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Veseljchak [2.6K]
3 years ago
10

Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm

al resistance for a wall

Engineering
1 answer:
Simora [160]3 years ago
5 0

Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow.  You can also see that R_{th} represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

  1. Wall Thickness.  More thickness, more thermal resistance.
  2. Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
  3. Cross-section Area. More cross-section area, less thermal resistance.

A expression to define  the thermal resistance for the wall is as follows:  R_{th} =\frac{l}{Ak}, where  l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

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3 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
3 years ago
Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req
Nataly_w [17]

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

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3 0
2 years ago
"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United
Paraphin [41]

Answer:

see explaination

Explanation:

Part a) Width of bay at Potomac River:

Given Data:

· Actual Width at Potomac River = 30 miles

· Bay Model Length Ratio Lr = 1/1000

In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.

Length of any part in model = Actual length x Lr

Hence,

Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles

Since 1 mile = 5280 ft

Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft

Part b) Model Length of bay bridge in model:

Given Data:

· Actual Length of bay bridge = 4.3 miles

· Bay Model Length Ratio Lr = 1/1000

Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles

Since 1 mile = 5280 ft

Model Length in feet = 0.0043 x 5280 = 22.704 ft

Part c) Model Length of bay bridge in model:

Given Data:

· Model Area = 8 acre

· Bay Model Length Ratio Lr = 1/1000

Model Area = Actual Area x Lr x Lr

8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000

Since 1 square mile = 640 acre,

Actual Area in square miles = 8,000,000/640 = 12,500 square miles

Part d) Average and maximum depth of model:

Given Data:

· Actual Average depth = 28 ft

· Actual Maximum depth = 174 ft

· Bay Model Length Ratio Lr = 1/1000

Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet

Since 1 ft = 12 inch

Model average depth in inch = 0.028 x 12 = 0.336 in

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Since 1 ft = 12 inch

Model maximum depth in inch = 0.174 x 12 = 2.088 in

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3 years ago
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.
Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761

Now, from the Probability table of standard normal distribution:

P(z\leq 1.761) = 0.9608

Now,

P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%

Now, for the expected no. of readings greater than 2.70 V:

P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}

No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)\times Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392\times 60 = 2.352 ≈ 2

7 0
3 years ago
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