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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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Sam decides to remove his oversized hooded jacket before he works near the pto driveline is it safe or unsafe
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Its safe because it isn't something with electricity
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Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3
viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

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A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

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<u>the liquid limit</u>

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LL=-0.8078(25)+49.503

LL=29.308

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PL=11.808

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

- Cutiepatutie ☺❀❤

6 0
3 years ago
Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

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