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Vikki [24]
3 years ago
14

An inventor tries to sell you his new heat engine that takes in 40 J of heat at 87°C on each cycle, expels 30 J at 27°C, and doe

s 10 J of work. Would it be wise to invest in this engine?
Engineering
1 answer:
algol133 years ago
5 0

Answer:

NOT WISE

Explanation:

given,

T₁ = 87⁰C  =  273 + 87 = 360 K

T₂ = 27⁰C  =  273 + 27 = 300 K

work output given = 10 J

efficiency will be equal to :

   \eta = 1 - \dfrac{T_2}{T_1}\\\eta = 1 - \dfrac{300}{360}\\\eta = 0.166

output delivery =  η ₓ input

                          =  0.166 ₓ 40  

                          = 6.66 J

hence, the output of new engine  is less than the old one so, investing in new engine will not be feasible.

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A cylindrical rod 100 mm long and having a diameter of 10.0 mm is to be deformed using a tensile load of 27,500 N. It must not e
mash [69]

Answer:

The steel is a candidate.

Explanation:

Given

P = 27,500 N

d₀ = 10.0 mm = 0.01 m

Δd = 7.5×10 ⁻³ mm (maximum value)

This problem asks that we assess the four alloys relative to the two criteria presented. The first  criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is  applied; this means that the stress corresponding to this load not exceed the yield strength of the material.

Upon computing the stress

σ = P/A₀ ⇒ σ = P/(π*d₀²/4) = 27,500 N/(π*(0.01 m)²/4) = 350*10⁶ N/m²

⇒ σ = 350 MPa

Of the alloys listed, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, (i.e., that Δd be less than 7.5 × 10 ⁻³  mm), it is necessary to  calculate the change in diameter Δd for these four alloys.

Then we use the equation   υ = - εx / εz = - (Δd/d₀)/(σ/E)

⇒  υ = - (E*Δd)/(σ*d₀)

Now, solving for ∆d from this expression,

∆d = - υ*σ*d₀/E

For the Aluminum alloy

∆d = - (0.33)*(350 MPa)*(10 mm)/(70*10³MPa) = - 0.0165 mm

0.0165 mm > 7.5×10 ⁻³ mm

Hence, the Aluminum alloy is not a candidate.

For the Brass alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(101*10³MPa) = - 0.0118 mm

0.0118 mm > 7.5×10 ⁻³ mm

Hence, the Brass alloy is not a candidate.

For the Steel alloy

∆d = - (0.3)*(350 MPa)*(10 mm)/(207*10³MPa) = - 0.005 mm

0.005 mm < 7.5×10 ⁻³ mm

Therefore, the steel is a candidate.

For the Titanium alloy

∆d = - (0.34)*(350 MPa)*(10 mm)/(107*10³MPa) = - 0.0111 mm

0.0111 mm > 7.5×10 ⁻³ mm

Hence, the Titanium alloy is not a candidate.

7 0
3 years ago
Oil enters a counterflow heat exchanger at 600 K with a mass flow rate of 10 kg/s and exits at 200 K. A separate stream of liqui
Elis [28]

Answer:

The minimum mass flow rate for the water is 14.39kg/s

Explanation:

In this question, we are asked to calculate the minimum mass flow rate for the water in kg/s.

Please check attachment for complete solution and step by step explanation

7 0
3 years ago
When you want to make a right turn your car must be
Eduardwww [97]

Answer:

(Have a great day)

¶Emma Jess¶

Explanation:

Get as far to the right side of the road as possible. Do not cut the across lanes of traffic to performed any turn..

8 0
3 years ago
How to make text take shape of object in affinity designer
Alina [70]

Answer:

To fit text to a shape in Affinity Designer, make sure you have your text selected. Then, grab the Frame Text Tool and click on the shape. A blinking cursor will appear within the shape, indicating that you can begin typing. The text you type will be confined to the boundaries of the shape.

Explanation:

6 0
3 years ago
An inventor claims to have developed a power cycle operating between hot and cold reservoirs at 1350 K and 295 K, respectively,
Art [367]

Answer:

efficiency = 0.71999  <    0.78148  

efficiency = 0.7999   ≈     0.78148  

Explanation:

given data

Hot reservoir Temp Th = 1350 K

Cold reservoir Temp Tc = 295 K

heat transfer rate  = 150,000 kJ/h

solution

first we get here maximum possible efficiency that is express as

maximum possible efficiency = 1 -  \frac{Tc}{Th}   ..........1

put here value  and we get

maximum possible efficiency = 1 - \frac{295}{1350}

maximum possible efficiency = 0.78148  

and

efficiency = \frac{Wout}{Qin}   .................2

here Qin = \frac{150000}{3600}  = 41.667

so put here value in equation 2 we get

efficiency = \frac{30}{41.667}  

efficiency = 0.71999  <    0.78148  

and

efficiency = \frac{33.33}{41.667}  

efficiency = 0.7999   ≈     0.78148  

claim might be true or not because there always be looser

5 0
4 years ago
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