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Gala2k [10]
3 years ago
11

A strong base (caustic or alkali) is added to oils to test for: b. entrained air a)- alkalinity b)- acidity c)- contamination. d

)- stability
Engineering
1 answer:
fredd [130]3 years ago
7 0

Answer:

A strong base(caustic or alkali) is added to oils test for stability

option (d) is correct

Explanation:

Formation of emulsion involves mixing together of measured volumes of specific concentrations of crude oil and sodium hydroxide (NaOH) solution. the viscosity of the emulsion and its type is determined and then this emulsion is made to rest for a long time in order to determine its stability.

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If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:
kirill [66]

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

Therefore, When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Option C is correct.

Learn more about creep:

brainly.com/question/10565749

#SPJ1

8 0
2 years ago
When the psychologist simply records the relationship between two variables...
Wewaii [24]
When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.
8 0
3 years ago
An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h
EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell (\eta), dimensionless, as:

\eta = \frac{\dot W_{out}}{\dot W_{in}} (Eq. 1)

Where:

\dot W_{in} - Maximum power possible from hydrogen flow, measured in kilowatts.

\dot W_{out} - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

\dot W_{in} = \dot V\cdot \rho \cdot L_{c} (Eq. 2)

Where:

\dot V - Volume flow rate, measured in cubic meters per second.

\rho - Density of hydrogen, measured in kilograms per cubic meter.

L_{c} - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that \dot V = \frac{28}{3600}\,\frac{m^{3}}{s}, \rho = 0.0899\,\frac{kg}{m^{3}}, L_{c} = 141790\,\frac{kJ}{kg} and \dot W_{out} = 80\,kW, then the efficiency of this fuel cell is:

(Eq. 1)

\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)

\dot W_{in} = 99.143\,kW

(Eq. 2)

\eta = \frac{80\,kW}{99.143\,kW}

\eta = 0.807

The efficiency of this fuel cell is 80.69 percent.

3 0
3 years ago
What are employers required to do to keep employees safe from caught-in and -between hazards from hand-held power tools?
nika2105 [10]

Answer:

Employees who use hand and power tools and who are exposed to the hazards of falling, flying, abrasive and splashing objects, or exposed to harmful dusts, fumes, mists, vapors, or gases must be provided with the appropriate equipment needed, including Personal Protective Equipment, to protect them from the hazard.

Explanation:

8 0
3 years ago
A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What a
statuscvo [17]

Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

3 0
3 years ago
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