Diesel Power Plant

A diesel power plant is a generating station in which a diesel (compression-ignition) engine is the prime mover driving an alternator. It is used for standby, peak load and small isolated loads.

Schematic of a diesel power plant

 Air filter -> Intake    Exhaust -> Silencer -> Stack
                 |          ^
 Fuel tank  +----v----------+----+   coupling  +-----+
 -> filter->| DIESEL ENGINE      |============| Alt |->Bus
 -> pump -> | (injectors)        |            +-----+
 -> injector+--+-----------+-----+               |
               |           |                  Exciter
     Cooling water     Lube oil
     pump <-> cooling  pump <-> oil
     tower/radiator    cooler, filter
     Starting: compressed air bottles / electric motor

Systems of a diesel power plant

1. Engine and alternator: the diesel engine (2- or 4-stroke) converts the chemical energy of fuel into mechanical work; it is coupled to an alternator with its exciter on a common bed plate.

2. Fuel supply system: storage tank → transfer pump → day tank → filters → fuel (booster) pump → fuel injection pump → injectors. It stores, cleans and injects the correct quantity of fuel at high pressure at the right time.

3. Air intake system: air filter, intake ducts and (often) a turbo-charger/supercharger with intercooler. It supplies clean air for combustion.

4. Exhaust system: exhaust manifold, silencer (muffler) and chimney. It removes exhaust gases and reduces noise; exhaust heat may be recovered in a waste-heat boiler or to heat oil/water.

5. Cooling system: water pump, jacket, heat exchanger, radiator or cooling tower. It removes about 30 % of fuel heat to keep cylinder, piston and valves within safe temperature.

6. Lubrication system: oil sump, pump, filter and oil cooler. It reduces friction and wear and also cools pistons and bearings.

7. Starting system: compressed-air bottles with compressor (large engines), electric starter motor with battery (small engines), or an auxiliary petrol engine. The engine cannot start by itself.

8. Governing system: governor that changes the fuel supply to keep speed (frequency) constant as load varies.

9. Electrical system: alternator, exciter, switchgear, protection, and control panel.

Need of lubricating system

Lubrication puts a thin oil film between moving surfaces of the engine. It is needed to:

• reduce friction and wear between piston, rings, liners, bearings and gears;
• carry away heat from pistons and bearings (cooling action);
• seal the gap between piston rings and liner, so compression and power are not lost;
• clean the parts by carrying away carbon and metal particles to the filter;
• prevent corrosion and reduce noise.

Without lubrication the parts would overheat and seize in a few minutes.

Cooling systems of a diesel power plant

About 25–35 % of fuel heat goes to the cylinder walls. If not removed, the lubricating oil breaks down, pistons seize and valves warp. Two methods are used.

1. Open (once-through) cooling system

 River/well -> pump -> filter -> engine jacket
                                    |
           river (downstream) <-----+ hot water

Water is taken from a natural source, passed through the jacket once and discharged. Simple and cheap, but needs plenty of clean water; scaling from impurities.

2. Closed (recirculating) cooling system - used in most plants.

  +-> engine jacket -> hot water -> heat exchanger -+
  |                                  (soft water)   |
  +---- pump <---- expansion tank <-----------------+
               raw water side:
  cooling tower/pond -> pump -> HX -> tower (spray)

Soft (treated) water circulates in a closed loop through the engine jacket and a heat exchanger. The heat exchanger is cooled by raw water from a cooling tower or spray pond (or by air in a radiator for small sets). No scaling in the jacket, little water loss, so it is preferred.

3. Air cooling (small engines only): fins on the cylinder and a fan blow air over them. Simple, but cooling is not uniform and suits only small sets.

The fuel supply system of a diesel power plant receives, stores, cleans and delivers fuel oil to the engine cylinders in the right quantity, at high pressure and at the correct time.

 Tanker/rail -> unloading pump -> Main storage tanks
                                      |
                                transfer pump
                                      |
                                filter (strainer)
                                      |
                                  Day tank
                                      |
                               fine filter
                                      |
                            fuel (booster) pump
                                      |
                           fuel injection pump
                         (high pressure, metered)
                                      |
                       injectors -> engine cylinders

Components

1. Storage tanks: large tanks (above or below ground) holding about a month's fuel, filled from tankers through unloading pumps and strainers.
2. Transfer pump and strainer: move fuel from storage to the day tank after coarse filtering.
3. Day tank: small tank near the engine, holding about 8 hours' supply; often placed high so fuel flows by gravity.
4. Fine filters: remove dirt and water; dirty fuel blocks injector holes and wears pump plungers.
5. Fuel (booster) pump: feeds fuel to the injection pump at low pressure.
6. Fuel injection pump: plunger-type pump that raises pressure to about 100–200 bar (much higher in modern engines) and meters exact quantity per cycle as set by the governor.
7. Injectors (atomisers): spray fine fuel droplets into hot compressed air in the cylinder at the end of the compression stroke; the fuel ignites by itself.
8. Heaters (for heavy oil) and return lines from injectors back to the day tank.

Types of fuel injection systems

• Common rail system: one high-pressure pump keeps a common header at constant pressure; injectors open by timing.
• Individual pump system: separate pump for each cylinder; most common.
• Distributor system: one pump meters fuel and a distributor sends it to each cylinder in turn.

Requirements

• Accurate metering and timing; fine atomisation; quick start and stop of injection without dribble.

Need of cooling system

During combustion the gas temperature in a diesel engine reaches about 2000 °C. About 25–35 % of this heat flows into the cylinder walls, head, piston and valves. Cooling is needed to:

• keep the cylinder wall temperature low (about 150–200 °C) so the lubricating oil film is not burnt;
• prevent seizure of the piston from uneven expansion;
• avoid cracking of the head and warping of valves from thermal stress;
• keep volumetric efficiency and strength of the materials.

Over-cooling is also bad (more heat loss, poor combustion, corrosion), so temperature is controlled with a thermostat.

Cooling systems

1. Air cooling (small sets)

Fins on the cylinder and head increase surface area; a fan blows air over them.

• Simple, light, no water needed.
• Uneven cooling, noisy; only for small engines.

2. Water cooling

(a) Open (once-through) system

 River/lake -> pump -> filter -> engine jacket
                                    |
            back to river <---------+ (hot)

Water from a natural source passes once through the jacket and is discharged. Cheap and simple, but needs a large water source; impurities cause scaling.

(b) Natural circulation (thermosyphon) system

   +--> radiator/cooling tank --+
   |   (hot water rises)        | (cool water sinks)
   +---- engine jacket <--------+

Circulation is by density difference only; no pump. Slow circulation; suits small engines.

(c) Forced circulation, closed system - used in diesel power plants.

 +-> engine jacket -> thermostat -> heat exchanger -+
 |                                     ^ raw water  |
 +-- pump <------- expansion tank <----+------------+
          cooling tower or spray pond
          cools the raw water

A pump circulates soft (treated) water through the jacket and a heat exchanger; raw water from a cooling tower or spray pond takes the heat away. A thermostat bypasses the cooler until the engine warms up. Water loss and scaling are small, and temperature is closely controlled.

Fuel supply system

The fuel supply system stores, cleans and delivers fuel to the engine in the correct quantity, at high pressure and at the right time.

 Tanker -> unloading pump -> Storage tank
                                  |
                            transfer pump + strainer
                                  |
                              Day tank
                                  |
                            fine filter
                                  |
                         booster (fuel) pump
                                  |
                        fuel injection pump
                                  |
                     injectors -> cylinders
       (surplus fuel returns to the day tank)

1. Storage tank: holds bulk fuel (about a month's need).
2. Transfer pump and strainer: send coarse-filtered fuel to the day tank.
3. Day tank: small tank holding a few hours' supply near the engine.
4. Fine filter: removes dirt and water that would block injector nozzles.
5. Booster pump: feeds the injection pump at low pressure.
6. Fuel injection pump: raises pressure to about 100–200 bar and meters fuel as set by the governor.
7. Injector: atomises fuel into the hot compressed air, where it ignites by itself.

Injection arrangements: common rail, individual pump (most common), and distributor systems.

Main functions of the lubricating system

• Reduce friction and wear between moving parts (piston, rings, bearings, cams).
• Cool pistons and bearings by carrying heat to the oil cooler.
• Seal piston rings against the liner to keep compression.
• Clean parts by carrying carbon and metal particles to the filter.
• Prevent corrosion and reduce noise.

Data: 4 cylinders ($k = 4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1820$ rpm, $\dot m_f = 0.2$ kg/min, $CV = 44100$ kJ/kg, net brake load $(W-S) = 40$ kg, brake circumference $\pi D_b = 3$ m, $\eta_m = 0.9$. Take $g = 9.81\ \text{m/s}^2$.

Brake power

$$\begin{aligned} (W-S) &= 40\times9.81 = 392.4\ \text{N} \\ BP &= \frac{(W-S)\,\pi D_b\,N}{60} = \frac{392.4\times3\times1820}{60} = 35.71\ \text{kW} \end{aligned}$$

Heat supplied by fuel

$$\dot m_f = \frac{0.2}{60} = 0.003333\ \text{kg/s},\qquad Q = \dot m_f\times CV = 0.003333\times44100 = 147.0\ \text{kW}$$

i) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{Q} = \frac{35.71}{147.0} = 24.29\ \%$$

ii) Indicated thermal efficiency

$$\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{35.71}{0.9} = 39.68\ \text{kW} \\ \eta_{ith} &= \frac{IP}{Q} = \frac{39.68}{147.0} = 26.99\ \% \end{aligned}$$

(Check: $\eta_{ith} = \eta_{bth}/\eta_m$.)

iii) Indicated mean effective pressure

For a four-stroke engine, power strokes per second per cylinder $n = N/(2\times60) = 15.167$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ IP &= p_{mi}\,L\,A\,n\,k \\ p_{mi} &= \frac{39.68\times10^3}{0.12\times7.854\times10^{-3}\times15.167\times4} = 6.939\times10^5\ \text{N/m}^2 = 6.94\ \text{bar} \end{aligned}$$

iv) Brake specific fuel consumption

$$BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{35.71} = \frac{12}{35.71} = 0.336\ \text{kg/kWh}$$

Answer: $\eta_{bth}$ = 24.29 %, $\eta_{ith}$ = 26.99 %, IMEP = 6.94 bar, BSFC = 0.336 kg/kWh (BP = 35.71 kW).

A diesel power plant uses a compression-ignition engine as prime mover to drive an alternator. Its main components are shown below.

 Air filter -> Intake    Exhaust -> Silencer -> Stack
                 |          ^
 Fuel system +---v----------+---+  coupling  +-----+
 ----------->| DIESEL ENGINE    |===========| Alt |->Bus
             +--+-----------+---+           +-----+
                |           |                  |
          Cooling system  Lube system       Exciter
       (pump, HX, tower)  (pump, filter, cooler)
     Starting system: compressed air / battery-motor
     Governor controls fuel to hold speed

Main components

1. Diesel engine: prime mover (2- or 4-stroke).
2. Alternator and exciter: convert mechanical power into electrical power.
3. Fuel supply system: stores, filters and injects fuel.
4. Air intake system: air filter, ducts, supercharger.
5. Exhaust system: manifold, silencer, chimney.
6. Cooling system: pumps, heat exchanger, cooling tower/radiator.
7. Lubrication system: oil pump, filter, cooler.
8. Starting system: compressed air or electric starter.
9. Governor and control panel.

Fuel supply system

 Storage tank -> transfer pump -> strainer -> Day tank
   -> fine filter -> booster pump -> injection pump
   -> injectors -> cylinders   (surplus back to day tank)

1. Storage tanks hold bulk fuel received from tankers through an unloading pump.
2. Transfer pump and strainer move fuel to the day tank, which holds a few hours' supply close to the engine.
3. Fine filters remove dirt and water to protect the injection pump and nozzles.
4. Booster pump feeds fuel at low pressure to the fuel injection pump.
5. The injection pump raises the pressure (about 100–200 bar) and meters the quantity set by the governor.
6. Injectors atomise the fuel into hot compressed air at the end of the compression stroke, where it self-ignites.

Injection arrangements: common rail (one pump, common high-pressure header), individual pump for each cylinder (most common), and distributor system.

Data: two-stroke engine; motoring power (friction power) $FP = 5$ kW; $T = 300$ N·m; $N = 1200$ rpm; fuel used = 6.5 kg in 1 h; $CV = 40$ MJ/kg.

In a motoring test the engine is driven by a motor with fuel cut off; the power drawn equals the friction power, so $IP = BP + FP$.

Brake power

$$BP = \frac{2\pi NT}{60} = \frac{2\pi\times1200\times300}{60} = 37{,}699\ \text{W} = 37.70\ \text{kW}$$

Indicated power

$$IP = BP + FP = 37.70 + 5 = 42.70\ \text{kW}$$

Heat supplied

$$Q = \dot m_f\times CV = \frac{6.5}{3600}\times40{,}000 = 72.22\ \text{kW}$$

i) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{37.70}{42.70} = 0.8829 = 88.29\ \%$$

ii) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{Q} = \frac{42.70}{72.22} = 0.5912 = 59.12\ \%$$

iii) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{Q} = \frac{37.70}{72.22} = 0.5220 = 52.20\ \%$$

(Check: $\eta_{bth} = \eta_m\times\eta_{ith} = 0.8829\times0.5912 = 0.522$.)

Answer: $\eta_m$ = 88.29 %, $\eta_{ith}$ = 59.12 %, $\eta_{bth}$ = 52.20 %. (These efficiencies are higher than in a real diesel engine, about 30–40 %; they follow from the given data.)

Data: 4 cylinders ($k = 4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, $CV = 44000$ kJ/kg, net brake load $(W-S) = 40$ kg, brake circumference $\pi D_b = 3$ m, $\eta_m = 0.9$. Take $g = 9.81\ \text{m/s}^2$.

Brake power

$$\begin{aligned} (W-S) &= 40\times9.81 = 392.4\ \text{N} \\ BP &= \frac{(W-S)\,\pi D_b\,N}{60} = \frac{392.4\times3\times1800}{60} = 35.32\ \text{kW} \end{aligned}$$

Heat supplied by fuel

$$\dot m_f = \frac{0.2}{60} = 0.003333\ \text{kg/s},\qquad Q = \dot m_f\times CV = 0.003333\times44000 = 146.67\ \text{kW}$$

i) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{Q} = \frac{35.32}{146.67} = 24.08\ \%$$

ii) Indicated thermal efficiency

$$\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{35.32}{0.9} = 39.24\ \text{kW} \\ \eta_{ith} &= \frac{IP}{Q} = \frac{39.24}{146.67} = 26.75\ \% \end{aligned}$$

(Check: $\eta_{ith} = \eta_{bth}/\eta_m$.)

iii) Indicated mean effective pressure

For a four-stroke engine, power strokes per second per cylinder $n = N/(2\times60) = 15$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ IP &= p_{mi}\,L\,A\,n\,k \\ p_{mi} &= \frac{39.24\times10^3}{0.12\times7.854\times10^{-3}\times15\times4} = 6.939\times10^5\ \text{N/m}^2 = 6.94\ \text{bar} \end{aligned}$$

iv) Brake specific fuel consumption

$$BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{35.32} = \frac{12}{35.32} = 0.340\ \text{kg/kWh}$$

Answer: $\eta_{bth}$ = 24.08 %, $\eta_{ith}$ = 26.75 %, IMEP = 6.94 bar, BSFC = 0.340 kg/kWh (BP = 35.32 kW).

Data: 4 cylinders ($k=4$), four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1600$ rpm, $\dot m_f = 0.2$ kg/min, $BP = 31.4$ kW, $\eta_m = 0.8$, $CV = 44{,}000$ kJ/kg.

i) Brake specific fuel consumption

$$BSFC = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{0.2\times60}{31.4} = \frac{12}{31.4} = 0.382\ \text{kg/kWh}$$

ii) Indicated mean effective pressure

$$\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{31.4}{0.8} = 39.25\ \text{kW} \\ n &= \frac{N}{2\times60} = \frac{1600}{120} = 13.333\ \text{power strokes/s per cylinder} \\ A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ p_{mi} &= \frac{IP}{L\,A\,n\,k} = \frac{39.25\times10^3}{0.12\times7.854\times10^{-3}\times13.333\times4} \\ &= 7.809\times10^5\ \text{N/m}^2 = 7.81\ \text{bar} \end{aligned}$$

iii) Brake thermal efficiency

$$\begin{aligned} Q &= \frac{0.2}{60}\times44{,}000 = 146.67\ \text{kW} \\ \eta_{bth} &= \frac{BP}{Q} = \frac{31.4}{146.67} = 0.2141 = 21.41\ \% \end{aligned}$$

Answer: BSFC = 0.382 kg/kWh, IMEP = 7.81 bar, $\eta_{bth}$ = 21.41 %.

Diesel cycle

The diesel cycle is the ideal (air-standard) cycle of a compression-ignition engine, in which heat is added at constant pressure. It has four processes:

  p
  |  2 _______ 3
  |   |        \
  |   |         \        1-2 isentropic compression
  |    \         \       2-3 heat added at const. p
  |     \         + 4    3-4 isentropic expansion
  |      `.       |      4-1 heat rejected at const. V
  |        `-._   |
  |            `--+ 1
  +--------------------- V

1. 1–2 Isentropic compression: air alone is compressed (ratio $r$ = 14 to 22), raising its temperature above the fuel's self-ignition point.
2. 2–3 Constant-pressure heat addition: fuel is injected and burns as the piston moves out.
3. 3–4 Isentropic expansion: power stroke.
4. 4–1 Constant-volume heat rejection: exhaust.

Efficiency, with cut-off ratio $\rho = V_3/V_2$:

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

How the engine is cooled

About 30 % of the fuel heat goes to the cylinder walls, so cooling keeps the lubricating oil film and metal parts safe:

• Air cooling: fins and a fan; for small sets only.
• Open water cooling: river/well water passes once through the jacket and is discharged.
• Closed (forced circulation) cooling: soft water is pumped through the jacket and a heat exchanger (or radiator); raw water from a cooling tower or spray pond removes the heat. A thermostat holds the jacket temperature. This is the usual method in diesel power plants.
• The lubricating oil also cools pistons and bearings and is itself cooled in an oil cooler.

Advantages of diesel power plant

• Simple design and layout; small space; can be built quickly.
• Can be located near the load; no large water need.
• Quick starting and loading; good for standby and peak loads.
• Low standby losses; high thermal efficiency at full load (about 35–40 %).
• Low capital cost per kW for small capacities; fewer operating staff.

Disadvantages

• High fuel cost (imported oil in Nepal) and high operating cost.
• Limited unit size (a few MW) and short life.
• High maintenance and lubrication cost.
• Noise, vibration and exhaust pollution.
• Cannot run on sustained overload.

Data: $D = 0.2$ m, $L = 0.28$ m, $p_m = 3.1$ bar, $N = 350$ rpm, net brake load $W = 640$ N, drum diameter 1 m (no rope diameter given, so effective radius $R = 0.5$ m), fuel 1.52 kg in 20 min, CV = 43 900 kJ/kg.

For a four-stroke engine the number of power strokes per minute is $n = N/2 = 175$.

(i) Indicated power

$$\begin{aligned} A &= \frac{\pi}{4}D^2 = \frac{\pi}{4}(0.2)^2 = 0.031416\ \text{m}^2 \\ IP &= \frac{p_m L A n}{60} = \frac{3.1\times10^5 \times 0.28 \times 0.031416 \times 175}{60} \\ &= 7953\ \text{W} = 7.953\ \text{kW} \end{aligned}$$

(ii) Brake power

$$\begin{aligned} T &= W R = 640 \times 0.5 = 320\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 350 \times 320}{60} = 11\,729\ \text{W} = 11.729\ \text{kW} \end{aligned}$$

(iii) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{11.729}{7.953} = 1.475 = 147.5\%$$

(iv) Indicated thermal efficiency

$$\begin{aligned} \dot m_f &= \frac{1.52}{20\times60} = 0.0012667\ \text{kg/s}\ (4.56\ \text{kg/h}) \\ \text{Heat supplied} &= \dot m_f \times CV = 0.0012667 \times 43\,900 = 55.61\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot m_f\,CV} = \frac{7.953}{55.61} = 0.1430 = 14.30\% \end{aligned}$$

Answer: IP = 7.953 kW, BP = 11.729 kW, $\eta_m$ = 147.5 %, $\eta_{ith}$ = 14.30 %.

Note on the data: a mechanical efficiency above 100 % is impossible (BP can never exceed IP), so the figures as printed are inconsistent. If the engine is taken as two-stroke ($n = N = 350$), IP doubles to 15.906 kW, giving $\eta_m = 11.729/15.906 = 73.7\%$ and $\eta_{ith} = 15.906/55.61 = 28.6\%$, which are realistic values. In the exam, show the working as above and point out the inconsistency.

Data: $\dot m_f = 5.5$ g/s $= 0.0055$ kg/s, $BP = 75$ kW, $\eta_m = 0.85$, LHV $= 44$ MJ/kg $= 44\,000$ kJ/kg.

Indicated power:

$$IP = \frac{BP}{\eta_m} = \frac{75}{0.85} = 88.24\ \text{kW}$$

Fuel used per hour: $0.0055 \times 3600 = 19.8$ kg/h.

a) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f\ (\text{kg/h})}{BP} = \frac{19.8}{75} = 0.264\ \text{kg/kWh}$$

b) Indicated specific fuel consumption

$$isfc = \frac{19.8}{88.24} = 0.2244\ \text{kg/kWh}$$

c) Brake thermal efficiency

Heat supplied by fuel: $\dot m_f \times LHV = 0.0055 \times 44\,000 = 242$ kW.

$$\eta_{bth} = \frac{BP}{\dot m_f\,LHV} = \frac{75}{242} = 0.3099 = 30.99\%$$

d) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{\dot m_f\,LHV} = \frac{88.24}{242} = 0.3646 = 36.46\%$$

Check: $\eta_{bth} = \eta_m \times \eta_{ith} = 0.85 \times 36.46 = 30.99\%$.

Answer: bsfc = 0.264 kg/kWh, isfc = 0.2244 kg/kWh, $\eta_{bth}$ = 30.99 %, $\eta_{ith}$ = 36.46 %.

A diesel power plant uses a compression-ignition (CI) engine coupled to an alternator. Air compressed in the cylinder becomes hot enough to ignite the injected fuel; the engine's shaft power is converted to electrical power by the alternator. Plants are usually 2–50 MW and are used for standby, peak load and isolated supply.

General layout

 Bulk tank -> Filter -> Day tank -> Inj. pump
                                       |
                                       v
 Air filter -> Supercharger -> +--------+
                               | DIESEL |==> ALTERNATOR
 Comp. -> Air bottle --------> | ENGINE |
                               +--------+
                                |  |  |
      Lube tank, filter, cooler-+  |  +-> Silencer->Stack
                                   v
      Cooling tower <- Heat exch <- Jacket pump

Main components: engine, alternator, air intake system (filter, supercharger), fuel system, exhaust system (silencer, stack), cooling system, lubricating system, starting system and governing system.

Starting system

A diesel engine cannot start by itself because the high compression needs an external torque to turn it to a speed at which compression temperature ignites the fuel. Methods:

1. Compressed-air starting (most common for large plants): an air compressor charges air bottles to about 17–25 bar. To start, a starting valve admits this air to some cylinders in firing order; the air pushes the pistons and rotates the crankshaft. When the engine reaches firing speed, fuel is injected and air supply is cut off.
2. Electric starting: a battery-driven DC motor turns the engine through a gear on the flywheel; used for small engines.
3. Auxiliary petrol engine: a small engine cranks the main engine through a clutch; rarely used now.

Fuel supply system

1. Fuel oil arrives by tanker and is pumped through a strainer into the bulk storage tank.
2. A transfer pump sends it through filters to a day tank (holding about one day's need), placed higher so oil flows by gravity.
3. From the day tank oil passes through fine filters to the fuel injection pump, which raises the pressure (about 100–200 bar or more) and meters the correct quantity according to the governor setting.
4. The injector (atomiser) sprays fine droplets into the hot compressed air at the right instant, where it ignites.
5. Excess oil from the injectors returns to the day tank. Heaters are used for heavy oils to reduce viscosity.

The fuel system may be a common-rail, individual pump or distributor type injection system.

A diesel power plant consists of a CI engine driving an alternator, together with auxiliary systems that supply fuel, air, cooling, lubrication and starting, and remove the exhaust.

Main components (sketch)

  Air filter -> Supercharger -+
                              v
 Fuel tank -> pump -> +--------------+
                      | DIESEL ENGINE|==> ALTERNATOR
 Air bottle --------> +--------------+
 (starting)             |    |    |
      Lube oil tank <---+    |    +--> Silencer -> Stack
      filter, cooler         v
                 Jacket water -> Heat exch -> Cooling tower

1. Engine – converts fuel energy to shaft work (2- or 4-stroke).
2. Air intake system – air filters, ducts and supercharger.
3. Exhaust system – exhaust manifold, silencer, stack; sometimes waste-heat boiler.
4. Fuel system – storage tank, day tank, filters, injection pump, injectors.
5. Cooling system – pumps, heat exchanger, cooling tower/radiator.
6. Lubricating system – oil tank, pump, filter, oil cooler.
7. Starting system – air compressor and air bottles (or battery motor).
8. Governing system – controls fuel supply to keep speed constant.

Functions of the lubricating system

• Reduces friction and wear between moving parts (piston and liner, bearings, gears) by keeping an oil film.
• Cools the parts: oil carries heat away from bearings and piston crowns.
• Seals the gap between piston rings and liner, preventing gas leakage.
• Cleans the engine by carrying away carbon and metal particles to the filter.
• Protects surfaces against corrosion and absorbs shocks in bearings.

Functions of the cooling system

• About 25–35 % of fuel heat goes into the cylinder walls, head and piston. The cooling system removes this heat so that metal temperatures stay within safe limits.
• Prevents thermal stresses, cracking and distortion of cylinder head and liner.
• Prevents breakdown of the lubricating oil film at high temperature, which would cause seizure.
• Keeps the engine at a suitable working temperature for good combustion and efficiency (over-cooling is also avoided).

Assumptions: the engine works on the air-standard Diesel cycle (isentropic compression 1–2, constant-pressure heat addition 2–3, isentropic expansion 3–4, constant-volume heat rejection 4–1). The heat added is taken as 1675 kJ/kg (1675 MJ/kg is clearly a misprint), $R = 0.287$ kJ/kg K, $c_p = 1.005$, $c_v = 0.718$ kJ/kg K, $\gamma = 1.4$.

Data: $r = 15$, $p_1 = 0.1$ MPa, $T_1 = 40 + 273 = 313$ K.

(i) Maximum temperature and pressure

Process 1–2 (isentropic compression):

$$\begin{aligned} T_2 &= T_1 r^{\gamma-1} = 313 \times 15^{0.4} = 924.66\ \text{K} \\ p_2 &= p_1 r^{\gamma} = 0.1 \times 15^{1.4} = 4.431\ \text{MPa} \end{aligned}$$

Process 2–3 (heat added at constant pressure):

$$T_3 = T_2 + \frac{q_s}{c_p} = 924.66 + \frac{1675}{1.005} = 2591.32\ \text{K}$$

Maximum temperature $T_3 = 2591.3$ K; maximum pressure $p_3 = p_2 = 4.431$ MPa.

(ii) Cut-off ratio

$$\rho = \frac{v_3}{v_2} = \frac{T_3}{T_2} = \frac{2591.32}{924.66} = 2.802$$

(iii) Net work per kg of air

$$\begin{aligned} T_4 &= T_3\left(\frac{\rho}{r}\right)^{\gamma-1} = 2591.32\left(\frac{2.802}{15}\right)^{0.4} = 1324.65\ \text{K} \\ q_r &= c_v (T_4 - T_1) = 0.718(1324.65 - 313) = 726.37\ \text{kJ/kg} \\ W_{net} &= q_s - q_r = 1675 - 726.37 = 948.63\ \text{kJ/kg} \end{aligned}$$

(iv) Thermal efficiency

$$\eta = \frac{W_{net}}{q_s} = \frac{948.63}{1675} = 0.5664 = 56.64\%$$

Check with the formula $\eta = 1 - \dfrac{1}{r^{\gamma-1}}\cdot\dfrac{\rho^{\gamma}-1}{\gamma(\rho-1)} = 56.64\%$.

(v) Mean effective pressure

$$\begin{aligned} v_1 &= \frac{R T_1}{p_1} = \frac{0.287 \times 313}{100} = 0.8983\ \text{m}^3/\text{kg} \\ v_2 &= \frac{v_1}{r} = 0.05989\ \text{m}^3/\text{kg} \\ p_m &= \frac{W_{net}}{v_1 - v_2} = \frac{948.63}{0.8983 - 0.05989} = 1131.5\ \text{kPa} \end{aligned}$$

Answer: $T_{max}$ = 2591.3 K, $p_{max}$ = 4.431 MPa, cut-off ratio = 2.802, $W_{net}$ = 948.6 kJ/kg, $\eta$ = 56.64 %, MEP = 1131.5 kPa ≈ 11.31 bar.

Data: 4 cylinders, four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1800$ rpm, $\dot m_f = 0.2$ kg/min, CV = 44 000 kJ/kg, net rope tension $(W - S) = 40$ kgf, brake circumference = 3.0 m, rope radius = 25 mm, $\eta_m = 0.90$.

Brake power

Drum radius from circumference: $R_d = \dfrac{3.0}{2\pi} = 0.4775$ m. Effective radius (to rope centre): $R_e = 0.4775 + 0.025 = 0.5025$ m. Net force: $F = 40 \times 9.81 = 392.4$ N.

$$\begin{aligned} T &= F R_e = 392.4 \times 0.5025 = 197.17\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 197.17}{60} = 37.165\ \text{kW} \\ IP &= \frac{BP}{\eta_m} = \frac{37.165}{0.9} = 41.295\ \text{kW} \end{aligned}$$

Heat supplied: $\dot m_f\, CV = \dfrac{0.2}{60} \times 44\,000 = 146.67$ kW.

(i) Brake thermal efficiency

$$\eta_{bth} = \frac{37.165}{146.67} = 0.2534 = 25.34\%$$

(ii) Indicated thermal efficiency

$$\eta_{ith} = \frac{41.295}{146.67} = 0.2816 = 28.16\%$$

(iii) Indicated mean effective pressure

For a four-stroke, $k$-cylinder engine: $IP = \dfrac{p_m L A (N/2) k}{60}$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2 \\ p_m &= \frac{60\, IP}{L A (N/2) k} = \frac{60 \times 41\,295}{0.12 \times 7.854\times10^{-3} \times 900 \times 4} \\ &= 7.302\times10^5\ \text{N/m}^2 = 7.30\ \text{bar} \end{aligned}$$

(iv) Brake specific fuel consumption

$$bsfc = \frac{0.2 \times 60}{37.165} = \frac{12}{37.165} = 0.3229\ \text{kg/kWh}$$

Answer: $\eta_{bth}$ = 25.34 %, $\eta_{ith}$ = 28.16 %, imep = 7.30 bar, bsfc = 0.323 kg/kWh (BP = 37.17 kW, IP = 41.30 kW).

A closed cooling system is a water (liquid) cooling system in which the same treated water (soft water or water + anti-freeze/inhibitor) circulates continuously through the engine jackets and a heat exchanger. The jacket water never comes in contact with air or raw water, so it stays clean; a separate raw-water circuit removes the heat from the heat exchanger.

Layout

        Primary (closed) circuit
   +--------------------------------+
   |                                |
 ENGINE  -->  Heat  ---> Pump  ---> +
 jackets      exchanger    ^
   ^            |  ^       |
   |            |  |    Expansion /
   +-- (return) |  |    surge tank
                v  |
       Raw water out  Raw water in
       (to cooling    (from river,
        tower/pond)    tower)
          Secondary (open) circuit

Working

1. A circulating pump forces soft water through the cylinder jackets and cylinder heads, where it absorbs heat (outlet about 70–85 °C).
2. Hot water flows to a heat exchanger (or radiator). In the heat exchanger, raw water from a river, pond or cooling tower passes through tubes and takes the heat away.
3. The cooled jacket water returns to the pump and engine, completing the closed loop.
4. A surge/expansion tank takes up volume changes and allows make-up water; a thermostat bypasses the heat exchanger when the engine is cold so that it warms up quickly.
5. The raw water is cooled in a cooling tower or spray pond and reused, or discharged.
6. Lubricating oil coolers are often placed in the same secondary circuit.

For small engines the heat exchanger is replaced by an air-cooled radiator with a fan (radiator cooling is also a closed system).

Advantages

• Jacket water is soft and clean, so there is no scale formation or corrosion in the jackets; heat transfer stays good.
• Water loss is very small; little make-up is needed.
• Engine temperature can be controlled accurately with a thermostat.
• Anti-freeze can be used in cold places.
• Raw water of poor quality (river water) can be used in the secondary circuit.
• Long life of engine parts and less maintenance of jackets.

Disadvantages

• Higher initial cost: heat exchanger, extra pumps, tanks and piping.
• More pumping power is needed (two circuits).
• Heat exchanger tubes on the raw-water side can still foul and need cleaning.
• More complicated than open (once-through) cooling; more space required.
• Leakage in the heat exchanger can contaminate the closed circuit.

In a motoring test, the engine (without fuel) is driven by the dynamometer acting as a motor. The power needed to turn it at the test speed equals the friction power of the engine. So $FP = 4.5$ kW.

Data: $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg in 1 h, CV = 40 MJ/kg.

Brake power:

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39\,270\ \text{W} = 39.27\ \text{kW}$$

Indicated power:

$$IP = BP + FP = 39.27 + 4.5 = 43.77\ \text{kW}$$

Heat supplied by fuel:

$$\dot Q = \frac{5}{3600} \times 40\,000 = 55.56\ \text{kW}$$

a) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{39.27}{43.77} = 0.8972 = 89.72\%$$

b) Indicated thermal efficiency

$$\eta_{ith} = \frac{IP}{\dot Q} = \frac{43.77}{55.56} = 0.7879 = 78.79\%$$

c) Brake thermal efficiency

$$\eta_{bth} = \frac{BP}{\dot Q} = \frac{39.27}{55.56} = 0.7069 = 70.69\%$$

Answer: $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %.

(These thermal efficiencies are far higher than any real diesel engine, about 30–40 %, because the given fuel consumption is low; the method is what matters.)

In water (liquid) cooling, water circulates through jackets around the cylinders and cylinder head, absorbs the heat conducted through the walls and gives it up in a cooler. The main types used in diesel power plants are:

1. Open (once-through) cooling system

Water is taken from a river, lake or well, pumped through the engine jackets once and then discharged to waste.

• Advantages: simplest and cheapest; no cooling tower or heat exchanger.
• Disadvantages: needs a large, continuous water supply; raw water causes scale and corrosion in jackets; temperature control is poor. Used only where plenty of clean water is available.

2. Natural (thermosiphon) circulation system

Hot water, being lighter, rises from the jackets to the top of a tank/radiator, cools and falls back to the engine. No pump is used.

• Advantages: no pump, simple, no power for circulation.
• Disadvantages: slow circulation, so suitable only for small engines; radiator must be above the engine; cooling depends on load.

3. Forced circulation system

A pump forces water through the jackets and then through a radiator or cooling tower. A thermostat controls the flow.

• Advantages: fast, uniform cooling; suitable for all sizes; good temperature control.
• Disadvantages: pump consumes power; cooling stops if the pump fails.

4. Closed (double-circuit) cooling system

Soft treated water circulates in a closed primary circuit through jackets and a heat exchanger; raw water in a secondary circuit removes the heat and is cooled in a cooling tower or pond.

• Advantages: no scale or corrosion in jackets, very low water loss, accurate temperature control, long engine life.
• Disadvantages: higher cost, more pumps and piping, heat-exchanger maintenance.

5. Evaporative cooling

Water is allowed to boil in the jackets; the steam is condensed and returned. Large heat removal per kg of water.

• Advantage: low water requirement.
• Disadvantage: engine runs hot; local hot spots; rarely used in power plants.

Closed system:
ENGINE -> Heat exch -> Pump -> ENGINE   (soft water)
            ^   |
  raw water in  out -> Cooling tower -> back

Assumptions: single-cylinder engine; the 5 kW is the power developed with the given mean effective pressure (brake mep basis), so $P = \dfrac{p_m L A n}{60}$ with $n = N/2$ for a four-stroke engine.

Data: $P = 5$ kW $= 5000$ W, $N = 2000$ rpm, $n = 1000$ per min, $p_m = 7.5$ bar $= 7.5\times10^5$ N/m², $L = 1.25D$.

$$\begin{aligned} P &= \frac{p_m \, L \, \frac{\pi}{4}D^2 \, n}{60} = \frac{p_m (1.25D)\frac{\pi}{4}D^2 n}{60} \\ 5000 &= \frac{7.5\times10^5 \times 1.25 \times 0.7854 \times D^3 \times 1000}{60} \\ D^3 &= \frac{5000 \times 60}{7.5\times10^5 \times 1.25 \times 0.7854 \times 1000} = 4.074\times10^{-4}\ \text{m}^3 \\ D &= 0.07413\ \text{m} = 74.1\ \text{mm} \\ L &= 1.25 \times 74.13 = 92.7\ \text{mm} \end{aligned}$$

Check: $\dfrac{7.5\times10^5 \times 0.09267 \times \frac{\pi}{4}(0.07413)^2 \times 1000}{60} = 5000$ W.

Answer: Bore $D \approx 74.1$ mm, stroke $L \approx 92.7$ mm.

A diesel power plant is a CI engine–alternator set supported by five essential auxiliary systems: air intake, cooling, fuel, lubrication and exhaust (plus starting and governing).

Layout

       Air filter --> Supercharger
                         |
 Bulk tank -> Filter -> Day tank -> Inj. pump
                         |            |
                         v            v
                    +-------------------+
 Lube tank->pump--> |   DIESEL ENGINE   |==> ALTERNATOR
 ->filter->cooler   +-------------------+
        ^  return     |              |
        +-------------+              v
                      |          Silencer -> Stack
         Jacket water v
   Cooling tower <- Heat exchanger <- Pump

(i) Air intake system

Supplies clean air for combustion. Air passes through air filters (oil-bath or dry type) that remove dust, then through ducts and often a supercharger/turbocharger that raises air density so more fuel can be burnt per stroke. Intake is placed outside the engine room, away from exhaust, and a silencer reduces suction noise. Dirty air would cause rapid wear of liners and rings.

(ii) Cooling system

Removes 25–35 % of the fuel heat that passes to cylinder walls, head and pistons. Water is pumped through the jackets, then to a heat exchanger or radiator; the heat is finally rejected in a cooling tower or spray pond. Closed (soft-water) systems are preferred to prevent scaling. Proper cooling keeps metal temperatures and lube-oil temperatures safe.

(iii) Fuel supply system

Oil from tankers goes to bulk storage tanks, then is transferred through filters to a day tank. From there the fuel injection pump pressurises and meters the oil, and injectors atomise it into the hot compressed air. Heaters are used for heavy oil. Governor controls the quantity injected according to load.

(iv) Lubrication system

An oil pump draws oil from the sump/tank and supplies it under pressure through filters and an oil cooler to main bearings, crank pins, gudgeon pins, camshaft and cylinder walls. Oil reduces friction and wear, cools parts, seals the piston rings and carries away dirt. Oil drains back to the sump and is recirculated.

(v) Exhaust system

Discharges hot exhaust gases from the cylinders to the atmosphere through the exhaust manifold, silencer (muffler) and stack. The silencer reduces noise; flexible joints isolate vibration and allow expansion. The exhaust heat (about 30 % of fuel energy) may be recovered in a waste-heat boiler or used to drive the turbocharger. The stack must be high enough to disperse fumes.

Data: $N = 1800$ rpm, $T = 8$ N m, $IP = 1.8$ kW.

Brake power:

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 1800 \times 8}{60} = 1508\ \text{W} = 1.508\ \text{kW}$$

Friction power:

$$FP = IP - BP = 1.8 - 1.508 = 0.292\ \text{kW}$$

Friction loss as a percentage of IP:

$$\frac{FP}{IP}\times 100 = \frac{0.292}{1.8}\times100 = 16.22\%$$

Answer: Friction power = 0.292 kW, which is 16.22 % of the indicated power (mechanical efficiency = 83.78 %).

In the motoring test the engine is turned by the dynamometer without fuel; the power required equals the friction power, so $FP = 2.2$ kW.

Data: $T = 150$ N m, $N = 800$ rpm, fuel = 2.5 kg in 1 h, CV = 40 MJ/kg = 40 000 kJ/kg.

(a) Brake power

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 800 \times 150}{60} = 12\,566\ \text{W} = 12.566\ \text{kW}$$

(b) Indicated power

$$IP = BP + FP = 12.566 + 2.2 = 14.766\ \text{kW}$$

(c) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{12.566}{14.766} = 0.8510 = 85.10\%$$

(d) Indicated thermal efficiency

Heat supplied:

$$\dot Q = \frac{2.5}{3600}\times 40\,000 = 27.78\ \text{kW}$$ $$\eta_{ith} = \frac{IP}{\dot Q} = \frac{14.766}{27.78} = 0.5316 = 53.16\%$$

Answer: BP = 12.57 kW, IP = 14.77 kW, $\eta_m$ = 85.10 %, $\eta_{ith}$ = 53.16 % (brake thermal efficiency would be 45.24 %).

The fuel storage and supply system receives, stores, cleans and delivers fuel oil to the engine cylinders in the correct quantity, at the correct time and in atomised form.

Sketch

 Tanker -> Strainer -> BULK STORAGE TANK
                          |
                   Transfer pump
                          |
                       Filter
                          v
                      DAY TANK (overhead)
                          | gravity
                     Fine filter
                          v
                 FUEL INJECTION PUMP <- Governor
                          | high pressure
                          v
                 INJECTORS (in cylinders)
                          |
        leak-off oil -----+--> back to day tank

Working

1. Unloading and storage: oil brought by tankers or rail wagons is pumped through a strainer into large bulk storage tanks (steel, above or below ground), holding one to several months' supply. Tanks have vents, level gauges and drain valves for water and sludge.
2. Transfer: a transfer pump moves oil from the bulk tank through filters to the day tank, which holds about 8–24 hours of consumption and is placed above the engine so oil flows by gravity.
3. Filtration and heating: fine filters (and centrifuges for heavy oil) remove dirt and water. Heavy oils are heated to reduce viscosity.
4. Injection pump: a plunger-type pump raises the pressure (about 100–200 bar or more) and delivers a metered quantity, controlled by the governor according to load.
5. Injector: sprays finely atomised oil into the hot compressed air near the end of compression, where it self-ignites.
6. Surplus/leak-off oil returns to the day tank.

Injection may be by common-rail, individual pump or distributor system.

Applications of diesel power plants

• Standby (emergency) plants for hospitals, telephone exchanges, data centres, airports, industries.
• Peak-load plants working with hydro or steam stations.
• Base-load plants in small isolated areas where hydro or grid supply is not available.
• Mobile plants (trailer-mounted) for construction sites, mines, temporary loads and relief work.
• Starting stations (black-start) for large steam and gas stations.
• Nursery stations in growing towns before grid extension, and supply on islands, hilly and remote areas.
• Captive power for industries where load-shedding occurs.

Data: $N = 650$ rpm, drum diameter $D = 0.6$ m, rope diameter $d = 0.05$ m, dead load $W = 32$ kg, spring balance $S = 4.75$ kg, $\eta_m = 0.80$.

For a rope brake dynamometer the effective radius is measured to the rope centre:

$$R_e = \frac{D + d}{2} = \frac{0.6 + 0.05}{2} = 0.325\ \text{m}$$

Net brake load:

$$(W - S)g = (32 - 4.75)\times 9.81 = 267.32\ \text{N}$$

Brake power

$$\begin{aligned} T &= (W-S)g\,R_e = 267.32 \times 0.325 = 86.88\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 650 \times 86.88}{60} = 5914\ \text{W} = 5.914\ \text{kW} \end{aligned}$$

Indicated power

$$IP = \frac{BP}{\eta_m} = \frac{5.914}{0.80} = 7.392\ \text{kW}$$

Answer: BP = 5.91 kW, IP = 7.39 kW (friction power = 1.48 kW).

A diesel power plant generates electricity using a compression-ignition engine coupled to an alternator. It is mainly suited to small capacities, standby and peak duties, and isolated areas.

Advantages

1. Simple design and layout; few auxiliaries compared with a steam plant (no boiler, condenser or ash handling).
2. Quick starting and loading: can be started and put on full load within a few minutes, so it is ideal for standby and peak loads.
3. Small space required; can be located near the load centre, reducing transmission cost.
4. Low initial cost per kW for small capacities; short construction time.
5. High thermal efficiency (about 30–40 %), better than small steam plants; good part-load efficiency.
6. Less water is needed (only for cooling).
7. No standby losses: fuel is used only when the engine runs.
8. Fewer operating staff; easy operation and automatic control.
9. Fuel is easy to handle and store; no ash disposal problem.
10. Can be mobile (skid or trailer mounted) and works independently of weather.

Disadvantages

1. High fuel cost: diesel oil is costly and in Nepal must be imported, so running cost is high.
2. Limited capacity: unit size usually up to about 50 MW; not economical for large base-load stations.
3. High maintenance and lubrication cost; frequent overhauls of injectors, valves and liners.
4. Noise and vibration are severe; heavy foundations and silencers are needed.
5. Short life (about 10–15 years) compared with hydro (50 years or more) and steam stations (25–30 years).
6. Overload capacity is poor: engine cannot run long above rated load.
7. Exhaust pollution (NOx, smoke, particulates).
8. Cost per kW rises sharply with size.

Comparison summary

Data: single cylinder, four-stroke, $D = 0.3$ m, $L = 0.45$ m, fuel 11.4 kg in 60 min, CV = 42 000 kJ/kg, $p_m = 6.0$ bar, net brake load $W = 1500$ N, $N = 300$ rpm, drum diameter 1.8 m, rope diameter 20 mm.

Power strokes per minute: $n = N/2 = 150$.

(a) Indicated power

$$\begin{aligned} A &= \frac{\pi}{4}(0.3)^2 = 0.070686\ \text{m}^2 \\ IP &= \frac{p_m L A n}{60} = \frac{6\times10^5 \times 0.45 \times 0.070686 \times 150}{60} \\ &= 47\,713\ \text{W} = 47.71\ \text{kW} \end{aligned}$$

(b) Brake power

Effective radius $R_e = \dfrac{1.8 + 0.02}{2} = 0.91$ m.

$$\begin{aligned} T &= W R_e = 1500 \times 0.91 = 1365\ \text{N m} \\ BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 300 \times 1365}{60} = 42\,883\ \text{W} = 42.88\ \text{kW} \end{aligned}$$

(c) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{42.88}{47.71} = 0.8988 = 89.88\%$$

(d) Indicated thermal efficiency

$$\begin{aligned} \dot Q &= \frac{11.4}{3600}\times 42\,000 = 133.0\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot Q} = \frac{47.71}{133.0} = 0.3587 = 35.87\% \end{aligned}$$

Answer: IP = 47.71 kW, BP = 42.88 kW, $\eta_m$ = 89.88 %, $\eta_{ith}$ = 35.87 %.

Diesel engines are cooled to remove the heat (about 25–35 % of fuel energy) that flows into cylinder walls, head and pistons, so that metal and lubricating oil stay within safe temperatures.

Methods of cooling diesel engines

1. Air cooling: fins on the cylinder and head increase surface area; air from a fan or vehicle motion carries heat away. Used only for small engines (a few kW) because air has low heat capacity.
2. Liquid (water) cooling: water circulates through jackets; used for all power-plant engines. Types:
   • Open (once-through) system – raw water used once and discharged.
   • Natural (thermosiphon) circulation – circulation by density difference, no pump; small engines.
   • Forced circulation – pump drives water through jackets and radiator/cooling tower.
   • Closed (double-circuit) system – soft water in a closed loop through jackets and a heat exchanger; raw water in the secondary loop goes to a cooling tower or spray pond.
   • Evaporative cooling – water allowed to boil in jackets, steam condensed and returned.
3. Oil cooling of pistons in large engines (oil sprayed under the piston crown).

Function of a cooling tower

A cooling tower cools the hot water coming from the engine jackets or heat exchanger so the same water can be reused, which saves water where a river or lake is not available.

      Air + vapour out
          ^  ^  ^
   +----------------+
   |  Spray nozzles | <-- hot water in
   |  ~~~~~~~~~~~~  |
   |  Fill/packing  |
   |  ~~~~~~~~~~~~  |
 ->| air in   air in|<-
   +----------------+
   |  Cold water    | --> pump -> engine/HX
   +----------------+

• Hot water (about 50–60 °C) is sprayed at the top and trickles down over fill/packing that breaks it into thin films and drops.
• Air flows upward (by natural draught or fans – induced or forced draught) through the falling water.
• A small part of the water evaporates; the latent heat for evaporation is taken from the rest of the water, which cools by 8–12 °C. Some sensible heat also passes to the air.
• Cold water collects in the basin and is pumped back. Make-up water replaces evaporation, drift and blow-down losses (about 1–3 %).

Thus the cooling tower rejects engine heat to the atmosphere with very small water consumption.

In a motoring test the engine is driven by the dynamometer without fuel; the motoring power equals the friction power: $FP = 4.5$ kW.

Data: $T = 250$ N m, $N = 1500$ rpm, fuel = 5 kg/h, CV = 40 MJ/kg = 40 000 kJ/kg.

Brake power and indicated power:

$$\begin{aligned} BP &= \frac{2\pi N T}{60} = \frac{2\pi \times 1500 \times 250}{60} = 39.27\ \text{kW} \\ IP &= BP + FP = 39.27 + 4.5 = 43.77\ \text{kW} \end{aligned}$$

Heat supplied:

$$\dot Q = \frac{5}{3600}\times 40\,000 = 55.56\ \text{kW}$$

i) Mechanical efficiency

$$\eta_m = \frac{39.27}{43.77} = 0.8972 = 89.72\%$$

ii) Indicated thermal efficiency

$$\eta_{ith} = \frac{43.77}{55.56} = 0.7879 = 78.79\%$$

iii) Brake thermal efficiency

$$\eta_{bth} = \frac{39.27}{55.56} = 0.7069 = 70.69\%$$

iv) Indicated specific fuel consumption

$$isfc = \frac{\dot m_f}{IP} = \frac{5}{43.77} = 0.1142\ \text{kg/kWh}$$

v) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f}{BP} = \frac{5}{39.27} = 0.1273\ \text{kg/kWh}$$

Answer: $\eta_m$ = 89.72 %, $\eta_{ith}$ = 78.79 %, $\eta_{bth}$ = 70.69 %, isfc = 0.114 kg/kWh, bsfc = 0.127 kg/kWh.

(The efficiencies are unrealistically high for a real engine because the given fuel rate is low; the method is the point of the problem.)

Data: 4 cylinders, four-stroke, $D = 0.12$ m, $L = 0.14$ m, $N = 2000$ rpm, $\dot m_f = 15$ kg/h, CV = 42 000 kJ/kg, net tension $= 40$ kgf, brake circumference $= 3.0$ m, $\eta_m = 0.80$.

Brake drum radius: $R = \dfrac{3.0}{2\pi} = 0.4775$ m (no rope size given, so this is taken as effective radius). Net force: $F = 40 \times 9.81 = 392.4$ N.

Brake power:

$$BP = \frac{2\pi N F R}{60} = F \times \frac{\text{circumference} \times N}{60} = 392.4 \times \frac{3.0 \times 2000}{60} = 39\,240\ \text{W} = 39.24\ \text{kW}$$

i) Indicated horse power

$$IP = \frac{BP}{\eta_m} = \frac{39.24}{0.80} = 49.05\ \text{kW}$$

With 1 hp = 0.746 kW: $IHP = 49.05/0.746 = 65.75$ hp (66.69 metric hp if 1 hp = 735.5 W).

ii) Brake thermal efficiency

$$\begin{aligned} \dot Q &= \frac{15}{3600}\times 42\,000 = 175\ \text{kW} \\ \eta_{bth} &= \frac{39.24}{175} = 0.2242 = 22.42\% \end{aligned}$$

iii) Indicated mean effective pressure

$$\begin{aligned} A &= \frac{\pi}{4}(0.12)^2 = 0.011310\ \text{m}^2,\quad n = \frac{N}{2} = 1000,\ k = 4 \\ p_m &= \frac{60\,IP}{L A n k} = \frac{60 \times 49\,050}{0.14 \times 0.011310 \times 1000 \times 4} \\ &= 4.647\times10^5\ \text{N/m}^2 = 4.65\ \text{bar} \end{aligned}$$

iv) Brake specific fuel consumption

$$bsfc = \frac{15}{39.24} = 0.3823\ \text{kg/kWh}$$

Answer: IHP = 49.05 kW (65.75 hp), $\eta_{bth}$ = 22.42 %, imep = 4.65 bar, bsfc = 0.382 kg/kWh.

Diesel engines produce high noise (from combustion knock, air intake, exhaust and mechanical parts) and vibration (from unbalanced reciprocating masses and fluctuating torque). Both must be controlled for the safety of the plant, operators and neighbours.

Methods to reduce noise

1. Exhaust silencers (mufflers): reactive (chambers and baffles) and absorptive types reduce exhaust noise, the biggest source.
2. Intake silencers / air filters on the air intake to damp suction noise.
3. Acoustic enclosures and canopies lined with sound-absorbing material (mineral wool, foam) around the engine–alternator set.
4. Sound-insulated engine house: thick masonry walls, double doors, acoustic louvres for ventilation air.
5. Locating the plant away from residential areas; exhaust stack pointing upward, away from buildings.
6. Good combustion (correct injection timing, quality fuel) to reduce diesel knock.
7. Proper maintenance: tight joints, balanced fans, lubricated gears.

Methods to reduce vibration

1. Heavy concrete foundation (mass several times engine weight) isolated from the building floor by an air gap or sand.
2. Anti-vibration mounts: springs, rubber pads or cork placed between engine skid and foundation.
3. Balancing of rotating and reciprocating masses; multi-cylinder engines with suitable firing order; balance weights on the crankshaft.
4. Flywheel to smooth torque fluctuation; vibration dampers on the crankshaft for torsional vibration.
5. Flexible connections (bellows, flexible hoses) in exhaust, fuel and water pipes so vibration is not carried to the building.
6. Proper alignment of engine and alternator shafts; flexible couplings.

Cooling system of a diesel power plant

About 25–35 % of the fuel heat flows to the cylinder walls, head, valves and pistons. The cooling system removes it to prevent overheating, thermal stresses, distortion and breakdown of the lube-oil film, while keeping the engine warm enough for good combustion.

Methods: air cooling (small engines, fins and fan) and water cooling (all power-plant engines): open once-through, thermosiphon, forced circulation and closed (double-circuit) systems.

Closed forced-circulation system (common in power plants):

 ENGINE jackets --> Heat exchanger --> Pump --+
      ^          (soft water loop)            |
      +---------------------------------------+
                     |   ^
          hot raw    v   |  cold raw water
              Cooling tower / spray pond

1. A pump circulates soft (treated) water through the cylinder jackets and heads; water leaves at about 70–85 °C.
2. In the heat exchanger, raw water takes this heat; the soft water returns to the engine.
3. The raw water is cooled in a cooling tower or spray pond and reused.
4. A thermostat bypasses the cooler during warm-up; an expansion tank allows for volume change and make-up.
5. Lube-oil coolers are placed in the raw-water circuit.

Soft water in the closed loop prevents scale and corrosion; the raw-water loop with a cooling tower reduces water consumption.

Data: 4 cylinders, four-stroke, $D = 0.1$ m, $L = 0.12$ m, $N = 1600$ rpm, fuel 0.27 L/min, specific gravity 0.74, $BP = 31.4$ kW, $\eta_m = 0.80$, CV = 44 000 kJ/kg.

Mass of fuel:

$$\dot m_f = 0.27\ \text{L/min} \times 0.74\ \text{kg/L} \times 60 = 11.988\ \text{kg/h}\ (0.00333\ \text{kg/s})$$

i) Brake specific fuel consumption

$$bsfc = \frac{\dot m_f}{BP} = \frac{11.988}{31.4} = 0.3818\ \text{kg/kWh}$$

ii) Indicated mean effective pressure

$$\begin{aligned} IP &= \frac{BP}{\eta_m} = \frac{31.4}{0.80} = 39.25\ \text{kW} \\ A &= \frac{\pi}{4}(0.1)^2 = 7.854\times10^{-3}\ \text{m}^2,\quad n = \frac{N}{2} = 800,\ k = 4 \\ p_{mi} &= \frac{60\,IP}{L A n k} = \frac{60 \times 39\,250}{0.12 \times 7.854\times10^{-3} \times 800 \times 4} \\ &= 7.809\times10^5\ \text{N/m}^2 = 7.81\ \text{bar} \end{aligned}$$

iii) Brake thermal efficiency

$$\begin{aligned} \dot Q &= \frac{11.988}{3600}\times 44\,000 = 146.52\ \text{kW} \\ \eta_{bth} &= \frac{31.4}{146.52} = 0.2143 = 21.43\% \end{aligned}$$

Answer: bsfc = 0.382 kg/kWh, imep = 7.81 bar, $\eta_{bth}$ = 21.43 %.

Lubrication in a diesel engine means supplying a film of oil between surfaces that move relative to each other, so that metal-to-metal contact is avoided.

Functions of lubricating oil

1. Reduces friction and wear of pistons, rings, liners, bearings, cams and gears; this also reduces friction power loss.
2. Cooling: carries heat away from bearings and from the underside of piston crowns.
3. Sealing: forms a seal between piston rings and cylinder liner, preventing blow-by of gas.
4. Cleaning: washes away carbon, metal particles and dirt to the filters.
5. Corrosion protection of metal surfaces against acids formed by combustion.
6. Shock absorption / noise reduction in bearings and gears.

Sketch: forced-feed (pressure) lubrication system

   +--------------- Engine ---------------+
   | main brgs, crank pins, gudgeon pins, |
   | camshaft, rocker arms, cyl. walls    |
   +--------------------------------------+
        ^                       |  drains
        | oil gallery           v
   Oil cooler <- Fine filter  SUMP / oil tank
        ^              ^        |
        |              |   strainer
        +---- Relief --+-- PUMP <+
              valve      (gear)
       raw cooling water through cooler

Working

1. A gear pump driven by the engine draws oil from the sump (or a separate oil tank in dry-sump systems) through a coarse strainer.
2. The oil is pumped at about 2–4 bar through a fine filter and an oil cooler (cooled by water from the cooling system). A pressure relief valve returns excess oil to the sump.
3. Oil enters the main oil gallery and is distributed through drilled passages to the main bearings, then through holes in the crankshaft to the crank-pin bearings and up the connecting rod to the gudgeon pin.
4. Branches feed the camshaft bearings, valve gear and rocker arms. Cylinder walls are lubricated by oil splash from the crank or by separate cylinder lubricators in large engines.
5. Oil drains back by gravity to the sump; in large plants it is periodically purified in a centrifuge to remove water and sludge.
6. A pressure gauge and temperature gauge (with low-pressure alarm/trip) protect the engine.

Other lubrication methods are mist (oil mixed with fuel, small two-strokes), wet sump and dry sump systems; power plants use forced-feed systems.

In a motoring test the engine is driven by the dynamometer without fuel; the motoring power equals the friction power: $FP = 2$ kW.

Data: $T = 150$ N m, $N = 900$ rpm, fuel = 3 kg in 1 h, CV = 40 MJ/kg = 40 000 kJ/kg.

(a) Brake power

$$BP = \frac{2\pi N T}{60} = \frac{2\pi \times 900 \times 150}{60} = 14\,137\ \text{W} = 14.137\ \text{kW}$$

(b) Indicated power

$$IP = BP + FP = 14.137 + 2 = 16.137\ \text{kW}$$

(c) Mechanical efficiency

$$\eta_m = \frac{BP}{IP} = \frac{14.137}{16.137} = 0.8761 = 87.61\%$$

(d) Indicated thermal efficiency

$$\begin{aligned} \dot Q &= \frac{3}{3600}\times 40\,000 = 33.33\ \text{kW} \\ \eta_{ith} &= \frac{IP}{\dot Q} = \frac{16.137}{33.33} = 0.4841 = 48.41\% \end{aligned}$$

Answer: BP = 14.14 kW, IP = 16.14 kW, $\eta_m$ = 87.61 %, $\eta_{ith}$ = 48.41 % (brake thermal efficiency = 42.41 %).