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jekas [21]
3 years ago
15

Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl

ied energy, assuming an environment temperature of 25°C.
Engineering
1 answer:
ANEK [815]3 years ago
5 0

Answer:

exergy = 33.39 kW

Explanation:

given data

thermal energy reservoir T2 = 1500 K

heat at a rate = 150,000 kJ/h = \frac{15000}{3600} kW =  41.67 kW

environment temperature T1 = 25°C = 298 K

solution

we get here maximum efficiency that is  reversible efficiency is express as

reversible efficiency = 1 - \frac{T1}{T2}    ...............1

reversible efficiency = 1 - \frac{298}{1500}  

reversible efficiency =  0.80133

and

the exergy of this supplied energy that is

exergy  = efficiency × hat supply   ................2

exergy = 0.80133 × 41.67 kW

exergy = 33.39 kW

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Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values
Nataliya [291]

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating  pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating  pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

7 0
3 years ago
5. One of the major road blocks for direct DC power in residences is the ___. A) generation of DC B) lack of standardization C)
Montano1993 [528]

Answer:

D) quantity of components required for this type of system

Explanation:

Electricity can be transmitted using the alternating and direct currents. The alternating current is one in which the flow of the current diverts at certain time intervals whereas, the direct current is one in which there is a constant one-directional flow of current. The DC is used in batteries and solar panels. Residential areas and business places make use of the AC current.

One of the several reasons why the DC is not used in homes is because unlike the AC it is not easy to build and sustain. Moreso, it has more components compared to the AC. For example, its motor consists of brushes and commutators . The components, example, switches are also large compared to the AC components.

6 0
3 years ago
What is a system that performs different functions according to a fixed schedule?
geniusboy [140]

<u>I'm pretty sure your answer is B, because Sequential Control operates during order like a schedule</u>

Sequential Control=A control system in which the individual steps are processed in a predetermined order, progression from one sequence step to the next being dependent on defined conditions being satisfied.

Tell me if I'm incorrect but, Hope this helps!

4 0
3 years ago
Read 2 more answers
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
ella [17]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

7 0
2 years ago
Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

ΔS = -0.544 KJ/Kg.K

3 0
3 years ago
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