(a) A four-cylinder gasoline engine has a bore of 75 mm and a stroke length of 100 mm. It is operated at 3000 rpm and tested at…

(a) The engine is treated as a four-stroke spark-ignition engine; the four-stroke cycle relation is used for mean effective pressures. Given D=75 mm=0.075 m, L=100 mm=0.100 m, n=4, N=3000 rpm, brake arm R=40 cm=0.40 m, and net brake load W=150 N. The net brake load is already corrected for tare, so it gives the net torque. The brake arm is the perpendicular distance from the shaft axis to the line of action of the net load. The brake torque is T=WR=150 N×0.40 m=60 N m. The angular speed is ω=2πN/60=2π×3000/60=100π rad/s. (i) Brake power By the rotational power theorem, BP=Tω. Hence BP=60 N m×100π rad/s=6000π W. In kilowatts, BP=6000π/1000=6π kW=18.85 kW. The result is exact while π is retained. BP=6000π W=18.85 kW.

(ii) bmep Mean effective pressure is the constant pressure that would produce the same work per cycle as the actual pressure-volume work. Swept volume of one cylinder is Vs=πD²L/4=π×(0.075 m)²×0.100 m/4=0.000441786 m³. Total swept volume is Vd=nVs=4×0.000441786 m³=0.001767146 m³=0.0005625π m³. The total displacement is also πD²L because the four cylinders cancel the factor 4. For a four-stroke engine, power strokes occur at N/120 per second; the factor 120 comes from 60 s/min divided by 2 revolutions per cycle. Hence BP=bmep×Vd×N/120. Therefore bmep=120BP/(VdN)=120×6000π W/(0.0005625π m³×3000)=426666.7 Pa. The pressure is a mean pressure difference, not an absolute gas pressure. bmep=1280/3 kPa=426.7 kPa=4.27 bar.

(iii) bsfc Brake specific fuel consumption is fuel mass flow divided by brake power. The unit kg/kWh is mass of fuel per unit brake energy output. Specific gravity 0.79 is relative to water, taken as 1000 kg/m³, so fuel density ρf=0.79×1000 kg/m³=790 kg/m³. Volume flow is 7.8 l/h=7.8×10⁻³ m³/h. Mass flow mf=7.8×10⁻³ m³/h×790 kg/m³=6.162 kg/h. Since BP=6π kW, bsfc=mf/BP=6.162/(6π)=1027/(1000π)=0.3269 kg/kWh. The calorific value 44000 kJ/kg is not needed for the requested quantities; it would be used for indicated thermal efficiency. bsfc=1027/(1000π) kg/kWh=0.327 kg/kWh.

(iv) Indicated power Morse test theorem: at the same speed and with friction unchanged, the fall in brake power when one cylinder is cut out equals the indicated power of that cylinder. Morse test assumes that cutting out one cylinder does not materially change the friction of the remaining engine. Since speed and brake arm are unchanged, brake power is proportional to brake load. Power per newton of brake load is 2πNR/60=2π×3000×0.40/60=40π W/N. The equivalent load losses are 40, 42, 44 and 46 N. Thus IP1=(150−110)×40π=1600π W, IP2=(150−108)×40π=1680π W, IP3=(150−106)×40π=1760π W, IP4=(150−104)×40π=1840π W. Total IP=1600π+1680π+1760π+1840π=6880π W=21.61 kW. The corresponding friction power is IP−BP=880π W=2.76 kW, but it is not separately asked. IP=6880π W=21.61 kW.

(v) Mechanical efficiency Mechanical efficiency is brake power divided by indicated power. ηm=BP/IP=6000π/6880π=6000/6880=75/86=0.87209. This value is below 100%, as required for a real engine. ηm=75/86=87.2%.

(vi) imep Indicated mean effective pressure is defined in the same way as bmep, but using indicated power. Hence IP=imep×Vd×N/120, so imep=120IP/(VdN)=120×6880π W/(0.0005625π m³×3000)=489244.4 Pa. The ratio imep/bmep equals IP/BP, giving the same check: 426.667×6880/6000=489.244 kPa. The imep is larger than bmep because it includes the work lost in friction. imep=22016/45 kPa=489.2 kPa=4.89 bar.

(b) Ideal vapour-compression heat pump with R-134a; table values (same convention as Q5(c)).

• State 1 (evaporator, 320 kPa, saturated vapour): interpolating the table (0 °C ↔ 294.0 kPa, 5 °C ↔ 350.9 kPa) gives Tsat ≈ 2.5 °C and h1 = hg ≈ 399.8 kJ/kg.
• State 2 (compressor outlet, 800 kPa): h2 = 420 kJ/kg (given).
• State 3 (condenser outlet, saturated liquid, 800 kPa): Tsat ≈ 31.3 °C and hf ≈ 243.7 kJ/kg (30 °C: 241.79; 35 °C: 249.10).
• State 4: isenthalpic throttling, so h4 = h3 = 243.7 kJ/kg.

The house loses heat at 75 MJ/h = 20.83 kW, so the condenser must supply QH = 20.83 kW.

Per kg: qH = h2 − h3 = 420 − 243.7 = 176.3 kJ/kg ⇒ mass flow ṁ = QH/qH = 20.83/176.3 = 0.1182 kg/s.

Compressor work: w = h2 − h1 = 420 − 399.8 = 20.2 kJ/kg ⇒ W_in = ṁw = 0.1182 × 20.2 = 2.39 kW.

Heat-pump COP = QH/W_in = 20.83/2.39 ≈ 8.7.

Evaporator heat pick-up: QL = ṁ(h1 − h4) = 0.1182(399.8 − 243.7) = 0.1182 × 156.1 = 18.45 kW — and QL + W_in = 18.45 + 2.39 = 20.84 kW = QH, so the energy balance closes.

A resistance heater would need 20.83 kW of electricity to deliver the same heat, so the electric power saved = 20.83 − 2.39 = 18.45 kW ≈ 18.5 kW (numerically equal to QL).

Final: W_in ≈ 2.39 kW; COP ≈ 8.7; power saved vs a resistance heater ≈ 18.5 kW. Assumptions: ideal cycle with saturated vapour at compressor inlet, isenthalpic expansion, no pressure drops or extra subcooling/superheat beyond the given h2.

(c) A steam turbine governor compares actual speed with desired speed and changes steam admission to hold speed nearly constant. The governor must be self-acting: a speed increase closes steam admission, and a speed decrease opens it. Two common types are:

• Throttle governing: In the sketch, draw a horizontal turbine casing. At the left, show a steam pipe from the stop valve to a throttle valve, then to a nozzle block. The nozzle block contains all nozzles of the first stage and feeds the rotor; show the following stages to the right. Label the throttle valve, nozzle block, rotor, and governor link. Connect the turbine shaft to a flyball or hydraulic governor by a spindle, and draw a link from the governor to the throttle valve stem. The throttle valve is usually a butterfly or globe valve in the steam line. When load falls, speed rises, and the governor moves the throttle valve partly closed. The pressure and density of steam in the nozzle block fall, so the mass flow through the unchanged nozzle area falls and power is reduced. When load rises, the valve opens. All nozzles remain open; regulation is by inlet pressure. Because the nozzle pressure falls, the nozzle exit velocity may fall slightly; the method is simple but less efficient at part load. It is used in impulse turbines, especially small or single-stage machines.
• Nozzle governing: In the sketch, draw the first-stage nozzle block divided into several nozzle groups, for example four groups, each with its own nozzle valve on top. Label each nozzle group and its valve. Draw the turbine shaft connected to a governor, and draw separate governor links to each nozzle valve. At full load all nozzle groups are open. When load falls, the governor closes one or more groups in a predetermined sequence, reducing the effective nozzle area while the pressure in the open groups remains nearly constant. The closed groups are isolated by the nozzle valves, so the open groups operate at nearly full boiler pressure. When load rises, groups are opened in reverse sequence. This method avoids a large throttling pressure drop and gives better part-load efficiency because the pressure drop across the open nozzles remains large. It is used in impulse turbines and for the first (impulse) stage of large turbines where nozzle groups can be controlled.