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leonid [27]
3 years ago
15

Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi

n at 65°C. Which source (1 or 2) would you choose to supply energy to an ideal reversible heat engine that is to produce the larger amount of power if the temperature of the surroundings is 38°C?
Engineering
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

\eta_{th} = 1 - \frac{T_{L}}{T_{H}}

Where:

T_{L} - Temperature of the cold reservoir, in K.

T_{H} - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}

\eta_{th} = 0.466 \,(46.6\,\%)

Source 2

\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}

\eta_{th} = 0.0798 \,(7.98\,\%)

The power produced by each device is presented below:

Source 1

\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 85.433\,kW

Source 2

\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 146.3\,kW

The source 2 produces the largest amount of power.

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n200080 [17]

Answer:

Advantages

The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights. In other words, it reduces the amount of actual force required to lift heavy objects. It also changes the direction of the force applied. These two advantages in the use pulleys make them an important tool for heavy lifting. It also provides a mechanical advantage.

The other advantage in the use of pulleys is that the distance between the operator and weight. There is a safe distance between them which avoids any disaster. Pulleys are easy to assemble and cost-effective. The combination of different directional pulleys can change the position of the load with little effort. Though there are moving parts in the pulley system they require less or no lubrication after installation.

Disadvantages

Apart from the above-said advantages while using pulley systems, there are several disadvantages in their use. The main disadvantage in the use of the pulley system is that it requires large space to install and operate. The mechanical advantage of pulleys can go to higher values but need more space to install them.

In some cases, the ropes/belts move over the wheel with no grooves, the chances of the slip of ropes/belts from the wheel are inevitable. If the system is installed to use for a long time, they require maintenance and regular check-up of ropes/cables as the friction between the wheels and cables/ropes occur causing wear and tear to them. Continuous use of the system makes the ropes weak. The rope may break while using the system causing damages to the operator, surrounding place and the load which is being lifted.

5 0
3 years ago
Based on the pattern, what are the next two terms of the sequence? 9,94,916,964,9256,... A. 91024,94096 B. 9260,91028 C. 9260,92
Hatshy [7]

Answer:

The answer is "Option A".

Explanation:

Series:

9, 94, 916, 964, 9256, ........

Solving the above series:

\to  9\\ \to  9(4) =94\\\to 9 (4^2) = 9(16) =916\\\to 9 (4^3) = 9(64) =964\\\to 9 (4^4) = 9(256) =9256\\\to 9 (4^5) = 9(1024) =91024\\\to 9 (4^6) = 9(4096) =94096\\

So, the series is:  9, 94, 916, 964, 9256, 91024, 94096, .................

6 0
3 years ago
Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find
Igoryamba

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

please kindly see attachment for the step by step solution of the given problem.

4 0
3 years ago
An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir
Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

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3 years ago
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