\)

\(WR = 1-\frac{{{}{{} {}} {}\left[ {{T_2} - {T_1}} \right]}}{{{}\left[ {{T_3} - {T_4}} \right]}}\)

Efficiency is defined as the ratio of net work output to the energy supplied to the cycle.

Heat added in system \( = {C_p}\left[ {{T_3} - {T_2'}} \right]\)

Head rejected = C_p [T_4' – T₁]

For ideal Regenerator

T₂ = T_4' and T2' = T4 

\(\eta = 1 - \frac{{{Q_{Rejected}}}}{{{Q_{added}}}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_4'} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_2'}} \right]}}\)

\(\eta = 1 - \frac{{{C_p}\left[ {{T_2} - {T_1}} \right]}}{{{C_p}\left[ {{T_3} - {T_4}} \right]}}\)

\(\eta = 1 - \frac{{\left[ {{T_2} - {T_1}} \right]}}{{\left[ {{T_3} - {T_4}} \right]}}\)

Hence, Efficiency is equal to work ratio

Concept:

Turbine power output (P) = ṁ × (h₁ - h₂)

ṁ = mass flow rate of steam, h₁ = enthalpy at the inlet, h₂ = enthalpy at the outlet

Calculation:

Given:

ṁ = 0.5 kg/s, h₁ = 3140 kJ/kg, h₂ = 2640 kJ/kg

Concept:

Work done by the turbine = Enthalpy at the inlet - Enthalpy at the outlet of turbine,

W_T = h₃ - h₄

Work input to the pump,

W_p = Enthalpy at outlet – Enthalpy at the inlet of the pump, 

W_p = h₂ – h₁

Net work output, W_net = W_T - W_P

Specific Steam consumption = \(\frac{{3600}}{{{W_{net}}}}\) kg/kW-hr

Calculation:

Given:

Enthalpy at the inlet of the turbine, h₃ = 2800 kJ/kg;

Enthalpy at the outlet of turbine, h₄ = 1800 kJ/kg, 

Pump work is negligible, W_p = 0 kJ/kg.

W_net = WT - WP = (h₃ - h₄) - 0 = 2800 - 1800 = 1000 kJ/kg.

Specific steam consumption = \(\frac{{3600}}{{{W_{net}}}}\) \(= \frac{1}{{1000}} \times 3600 = 3.6\) kg/kW-hr.

The specific steam consumption is 3.6 kg/kW-hour.

Concept:

Rankine cycle:

Rankine cycle is a reversible cycle. It is used for power generation through the steam turbine. It consists of two constant pressure and two isentropic processes. These are the four processes in the Rankine cycle:

Process 1 – 2: Isentropic compression

Process 2 – 3: Isobaric heat addition

Process 3 – 4: Isentropic expansion

Process 4 – 1: Isobaric heat rejection

The work ratio is defined as the ratio of net-work to the work done in the turbine.

\(r_w=\frac{W_{net}}{W_T}=\frac{W_T\;-\;W_C}{W_T}\)

Where W_T = Work done by turbine during the expansion of steam, W_c = Work required for the pump to feedback the water into the boiler.

Calculation:

Given:

WT =  8 kJ, Wc = 6 kJ

\(r_w=\frac{W_{net}}{W_T}=\frac{W_T\;-\;W_C}{W_T}\)

\(r_w=\frac{8\;-\;6}{8} = \frac{2}{8}=0.25\)

∴ The work ratio is 0.25.

Explanation:

Concept of Regeneration in the Rankine cycle:

• Regeneration means to take some part of the heat from the expanding steam in the turbine to decrease the overall heat supplied in the process.
• The thermal efficiency of the Rankine cycle can be increased by the use of a regenerative heat exchanger.
• In the regenerative cycle, a portion of the partially expanded steam is drawn off between the high and low-pressure turbines.
• The steam is used to preheat the condensed liquid before it returned to the boiler.
• In this way, the amount of heat added at the low temperatures is reduced and the mean effective temperature of heat addition is increased, thus cycle efficiency is increased.
• At point 2 some amount of steam is taken out for the heating purpose of water.

Effect of Regeneration:

• Decrease in turbine work due to a decrease in the mass flow rate of steam.
• Decrease in heat rejection in condenser due to a reduction in mass flow rate.
• Increased of efficiency and mean temperature of heat addition (Tm).

Concept:

• Work done by the pump is given as,
• \(\int_{1}^{2}dW=v\int_{1}^{2}dP~\)
• \(\rm{W}=\rm v(P_2-P_1)~\)
• But we know that, \(ρ=\frac{m}{V}\)
• and for unit mass,\(ρ=\frac{1}{V}\) means, \(V=\frac{1}{ρ}\)
• therefore \(W=\frac{P_2\;-\;P_1}{ρ}\) kJ/kg

Calculation:

Given:

ρ = 990 kg/m³, P₂ = 30 bar = 3000 kPa, P₁ = 1 bar = 100 kPa

\(W=\frac{P_2\;-\;P_1}{ρ}=\frac{3000\;-\;100}{990}=2.93~kJ/kg~\)

Explanation:

Rankine Cycle:

• The Rankine cycle is an idealized thermodynamic cycle describing the process by which certain heat engines, such as steam turbines or reciprocating steam engines, allow mechanical work to be extracted from a fluid as it moves between a heat source and heat sink.

There are four processes in the Rankine cycle :

• Process 1-2 ⇒  Isentropic compression: The working fluid is pumped from low to high pressure. As the fluid is a liquid at this stage, the pump requires little input energy.
• Process 2-3 ⇒ Constant pressure heat addition in the boiler: The high-pressure liquid enters a boiler, where it is heated at constant pressure by an external heat source to become a dry saturated vapour. 
• Process 3-4 ⇒ Isentropic expansion: The dry saturated vapour expands through a turbine, generating power. This decreases the temperature and pressure of the vapour, and some condensation may occur.
• Process 4-1 ⇒ Constant pressure heat rejection in the condenser.

• ​The pump handles liquid water which is incompressible, i.e., its density or specific volume undergoes little change with an increase in pressure.
• For reversible adiabatic compression, by the use of the property relation:

​\(Tds~=~dh~-~vdp\)

Since, ds = 0,

∴ \(\int_{1}^{2}dh~=~\int_{1}^{2}vdp\)

⇒ Pump work (W_P) = h₂ - h₁ = v₁ × (P₂ - P₁)

Concept:

\(Specific\;Steam\;Consumption = \frac{{3600}}{{{W_{net}}}}\)

\(Specific\;Steam\;Consumption = \frac{{3600}}{{({W_T} - {W_{pump}}})}\)

Calculation:

Given:

h1 – h2 = 840 kJ/kg

h1 - h₄­ = 294 kJ/kg

pump work is negligible! hf4 ≈ hf3 = 191.81 kJ/kg

W_net = W_T - W_P = (h1 – h2) - (h₄ - hf3 ) = 840 - 0 = 840 kJ/kg

\(Specific\;Steam\;Consumption = \frac{{3600}}{{{W_{net}}}}= \frac{3600}{840}= 4.28\) kg/kWh

Explanation:

Let us consider the Rankine cycle.

1-2-3-4-1 is a normal Rankine cycle process

Process 1 - 2 → Isentropic expansion (Turbine)

Process 2 – 3 → Constant pressure heat rejection (Condenser)

Process 3 - 4 → Pump work

Process 4 – 1 → Constant pressure heat addition (Boiler)

1´- 2´- 3 - 4´- 1 is a Rankine cycle when the maximum steam temperature is fixed.

As the maximum steam temperature is fixed because of the metallurgical condition and the boiler pressure is increased.

The process 1 – 2 shifted to a process 1´ - 2´ and the boiler pressure changed from PB to P'B.

∴ we can conclude from the Rankine cycle that the quality of turbine exhaust will decrease i.e., from 2 to 2´.