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Molodets [167]
4 years ago
9

Liquid ethanol is a flammable fluid and can release vapors that form explosive mixtures at temperatures above its flashpoint at

16.6°C. In a chemical plant, liquid ethanol (c_p = 2.44 kJ/kg*K, rho = 789 kg/m^3) is being transported in a pipe with an inside diameter of 5 cm. The pipe is located in a hot area with the 20 kW of heat is added to the ethanol. Your task, as an engineer is to design a pumping system to transport the ethanol safely and to prevent fire hazard.
1. If the inlet temperature of the ethanol is 10°C, determine the volume flow rate that is necessary to keep the temperature of the ethanol in the pipe below its flashpoint.
Engineering
1 answer:
marta [7]4 years ago
6 0

Answer:

The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s

Explanation:

Q = MCp(T2 - T1)

Q (quantity of heat) = Power (P) × time (t)

Density (D) = Mass (M)/Volume (V)

M = DV

Therefore, Pt = DVCp(T2 - T1)

V/t (volume flow rate) = P/DCp(T2 - T1)

P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K

Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)

The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s

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The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat
Snowcat [4.5K]

Answer:

<em>Heat is lost at the rate of 750 J/s or W</em>

<em>The thermal resistance is 1 K/W</em>

Explanation:

interior temperature T_{2} = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature T_{1} = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference dt = T_{2} - T_{1} = 900 - 150 = 750 °C

note that dt is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak\frac{dt}{t}

substituting values, we have

Q = 1.5 x 0.4 x \frac{750}{0.6} = <em>750 J/s or W</em>

Thermal resistance R_{t} = \frac{dt}{Q}

R_{t} = 750/750 =<em> 1 K/W</em>

7 0
3 years ago
The type of current that flows from the electrode across the arc to the work is called what?
Scrat [10]

Answer:

Direct current.

Explanation:

5 0
3 years ago
Define and discuss the difference between micronutrients and macronutrients. Also, discuss their importance in the body at rest
almond37 [142]

Answer:

Macronutrients are simply nutrients the body needs in a very high amount e.g Carbohydrate.

MicroNutrients are simply nutrients the body needs but in little amount e.g  Minerals.

Explanation:

So for further breakdown:

What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.

Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.

Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins

In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.

In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.

5 0
3 years ago
What must engineers keep in mind so that their solutions will be appropriate?
vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0
3 years ago
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