Figure 1: Reheat Cycle.

T-s and h-s Diagrams of Reheat Cycle:

Figure 2: T-s Diagram.

Figure 3: h-s Diagram.

In a reheat cycle the steam in a superheated state at point- 1 expands isentropically while flowing through the turbine, as shown by the vertical line in figure 2. After expansion, the steam becomes wet. Which is reheated at a constant pressure (generally upto the same temperature as that at point- 1) shown by the point-3, where it is superheated. The steam again expands isentropically while flowing through the next stage of the turbine, as shown by the vertical line 3 to 4 in h-s diagram (see Figure 3).

Salient Points

1. Steam entering into the turbine.
2. Steam condition after partial expansion; (point where reheating starts).
3. Steam condition after reheating.
4. Steam condition after expansion in turbine.

Efficiency of Reheat Cycle

Efficiency of reheats Rankine cycle is given as.

\[{{\eta }_{\text{Reheat}}}=\frac{Workdone}{Total\text{ }heat\operatorname{s}upplied}\]

Workdone = Turbine work + Pump work

\[{{W}_{T}}=({{h}_{1}}-{{h}_{2}})+({{h}_{3}}-{{h}_{4}})\]

\[{{W}_{p}}=({{h}_{6}}-{{h}_{5}})\]

Heat Supplied,

\[{{Q}_{s}}=({{h}_{1}}-{{h}_{6}})+({{h}_{3}}-{{h}_{4}})\]

\[{{\eta }_{\text{Reheat}}}=\frac{{{W}_{r}}-{{W}_{P}}}{{{Q}_{s}}}\]

\[{{\eta }_{\text{Reheat}}}=\frac{({{h}_{1}}-{{h}_{2}})+({{h}_{3}}-{{h}_{4}})-({{h}_{6}}-{{h}_{5}})}{({{h}_{1}}-{{h}_{6}})+({{h}_{3}}-{{h}_{2}})}\]

if pump work is neglected,

\[{{h}_{6}}={{h}_{5}}\]

Then,

\[{{\eta }_{\text{Reheat}}}=\frac{({{h}_{1}}-{{h}_{2}})+({{h}_{3}}-{{h}_{4}})}{({{h}_{1}}-{{h}_{6}})+({{h}_{3}}-{{h}_{2}})}\]

Comparison between Efficiency of Reheat Cycle and  Rankine,

The efficiency of rankine cycle is given by.

\[{{\eta }_{\text{Renkine}}}=\frac{{{h}_{1}}-{{h}_{2}}}{{{h}_{1}}-{{h}_{3}}}\]

The efficiency of reheat cycle is given us,

\[{{\eta }_{\text{Reheat}}}=\frac{({{h}_{1}}-{{h}_{2}})+({{h}_{3}}-{{h}_{4}})-({{h}_{6}}-{{h}_{5}})}{({{h}_{1}}-{{h}_{6}})+({{h}_{3}}-{{h}_{2}})}\]

By comparing both the efficiencies, it can be concluded that the efficiency of rankine cycle with reheating is more than simple rankine cycle.

Advantages of Reheat Cycle

1. It increases the thermal efficiency and output of the turbine.
2. Reheating process improves the final dryness fraction or quality of steam.
3. It also increases the nozzle and blade efficiencies of turbine.
4. It eliminates the erosion and corrosion problems occurring in the steam turbine.

Disadvantages of Reheat Cycle

1. It requires more maintenance.
2. In comparison to the expenditure incurred in reheating, the increase in thermal efficiency is not reasonable.

The post What is Reheat Cycle? Process, Derivation, Diagram & Efficiency appeared first on ElectricalWorkbook.

Figure 1: Dual Cycle.

Dual combustion cycle consists of five processes. They are.

Process 1-2: Isentropic compression – It is a reversible adiabatic compression process during which pressure and temperature increases from P₁, T₁ to P₂, T₂ respectively.

Process 2-3: Reversible constant volume heat addition – During this process, heat is added to the air at constant volume by a suitable heating source such that the pressure and temperature increases from P₂, T₂, to P₃, T₃ respectively.

Process 3-4: Reversible constant pressure heat addition – During this process, heat is added at constant pressure by a suitable heating source such that the volume and temperature increases from V₃, T₃ to V₄, T₄  respectively.

Process 4-5: Isentropic expansion – It is a reversible adiabatic expansion process during which the pressure and temperature decreases from P₄ ,T₄ to P₅, T₅ respectively.

Process 5-1: Reversible constant volume heat rejection – During this process. heat is rejected at constant volume such that the pressure and temperature falls from P₅, T₅ to P₁, T₁ respectively.

Heat supplied,

Q₁ = mC_v (T₃ – T₂) + mC_p (T₄ – T₃)

Heat rejected,

Q₂ = mC_v (T₅ – T₁)

Work done =   Heat supplied  ˗ Heat rejected

\[W={{Q}_{1}}-{{Q}_{2}}\]

\[=m{{C}_{v}}({{T}_{3}}-{{T}_{2}})+m{{C}_{p}}({{T}_{4}}-{{T}_{3}})-m{{C}_{v}}({{T}_{5}}-{{T}_{1}})\]

Thermal efficiency

\[\eta =\frac{Work\text{ }done}{Heat\text{ }added}\]

\[=\frac{m{{C}_{v}}({{T}_{3}}-{{T}_{2}})+m{{C}_{p}}({{T}_{4}}-{{T}_{3}})-m{{C}_{v}}({{T}_{5}}-{{T}_{1}})}{m{{C}_{v}}({{T}_{3}}-{{T}_{2}})+m{{C}_{p}}({{T}_{4}}-{{T}_{3}})}\]

\[=1-\frac{m{{C}_{v}}({{T}_{5}}-{{T}_{1}})}{m{{C}_{v}}({{T}_{3}}-{{T}_{2}})+m{{C}_{p}}({{T}_{4}}-{{T}_{3}})}\]

Therefore,

\[ \eta =1-\frac{{{T}_{5}}-{{T}_{1}}}{({{T}_{3}}-{{T}_{2}})+({{T}_{4}}-{{T}_{3}})}\]

\[\text{  }\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\]

Now,

Compression ratio,

\[r=\frac{{{V}_{1}}}{{{V}_{2}}}\]

Expansion ratio,

\[{{r}_{e}}=\frac{{{V}_{5}}}{{{V}_{4}}}\]

Cut-off ratio,

\[\rho =\frac{{{V}_{4}}}{{{V}_{3}}}\]

Pressure ratio,

\[{{r}_{p}}=\frac{{{P}_{3}}}{{{P}_{2}}}\]

Consider process 1-2,

\[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}={{\left( r \right)}^{\gamma -1}}\]

\[{{T}_{2}}={{T}_{1}}\text{  }{{\text{r}}^{\gamma -1}}\]

Consider process 2-3,

\[\frac{{{T}_{3}}}{{{T}_{2}}}=\frac{{{P}_{3}}}{{{P}_{2}}}={{r}_{p}}\]

\[{{T}_{3}}={{T}_{2}}\text{ }{{r}_{p}}={{T}_{1.}}\text{ }{{\text{r}}^{\gamma -1}}.{{r}_{p}}.\rho \]

Consider process 3-4,

\[\frac{{{V}_{4}}}{{{T}_{4}}}=\frac{{{V}_{3}}}{{{T}_{3}}}\]

\[{{T}_{4}}=\frac{{{V}_{4}}}{{{V}_{3}}}{{T}_{3}}\]

\[=\rho {{T}_{3}}={{T}_{1.}}\text{ }{{\text{r}}^{\gamma -1}}{{r}_{p}}.\rho \]

Consider process 4-5,

\[\frac{{{T}_{5}}}{{{T}_{4}}}={{\left( \frac{{{V}_{4}}}{{{V}_{5}}} \right)}^{\gamma -1}}={{\left( \frac{{{V}_{4}}}{{{V}_{1}}} \right)}^{\gamma -1}}\]

\[={{\left[ \frac{{{V}_{4}}}{{{V}_{3}}}\times \frac{{{V}_{3}}}{{{V}_{2}}}\times \frac{{{V}_{2}}}{{{V}_{1}}} \right]}^{\gamma -1}}\]

\[{{V}_{3}}={{V}_{2}}\]

\[={{\left[ \frac{{{V}_{4}}}{{{V}_{3}}}\times \frac{{{V}_{2}}}{{{V}_{1}}} \right]}^{\gamma -1}}\text{ }\]

\[={{\left( \frac{\rho }{r} \right)}^{\gamma -1}}\]

\[{{T}_{5}}={{T}_{4}}.{{\left( \frac{\rho }{r} \right)}^{\gamma -1}}\]

\[={{T}_{1}}{{r}^{\gamma -1}}.{{r}_{p}}.\rho .{{\left( \frac{\rho }{r} \right)}^{\gamma -1}}\]

\[={{T}_{1}}{{r}_{p}}.{{\rho }^{\gamma }}\]

Substitute eq.(2)(3)(4)and(5)in eq.(1),

Then,

\[\eta =1-\frac{{{T}_{1}}\left( {{r}_{p}}{{\rho }^{\gamma }}-1 \right)}{{{T}_{1}}\left( {{r}^{\gamma -1}}{{r}_{p}}-{{r}^{\gamma -1}} \right)+\gamma {{T}_{1}}\left( {{r}^{\gamma -1}}{{r}_{p}}\rho -{{r}^{\gamma -1}}{{r}_{p}} \right)}\]

\[=1-\frac{\left( {{r}_{p}}{{\rho }^{\gamma }}-1 \right)}{{{r}^{\gamma -1}}\left( {{r}_{p}}-1 \right)+\gamma .{{r}^{\gamma -1}}{{r}_{p}}\left( \rho -1 \right)}\]

\[=1-\frac{\left( {{r}_{p}}{{\rho }^{\gamma }}-1 \right)}{{{r}^{\gamma -1}}\left[ \left( {{r}_{p}}-1 \right)+\gamma .{{r}_{p}}\left( \rho -1 \right) \right]}\]

\[=1-\frac{1}{{{r}^{\gamma -1}}}\left[ \frac{\left( {{r}_{p}}{{\rho }^{\gamma }}-1 \right)}{\left[ \left( {{r}_{p}}-1 \right)+\gamma .{{r}_{p}}\left( \rho -1 \right) \right]} \right]\]

The efficiency of dual cycle is function r, ρ, r_p and γ.

The post What is Dual Cycle? Process, Derivation, Diagram & Efficiency appeared first on ElectricalWorkbook.

1. Isentropic compression process
2. Constant pressure heat addition process
3. Isentropic expansion process
4. Constant volume heat rejection process.

The representation of diesel cycle on P-V and T-s diagram is shown in below figure 1,

Figure 1: Diesel Cycle.

Isentropic Compression Process: During this process, the air in the cylinder is compressed by moving the piston from B.D.C to T.D.C. Since, it is an isentropic process, there will be no heat transfer and hence the entropy remains constant. This process is represented by the line 1-2 in P-V and T-s diagram. Due to compression, the pressure and temperature in the cylinder increases. The ratio of initial volume to the volume after compression $\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)$ is known as compression ratio.

Constant Pressure Heat Addition Process: In this process, the compressed air which is at high pressure is heated with the help of constant addition. This process is represented by the line 2-3 in P-V and T-s diagram. After completion of this process, the temperature reaches its highest point in the cycle. Thus, entropy of air also increases. During this process the volume of air increases and the ratio of volume of air after heat addition to the volume of air before heat addition $\left( \frac{{{V}_{3}}}{{{V}_{2}}} \right)$ is known as cut-off ratio.

Isentropic Expansion Process: In this process, the high pressure and temperature air is expanded isentropically. Due to this, the pressure and temperature of air decreases. Expansion takes place till the piston reaches B.D.C work is produced during expansion and it is called the power stroke. The line ‘3-4’ represents this process on P-V and T-s diagram.

Constant Volume Heat Rejection Process: In the expansion process, the piston reaches to B.D.C and from there it can’t move further. So the hot air is the cylinder rejects heat at constant volume to achieve equilibrium with the surrounding, after rejecting heat the air reaches to its initial condition. The line ‘4-1’ in P-V and T-s diagram represent this process. As heat is rejected the entropy of the air also decreases. This completes the cycle and it is repeated.

Heat added in the cycle,

Q_S = Q_2-3 = C_P (T₃ – T₂)

Heat rejected in the cycle,

Q_R = Q_4-1 = C_v (T₄ – T₁)

Workdone per cycle,

W = Q_S – Q_R

= C_P (T₃ – T₂) – C_v (T₄ – T₁)

Thermal efficiency,

\[{{\eta }_{th}}=\frac{W}{{{Q}_{S}}}\]

\[=\frac{{{C}_{p}}({{T}_{3}}-{{T}_{2}})-{{C}_{v}}({{T}_{4}}-{{T}_{1}})}{{{C}_{p}}({{T}_{3}}-{{T}_{2}})}\]

\[{{\eta }_{th}}=1-\frac{1}{\gamma }\left( \frac{({{T}_{4}}-{{T}_{1}})}{({{T}_{3}}-{{T}_{2}})} \right)\]

Air standard efficiency of a Diesel Cycle

Diesel cycle is also known as constant pressure heat addition cycle. It consists of four reversible processes. They are,

1. Isentropic compression process.
2. Constant pressure heat addition process.
3. Isentropic expansion process.
4. Constant volume heat rejection process.

The P-V and T-s diagram of diesel cycle is shown below.

Heat supplied to the cycle,

Q_S = Q_2-3 = C_P (T₃ – T₂)

Heat rejected from the cycle.

Q_R = Q_4-1 = C_V (T₄ – T₁)

Workdone per cycle,

W = Q_S – Q_R

= C_P (T₃ – T₂) – C_V (T₄ – T₁).

Thermal efficiency of diesel cycle,

\[{{\eta }_{th}}=1-\frac{1}{\gamma }\left( \frac{({{T}_{4}}-{{T}_{1}})}{({{T}_{3}}-{{T}_{2}})} \right)\]

\[{{\eta }_{th}}=\frac{Work\text{ }done}{Heat\text{ }added}\]

\[=\frac{{{C}_{p}}({{T}_{3}}-{{T}_{2}})-{{C}_{v}}({{T}_{4}}-{{T}_{1}})}{{{C}_{p}}({{T}_{3}}-{{T}_{2}})}\]

Since,

\[\frac{{{C}_{p}}}{{{C}_{v}}}=\gamma \]

\[=1-\frac{1}{\gamma }\left( \frac{({{T}_{4}}-{{T}_{1}})}{({{T}_{3}}-{{T}_{2}})} \right)\]

From isentropic process ‘1-2’,

\[\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}={{\left( r \right)}^{\gamma -1}}\]

Where,

\[r=\frac{{{V}_{1}}}{{{V}_{2}}}\]

r known as as compression ratio

\[{{T}_{2}}={{T}_{1}}{{\left( r \right)}^{\gamma -1}}\]

From constant pressure process ‘2-3’,

\[\frac{{{T}_{3}}}{{{T}_{2}}}=\frac{{{V}_{3}}}{{{V}_{2}}}=\rho \]

Where, ρ is cut off ratio

Therefore,

\[{{T}_{3}}={{T}_{2}}\rho \]

\[={{T}_{1}}{{\left( r \right)}^{\gamma -1}}\rho \]

From isentropic process ‘3-4’,

\[\frac{{{T}_{4}}}{{{T}_{3}}}={{\left( \frac{{{V}_{3}}}{{{V}_{4}}} \right)}^{\gamma -1}}\]

\[{{T}_{4}}={{V}_{3}}{{\left( \frac{{{V}_{3}}}{{{V}_{2}}}\times \frac{{{V}_{2}}}{{{V}_{4}}} \right)}^{\gamma -1}}\]

But,

\[\frac{{{V}_{3}}}{{{V}_{2}}}=\rho \]

And

\[\frac{{{V}_{2}}}{{{V}_{4}}}=\frac{{{V}_{2}}}{{{V}_{1}}}=\frac{1}{\gamma }\]

Since,

\[\text{ }{{V}_{2}}={{V}_{4}}\]

\[{{T}_{4}}={{T}_{3}}{{\left( \frac{\rho }{r} \right)}^{\gamma -1}}\]

\[{{T}_{4}}={{T}_{1}}{{r}^{\gamma -1}}\rho {{\left( \frac{\rho }{r} \right)}^{\gamma -1}}\]

\[{{T}_{4}}={{T}_{1}}{{\rho }^{\gamma }}\]

Substitute equations (2),(3) and (4) in eq. (1),

\[{{\eta }_{th}}=1-\frac{1\left( {{T}_{1}}{{\rho }^{\gamma }}-{{T}_{1}} \right)}{\gamma \left( {{T}_{1}}{{r}^{\gamma -1}}\rho -{{r}^{\gamma -1}} \right)}\]

\[1-\frac{1\left( {{\rho }^{\gamma }}-1 \right)}{\gamma \left( {{r}^{\gamma -1}}\rho -{{r}^{\gamma -1}} \right)}\]

\[ {{\eta }_{th}}=1-\frac{1}{{{r}^{\gamma -1}}}\left[ \frac{{{\rho }^{\gamma }}-1}{\gamma \left( \rho -1 \right)} \right]\]

Therefore, the efficiency of diesel cycle depends on compression ratio and cut-off ratio. The efficiency can be increased by increasing the compression ratio and decreasing the cut-off ratio. For the same compression ratio, diesel cycle efficiency is less than otto cycle efficiency.

Q. Name the factors that affect the air standard efficiency of Diesel Cycle.

The following factors affect the air standard efficiency of diesel cycle.

1. Compression Ratio: Increase in compression ratio (r), increases the air standard efficiency of diesel cycle.
2. Cut-off Ratio: Increase in cut-off ratio (ρ), decreases the air standard efficiency of diesel cycle.
3. Adiabatic Index: For large values of adiabatic index (γ), air standard efficiency is high.

Q. When compression ratio is kept constant, what is the effect of cut-off ratio on the efficiency of diesel cycle.

Cut-off ratio (ρ) is defined as the ratio of initial and final volumes of air during constant pressure heat addition process. For same compression ratio (r) and adiabatic index (γ), the air standard efficiency of diesel cycle decreases with increase in cut-off ratio. The relation between cut-off ratio and air standard efficiency at constant r. γ values is shown in below figure,

Q. Explain with the help of suitable graphs the variation of the efficiency of the diesel cycle with compression ratio and cut-off ratio.

The equation relating thermal efficiency (η_th), compression ratio (r) and cut-off ratio (ρ) for diesel cycle is shown below,

\[{{\eta }_{th}}=\frac{1}{{{r}^{\gamma -1}}}\left[ \frac{{{\rho }^{\gamma -1}}}{\gamma \left( \rho -1 \right)} \right]\]

Graphs are drawn between compression ratio (r) and thermal efficiency (η_th) and also between cut-off ratio (ρ) and thermal efficiency (η_th).

Figure (1) is drawn between compression ratio (r) and thermal efficiency (η_th) for different values of cut-off ratio (ρ). From the graph, it can be seen that, for different values of cut-off ratio (ρ) the thermal efficiency (η_th) increases with the increase in compression ratio (r). When cut-off ratio (p) is unity, then the efficiency of diesel cycle is equal to the efficiency of Otto cycle.

Figure (2) is drawn between cut-off ratio (ρ) and thermal efficiency (η_th) for different values of compression ratio (r). As the value of cut-off ratio (ρ) increases, the thermal efficiency of diesel cycle decreases for different values of compression ratio (r). If diesel cycle has low cut-off ratio (ρ) its efficiency will increase but, if it has high cut-off ratio (ρ) its power output is more. Smoking occurs due to large cut-off ratio (ρ). Thus, for a diesel cycle to have good efficiency it should have high compression ratio (r) and low cut-off ratio (ρ).

The post What is Diesel Cycle? Process, Derivation, Diagram & Efficiency appeared first on ElectricalWorkbook.

1. Reversible adiabatic compression
2. Constant volume heat addition process
3. Reversible adiabatic expansion
4. Constant volume heat rejection process.

The P-V and T-s diagrams of the cycle (see Figure 1),

Figure 1: Otto Cycle.

Process 1-2: It is a reversible adiabatic compression process. During this process (lie air in the cylinder is compressed such that there is no heat transfer. Iii this process, the temperature increases from T₁ to T_2.

Process 2-3: It is a constant volume heat addition process. During this process, the compressed air in the cylinder is heated by means of a source. As the heat is supplied, the temperature increases to T₃.

Process 3-4: It is reversible adiabatic expansion process. Due to the supply of heat, the temperature of air increases and starts to expanding without any heat transfer. The change in entropy in this process is zero and the temperature reduces to T₄.

Process 4-1: This process is constant volume heat rejection process. After expanding, the hot air rejects heat to the sink at constant volume. This completes the cycle and it is repeated.

The workdone and efficiency of Otto cycle is calculated as follows.

Heat supplied during process ‘2 – 3’.

Q_S = C_V (T₃ – T₂)

Heat rejected during process  ‘4 – 1’.

Q_R= C_V (T₄ – T₁)

Workdone per kg of air,

\[W={{Q}_{S}}-{{Q}_{R}}\]

\[={{C}_{V}}({{T}_{3}}-{{T}_{2}})-{{C}_{V}}({{T}_{4}}-{{T}_{1}})\]

Air standard effciency,

\[\eta =\frac{Workdone}{Heat\text{ }Supplied}\]

\[=\frac{W}{{{Q}_{S}}}\]

\[=\frac{{{C}_{V}}({{T}_{3}}-{{T}_{2}})-{{C}_{V}}({{T}_{4}}-{{T}_{1}})}{{{C}_{V}}({{T}_{3}}-{{T}_{2}})}\]

\[=1-\frac{({{T}_{4}}-{{T}_{1}})}{({{T}_{3}}-{{T}_{2}})}\]

\[=1-\frac{{{T}_{1}}\left[ \frac{{{T}_{4}}}{{{T}_{1}}}-1 \right]}{{{T}_{2}}\left[ \frac{{{T}_{3}}}{{{T}_{2}}}-1 \right]}…(1)\]

Since process ‘1-2’ and ‘3-4’ are isentropic.

\[\frac{{{T}_{2}}}{{{T}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}={{r}^{\gamma -1}}…(2)\]

\[{{T}_{2}}={{T}_{1}}\text{  }{{r}^{\gamma -1}}\]

And

\[\frac{{{T}_{3}}}{{{T}_{4}}}={{\left( \frac{{{V}_{4}}}{{{V}_{3}}} \right)}^{\gamma -1}}={{r}^{\gamma -1}}…(3)\]

\[{{T}_{4}}=\frac{{{T}_{3}}}{{{r}^{\gamma -1}}}\]

Where,

\[r=\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{V}_{4}}}{{{V}_{3}}}\]

r known as as compression ratio

From equations (2) and (3),

\[\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{4}}}\frac{{{T}_{4}}}{{{T}_{1}}}=\frac{{{T}_{3}}}{{{T}_{2}}}\]

From equation (1),

\[{{\eta }_{a}}=1-\frac{{{T}_{1}}}{{{T}_{2}}}\]

Also,

\[\text{ }\frac{{{T}_{1}}}{{{T}_{2}}}={{r}^{\gamma -1}}\]

Hence,

\[{{\eta }_{a}}=1-\frac{1}{{{r}^{\gamma -1}}}\]

Since ‘γ’ is constant for air, the efficiency of Otto cycle depends on compression ratio and is independent of temperatures. The efficiency can be improved by increasing the compression ratio upto a certain limit. Beyond that limit, the increase in efficiency is considerably low and also leads to detonation.

Mean effective Pressure of Otto Cycle

The net workdone per kg in the otto cycle is given as,

W = Work during expansion – Work during compression.

\[={{W}_{3-4}}-{{W}_{1-2}}\]

\[=\frac{{{P}_{3}}{{V}_{3}}-{{P}_{4}}{{V}_{4}}}{\gamma -1}-\frac{{{P}_{2}}{{V}_{2}}-{{P}_{1}}{{V}_{1}}}{\gamma -1}\]

From isentropic relations,

\[\frac{{{P}_{3}}}{{{P}_{4}}}=\frac{{{P}_{2}}}{{{P}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}={{r}^{\gamma }}\]

And, pressure ratio

\[{{r}_{P}}=\frac{{{P}_{3}}}{{{P}_{2}}}=\frac{{{P}_{4}}}{{{P}_{1}}}\]

Since,

\[r=\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{V}_{4}}}{{{V}_{3}}}\]

\[{{V}_{1}}=r{{V}_{2}}={{V}_{4}}=r{{V}_{3}}\]

\[W=\frac{1}{\gamma -1}\left[ {{P}_{4}}{{V}_{4}}\left( \frac{{{P}_{3}}{{V}_{3}}}{{{P}_{4}}{{V}_{4}}}-1 \right)-{{P}_{1}}{{V}_{1}}\left( \frac{{{P}_{2}}{{V}_{2}}}{{{P}_{1}}{{V}_{1}}}-1 \right) \right]\]

\[=\frac{1}{\gamma -1}\left[ {{P}_{4}}{{V}_{4}}\left( \frac{{{P}_{3}}}{{{P}_{4}}}.\frac{1}{r}-1 \right)-{{P}_{1}}{{V}_{1}}\left( \frac{{{P}_{2}}}{{{P}_{1}}}.\frac{1}{r}-1 \right) \right]\]

\[=\frac{1}{\gamma -1}\left[ {{P}_{4}}{{V}_{4}}\left( {{r}^{\gamma }}.\frac{1}{r}-1 \right)-{{P}_{1}}{{V}_{1}}\left( {{r}^{\gamma }}.\frac{1}{r}-1 \right) \right]\]

\[=\frac{1}{\gamma -1}\left[ {{P}_{4}}{{V}_{4}}\left( {{r}^{\gamma -1}}-1 \right)-{{P}_{1}}{{V}_{1}}\left( {{r}^{\gamma -1}}-1 \right) \right]\]

Also,

\[{{V}_{1}}={{V}_{4}}\]

\[W=\frac{{{V}_{1}}}{\gamma -1}\left[ {{P}_{4}}\left( {{r}^{\gamma -1}}-1 \right)-{{P}_{1}}\left( {{r}^{\gamma -1}}-1 \right) \right]\]

\[=\frac{{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{P}_{4}}-{{P}_{1}} \right) \right]\]

\[=\frac{{{P}_{1}}{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( \frac{{{P}_{4}}}{{{P}_{1}}}-1 \right) \right]\]

\[=\frac{{{P}_{1}}{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{r}_{p}}-1 \right) \right]\]

Mean effective pressure is given as,

\[MEP=\frac{Work\text{ }done}{Swept\text{b}Volume}\]

\[=\frac{\frac{{{P}_{1}}{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{r}_{p}}-1 \right) \right]}{{{V}_{1}}-{{V}_{2}}}\]

\[=\frac{\frac{{{P}_{1}}{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{r}_{p}}-1 \right) \right]}{{{V}_{1}}\left[ 1-\frac{{{V}_{2}}}{{{V}_{1}}} \right]}\]

\[=\frac{\frac{{{P}_{1}}{{V}_{1}}}{\gamma -1}\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{r}_{p}}-1 \right) \right]}{{{V}_{1}}\left[ 1-\frac{1}{r} \right]}\]

Therefore,

\[ MEP=\frac{{{P}_{1}}r\left[ \left( {{r}^{\gamma -1}}-1 \right)\left( {{r}_{p}}-1 \right) \right]}{\left( \gamma -1 \right)\left( r-1 \right)}\]

Differences Between Otto cycle and Diesel cycle

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Vane Sealed Compressor

Figure 1: Vane Sealed Compressor.

The vanes are generally made of steel and sometimes of nonmetallic material such as fibre or carbon. Figure (1) shows a vane sealed compressor consisting of a rotor having slots mounted eccentrically in the body and vanes mounted on the rotor. During the rotation of rotor, the air from the inlet passage is drawn into the space between the rotor, the cylinder and two adjacent vanes. Further rotation of rotor causes the area between the rotor and the vanes to decrease and thereby resulting in compression of the air. As the rotor reaches the point just opposite to the eccentricity of the rotor, the suction ends and the compression proceeds and finally discharges the air through the discharge port. In this type of compressors, some or all of the compression is obtained before the air is delivered. It is also called as the sliding vane compressor. as the vanes slides in the rotor slots during its rotation. The vanes generally provided are 20 – 30 in number and as the number of vanes increases, the leakage losses decrease due to less pressure difference between the adjacent spaces. They are generally designed for the capacity upto 150 m³/min at a compression ratio upto 8.5. These compressors requires less work input compared to roots blower for the same pressure ratio and air delivery.

Figure 2: P-V Diagram for Vane Sealed Compressor.

The P-V diagram of a vane sealed compressor is shown in figure (2) below. From the figure, it is noticed that the air is compressed in the compressor upto a pressure of P and the remaining pressure rise occurs due to the back flow of air from the receiver.

Let, V_S – induced volume at pressure P₁ and temperature T₁.

The workdone per revolution with N vanes is given by,

\[W=\frac{N\gamma }{\gamma -1}{{P}_{1}}{{V}_{1}}\left[ {{\left( \frac{{{P}_{i}}}{{{P}_{i}}} \right)}^{\frac{\gamma }{\gamma -1}}}-1 \right]+N({{P}_{2}}-{{P}_{i}}){{V}_{i}}\]

The post What is Positive Displacement Compressor? Working, Diagram & Parts appeared first on ElectricalWorkbook.

Figure 1: Centrifugal Air Compressor.

A Centrifugal Air Compressor consists of a rotor (or impeller) to which a number of curved vanes are fined symmetrically. The rotor rotates in an air tight casing with inlet and outlet points.

The casing is so designed that the kinetic energy of the air is convened into pressure energy before it leaves the casing. The mechanical energy to the rotor is provided by some external source. As the rotor rotates, it sucks the air through its inlet, increasing its pressure due to centrifugal force and forces the air to flow over the diffuser. In the diffuser, the pressure of air is father increased and hence the air is passed onto the receiver.

Parts of Centrifugal Air Compressor

The essential components that are required for the operation of centrifugal compressor are,

1. Inlet pipe
2. Curved radial vanes
3. Impeller
4. Casing
5. Diffuser
6. Outlet pipe.

1. Inlet Pipe: The main purpose of inlet pipe is to introduce the fluid into impeller inlet.

2. Curved Radial Vanes: A series of curved vanes mounted on the shaft and they rotate with the shaft. The movement of air is guided by the radial vanes. The fluid flows radially outwards through impeller because of these radial vanes.

3. Impeller: The rotating member in centrifugal compressor is impeller. The fluid is rotated by the impeller and increases both pressure and velocity of the fluid.

4. Diffuser: It contains diverging passages formed by the diffuser blades. By passing through these passages, the velocity of air decreases and pressure increases still further.

5. Casing: The casing surrounds the totaling impeller. It collects the high pressure air from diffuser.

6. Outlet Pipe: Compressed air is taken out through outlet pipe.

Expression for Workdone in a Centrifugal Air Compressor

The work done by the impeller of a centrifugal compressor can be derived by considering the inlet and outlet velocity triangles as shown in the figures (a) and (b). Assuming that the air enters axially, hence the velocity of whirl at inlet is zero, therefore, V₁ = V_f1

Let mass of the fluid enters the impeller is one kg.

According to the principle of moment of momentum, theoretical torque supplied to the impeller must be equal to the rate of change of momentum.

Therefore,

\[\text{Workdone}=\text{Torque}\times \text{Angular Velocity}\]

\[\text{W}\text{.D}=T\times \omega \]

\[=({{V}_{{{w}_{2}}}}{{r}_{2}}-{{V}_{w1}}{{r}_{1}})\omega \]

\[={{V}_{{{w}_{2}}}}{{r}_{2}}\omega -{{V}_{w1}}{{r}_{1}}\omega \]

Also

\[{{r}_{1}}{{\omega }_{1}}={{u}_{1}}\text{ and }{{r}_{2}}{{\omega }_{2}}={{u}_{2}}\]

But,

\[\left( {{V}_{w2}}{{u}_{2}}-{{V}_{w1}}{{u}_{1}} \right)=\left( {{h}_{2}}-{{h}_{1}} \right)\]

\[{{h}_{2}}-{{h}_{1}}={{C}_{p}}({{T}_{2}}-{{T}_{1}})\]

Therefore,

\[\text{W}\text{.D}=\left( {{V}_{w2}}{{u}_{2}}-{{V}_{w1}}{{u}_{1}} \right)\]

\[={{h}_{2}}={{h}_{1}}={{C}_{p}}({{T}_{2}}-{{T}_{1}})\]

The above equation is known as Euler’s work.

If the fluid enters the impeller radially then V_w2 = 0. Work done = V_w2 u₂ J/kg.
From inlet and outlet velocity triangles, we have,

\[V_{r1}^{2}=u_{1}^{2}+V_{1}^{2}-2{{u}_{1}}{{V}_{w1}}\]

\[V_{r2}^{2}=u_{2}^{2}+V_{2}^{2}-2{{u}_{2}}{{V}_{w2}}\]

By substituting the values of V_w2u₂ and V_w1u₁ , from the equations (2) and (3) in equation (1).

\[\text{W}\text{.D}=\frac{V_{2}^{2}-V_{1}^{2}}{2}+\frac{V_{r1}^{2}-V_{r2}^{2}}{2}+\frac{u_{2}^{2}-u_{1}^{2}}{2}\]

In the above equation, the term $\left[ \frac{V_{2}^{2}-V_{1}^{2}}{2} \right]$ represents the increase in kinetic energy of fluid.

The term $\left[ \frac{V_{r1}^{2}-V_{r2}^{2}}{2} \right]$ represents the pressure rise in the impeller due to diffusion action. Whereas the term $\left[ \frac{u_{2}^{2}-u_{1}^{2}}{2} \right]$ represents the pressure rise in the impeller due to centrifugal action.

The post What is Centrifugal Air Compressor? Working, Diagram & Parts appeared first on ElectricalWorkbook.

Figure 1: Roots Blower.

The roots blower compressor consists of two rotors or lobes arranged on two separate axes as shown in figure 1. One of the rotors is directly connected to a drive and the other is driven by gear connected to the first. These rotors rotate in opposite direction to each other. The rotor or lobes are of cycloidal or involute profile and are connected in such a way that it seals the delivery side from the suction or inlet side. A small clearance is to be provided between the lobes, casing for reducing wear and friction. The provided clearance acts as a leakage passage and hence affects the performance of the compressor. From the intake port, the air enters into the casing and the air trapped between the rotors or lobes moves along the port and then discharged through the discharge port to the receiver. But, when the outlet port is open. The gas from the receiver flows back, as it is at higher pressure. The gas is compressed irreversibility to the delivery pressure and then delivery begins. This process is carried out four times in one revolution of the driving shaft. It is shown in P-V diagram by dotted lines.

Efficiency of Roots Blower

Let. P₁– Pressure of air at inlet

T₁– Temperature of air at inlet

V – volume between lobe face and the casing

V_S – Swept volume

P₂ – Pressure of air after compression.

Workdone per cycle = (P₂ – P₁)V

Workdone per revolution = 4(P₂ – P₁)V

Workdone/minute = (P₂ – P₁)V_S

For an ideal compression process i.e. reversible adiabatic process the work done per minute is given by,

\[\text{Workdone/Minute}=\frac{\gamma }{\gamma -1}{{P}_{1}}{{V}_{S}}\left[ {{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{\frac{\gamma -1 }{\gamma}}}-1 \right]\]

Roots blower efficiency is given as the ratio of isentropic workdone to the actual work done.

i.e.,

\[\text{roots blower efficiency}=\frac{\text{Isentropic workdone}}{\text{Actual workdone}}\]

\[{{\eta }_{roots}}=\frac{\frac{\gamma }{\gamma -1}{{P}_{1}}{{V}_{S}}\left[ {{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{\frac{\gamma -1 }{\gamma}}}-1 \right]}{{{V}_{S}}({{P}_{2}}-{{P}_{1}})}\]

\[{{\eta }_{roots}}=\frac{\frac{\gamma }{\gamma -1}{{P}_{1}}{{V}_{S}}\left[ {{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{\frac{\gamma -1 }{\gamma}}}-1 \right]}{{{P}_{1}}{{V}_{S}}\left[ \frac{{{P}_{2}}}{{{P}_{1}}}-1 \right]}\]

\[=\frac{\frac{\gamma }{\gamma -1}\left[ {{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{\frac{\gamma -1 }{\gamma}}}-1 \right]}{\left[ \frac{{{P}_{2}}}{{{P}_{1}}}-1 \right]}\]

\[\frac{{{P}_{2}}}{{{P}_{1}}}=r=compression\text{ }ratio\]

\[{{\eta }_{roots}}=\frac{\gamma }{\gamma -1}\frac{\left[ {{\left( r \right)}^{\frac{\gamma -1 }{\gamma}}}-1 \right]}{\left( r-1 \right)}\]

Since,

\[\frac{\gamma }{\gamma -1}=\frac{{{C}_{p}}}{R}\]

\[{{\eta }_{roots}}=\frac{{{C}_{p}}}{R}\left[ \frac{{{\left( r \right)}^{\frac{\gamma -1 }{\gamma}}}-1}{r-1} \right]\text{   }\]

From the above equation, it can be noticed that the roots efficiency decreases with increase in pressure ratio and can be used up to a pressure ratio of 2 for single stage and 3 for two- stage only. These compressors are generally designed for the capacities varying from 0.14 m³/min, to 1400 m³/min.

The post What is Roots Blower? Working, Diagram, Parts & Efficiency appeared first on ElectricalWorkbook.

Rope Brake Dynamometer

Figure 1: Rope Brake Dynamometer.

Consider a rope brake dynamometer consisting of a rope warped around the circumference of the flywheel of the engine as shown in figure 1.

The spring balance is attached to one end of the rope and the other end carries a load (W). The friction produced between the rim and the wheel absorbs the whole power developed by the engine. Due to the friction, heat is developed which inturn increase temperature of the brake wheel. Therefore, means of cooling (water) is provided to reduce the temperature of the wheel.

Let,

W – Load applied at the end of the rope in N

S – Spring balance reading in N

N – Speed of the engine in r.p.m

D – Diameter of the brake wheel in m

d – Rope diameter in m

Braking torque,

T = Friction force × Effective radius

\[=\text{(W}-\text{S})\left( \frac{\text{D}+\text{d}}{\text{2}} \right)\]

Power developed = Frictional torque × Angular rotation

\[=\text{(W}-\text{S)}\left( \frac{\text{D}+\text{d}}{\text{2}} \right)\left( \frac{\text{2 }\!\!\pi\!\!\text{}\!\!\omega\!\!\text{ }}{\text{60}} \right)\text{ W}\]

\[=\frac{\text{ }\!\!\pi\!\!\text{ N}(\text{W}-\text{S)}(\text{D}+\text{d)}}{\text{60}}\text{ W}\]

\[=\frac{\text{ }\!\!\pi\!\!\text{ N}(\text{W}-\text{S)}(\text{D}+\text{d)}}{\text{60 }\!\!\times\!\!\text{ 1000}}\text{ kW}\]

If d is neglected, then

\[\text{B}\text{.P}=\frac{\text{ }\!\!\pi\!\!\text{ DN}(\text{W}-\text{S)}}{\text{60 }\!\!\times\!\!\text{ 1000}}\text{ kW}\]

As the rope brakes are cheap and easily manufactured, they are extensively used for testing B.P of an engine.

Prony Brake Dynamometer

Figure 2: Prony Brake Dynamometer.

The simple arrangement of prony brake system is as shown in figure 2. The basic principle involved in the working of the prony brake is to convert the power into heat by means of dry friction. It consists of a wooden block or shoe clamped to the brake rim by means of bolts. When these blocks are pressed into contact by means of bolts, the power of the engine is dissipated in frictional resistance. The power thus absorbed is converted into heat. Due to high heat involved, these systems are provided with cooling water to cool the rim of the brake drum.

\[\text{B}\text{.P}=\frac{\text{2 }\!\!\pi\!\!\text{ NT}}{\text{60}}\text{ W}\]

Also

\[\text{B}\text{.P}=\frac{\text{2 }\!\!\pi\!\!\text{ NT}}{\text{60000}}\text{ kW}\]

Where,

T = (W × L)

W – Weight on load carrier

L – Distance from the point of load to the centre of shaft.

Prony Brake Dynamometer Advantage & Disadvantages

It has an advantage of being simple, easy to construct and inexpensive. Due to the above advantages, it is extensively used for testing of low speed engines. The main drawback of prony brake is that, it cannot be used for testing in varying conditions due to its constant torque at any one band pressure.

The post What is Brake Power in IC Engine? Definition, Formula & Unit (Rope Brake & Prony Brake Dynamometer) appeared first on ElectricalWorkbook.

Figure 1: Heat Pipe.

Working of Heat Pipe

The heat pipe is a type of heat exchanger that uses the principles of phase transformation and thermal conductivity, to transfer heat in between two interfaces. These pipes are also called as superconductors due to their impeccable heat transfer capabilities. This device is a sealed container which is usually a pipe-like structure, consists of two ends and is filled with a work in a fluid. One end of the heat pipe is connected to the heat source and the other end is connected to the device to which heat is to transferred and a porous material called a wick is arranged on the inner walls of the pipe. When the evaporator section experiences high temperature, the working fluid gets evaporated and the vapour exerts pressure, this rise in pressure causes the vapour to flow to the condenser region. In this region, the heat is rejected to the sink or the device need to be heated and therefore, the fluid in heat pipe condenses. This condensate is driven back to the evaporator end by the capillary action in the wick. Hence in a heat pipe, the heat transfer occurs through the capillary movement of fluid.

A heat pipe works efficiently when the evaporator end is below the condenser end or when both the ends are in same horizontal line, because the return of working fluid from condenser end is due to both capillary action and gravity. If the condenser end is below the evaporator end, the fluid must flow against the gravity which decreases the rate of fluid flow thereby decreasing the heat transfer rate.

Types of heat Pipes

In general, the shape of heat pipes is not confined to the circular cross-section, there are heat pipes with other shapes also. The different kinds of heat pipes are as follows.

Capillary Heat Pipe with Multiple Tubes

In this kind of heat pipes, there are multiple tubes between evaporator and condenser regions for the flow of vapours. While the flow of condensate solely depends on capillarity. In horizontal heat pipes, internal heat exchange of 10000 W/m²K is achieved whereas the external heat transfer coefficients being only 50W/m²K between fluid inside the heat pipe and external forced air stream. Due to this, surface areas of condenser and evaporator regions must be 30 and 20 times greater than the heat pipe’s cross-sectional area respectively. Heat pipes with circumferentially grooved wicks, are able to transfer heat tip to 80000 W/m²K with water as a working fluid.

Flexibility Heat Pipes

When vibrations are involved, it is difficult to install a heat pipe. To overcome this problem, heat pipes are made with flexible bellows in which the design of the wick is capable to transfer heat under vibrating conditions.

Rotating Heat Pipe

The heat pipes of this kind are designed in a shape of truncated cone. The wider end of pipe is a evaporator region and the other end is a condenser region. As the vapour condenses and reaches the condenser region. As the heat pipe is allowed to rotate on its own axis. Due to the centrifugal force, fluid flows towards the evaporator region. Thus, the rate of fluid flow is proportional to the angular momentum.

Heat Plates

These are not considered as a heat pipes, but their function is similar to heat pipes. The heat plates are arranged with a mesh of wicks and vapour passages alternatively. One end of heat plate act as a condenser region and the other as a evaporator region similar to conventional pipes. A flat heat plate works only when it is oriented horizontally.

Heat Pipes with Gravity Induced Fluid Flow

In this kind of heat pipes, the wick is designed relatively smaller and is only for providing thermal insulation between vapour and liquid. And. the flow of fluid from the condenser region to evaporator region is entirely due to gravity.

Heat Pipe with the Osmotic Flow

In this kind of heat pipe, the sugar solution is used as a working fluid. When the evaporator region is exposed to heat, the water in the sugar solution gets vapourised and travels towards the condenser region where it transfers its latent heat by condensation. A semi-permeable membrane (cellulose) is arranged in between vapour region and fluid flow as a result osmotic pressure is developed on the condensate, due to which water flows into the sugar solution and the cycle is repeated. This kind of heat pipes can be arranged with condenser end lower than the evaporator end, because the gravitational pull is smaller than the osmotic pressure where this arrangement is not possible with conventional capillary heat pipes. The major drawback is that the flow rate of fluid through the membrane is very slow and thus the capacity of this heat pipe is smaller.

Electro-osmotic Heat Pipes

An electric field is applied along the heat pipes with the osmotic flow, due to the electro-osmotic effect the speed of the fluid increases along the wick. This effect only occurs when fluids with high dielectric nature are used as a working fluid. In few designs, where fluids flowing up to 500 mm against gravity this phenomenon is observed and it is almost impossible to observe in conventional heat pipes which solely depend on capillarity.

Inverse thermosiphon heat pipes

The evaporator region of this kind of heat pipes is above the condenser region. A return tube with an auxiliary heater on one end is arranged between condenser and evaporator regions. The other end of the return tube is dipped in the condenser region. The condensed vapour enters into the return tube, where the auxiliary heater heats the liquid and vapour bubbles are produced. Due to the density difference between pure fluid surrounding the return tube and vapour or liquid column in return tube, the fluid flows from condenser region to evaporator region.

Applications of Heat Pipe

The following are the applications of heat pipe.

1. These pipes are used in industrial plants for heat recovery purpose.
2. In aerospace for temperature stabilisation and cooling of space craft.
3. These pipes are used for cooling of electronic components.
4. These pipes carry heat from ovens and furnaces.
5. Heat exchangers.
6. In solar thermal applications such as.

• Solar distillation.
• Solar cooking.

What are the characteristics of fluids used in heat pipes?

The fluids used in the heat pipes must possess the following characteristics.

1. The fluids must be chemically stable and does not react with the heat pipe material.
2. Purification and degasification of fluids must be easy.
3. The cost of fluid should be economical.

The post What is Heat Pipe? Working, Diagram, Types & Applications appeared first on ElectricalWorkbook.

This type of heat exchangers are most suitable for large capacities of heat transfer and are widely used in steel and glass melting furnaces. The following are the factors that affect heat transfer rate in regenerators are,

1. The time interval between the flow of hot and cold fluids over the thermal storage medium.
2. The conductivity of the thermal storage medium.
3. Regenerator size.

Types of regenerative heat exchangers

There are two types of regenerators.

Static or Furnace Regenerator

Figure 1: Furnace Regenerator.

This type of heat exchanger consists of two chambers arranged with a thermal storage medium (Example: refractory fire bricks) in the form of a matrix. The hot gas or fluid is allowed to pass through the chamber where the medium absorbs the heat. The cold fluid or gas is passed through that chamber which absorbs heat from the medium. Two chambers are used such that one chamber absorbs heat while the other chamber transfers the heat. The direction of fluid flow is changed after a certain time. These regenerators are generally used in glass furnaces, coke ovens, open-hearth furnaces, etc. for preheating the combustion air or preheating the boiler feed water, etc.

Dynamic or Rotary Regenerator

Figure 1: Rotary Regenerator.

In this type of regenerator, a porous packing material in the shape of a disc with high heat capacity is arranged in between the two ducts in which the cold and hot fluids flow. As the disc rotates through the hot fluid, it stores heat and this thermal energy is transferred to the cold fluid. The overall efficiency of rotary regenerators is up to 85

The post What is Regenerative Heat Exchanger? Working, Diagram & Types appeared first on ElectricalWorkbook.