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Setler [38]
3 years ago
14

A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi

ce of that of loose side, F1= 2F2. Please calculate the F1, Fe, and Fo.
Engineering
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

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1+1=2
Solution
3:932(2)=61
4 0
3 years ago
Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from
kykrilka [37]

Answer:

Given:

high temperature reservoir T_{H} =1000k

low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

The engines are said to  operate on Carnot cycle which is totally reversible.

To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

n_{2} =1-\frac{T_{L} }{T}

The temperature of intermediate reservoir can be defined as  

1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k

8 0
3 years ago
What is the maximum thermal efficiency possible for a power cycle operating between 600P'c and 110°C? a). 47% b). 56% c). 63% d)
Anastasy [175]

Answer:

(b) 56%

Explanation:

the maximum thermal efficiency is possible only when power cycle is reversible in nature and when power cycle is reversible in nature the thermal efficiency depends on the temperature

here we have given T₁ (Higher temperature)= 600+273=873

lower temperature T₂=110+273=383

Efficiency of power cycle is given by =1-\frac{T2}{T1}

=1-\frac{383}{873}

=1-0.43871

=.56

=56%

5 0
3 years ago
Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25
Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0
2 years ago
Two engineers are to solve an actual heat transfer problem in a manufacturing facility. Engineer A makes the necessary simplifyi
deff fn [24]

Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

5 0
3 years ago
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