These are the most widely used tools and most often abuse tool

Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th

Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0

4 years ago

Write the heat equation for each of the following cases:

jok3333 [9.3K]

Answer:

Explanation:

a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

$\dfrac{\partial^2T}{\partial x^2}= \ 0 \ ; \ if \ T = f(x) \\ \\ \dfrac{\partial^2T}{\partial y^2}= \ 0 \ ; \ if \ T = f(y) \\ \\ \dfrac{\partial^2T}{\partial z^2}= \ 0 \ ; \ if \ T = f(z)$

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

$\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}$

where;

$Q_g$ = heat generated per unit volume

$\alpha$ = Thermal diffusivity

c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

$\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0$

where;

The radial directional term = $\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r})$ and the axial directional term is $\dfrac{\partial^2 T}{\partial z^2 }$

d) The heat equation for a wire going through a furnace is:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$

since;

the steady-state is zero, Then:

$\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]$ '

e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

$\dfrac{1}{r} \times \dfrac{\partial}{\partial r} \Big ( r^2 \times \dfrac{\partial T}{\partial r} \Big ) + \dfrac{Q_q}{K} = \dfrac{1}{\alpha}\times \dfrac{\partial T}{\partial t}$

4 0

3 years ago

A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and th

Dennis_Churaev [7]

Answer:

A)

- Q ( kw ) for vapor = -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor = -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate - Q ( KJ )

- Q ( kj ) for vapor = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

7 0

3 years ago

Column arrays: Transpose a row array Construct a row array countValues with elements 1 to endValue, using the double colon opera

White raven [17]

Answer:

Matlab code with step by step explanation and output results are given below

Explanation:

We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).

function countValues = CreateArray(endValue)

% Here we construct a row vector countValues from 1:endValue

countValues = 1:endValue;

% then we transpose this row vector into column vector

countValues = countValues';

end

Output:

Calling this function with the endValue=11 returns following output

CreateArray(11)

ans =

1

2

3

4

5

6

7

8

9

10

11

Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.

7 0

3 years ago

You get invited to a party at a 1,200 sf apartment, where there are a total of 23 people. The apartment has ceilings that are 8

yawa3891 [41]

Answer:

concentration of air = 0.0000366 ppm is not sufficient to cause any eye irritation

Explanation:

Average amount of aldehyde generated by a cigarette = 1 mg = 10⁻⁶g

Room volume = Height * Base area

Room volume = 8ft * 1200 sq.ft = 9600 cubic feet

number of people in the room = 23

If 1 person smokes 2 cigarette in 1 hour,

23 people will smoke 23*2 cigarettes = 46 cigarettes in 1 hr

Mass of aldehyde generated by 1 cigarette = 10⁻⁶g

Mass of aldehyde generated by 46 cigarettes = 46 * 10⁻⁶g

Based on the law of mass conservation,

Mass of aldehyde coming in = mass of aldehyde going out = Rate of accumulation of aldehyde

Rate of accumulation of aldehyde in the steady state = 0

m*C*V= 0

m = 46 * 10⁻⁶g

C = concentration in the steady state

V = Volumetic flow of the fresh air

V = 600 ft³/min = 600 *60

V = 36000 ft³/hr

Density = Mass/volume

Density = 46 * 10⁻⁶g /36*10³

Density = 1.28 * 10⁻⁹ g/ft³

If the density of air is taken as 36 g/ft³

concentration = (1.28 * 10⁻⁹ g/ft³)/36 g/ft³

concentration = 3.66 * 10⁻¹¹g/g

1*10⁻⁶g/g of air = 1ppm(part per million)

concentration of air =( 3.66 * 10⁻¹¹)(/1*10⁻⁶)

concentration of air = 3.66 * 10⁻⁵ppm = 0.0000366 ppm

concentration of air = 0.0000366 ppm is not sufficient to cause any eye irritation

7 0

3 years ago

These are the most widely used tools and most often abuse tool​

These are the most widely used tools and most often abuse tool