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MArishka [77]
3 years ago
15

Two different fuels are being considered for a 2.5 MW (net output) heat engine which can operate between the highest temperature

produced during the burning of the fuels and an atmospheric temperature of 300 K. Fuel A burns at 2,500 K, delivering 50 MJ/kg (heating value) and costs $2.00/kg. Fuel B burns at 1,500 K, delivering 40 MJ/kg and costs $1.50/kg.
Compare the fuel costs per hour of fuel A and fuel B, assuming that the heat engine operates
(a) at Carnot efficiency
(b) at 40% of Carnot efficiency
Engineering
1 answer:
sveta [45]3 years ago
8 0

Answer:

If the heat engine operates for one hour:

a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.

b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.

In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.

Explanation:

The Carnot efficiency is obtained as:

\epsilon_{car}=1-\frac{T_c}{T_H}

Where T_c is the atmospheric temperature and T_H is the maximum burn temperature.

For the case (B), the efficiency we will use is:

\epsilon_{b}=0.4\epsilon_{car}

The work done by the engine can be calculated as:

W=\epsilon Q=\epsilon H_v\cdot m_{fuel} where Hv is the heat value.

If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.

m=\frac{P\cdot t}{\epsilon H_v}

If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.

TC=m\cdot c

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Levart [38]

Answer:

This is a for loop.

Explanation:

6 0
3 years ago
A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse
d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

    T = T_m + (T_0-T_m)e^{kt}

where,

T_0 = initial temp

T_m = temp of room

T = temp after t hours

k = how fast the temp is changing

t = time (hours)

T_0 = 31     because the body was initlally 31ºC when the police found it

T_m = 22   because that was the room temp

T = 30  because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

k=???   we don't know k so we must solve for this

rearrange the equation to solve for k

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k

plug in the numbers to solve for k

k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}

k = \frac{ln(\frac{30 - 22}{31-22})}{1}

k=ln(\frac{8}{9})

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

T_0 = 37     because the body was 37ºC right after being killed

T_m = 22   because that was the room temp

T = 31  because the body temp when the police found it

k=ln(\frac{8}{9})   we solved this earlier

t = ???   we don't know how long it took from the time of the murder to when the police found the body

Rearrange the equation to solve for t

T = T_m + (T_0-T_m)e^{kt}

T - T_m= (T_0-T_m)e^{kt}

\frac{T - T_m}{(T_0-T_m)}= e^{kt}

ln(\frac{T - T_m}{T_0-T_m})=kt

\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}=t

plug in the values

t=\frac{ln(\frac{T - T_m}{T_0-T_m})}{k}

t=\frac{ln(\frac{31 - 22}{37-22})}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t=\frac{ln(3/5)}{ln(8/9)}

t ≈ 4.337 hours from the time the body was killed to when the police found it.

The police found the body at 11:00PM so subtract 4.337 from that.

11 - 4.33 = 6.66 ≈ 6:30PM

7 0
3 years ago
You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC
aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

<h3>What is PMCS?</h3>

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

  • M1114
  • M1151
  • M1123

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

Read more on PMCS here: brainly.com/question/15720250

#SPJ1

4 0
2 years ago
Problem 2: Sieve Analysis and Soil Gradation
Marina CMI [18]

<u>Explanation:</u>

% Gravel = 100 - 72% = 28%

% Sand = 100 - 28 - 15 = 57%

% Fires = 100 - 0 - 28 - 57 = 15%

\begin{aligned}&D_{n} \cdot 0.03 \quad ; \quad D_{x}=1.1 \quad ; \quad D_{0}=3.3\\&C_{u}=\frac{D_{60}}{D_{10}}=\frac{3.3}{0.03}=110\\&C_{c}=\frac{D_{30}^{2}}{D_{0} \cdot D_{0}}=\frac{1.1^{2}}{3.3 \times 0.03}=12.22\end{aligned}

The soil is poorly graded soil.

6 0
3 years ago
Single-use earplugs require a professional fitting before they can be used.
Musya8 [376]
What are you asking? are you asking if it is true or false? sorry :(( i can help you tho!
8 0
3 years ago
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