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FinnZ [79.3K]
3 years ago
15

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K,

D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius?

Engineering
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

See attachment for complete solving

Explanation:

Given that: Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K, D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius

See attachment fir completed solving

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Question 2: (a) In your own words, clearly distinguish and differentiate between Ethics in Engineering and Ethics in Computing (
zlopas [31]

Engineering ethics is not without abstraction, but in contrast with computing, it is animated by a robust and active movement concerned with the seamless identification of ethics with practice.

<h3 /><h3>What is engineering?</h3>

This is a branch of science and technology concerned with the design, building, and use of engines, machines, and structures that uses scientific principles.

Comparing ethics in engineering and ethics in computing:

  • Engineering ethics are a set of rules and guidelines. While computing ethics deals with procedures, values and practices.
  • In engineering ethics, engineers must adhere to these rules as a moral obligation to their profession While in computing ethics, the ethics govern the process of consuming computer technology.
  • Following these ethics for the two professions will NOT cause damage, but disobeying them causes damage.

Some practical examples in the computing field:

  • Avoid using the computer to harm other people such as creating a bomb or destroying other people's work.
  • Users also should not use a computer for stealing activities like breaking into a bank or company.
  • Make sure a copy of the software had been paid for by the users before it is used.

Some practical examples in the engineering field:

  • Integrity for oneself.
  • Respect for one another.
  • Pursuit of excellence and accountability.

Hence, Engineering ethics is the field of system of moral principles that apply to the practice of engineering and following them is important to the profession.

Read more about <em>engineering</em> here:

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7 0
2 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can
Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a)   ln(T2/T1)=μβ         β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0
3 years ago
Q.14
Kay [80]

Answer:

A. optical isolation

Explanation:

well I can't really give a good explanation because I also saw the same question in my exams and option A was the correct answer

6 0
2 years ago
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