Question

In: Mechanical Engineering

Top fuel dragsters (right) as the quickest accelerating cars in the world, and the fastest competitors...

Top fuel dragsters (right) as the quickest accelerating cars in the world, and the fastest competitors reach speeds of 335mph, completing a 1000ft run in just 3.64s. The 0 to 100mph time during these runs is as little as 0.8 seconds. The 4-stroke engine geometry used in these dragsters can be seen in Table 1. These engines run on nearly pure nitromethane (properties in Table 2).

Table1: Top Fuel Engine Geometry

Displacement, V_d [cm³]	8190
Bore [cm]	10.6
Stroke [cm]	11.4
Compression Ratio [-]	7.0
Configuration	V8
Peak Power [bhP]	10,000 @ 8000rpm

Table2: Nitromethane Properties

Chemical Formula	CH₃NO₂
Density [kg/L]	1.140
LHV [MJ/kg]	11.2

A) (10pts) Calculate the BMEP in [bar] produced by a top fuel dragster at peak power.

(Answer: BMEP = 136.6bar)

B) (25pts) A top fuel dragster consumes about 11.2 gal/s of nitromethane at full throttle, and operates at an equivalence ratio, ϕ, of 5.0. Determine the air flow and fuel flow per cylinder at full throttle in [kg/s] (1gal = 3.785L). (Answer: ṁ_fuel= 6.04 kg/s per cylinder, ṁ_air= 2.052 kg/s per cylinder)

C) (15pts) Determine the brake fuel conversion efficiency of a top fuel engine operating at peak power. (Answer: η_f= 1.38%)

Expert Solution

A)

10,000 bhp = 7457 kW

Displacement V = 8190 cm³ = 8.19 L

P = 2*pi*N*T / 60......where P is Power, N is rpm and T is Torque

7457 * 10³ = 2 * 3.14 * 8000 * T / 60

T = 8905 Nm

For 4-stroke engine, Power = BMEP * V * N / 120

7457 = BMEP * 8.19*10^-3 * 8000 / 120

BMEP = 13657 kPa = 136.6 bar

B)

Fuel consumption = 11.2 gal/s = 11.2 * 3.785 L/s = 42.392 L/s

Fuel mass consumption = Volume consumption * Density

= 42.392 * 1.14

= 48.327 kg/s

Fuel mass per cylinder = 48.327 / 8 = 6.04 kg/s

From CH3NO2 property tables, stoichiometric Fuel-Air ratio = 0.588

Actual Fuel-air ratio = Stoich Fuel-Air ratio * Equivalence ratio

= 0.588 * 5

Actual Fuel-air ratio = 2.94

Mass of air per cylinder = 6.04 / 2.94

Mass of air per cylinder = 2.052 kg/s

C)

Heat obtained by burning fuel = Fuel mass * LHV

= 48.327 * 11.2

= 541.26 MW = 541260 kW

Efficiency = Power / Heat supplied

= 7457 / 541260

= 0.0138 or 1.38%

samet mamat answered 2 years ago

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