Answer:
A number of engine geometry properties and performance parameters obtained from dynamometer testing are given in Table below:
Please calculate:
i) Cylinder bore, bore to stroke ratio, connecting rod to crank radius ratio, and clearance volume in each engine cylinder.
ii) Brake power output at 3800 rpm and the torque output at 6000 rpm.
iii) Brake, indicated and friction mean effective pressures at both 3800 rpm and 6000 rpm.
iv) Brake specific fuel consumption in kg/(kW h) and brake thermal efficiency at 3800 rpm.
v) Volumetric efficiency at 3800 rpm.
vi) Air to fuel ratio and the relative air to fuel ratio at 3800 rpm. Is this a lean or rich mixture?
Engine type Displacement/ size Number of cylinders Stroke length Connecting rod length Compression ratio Maximum brake power6000 rpm Maximum [email protected] 3800 rpm Mass flow rate of fuel 3800 rpm Volumetric flow rate of air 3800 rpm Mechanical efficiency @ 3800 rpm Mechanical efficiency @ 6000 rpm Calorific value of fuel Ambient air temperature Ambient air pressure Specific gas constant for air Gasoline, naturally aspirated, four-stroke 1.329 L 4 (Straight) 85 mm 141 mm 11.5 73 kW 128 N m 3.2 g/s 37.5 L/s 0.88 0.72 46 MJ/kg 10 °C 101.3 kPa 287.1 J/(kg K)
Explanation:
Answer:
Massive destruction
Explanation:
If the asteroid collides with the ground, a massive volume of dust will be blasted into the environment. If it collides with water, the amount of water vapour in the atmosphere will rise. This would result in more rain, which would cause earthquakes and mudslides.
As the asteroid collided with the Earth, massive volumes of dust were ejected into to the atmosphere. The sun's light were stopped from entering the Earth's surface, which is terrible news for plants.
Answer:
This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.
I hope it will help you a lot.
Explanation:
1.25, because change the fractions into decimals and then add it from there so it would be add like, 0.75+0.5=1.25
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
<h3>PART (a)</h3>
For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
<h3>
PART (b)</h3>
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
<h3>PART(c)</h3>
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
<h3>
PART (d)</h3>
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)