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Bumek [7]
3 years ago
7

A Carnot heat engine absorbs 235 KW of heat from a heat source and rejects 164 KW to the atmosphere. Determine the thermal effic

iency of the heat engine.
Engineering
1 answer:
nevsk [136]3 years ago
5 0

Answer:

43.2%

Explanation:

Given that,

Heat absorbed by a carnot heat engine, Q_1=235\ kW

Heat rejected to the atmosphere, Q_2=164\ kW

We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :

\eta=1-\dfrac{Q_1}{Q_2}\\\\\eta=1-\dfrac{235}{164}\\\\\eta=-0.432

or

\eta=-43.2\%

The egative value of efficiency shows work is done by the engine.

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Find: factor of safety (n)for point A and B by using both MSS and DE (you can neglect shear stress due to shear force and also n
gladu [14]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

8 0
3 years ago
Marks
Lina20 [59]

Answer:

i think C

Explanation:

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3 0
2 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
Select the best answer for the question.
dalvyx [7]
I think the Acid level
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3 years ago
The Chinese had a toy helicopter some 1000 years ago. True Or False ?
Vadim26 [7]

Answer:

f

Explanation:

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