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blagie [28]
3 years ago
14

Two engineers are to solve an actual heat transfer problem in a manufacturing facility. Engineer A makes the necessary simplifyi

ng assumptions and solves the problem analytically, while engineer B solves it numerically using a powerful software package. Engineer A claims he solved the problem exactly and, thus, his results are better, while engineer B claims that he used a more realistic model and, thus, his results are better. Will the experiments prove engineer B right
Engineering
1 answer:
deff fn [24]3 years ago
5 0

Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

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Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
3 years ago
In contrast to the leading-trailing drum brake system, the duo-servo drum brake system will:
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Leading/trailing shoe type drum brake This is called the servo effect (self-boosting effect) which realizes the powerful braking forces of drum brakes. ... This is because drum brakes generate the same braking force in either direction. Generally, this type is used for the rear brakes of passenger cars.

6 0
4 years ago
Consider a sinusoidal oscillator consisting of an amplifier having a frequency-independent gain A (where A is positive) and a se
mafiozo [28]

Sinusoidal oscillator frequency of oscillation is given below.

Explanation:

The criterion for a stable oscillator is given in the equation

l A(jw)β(jw) l ≥ 1

In this task A represents the gain of the amplifier , and

β represents gain/attenuation of the second-order bandpass filter.

This sinusoidal oscillation is a special edge case where the product is equal to one.

So the condition is A-K=1

to obtain the sustained oscillations at the desired frequency of oscillations, the product of the voltage gain A and the feedback gain β must be one or greater than one. In this case, the amplifier gain A must be 3. Hence, to satisfy the product condition, feedback gain β must be 1/3.

4 0
3 years ago
Explain two ways that anthropometric data could be useful when designing a tennis racket.
12345 [234]

Answer:

I hope it helps :)

Explanation:

It is useful to measure Height and Arm Span in tennis players. Body fat can be measured using the skinfold method. If this is not available, monitoring body weight changes would give an indication of body fat changes, assuming no

3 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Nataly_w [17]

Answer:

The exit temperature of the gas = 32° C

Explanation:

Solution

Given that:

Inlet temperature T₁ = 27°C ≈ 300.15 K

Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa

Volume flow rate , V = 15 m/s³

Diameter of the deduct, D = 500 mm = 0.5 m

Electric heater power, W heater = 130 kW = 130 * 10^3 W

The heat lost Q = 80 kW =  80 * 10^3 W

Now,

From the ideal gas law, density of the air at the inlet is given as :

ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300

=0.6667 kg/m³

The mass flow rate through the duct is computed below:

m = ρ₁ V = 0.6667 * 15 = 10 kg/s

Thus

Applying the first law of thermodynamics to the process is shown below:

Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)

So,

If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:

Q + m (h₁) = m (h₂) + W

or

Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)

Thus

h₂ - h₁ = Cp T₂ - T₁

Now by method of substitution the known values are:

(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)

Note: The heat transfer is  taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas

So,

Solving for T₂,

T₂ = 32° C

Therefore the exit temperature of the gas = 32° C

7 0
4 years ago
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