32
Problem solving
Textbook: pp. 89–90
GoalSolve problems linking engine power, fuel consumption and efficiency.
New words
power · quvvatfuel consumption · yoqilg‘i sarfiuseful work A = Pt · foydali ish A = Ptheating efficiency · issiqlik berish koeffitsiyenti
Explanation
In engine problems useful work follows from power and time: A = P·t (P in watts, t in seconds). The heat released by the fuel is Q₁ = q·m. So η = A/Q₁ = Pt/(qm), hence the fuel mass m = Pt/(ηq). If the device (stove, heater) warms water with the heat, the useful heat is Q = cmΔt and η = cmΔt/(q·m_fuel). Convert to SI: hour → 3600 s, kW → 1000 W, MJ → 10⁶ J, km/h → m/s (÷3.6). Before answering check that the efficiency is below 100 %.
Worked examples
How long can a 40 kW engine of efficiency 35 % run on 12 kg of diesel fuel (q = 42 MJ/kg)? Q₁ = qm = 42·10⁶·12 = 5.04·10⁸ J; A = ηQ₁ = 0.35·5.04·10⁸ ≈ 1.76·10⁸ J; t = A/P = 1.76·10⁸/40 000 ≈ 4.4·10³ s, i.e. about 1.2 hours.
On a gas stove 2 kg of water is heated from 20 °C to 70 °C while 25 g of natural gas (q = 44 MJ/kg) burns. Useful heat Q = cmΔt = 4200·2·50 = 4.2·10⁵ J; heat of the fuel qm = 44·10⁶·0.025 = 1.1·10⁶ J; η = 4.2·10⁵/1.1·10⁶ ≈ 0.38 = 38 %.
Class activity
Calculation marathon: 3 problems (power–time, fuel mass, heating water); each group solves one on the board and the class checks units and η < 100 %. No fuel is burned.
Practice
1
An engine takes 5 MJ from the heater with efficiency 40 %. How many MJ is the useful work?
2
2
A generator engine uses 18 kg of diesel fuel (q = 42 MJ/kg) in 2 hours with efficiency 35 %. About what is its useful power?
P = ηqm/t = 0.35·42·10⁶·18/7200 ≈ 3.7·10⁴ W ≈ 37 kW
3
Carnot engine with T₁ = 600 K and T₂ = 300 K: efficiency (%)?
50
4
Why is a real engine’s efficiency lower than the Carnot efficiency?
Friction, heat losses and irreversible processes exist and the working gas does not follow the ideal cycle.