Answer: d) property tests
Explanation: Product service life can be referred as the life that define the service that can be provided by the product manufactured.The service life contains the testing and calculation of the product's quality, reliability, maintenance factor etc. These factors are known as the property of the product and so is calculated by the property test. Therefore option (d) is the correct option because other option does not define the factors for defining the product service life.
Answer:
b)10
Explanation:
We know that in Otto cycle there are four process .In first process mass of air fuel mixture enters in the cylinder.In second process that intake mass is compresses up to a specified limit.In general compression ratio in Otto cycle is about 9 or 10.In third process burning of fuel takes place where certain amount of work is obtained.Last process is exhaust process in which burning gas escape out from the cylinder.
Best compression ratio in Otto cycle is about 10.Beyond this value of compression ratio the cycle will not proved best work out put. So our option b is right.
Answer:
Explanation:
Kelvin's climbing represents the <em>absolute temperature</em>. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.
The Celsius scale has an <em>artificial zero</em>, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.
Answer:
The diameter of the test cylinder should be 7.65 meters.
Explanation:
The Hooke's law relation between stress and strain is mathematically represented as
Where 'E' is modulus of elasticity of the material
Now by definition of strain we have
Applying values to obtain strain we get
Thus the stress developed in the material equals
Now by definition of stress we have
Solving for 'D' we get
Answer:
The energy in kJ is 8558.16 kJ.
Explanation:
Data presented in the problem:
Water is heated from 70 (T1) to 200 °F (T2).
Volume (V) of the water is 1 ft3.
It is required for the specific heat of water(HW), which is 1 BTU/lb°F.
First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.
M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.
.After that, we can calculate the heat required (Q).
Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)
Q = 62.4 * 130 BTU = 8112 BTU.
Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ
Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.
Finally, the energy required is 8558.16 kJ.
