Engineering physics (A-level only)

3.11 Engineering physics (A-level only) — coverage pack

12 specification leaves · notes, questions, answers and worked methods

3.11.1.1 · Concept of moment of inertia

  • Moment of inertia is the rotational analogue of mass: it measures resistance to angular acceleration about a specified axis and has unit kg m2\text{kg m}^2.
  • For a point mass, I=mr2I=mr^2; for separated point masses, I=∑mr2I=\sum mr^2. Expressions for continuous extended objects are supplied when needed.
  • Moving the same mass farther from the axis increases II strongly because the perpendicular distance is squared; both the mass distribution and rotation axis must therefore be stated.
  • A common error is to use an object's diameter for rr. Use the perpendicular distance from each mass to the stated axis and square it before summing.

Tier 1 · Easy

  1. 1. Two small balancing masses rotate about the same shaft. A 0.30 kg0.30\,\text{kg} mass is 0.22 m0.22\,\text{m} from the shaft and a 0.45 kg0.45\,\text{kg} mass is 0.12 m0.12\,\text{m} from it. Calculate their combined moment of inertia.[2 marks]

    Answer

    • 2.1×10−2 kg m22.1\times10^{-2}\,\text{kg m}^2

    Method: Use I=∑mr2I=\sum mr^2. Hence I=(0.30)(0.22)2+(0.45)(0.12)2=0.01452+0.00648=0.02100 kg m2I=(0.30)(0.22)^2+(0.45)(0.12)^2=0.01452+0.00648=0.02100\,\text{kg m}^2, which is 2.1×10−2 kg m22.1\times10^{-2}\,\text{kg m}^2 to two significant figures.

Tier 2 · Standard

  1. 1. A training flywheel has moment of inertia 0.080 kg m20.080\,\text{kg m}^2 before four identical 0.25 kg0.25\,\text{kg} sliders are attached. Determine the total moment of inertia when every slider is fixed 0.30 m0.30\,\text{m} from the axis.[3 marks]

    Answer

    • 0.17 kg m20.17\,\text{kg m}^2

    Method: Each slider contributes mr2=(0.25)(0.30)2=0.0225 kg m2mr^2=(0.25)(0.30)^2=0.0225\,\text{kg m}^2. Four contribute 4(0.0225)=0.0900 kg m24(0.0225)=0.0900\,\text{kg m}^2. Add the original flywheel value: I=0.080+0.0900=0.170 kg m2I=0.080+0.0900=0.170\,\text{kg m}^2, which is 0.17 kg m20.17\,\text{kg m}^2 to two significant figures.

Tier 3 · Hard

  1. 1. A prototype rotor consists of a uniform disc of mass 6.0 kg6.0\,\text{kg} and radius 0.25 m0.25\,\text{m}, a shaft with moment of inertia 0.028 kg m20.028\,\text{kg m}^2, and three identical point masses fixed 0.35 m0.35\,\text{m} from the axis. The total moment of inertia is 0.456 kg m20.456\,\text{kg m}^2. The disc expression is I=12MR2I=\frac12MR^2. Determine the mass of each point mass.[5 marks]

    Answer

    • 0.65 kg0.65\,\text{kg}

    Method: The disc contributes Idisc=12(6.0)(0.25)2=0.1875 kg m2I_{\text{disc}}=\frac12(6.0)(0.25)^2=0.1875\,\text{kg m}^2. The three masses therefore contribute 0.456−0.1875−0.028=0.2405 kg m20.456-0.1875-0.028=0.2405\,\text{kg m}^2. With 0.2405=3m(0.35)20.2405=3m(0.35)^2, m=0.2405/[3(0.35)2]=0.654 kgm=0.2405/[3(0.35)^2]=0.654\,\text{kg}, which is 0.65 kg0.65\,\text{kg} to two significant figures.

3.11.1.2 · Rotational kinetic energy

  • The rotational kinetic energy of a rigid object is Ek=12Iω2E_k=\frac12I\omega^2, directly analogous to Ek=12mv2E_k=\frac12mv^2 for translation.
  • A flywheel stores more energy when its moment of inertia or angular speed is larger; doubling ω\omega makes the stored energy four times as large.
  • Flywheels smooth variations in torque and speed and can store energy in vehicles or production machinery, releasing it when demand rises.
  • Use II in kg m2\text{kg m}^2 and ω\omega in rad s−1\text{rad s}^{-1}. A common error is to substitute revolutions per second without multiplying by 2π2\pi.

Tier 1 · Easy

  1. 1. A small flywheel has moment of inertia 0.85 kg m20.85\,\text{kg m}^2 and angular speed 14 rad s−114\,\text{rad s}^{-1}. Calculate its rotational kinetic energy.[2 marks]

    Answer

    • 83 J83\,\text{J}

    Method: Use Ek=12Iω2E_k=\frac12I\omega^2. Thus Ek=12(0.85)(14)2=83.3 JE_k=\frac12(0.85)(14)^2=83.3\,\text{J}, giving 83 J83\,\text{J} to two significant figures.

Tier 2 · Standard

  1. 1. An energy-recovery flywheel of moment of inertia 40 kg m240\,\text{kg m}^2 stores 32 kJ32\,\text{kJ} as rotational kinetic energy. Determine its angular speed.[3 marks]

    Answer

    • 40 rad s−140\,\text{rad s}^{-1}

    Method: Convert the energy: Ek=3.2×104 JE_k=3.2\times10^4\,\text{J}. Rearranging Ek=12Iω2E_k=\frac12I\omega^2 gives ω=2Ek/I=2(3.2×104)/40=1600=40 rad s−1\omega=\sqrt{2E_k/I}=\sqrt{2(3.2\times10^4)/40}=\sqrt{1600}=40\,\text{rad s}^{-1}.

Tier 3 · Hard

  1. 1. A machine flywheel with moment of inertia 18 kg m218\,\text{kg m}^2 slows from 75 rad s−175\,\text{rad s}^{-1} to 45 rad s−145\,\text{rad s}^{-1} in 5.5 s5.5\,\text{s}. During this interval, 72%72\% of the decrease in rotational kinetic energy is transferred usefully. Calculate the mean useful power.[5 marks]

    Answer

    • 4.2 kW4.2\,\text{kW}

    Method: The energy decrease is ΔEk=12I(ω12−ω22)=12(18)(752−452)=32400 J\Delta E_k=\frac12I(\omega_1^2-\omega_2^2)=\frac12(18)(75^2-45^2)=32400\,\text{J}. The useful energy is 0.72(32400)=23328 J0.72(32400)=23328\,\text{J}. Hence the mean useful power is P=23328/5.5=4.24×103 W=4.2 kWP=23328/5.5=4.24\times10^3\,\text{W}=4.2\,\text{kW} to two significant figures.

3.11.1.3 · Rotational motion

  • Angular displacement θ\theta is measured in radians. Angular velocity ω=Δθ/Δt\omega=\Delta\theta/\Delta t is signed for a chosen positive rotation direction; angular speed is its magnitude, and α=Δω/Δt\alpha=\Delta\omega/\Delta t.
  • For uniform angular acceleration, use the linear-motion analogues: ω2=ω1+αt\omega_2=\omega_1+\alpha t, θ=12(ω1+ω2)t\theta=\frac12(\omega_1+\omega_2)t, θ=ω1t+12αt2\theta=\omega_1t+\frac12\alpha t^2, and ω22=ω12+2αθ\omega_2^2=\omega_1^2+2\alpha\theta.
  • On an ω\omega-tt graph, gradient is angular acceleration and area is angular displacement; curved sections represent non-uniform angular acceleration.
  • Convert revolutions using 1 revolution=2π rad1\text{ revolution}=2\pi\,\text{rad}. A common error is to use rotational frequency in place of angular speed.

Tier 1 · Easy

  1. 1. The angular speed of a mixer rotor rises uniformly from 5.0 rad s−15.0\,\text{rad s}^{-1} to 17 rad s−117\,\text{rad s}^{-1} in 4.0 s4.0\,\text{s}. Calculate its angular acceleration.[2 marks]

    Answer

    • 3.0 rad s−23.0\,\text{rad s}^{-2}

    Method: Use α=Δω/Δt=(17−5.0)/4.0=3.0 rad s−2\alpha=\Delta\omega/\Delta t=(17-5.0)/4.0=3.0\,\text{rad s}^{-2}.

Tier 2 · Standard

  1. 1. A rotor starts from rest and has constant angular acceleration 4.2 rad s−24.2\,\text{rad s}^{-2} for 6.0 s6.0\,\text{s}. Determine the number of revolutions completed.[3 marks]

    Answer

    • 1212 revolutions

    Method: The angular displacement is θ=ω1t+12αt2=0+12(4.2)(6.0)2=75.6 rad\theta=\omega_1t+\frac12\alpha t^2=0+\frac12(4.2)(6.0)^2=75.6\,\text{rad}. Divide by 2π2\pi: N=75.6/(2π)=12.0N=75.6/(2\pi)=12.0 revolutions.

Tier 3 · Hard

  1. 1. A flywheel's angular speed rises uniformly from 12 rad s−112\,\text{rad s}^{-1} to 48 rad s−148\,\text{rad s}^{-1} in 6.0 s6.0\,\text{s}, remains at 48 rad s−148\,\text{rad s}^{-1} for 5.0 s5.0\,\text{s}, then falls uniformly to 18 rad s−118\,\text{rad s}^{-1} in 10 s10\,\text{s}. Determine its total angular displacement and the corresponding number of revolutions.[5 marks]

    Answer

    • 7.5×102 rad7.5\times10^2\,\text{rad}, or 1.2×1021.2\times10^2 revolutions

    Method: Angular displacement is the area under the ω\omega-tt graph. During acceleration, θ1=12(12+48)(6.0)=180 rad\theta_1=\frac12(12+48)(6.0)=180\,\text{rad}. At constant speed, θ2=(48)(5.0)=240 rad\theta_2=(48)(5.0)=240\,\text{rad}. During slowing, θ3=12(48+18)(10)=330 rad\theta_3=\frac12(48+18)(10)=330\,\text{rad}. Therefore θ=180+240+330=750 rad\theta=180+240+330=750\,\text{rad} and N=750/(2π)=119.4N=750/(2\pi)=119.4. To two significant figures these are 7.5×102 rad7.5\times10^2\,\text{rad} and 1.2×1021.2\times10^2 revolutions.

3.11.1.4 · Torque and angular acceleration

  • For a force perpendicular to the radius, torque is T=FrT=Fr and is measured in N m\text{N m}; use the perpendicular distance from the axis to the force's line of action.
  • The rotational form of Newton's second law is T=IαT=I\alpha, analogous to F=maF=ma; TT is the resultant torque, not automatically the driving torque.
  • Opposing torques such as bearing friction must be assigned the opposite sign before calculating α\alpha.
  • A common error is to use the applied force itself in T=IαT=I\alpha. First convert each force to a torque and then find the resultant.

Tier 1 · Easy

  1. 1. A tangential force of 85 N85\,\text{N} acts at the rim of a wheel of radius 0.16 m0.16\,\text{m}. Calculate the torque about the axle.[2 marks]

    Answer

    • 14 N m14\,\text{N m}

    Method: The force is perpendicular to the radius, so T=Fr=(85)(0.16)=13.6 N mT=Fr=(85)(0.16)=13.6\,\text{N m}, which is 14 N m14\,\text{N m} to two significant figures.

Tier 2 · Standard

  1. 1. A constant resultant torque of 24 N m24\,\text{N m} acts on a rotor with moment of inertia 3.2 kg m23.2\,\text{kg m}^2. Its initial angular speed is 6.0 rad s−16.0\,\text{rad s}^{-1}. Determine its angular speed after 3.0 s3.0\,\text{s}.[3 marks]

    Answer

    • 29 rad s−129\,\text{rad s}^{-1}

    Method: From T=IαT=I\alpha, α=T/I=24/3.2=7.5 rad s−2\alpha=T/I=24/3.2=7.5\,\text{rad s}^{-2}. Then ω=ω1+αt=6.0+(7.5)(3.0)=28.5 rad s−1\omega=\omega_1+\alpha t=6.0+(7.5)(3.0)=28.5\,\text{rad s}^{-1}, giving 29 rad s−129\,\text{rad s}^{-1} to two significant figures.

Tier 3 · Hard

  1. 1. A drive supplies a constant torque of 54 N m54\,\text{N m} to a flywheel of moment of inertia 6.2 kg m26.2\,\text{kg m}^2. Bearing friction provides a constant opposing torque of 7.5 N m7.5\,\text{N m}. The initial angular speed is 10 rad s−110\,\text{rad s}^{-1}. Calculate the angular speed after 4.5 s4.5\,\text{s}.[5 marks]

    Answer

    • 44 rad s−144\,\text{rad s}^{-1}

    Method: The resultant torque is T=54−7.5=46.5 N mT=54-7.5=46.5\,\text{N m}. Hence α=T/I=46.5/6.2=7.50 rad s−2\alpha=T/I=46.5/6.2=7.50\,\text{rad s}^{-2}. Using ω=ω1+αt\omega=\omega_1+\alpha t gives ω=10+(7.50)(4.5)=43.75 rad s−1\omega=10+(7.50)(4.5)=43.75\,\text{rad s}^{-1}, which is 44 rad s−144\,\text{rad s}^{-1} to two significant figures.

3.11.1.5 · Angular momentum

  • For a rigid body rotating about a fixed axis, angular momentum is L=IωL=I\omega with unit kg m2 s−1\text{kg m}^2\text{ s}^{-1}.
  • Angular momentum is conserved when the resultant external torque is zero; if II changes, ω\omega changes so that total IωI\omega remains constant.
  • A constant torque acting for time Δt\Delta t gives angular impulse TΔt=Δ(Iω)T\Delta t=\Delta(I\omega), analogous to FΔt=Δ(mv)F\Delta t=\Delta(mv).
  • Choose a positive rotation direction and retain signs. A common error is to conserve rotational kinetic energy when two rotating objects lock together; angular momentum, not kinetic energy, is conserved in that collision.

Tier 1 · Easy

  1. 1. A turntable has moment of inertia 2.4 kg m22.4\,\text{kg m}^2 and rotates at 18 rad s−118\,\text{rad s}^{-1}. Calculate its angular momentum.[2 marks]

    Answer

    • 43 kg m2 s−143\,\text{kg m}^2\text{ s}^{-1}

    Method: Use L=Iω=(2.4)(18)=43.2 kg m2 s−1L=I\omega=(2.4)(18)=43.2\,\text{kg m}^2\text{ s}^{-1}, giving 43 kg m2 s−143\,\text{kg m}^2\text{ s}^{-1} to two significant figures.

Tier 2 · Standard

  1. 1. A flywheel with moment of inertia 0.80 kg m20.80\,\text{kg m}^2 rotates at 30 rad s−130\,\text{rad s}^{-1}. It is coupled to a stationary coaxial wheel of moment of inertia 1.20 kg m21.20\,\text{kg m}^2, and the pair then rotate together. Determine their common angular speed.[3 marks]

    Answer

    • 12 rad s−112\,\text{rad s}^{-1}

    Method: With negligible external torque, angular momentum is conserved: (0.80)(30)+(1.20)(0)=(0.80+1.20)ω(0.80)(30)+(1.20)(0)=(0.80+1.20)\omega. Therefore 24=2.00ω24=2.00\omega, so ω=12 rad s−1\omega=12\,\text{rad s}^{-1}.

Tier 3 · Hard

  1. 1. A rotating platform and athlete initially have moment of inertia 4.8 kg m24.8\,\text{kg m}^2 and angular speed 9.0 rad s−19.0\,\text{rad s}^{-1}. An external constant opposing torque of 3.2 N m3.2\,\text{N m} acts for 1.5 s1.5\,\text{s}. The athlete then changes position so that the combined moment of inertia is 3.0 kg m23.0\,\text{kg m}^2. Determine the final angular speed, assuming the position change itself involves negligible external torque.[5 marks]

    Answer

    • 13 rad s−113\,\text{rad s}^{-1}

    Method: Initially Li=Iω=(4.8)(9.0)=43.2 kg m2 s−1L_i=I\omega=(4.8)(9.0)=43.2\,\text{kg m}^2\text{ s}^{-1}. The opposing angular impulse is TΔt=−(3.2)(1.5)=−4.8 kg m2 s−1T\Delta t=-(3.2)(1.5)=-4.8\,\text{kg m}^2\text{ s}^{-1}, so the angular momentum becomes L=43.2−4.8=38.4 kg m2 s−1L=43.2-4.8=38.4\,\text{kg m}^2\text{ s}^{-1}. During the position change this is conserved, hence ω=L/I=38.4/3.0=12.8 rad s−1\omega=L/I=38.4/3.0=12.8\,\text{rad s}^{-1}, or 13 rad s−113\,\text{rad s}^{-1} to two significant figures.

3.11.1.6 · Work and power

  • Work done by a constant torque through angular displacement is W=TθW=T\theta, analogous to W=FsW=Fs; θ\theta must be in radians.
  • Rotational power is P=TωP=T\omega, analogous to P=FvP=Fv. Use the torque and angular speed at the same instant when either changes.
  • Frictional torque dissipates energy and must be included in power and energy balances for rotating machinery.
  • At steady angular speed the resultant torque is zero, but the driving torque can still do work against friction and a load. A common error is to conclude that the power is zero.

Tier 1 · Easy

  1. 1. A constant torque of 18 N m18\,\text{N m} turns a shaft through 3.5 rad3.5\,\text{rad}. Calculate the work done by the torque.[2 marks]

    Answer

    • 63 J63\,\text{J}

    Method: Use W=Tθ=(18)(3.5)=63 JW=T\theta=(18)(3.5)=63\,\text{J}.

Tier 2 · Standard

  1. 1. A motor provides a torque of 42 N m42\,\text{N m} while its shaft rotates at 1200 rev min−11200\,\text{rev min}^{-1}. Determine the power transferred by the shaft.[3 marks]

    Answer

    • 5.3 kW5.3\,\text{kW}

    Method: Convert the rotation rate: 1200 rev min−1=20 rev s−11200\,\text{rev min}^{-1}=20\,\text{rev s}^{-1}, so ω=2π(20)=125.7 rad s−1\omega=2\pi(20)=125.7\,\text{rad s}^{-1}. Then P=Tω=(42)(125.7)=5.28×103 W=5.3 kWP=T\omega=(42)(125.7)=5.28\times10^3\,\text{W}=5.3\,\text{kW}.

Tier 3 · Hard

  1. 1. A motor maintains a constant angular speed of 32 rad s−132\,\text{rad s}^{-1} for 25 s25\,\text{s} while supplying torque 84 N m84\,\text{N m}. Bearing friction opposes the motion with torque 9.0 N m9.0\,\text{N m}; the remaining torque acts on the load. Calculate the useful power and the useful energy transferred to the load.[5 marks]

    Answer

    • 2.4 kW2.4\,\text{kW} and 6.0×104 J6.0\times10^4\,\text{J}

    Method: The load torque is Tload=84−9.0=75 N mT_{\text{load}}=84-9.0=75\,\text{N m} because the speed is constant. The useful power is P=Tloadω=(75)(32)=2400 W=2.4 kWP=T_{\text{load}}\omega=(75)(32)=2400\,\text{W}=2.4\,\text{kW}. The useful energy is E=Pt=(2400)(25)=60000 J=6.0×104 JE=Pt=(2400)(25)=60000\,\text{J}=6.0\times10^4\,\text{J}.

3.11.2.1 · First law of thermodynamics

  • AQA uses Q=ΔU+WQ=\Delta U+W, where QQ is energy transferred to the gas by heating, ΔU\Delta U is the increase in internal energy, and WW is work done by the gas.
  • With this convention, heating gives Q>0Q>0, cooling gives Q<0Q<0, expansion normally gives W>0W>0, and compression gives W<0W<0 because work is done on the gas.
  • The equivalent rearrangement ΔU=Q−W\Delta U=Q-W is often safest when substituting signed values.
  • Internal energy is a state function, so ΔU=0\Delta U=0 over a complete cycle even though the net heat transfer and net work need not be zero.
  • A common error is to make work done on the gas positive in AQA's formula. Translate the wording into the stated sign convention before calculating.

Tier 1 · Easy

  1. 1. A gas receives 520 J520\,\text{J} by heating and does 180 J180\,\text{J} of work. Calculate its change in internal energy using AQA's convention Q=ΔU+WQ=\Delta U+W.[2 marks]

    Answer

    • +340 J+340\,\text{J}

    Method: Here Q=+520 JQ=+520\,\text{J} and W=+180 JW=+180\,\text{J}. Therefore ΔU=Q−W=520−180=+340 J\Delta U=Q-W=520-180=+340\,\text{J}.

Tier 2 · Standard

  1. 1. A gas is compressed so that 240 J240\,\text{J} of work is done on it. During the compression, 90 J90\,\text{J} of energy leaves the gas by heating. Using Q=ΔU+WQ=\Delta U+W and taking work output from the gas as positive, determine ΔU\Delta U.[3 marks]

    Answer

    • +150 J+150\,\text{J}

    Method: Energy leaving by heating gives Q=−90 JQ=-90\,\text{J}. Work done on the gas means work done by the gas is W=−240 JW=-240\,\text{J}. Hence ΔU=Q−W=−90−(−240)=+150 J\Delta U=Q-W=-90-(-240)=+150\,\text{J}.

Tier 3 · Hard

  1. 1. A gas completes a two-process cycle. In process A it receives 3.2 kJ3.2\,\text{kJ} by heating and its internal energy increases by 1.1 kJ1.1\,\text{kJ}. In process B the gas returns to its initial state while 1.4 kJ1.4\,\text{kJ} leaves it by heating. Determine the net work done by the gas during the cycle. State the signs used in Q=ΔU+WQ=\Delta U+W.[5 marks]

    Answer

    • +1.8 kJ+1.8\,\text{kJ}, net work done by the gas

    Method: For A, QA=+3.2 kJQ_A=+3.2\,\text{kJ} and ΔUA=+1.1 kJ\Delta U_A=+1.1\,\text{kJ}, so WA=QA−ΔUA=+2.1 kJW_A=Q_A-\Delta U_A=+2.1\,\text{kJ}. Returning to the initial state requires ΔUB=−1.1 kJ\Delta U_B=-1.1\,\text{kJ}. For B, QB=−1.4 kJQ_B=-1.4\,\text{kJ}, so WB=QB−ΔUB=−1.4−(−1.1)=−0.3 kJW_B=Q_B-\Delta U_B=-1.4-(-1.1)=-0.3\,\text{kJ}. Thus Wnet=2.1−0.3=+1.8 kJW_{\text{net}}=2.1-0.3=+1.8\,\text{kJ}.

3.11.2.2 · Non-flow processes

  • The named non-flow processes are isothermal (constant temperature), adiabatic (no heat transfer), constant pressure, and constant volume.
  • For an ideal gas, pV=nRTpV=nRT. Isothermal change obeys pV=constantpV=\text{constant}, while adiabatic change obeys pVγ=constantpV^\gamma=\text{constant}.
  • At constant pressure, W=pΔVW=p\Delta V. At constant volume, ΔV=0\Delta V=0 so W=0W=0; apply Q=ΔU+WQ=\Delta U+W with AQA's work-by-the-gas sign convention.
  • For an ideal gas in an isothermal process, internal energy does not change, so ΔU=0\Delta U=0 and Q=WQ=W. In an adiabatic process, Q=0Q=0 so ΔU=−W\Delta U=-W.
  • A common error is to call a rapid change isothermal. Rapid compression or expansion is more nearly adiabatic because there is little time for heat transfer.

Tier 1 · Easy

  1. 1. A rigid sealed vessel receives 450 J450\,\text{J} by heating. State the work done by the gas and calculate its increase in internal energy.[2 marks]

    Answer

    • W=0 JW=0\,\text{J} and ΔU=+450 J\Delta U=+450\,\text{J}

    Method: The vessel is rigid, so the volume is constant and W=0W=0. From Q=ΔU+WQ=\Delta U+W, ΔU=Q−W=450−0=+450 J\Delta U=Q-W=450-0=+450\,\text{J}.

Tier 2 · Standard

  1. 1. An ideal gas expands isothermally from pressure 240 kPa240\,\text{kPa} and volume 2.0×10−3 m32.0\times10^{-3}\,\text{m}^3 to volume 5.0×10−3 m35.0\times10^{-3}\,\text{m}^3. Determine its final pressure and state its change in internal energy.[3 marks]

    Answer

    • 96 kPa96\,\text{kPa} and ΔU=0\Delta U=0

    Method: For an isothermal change, p1V1=p2V2p_1V_1=p_2V_2. Hence p2=p1V1/V2=(240)(2.0/5.0)=96 kPap_2=p_1V_1/V_2=(240)(2.0/5.0)=96\,\text{kPa}. The temperature of an ideal gas is unchanged, so its internal energy is unchanged and ΔU=0\Delta U=0.

Tier 3 · Hard

  1. 1. A gas is compressed adiabatically from p1=100 kPap_1=100\,\text{kPa} and V1=4.8×10−3 m3V_1=4.8\times10^{-3}\,\text{m}^3 to V2=1.8×10−3 m3V_2=1.8\times10^{-3}\,\text{m}^3. Take γ=1.40\gamma=1.40. During the compression, 720 J720\,\text{J} of work is done on the gas. Calculate the final pressure and the change in internal energy.[5 marks]

    Answer

    • 3.9×102 kPa3.9\times10^2\,\text{kPa} and ΔU=+720 J\Delta U=+720\,\text{J}

    Method: For an adiabatic change, p1V1γ=p2V2γp_1V_1^\gamma=p_2V_2^\gamma, so p2=p1(V1/V2)γ=100(4.8/1.8)1.40=394.8 kPap_2=p_1(V_1/V_2)^\gamma=100(4.8/1.8)^{1.40}=394.8\,\text{kPa}, or 3.9×102 kPa3.9\times10^2\,\text{kPa} to two significant figures. Adiabatic means Q=0Q=0. Work done on the gas makes work done by the gas W=−720 JW=-720\,\text{J}. Thus ΔU=Q−W=0−(−720)=+720 J\Delta U=Q-W=0-(-720)=+720\,\text{J}.

3.11.2.3 · The p-V diagram

  • A process is represented as a path on a pp-VV diagram; movement to larger VV is expansion and movement to smaller VV is compression.
  • The work done by the gas is the area under the process curve. For constant pressure, W=pΔVW=p\Delta V; expansion gives positive work and compression negative work.
  • For a cyclic process, the enclosed loop area is the net work per cycle. A clockwise loop gives positive net work by the gas; an anticlockwise loop requires net work input.
  • Convert pressure to pascals and volume to m3\text{m}^3 so that the area has unit joules. A common error is to use the whole area under one branch instead of the area enclosed by a cycle.

Tier 1 · Easy

  1. 1. A gas expands at constant pressure 180 kPa180\,\text{kPa} from volume 1.2×10−3 m31.2\times10^{-3}\,\text{m}^3 to 3.7×10−3 m33.7\times10^{-3}\,\text{m}^3. Calculate the work done by the gas.[2 marks]

    Answer

    • 450 J450\,\text{J}

    Method: Use W=pΔVW=p\Delta V. Here p=1.80×105 Pap=1.80\times10^5\,\text{Pa} and ΔV=(3.7−1.2)×10−3=2.5×10−3 m3\Delta V=(3.7-1.2)\times10^{-3}=2.5\times10^{-3}\,\text{m}^3. Hence W=(1.80×105)(2.5×10−3)=450 JW=(1.80\times10^5)(2.5\times10^{-3})=450\,\text{J}.

Tier 2 · Standard

  1. 1. A gas follows a clockwise rectangular cycle on a pp-VV diagram between pressures 150 kPa150\,\text{kPa} and 400 kPa400\,\text{kPa} and between volumes 1.0×10−3 m31.0\times10^{-3}\,\text{m}^3 and 3.5×10−3 m33.5\times10^{-3}\,\text{m}^3. Determine the net work done per cycle.[3 marks]

    Answer

    • 6.3×102 J6.3\times10^2\,\text{J}

    Method: The net work is the rectangular loop area: W=(phigh−plow)(Vhigh−Vlow)=(400−150)×103(3.5−1.0)×10−3=625 JW=(p_{\text{high}}-p_{\text{low}})(V_{\text{high}}-V_{\text{low}})=(400-150)\times10^3(3.5-1.0)\times10^{-3}=625\,\text{J}. To two significant figures this is 6.3×102 J6.3\times10^2\,\text{J}. The clockwise direction makes this work positive, done by the gas.

Tier 3 · Hard

  1. 1. On a pp-VV diagram, a gas expands along a straight line from (0.8×10−3 m3,600 kPa)(0.8\times10^{-3}\,\text{m}^3,600\,\text{kPa}) to (3.2×10−3 m3,200 kPa)(3.2\times10^{-3}\,\text{m}^3,200\,\text{kPa}). It then compresses at constant pressure 200 kPa200\,\text{kPa} to its initial volume before returning at constant volume to its initial state. Calculate the net work done by the gas in the cycle.[5 marks]

    Answer

    • 480 J480\,\text{J}

    Method: For the straight expansion path, the area under the graph is average pressure times volume change: W1=12(600+200)×103(3.2−0.8)×10−3=960 JW_1=\frac12(600+200)\times10^3(3.2-0.8)\times10^{-3}=960\,\text{J}. On the constant-pressure compression, W2=pΔV=(200×103)(0.8−3.2)×10−3=−480 JW_2=p\Delta V=(200\times10^3)(0.8-3.2)\times10^{-3}=-480\,\text{J}. The final constant-volume path does no work. Therefore Wnet=960−480=480 JW_{\text{net}}=960-480=480\,\text{J}, positive because the loop is clockwise.

3.11.2.4 · Engine cycles

  • A four-stroke engine completes induction, compression, power and exhaust strokes. A petrol engine compresses a fuel-air mixture before spark ignition, whereas a diesel engine compresses air before injected fuel ignites; their indicator diagrams show the corresponding pp-VV paths without requiring constructional detail.
  • Indicated power is (loop area)(cycles per second per cylinder)(number of cylinders)(\text{loop area})(\text{cycles per second per cylinder})(\text{number of cylinders}). A four-stroke cylinder completes one cycle per two crankshaft revolutions.
  • Input power is calorific value multiplied by fuel mass-flow rate. Brake power is P=TωP=T\omega, and friction power is indicated power minus brake power.
  • Overall efficiency is brake power divided by input power, thermal efficiency is indicated power divided by input power, and mechanical efficiency is brake power divided by indicated power.
  • A common error is to use revolutions per second as cycles per second for a four-stroke engine, doubling the indicated power.

Tier 1 · Easy

  1. 1. An engine produces a steady crankshaft torque of 140 N m140\,\text{N m} at 3000 rev min−13000\,\text{rev min}^{-1}. Calculate its brake power.[2 marks]

    Answer

    • 44 kW44\,\text{kW}

    Method: 3000 rev min−1=50 rev s−13000\,\text{rev min}^{-1}=50\,\text{rev s}^{-1}, so ω=2π(50)=314 rad s−1\omega=2\pi(50)=314\,\text{rad s}^{-1}. Brake power is P=Tω=(140)(314)=4.40×104 W=44 kWP=T\omega=(140)(314)=4.40\times10^4\,\text{W}=44\,\text{kW}.

Tier 2 · Standard

  1. 1. The area enclosed by the indicator diagram for one cylinder of a four-stroke engine is 620 J620\,\text{J} per cycle. The four-cylinder engine runs at 2400 rev min−12400\,\text{rev min}^{-1}. Determine the indicated power.[3 marks]

    Answer

    • 49.6 kW49.6\,\text{kW}

    Method: The crankshaft speed is 2400/60=40 rev s−12400/60=40\,\text{rev s}^{-1}. A four-stroke cylinder completes one cycle every two revolutions, so each cylinder completes 20 cycles s−120\,\text{cycles s}^{-1}. Thus Pi=(620)(20)(4)=49600 W=49.6 kWP_i=(620)(20)(4)=49600\,\text{W}=49.6\,\text{kW}.

Tier 3 · Hard

  1. 1. A four-cylinder, four-stroke engine runs at 3000 rev min−13000\,\text{rev min}^{-1}. Each cylinder's indicator-loop area is 380 J380\,\text{J}. The brake torque is 105 N m105\,\text{N m}. Fuel of calorific value 44 MJ kg−144\,\text{MJ kg}^{-1} is supplied at 1.60×10−3 kg s−11.60\times10^{-3}\,\text{kg s}^{-1}. Determine the indicated power, brake power, friction power, and thermal efficiency.[6 marks]

    Answer

    • Pi=38.0 kWP_i=38.0\,\text{kW}, Pb=33.0 kWP_b=33.0\,\text{kW}, Pf=5.0 kWP_f=5.0\,\text{kW}, thermal efficiency =54.0%=54.0\%

    Method: The speed is 3000/60=50 rev s−13000/60=50\,\text{rev s}^{-1}, so a four-stroke cylinder completes 25 cycles s−125\,\text{cycles s}^{-1}. Hence Pi=(380)(25)(4)=38000 W=38.0 kWP_i=(380)(25)(4)=38000\,\text{W}=38.0\,\text{kW}. Also ω=2π(50)=314.16 rad s−1\omega=2\pi(50)=314.16\,\text{rad s}^{-1}, so Pb=Tω=(105)(314.16)=32.99 kW=33.0 kWP_b=T\omega=(105)(314.16)=32.99\,\text{kW}=33.0\,\text{kW}. Friction power is Pf=Pi−Pb=38.0−32.99=5.01 kW=5.0 kWP_f=P_i-P_b=38.0-32.99=5.01\,\text{kW}=5.0\,\text{kW}. Input power is (44×106)(1.60×10−3)=70.4 kW(44\times10^6)(1.60\times10^{-3})=70.4\,\text{kW}. Therefore thermal efficiency is Pi/Pinput=38.0/70.4=0.540=54.0%P_i/P_{\text{input}}=38.0/70.4=0.540=54.0\%.

3.11.2.5 · Second Law and engines

  • The First Law alone would allow complete conversion of heat to work, but the Second Law requires a heat engine to operate between a hot source and a colder sink.
  • For each cycle, QH=W+QCQ_H=W+Q_C and efficiency is η=W/QH=(QH−QC)/QH\eta=W/Q_H=(Q_H-Q_C)/Q_H.
  • The maximum theoretical efficiency is ηmax⁡=(TH−TC)/TH\eta_{\max}=(T_H-T_C)/T_H, with both temperatures in kelvin; it is below 100%100\% whenever TC>0T_C>0.
  • Practical engines are less efficient than the theoretical maximum because of friction, heat transfer to surroundings, finite-rate processes, and incomplete combustion.
  • Combined heat and power schemes use both WW and otherwise-wasted heat. A common error is to call the sum an engine efficiency; it is an energy-utilisation fraction.

Tier 1 · Easy

  1. 1. A heat engine receives 5.0 kJ5.0\,\text{kJ} from its hot source and rejects 3.2 kJ3.2\,\text{kJ} to its sink each cycle. Calculate its efficiency.[2 marks]

    Answer

    • 0.360.36, or 36%36\%

    Method: The work output is W=QH−QC=5.0−3.2=1.8 kJW=Q_H-Q_C=5.0-3.2=1.8\,\text{kJ}. Therefore η=W/QH=1.8/5.0=0.36=36%\eta=W/Q_H=1.8/5.0=0.36=36\%.

Tier 2 · Standard

  1. 1. An engine operates between a source at 720 K720\,\text{K} and a sink at 310 K310\,\text{K}. Determine its maximum theoretical efficiency and explain why a practical engine operating at these temperatures has a lower efficiency.[3 marks]

    Answer

    • 0.5690.569, or 56.9%56.9\%; practical irreversibilities such as friction and unwanted heat transfer make the actual efficiency lower.

    Method: Use ηmax⁡=(TH−TC)/TH=(720−310)/720=0.569=56.9%\eta_{\max}=(T_H-T_C)/T_H=(720-310)/720=0.569=56.9\%. This limit assumes an ideal reversible engine. A real engine has friction and transfers heat across finite temperature differences and to its surroundings, so less of QHQ_H becomes useful work.

Tier 3 · Hard

  1. 1. A combined heat and power plant receives thermal energy at 120 MW120\,\text{MW} and produces 45 MW45\,\text{MW} of electrical work. It supplies 60 MW60\,\text{MW} of otherwise rejected heat to nearby buildings. The engine source and sink temperatures are 850 K850\,\text{K} and 300 K300\,\text{K}. Calculate the engine efficiency, the total useful-energy fraction, and the maximum theoretical efficiency. Explain the difference between the last two quantities.[5 marks]

    Answer

    • Engine efficiency =37.5%=37.5\%, useful-energy fraction =87.5%=87.5\%, maximum theoretical efficiency =64.7%=64.7\%

    Method: The engine efficiency counts work output: η=W/QH=45/120=0.375=37.5%\eta=W/Q_H=45/120=0.375=37.5\%. The CHP useful-energy fraction counts electrical work plus useful heating: (45+60)/120=0.875=87.5%(45+60)/120=0.875=87.5\%. The maximum theoretical engine efficiency is (850−300)/850=0.647=64.7%(850-300)/850=0.647=64.7\%. The 87.5%87.5\% figure is not a heat-to-work efficiency: it includes lower-grade heat that the Second Law requires the engine to reject, while the Carnot figure limits only the fraction convertible to work.

3.11.2.6 · Reversed heat engines

  • A reversed heat engine uses work input WW to transfer energy QCQ_C from a cold region to a hot region, with QH=QC+WQ_H=Q_C+W.
  • For a refrigerator, COPref=QC/W=QC/(QH−QC)\mathrm{COP}_{\text{ref}}=Q_C/W=Q_C/(Q_H-Q_C) because the desired transfer is out of the cold space.
  • For a heat pump, COPhp=QH/W=QH/(QH−QC)\mathrm{COP}_{\text{hp}}=Q_H/W=Q_H/(Q_H-Q_C) because the desired transfer is into the warm space; therefore COPhp=COPref+1\mathrm{COP}_{\text{hp}}=\mathrm{COP}_{\text{ref}}+1.
  • For ideal devices, COPref=TC/(TH−TC)\mathrm{COP}_{\text{ref}}=T_C/(T_H-T_C) and COPhp=TH/(TH−TC)\mathrm{COP}_{\text{hp}}=T_H/(T_H-T_C), using kelvin.
  • COP can exceed 11 because it measures heat transferred per unit work input, not conversion efficiency. A common error is to use QCQ_C for a heat pump or QHQ_H for a refrigerator.

Tier 1 · Easy

  1. 1. A refrigerator removes 360 J360\,\text{J} from its cold compartment for every 120 J120\,\text{J} of electrical work supplied. Calculate its coefficient of performance.[2 marks]

    Answer

    • COPref=3.0\mathrm{COP}_{\text{ref}}=3.0

    Method: For a refrigerator, the desired transfer is QCQ_C. Hence COPref=QC/W=360/120=3.0\mathrm{COP}_{\text{ref}}=Q_C/W=360/120=3.0.

Tier 2 · Standard

  1. 1. A heat pump delivers 14 kW14\,\text{kW} to a building while consuming 4.0 kW4.0\,\text{kW} of electrical power. Determine its heat-pump coefficient of performance and the rate at which it extracts energy from outside.[3 marks]

    Answer

    • COPhp=3.5\mathrm{COP}_{\text{hp}}=3.5 and 10 kW10\,\text{kW} extracted

    Method: For a heat pump, COPhp=QH/W=14/4.0=3.5\mathrm{COP}_{\text{hp}}=Q_H/W=14/4.0=3.5. Energy conservation gives QC=QH−W=14−4.0=10 kWQ_C=Q_H-W=14-4.0=10\,\text{kW} extracted from outside.

Tier 3 · Hard

  1. 1. A heat pump keeps a building at 293 K293\,\text{K} when the outside temperature is 268 K268\,\text{K}. Its actual coefficient of performance is 42%42\% of the maximum theoretical value. Calculate the electrical power required when the heat delivered to the building is 18 kW18\,\text{kW}.[5 marks]

    Answer

    • 3.7 kW3.7\,\text{kW}

    Method: For an ideal heat pump, COPhp,max=TH/(TH−TC)=293/(293−268)=11.72\mathrm{COP}_{\text{hp,max}}=T_H/(T_H-T_C)=293/(293-268)=11.72. The actual COP is 0.42(11.72)=4.9220.42(11.72)=4.922. Since COPhp=QH/W\mathrm{COP}_{\text{hp}}=Q_H/W, the electrical input is W=QH/COPhp=18/4.922=3.66 kWW=Q_H/\mathrm{COP}_{\text{hp}}=18/4.922=3.66\,\text{kW}, giving 3.7 kW3.7\,\text{kW} to two significant figures.