Answer:
Explanation:
Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.
1. Exhaust clamp
2. Exhaust putty
3. Exhaust tape
4. New exhaust
Exhaust clamp is best used for holes on straight pipes.
Putty is best used on welds or small holes like on exhaust manifolds or welds connecting various pieces like catalytic converters, mufflers, or resonators.
Tape will work best on straight pipes with holes.
New exhaust is for when the thig is beyond repair, like rust.
Now good luck because working on exhausts is a pain.
Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
Answer:
A. National Highway Safety Act
Explanation:
The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965
National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers by ensuring national safety standards and issuance recalls for defective vehicles
Uniform Traffic Control Devices Act defines shapes, colors and locations for road signs, traffic signals, and road markings
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
