(a) Show in the form of a table, how the increase in the following variables affects (increase or decrease) the ignition delay…

(a) The effect of an increase in each variable on the ignition delay period of a CI engine, assuming all other parameters remain constant, is:

• (i) Self-ignition temperature increases → ignition delay increases.
• (ii) Cetane number increases → ignition delay decreases.
• (iii) Compression ratio increases → ignition delay decreases.
• (iv) Intake pressure increases → ignition delay decreases.
• (v) Intake temperature increases → ignition delay decreases.
• (vi) Air-fuel ratio increases, i.e. mixture becomes leaner → ignition delay increases.
• (vii) Exhaust gas recirculation increases → ignition delay increases.

(b) The desirable characteristics of an ideal working fluid for a vapour power cycle are:

• It should have a high latent heat of vaporisation, so that a large amount of energy can be transported per unit mass.
• It should have a high critical temperature so that a high boiler temperature can be used without going supercritical.
• Its critical pressure should be moderate, and its saturation pressure at the boiler temperature should not be excessively high.
• Its saturation pressure at the condenser temperature should not be too low, otherwise deep vacuum is required and air leakage becomes serious.
• It should have a high vapour density and low specific volume, so that turbine, piping and heat-exchanger sizes are small.
• It should have a low liquid specific volume, so that pump work is small.
• It should have a low liquid specific heat, so that less heat is required to raise the liquid to the boiling temperature.
• It should have high thermal conductivity, high heat-transfer coefficient and low viscosity.
• It should be chemically stable over the entire cycle temperature range.
• It should be non-corrosive, non-toxic, non-flammable and safe to handle.
• Its expansion in the turbine should remain dry or only slightly wet, to avoid blade erosion.
• It should be cheap, readily available and environmentally benign.

(c) The reheat factor of a steam turbine is defined as the ratio of the sum of the isentropic heat drops of all individual stages to the overall isentropic heat drop between the initial pressure and the final pressure. Thus, if H₁, H₂, ..., Hₙ are the stage isentropic drops and H is the overall isentropic drop, then

R.F. = (H₁ + H₂ + ... + Hₙ)/H.

Consider an infinitesimal expansion through pressure difference dp. For an isentropic process, T ds = dh − v dp = 0, so dh = v dp. Since pressure decreases, dp is negative, and the isentropic enthalpy drop is dH = −v dp.

Let p₁ be the initial pressure and p₂ the final pressure, with p₁ > p₂. Let s₀ be the initial entropy. The overall isentropic drop is H = ∫_p₁^p₂ −v(p, s₀) dp = ∫_p₂^p₁ v(p, s₀) dp.

In an actual multistage turbine, friction and other irreversibilities increase the entropy. Therefore, at any pressure p, the actual entropy s(p) is greater than s₀. The sum of the stage isentropic drops is ΣHᵢ = ∫_p₁^p₂ −v(p, s(p)) dp = ∫_p₂^p₁ v(p, s(p)) dp.

For steam, at constant pressure, specific volume increases with entropy: (∂v/∂s)_p > 0. This is true in the wet region, where higher entropy means higher dryness fraction, and in the superheated region, where higher entropy means higher temperature. Hence v(p, s(p)) > v(p, s₀). Therefore, the integrand in ΣHᵢ is greater than that in H. It follows that H₁ + H₂ + ... + Hₙ > H. Hence R.F. = (ΣHᵢ)/H > 1.

Thus the reheat factor is always greater than unity. Typical values lie between about 1.02 and 1.06.

(d) Comparison of vapour compression refrigeration (VCR) and vapour absorption refrigeration (VAR) systems:

• Energy input: VCR is driven mainly by mechanical work supplied to the compressor. VAR is driven mainly by thermal energy supplied to the generator, with a small amount of pump work.
• Pressure rise: In VCR, pressure is raised by the compressor. In VAR, pressure is raised by the combined absorber-generator-pump arrangement, often called a thermal compressor.
• Refrigerant: VCR uses a single pure refrigerant. VAR uses a binary mixture of refrigerant and absorbent, such as NH₃-H₂O or LiBr-H₂O.
• Main components: VCR has compressor, condenser, expansion valve and evaporator. VAR has generator, absorber, pump, rectifier, condenser, expansion valve and evaporator.
• COP: For VCR, COP = Q_evaporator/W_compressor. For VAR, COP = Q_evaporator/(Q_generator + W_pump). VCR generally has a much higher COP.
• Moving parts: VCR has many moving parts because of the compressor, so it is noisier and has more vibration. VAR has very few moving parts, so it is quieter and smoother.
• Maintenance: VCR requires more maintenance because of the compressor. VAR requires less maintenance.
• Size and capacity: VCR is compact and economical for small and medium capacities. VAR is bulky, but becomes economical for large capacities where waste heat, solar heat or cheap heat is available.
• Temperature range: VCR can achieve very low temperatures. LiBr-H₂O VAR is limited to temperatures above 0 °C, while NH₃-H₂O VAR can reach lower temperatures but ammonia is toxic.
• Applications: VCR is used in domestic refrigerators, air conditioners, automobiles and cold storage. VAR is used in industrial refrigeration, large air-conditioning plants, waste-heat recovery and solar cooling.
• Initial and running cost: VCR usually has lower initial cost for small units. VAR may have higher initial cost, but running cost can be low if waste heat is available.
• Environmental aspect: VCR may use synthetic refrigerants with global warming or ozone-depletion potential. VAR using water-LiBr is environmentally benign, while NH₃ is toxic but environmentally friendly.

(e) Given: DBT t = 35 °C, WBT t′ = 23 °C, barometer p = 755 mm Hg. Convert pressure to bar: p = (755/760) × 1.01325 = 1.00658 bar.

For WBT 23 °C, interpolate between 22 °C and 24 °C: p′_v = 0.026431 + (23 − 22)/(24 − 22) × (0.029832 − 0.026431) = 0.026431 + 0.5 × 0.003401 = 0.0281315 bar.

Using the modified Apjohn equation, p_v = p′_v − (1.8 p(t − t′))/2700 = 0.0281315 − (1.8 × 1.00658 × 12)/2700 = 0.0281315 − 0.0080527 = 0.0200788 bar.

(e)(i) Relative humidity At DBT 35 °C, interpolate between 34 °C and 36 °C: p_sat,35 = (0.053201 + 0.059423)/2 = 0.056312 bar.

Relative humidity φ = p_v/p_sat,35 = 0.0200788/0.056312 = 0.3566. Therefore, φ = 35.66 %.

(e)(ii) Humidity ratio ω = 0.622 × p_v/(p − p_v) = 0.622 × 0.0200788/(1.00658 − 0.0200788) = 0.622 × 0.0200788/0.9865012 = 0.01266 kg water vapour/kg dry air. Therefore, ω = 0.01266 kg/kg dry air = 12.66 g/kg dry air.

(e)(iii) Dew point temperature The dew point is the saturation temperature corresponding to p_v = 0.0200788 bar. From the table, this lies between 16 °C and 18 °C.

At 16 °C, p = 0.018173 bar. At 18 °C, p = 0.020630 bar.

Interpolation: t_dp = 16 + [(0.0200788 − 0.018173)/(0.020630 − 0.018173)] × (18 − 16) = 16 + [0.0019058/0.002457] × 2 = 16 + 1.5514 = 17.55 °C. Therefore, Dew point temperature = 17.55 °C.

(e)(iv) Enthalpy of moist air Using h = 1.005 t + ω(2501 + 1.86 t) kJ/kg dry air, h = 1.005 × 35 + 0.01266 × (2501 + 1.86 × 35) = 35.175 + 0.01266 × (2501 + 65.1) = 35.175 + 0.01266 × 2566.1 = 35.175 + 32.486 = 67.66 kJ/kg dry air. Therefore, Enthalpy of moist air = 67.66 kJ/kg dry air.