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3 years ago

16 engineering design process

Engineering

1 answer:

Nesterboy [21]3 years ago

6 0

Answer:

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Explanation:

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PLEASE HELP, THANK YOU

Elena L [17]

Answer:

6,3,2,5,1,4 because they jst are

Explanation:

3 0

3 years ago

A car engine with a thermal efficiency of 33% drives the air-conditioner unit (a refrigerator) besides powering the car and othe

fgiga [73]

Answer:

The rate of fuel required to drive the air conditioner $Q_h = 6.061 kW$

The flow rate of the cold air is $\r m = 0.30765 kg/s$

Explanation:

From this question we are told that

The efficiency is $\eta = 33$ % = 0.33

Temperature for the hot day is $T_h = 35^oC = 308 K \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (35+273)$

Temperature after cooling is $T_c = 5^oC = 278K$

The input power is $P_{in} = 2kW$

The rate of fuel required to drive the air conditioner can be mathematically represented as

$Q_h = \frac{P_{in}}{\eta}$

$= \frac{2}{0.33} = 6.061 kW$

From the question the air condition is assumed to be half as a Carnot refrigeration unit

This can be Mathematically interpreted in terms of COP(coefficient of performance) as

$\beta_{air} = 0.5 \beta$

where $\beta$ denotes COP and is mathematically represented as

$\beta = \frac{Q_c}{P_{in}}$

= > $Q_c = \beta P_{in}$

Where $Q_c$ is the rate of flue being burned for cold air to flow

Now if the COP of a Carnot refrigerator is having this value

$\beta_{Carnot } = \frac{T_c}{T_h - T_c}$

$= \frac{278}{308-278}$

$\beta_{Carnot} = 9.267\\$

Then

$\beta_{air} = 0.5 * 9.2667$

$= 4.6333$

Now substituting the value of $\beta$ to solve for $Q_c$

$Q_c = \beta P_{in}$

$= 4.6333 *2$

$9.2667kW$

The equation for the rate of fuel being burned for the cold air to flow

$Q_c = \r mc_p \Delta T$

Making the flow rate of the cold air

$\r m = \frac{Q_c}{c_p \Delta T}$

$= \frac{9.2667}{1.004}* (308 - 278)$

$= 0.30765 kg/s$

4 0

3 years ago

Which of the following is a way to heat or cool a building without using electricity or another power source?

zzz [600]

I believe the answer is: A. Passive heating and cooling.

8 0

3 years ago

Read 2 more answers

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn

Ostrovityanka [42]

Answer: a) 150 rev. b) 2105 rev.

Explanation:

a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.

In order to get the number of revolutions executed, we can use this:

ωf² - ω₀² = 2 γ Δθ (1)

For the first part, we know that ω₀ = 0 (as it starts from rest).

We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:

γ = (ωf - ω₀) / Δt (2)

As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.

3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec

Replacing in (2), we get γ:

γ = 120 π / 5 rad/sec² = 24 π rad/sec²

Replacing in (1) and solving for Δθ:

Δθ = 120² π² / 2. 24 π = 300 π rad

As 1 rev = 2π rad, Δθ = 150 rev

b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.

The new value for γ is as follows:

γ = -120π / 70 rad/sec² = -1.71 rad/sec²

Replacing in (1) and solving for Δθ, we get:

Δθ = -120² π² / 2. (-1.71) π = 4210 π rad

As 1 rev = 2π rad, Δθ = 2105 rev

7 0

4 years ago

An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh

Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in

B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

(ii) Average shear stress at the web = 18.289 * 10^5 psi

(iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 * 10^6 * 2.375}{53.54}$ = 0.287 * 10^6 Ib-in