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kiruha [24]
2 years ago
11

The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer

, TV, and lights at a rate of 1.2 kW. What is the net amount of energy change of the air in the room during a 30-min period?

Engineering
2 answers:
Mkey [24]2 years ago
8 0

Answer:

660KJ

Explanation:

Given

Let Q = Heat Loss from room = 50kj/min

Let W = Work Supplied to room = 1.2KW

1 kilowatt = 1 kilojoules per second

So, W = 1.2KJ/s

In heat and work (Sign Convention)

We know that

1. Heat takes positive sign when it is added to the system

2. Heat takes negative sign when it is removed from the system.

3. Work done is considered positive when work is done by the system

4. Work done is considered negative when work is done on the system.

From the above illustration, heat loss (Q) = -50KJ/Min

In 30 minutes time, Q = -50Kj/Min * 30 Min

Work done in 30 minutes = -1500 KJ

Also, work supplied = -1.2Kj/s

Work supplied to the system in 30 minutes = -1.2Kj/s * 30 minutes

W = -1.2 KJ/s * 30 * 60 seconds

W = -2160KJ

In thermodynamics (First Law)

Q = W + ΔU

-1500 = -2160 + ΔU

∆U = 2160 - 1500

∆U = 660KJ

Artemon [7]2 years ago
4 0

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

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What is the damped natural frequency (in rad/s) of a second order system whose undamped natural frequency is 25 rad/s and has a
TiliK225 [7]

Answer:

damped natural frequency = 23.84 rad/s

Explanation:

given data

damping ratio = 0.3

undamped natural frequency = 25 rad/s

to find out

damped natural frequency of a second order system

solution

we know that if damping ratio is = 0

then it is undamped system

and if damping ratio is > 1

then it is overdamped system

and  and if damping ratio is ≈ 1

then it is critical damped system

so damped natural frequency of a second order system formula is

damped natural frequency = wn × \sqrt{1-r^2}

here wn is undamped natural frequency and r is damping ratio

damped natural frequency = 25 × \sqrt{1-0.3^2}

damped natural frequency = 23.84 rad/s

3 0
2 years ago
Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i
leonid [27]

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

5 0
2 years ago
Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi
leva [86]

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

3 0
2 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
2 years ago
. A cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of 50.80 mm (2.000 in.) is pu
kogti [31]

Answer:

% reduction in area==PR=0.734=73.4%

% elongation=EL=0.42=42%

Explanation:

given do=12.8 mm

df=6.60

Lf=72.4 mm

Lo=50.8 mm

% reduction in area=((\pi*(do/2)^2)-(\pi*(df/2)^2)))/\pi*(do/2)^2

substitute values

% reduction in area=73.4%

% elongation=EL=((Lf-Lo)/Lo))*100

% elongation=((72.4-50. 8)/50.8)*100=42%

6 0
2 years ago
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