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Mariulka [41]
3 years ago
13

Two Carnot engines operate in series such that the heat rejected from one is the heat input to the other. The heat transfer from

the high-temperature reservoir is 500 kJ. The overall temperature limits are 1000K and 400K. Both engines produce equal work. What is most nearly the intermediate temperature between the two engines?
Engineering
1 answer:
kykrilka [37]3 years ago
8 0

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

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MakcuM [25]

Answer: π= G[√(u.V/W)]

STEP 1

Given parameters:

Power Input W= FL/T,

Absolute Viscosity u= FT/L²

Basin volume V= V/L³

Velocity gradient G= V/L³

STEP 2

We start by expressing the velocity gradient G as a function of W, u, V

G= G(W,u,V)

To get the pii terms, we use the dimension number formula n=k - r

where n and k are natural numbers representing number of fundamental dimensions and variable present respectively.

n= 4-3=1

STEP 3:

We expressed the pii terms as

π= G.W^a.u^b.V^c

The three fundamental F L T

We can write as

Fⁿ.Lⁿ.Tⁿ= 1/T. (FL/T)^a.(FT/L²)^b.(L³)

Using the exponential rule and by comparing coefficient on both sides;

Fⁿ.Lⁿ.Tⁿ= F^a+b. L^a-2b+3c. T^-a+b-1

Fⁿ= F^a+b = a+b= 0..............I

Lⁿ= L^a-2b+3c=0 = a-2b+3c=0...........ii

Tⁿ=L^-a+b-1=0. -a+b-1=0............iii

From the above equations we have,

a+b =0: b=-a...........iv

putting eq. iv into iii , we have

-a-a-1=0: -2a-1=0: a= -1/2

substituting the above value of a into eq iv, we have

b= 1/2

substituting the value of b above into eq 2, we have,

-1/2-2(1/2)+3c=0

c=1/2.

Lastly, from the pii terms given above we can obtain dimensionless relationship,

π=G(W^-1/2.u^1/2.V^1/2)

We can write this as

π= G[ √1/W.√u. √1/2] = G[(√u.V/√W)] or G[√(u.V/W)].... final answer.

5 0
3 years ago
Isormophous phase diagram
shusha [124]

Answer:

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Hope it help you friend

6 0
3 years ago
can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

6 0
3 years ago
Can some help me with this !!! Is 26 points!!
Aleonysh [2.5K]
Third one
15,000,000 ohms because M=10^6
8 0
3 years ago
A company purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate
olga2289 [7]

Answer:

1) The probability of at least 1 defective is approximately 45.621%

2) The probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is approximately 16.0212%

Explanation:

The given parameters are;

The defective rate of the device = 3%

Therefore, the probability that a selected device will be defective, p = 3/100

The probability of at least one defective item in 20 items inspected is given by binomial theorem as follows;

The probability that a device is mot defective, q = 1 - p = 1 - 3/100 = 97/100 = 0.97

The probability of 0 defective in 20 = ₂₀C₀(0.03)⁰·(0.97)²⁰ ≈ 0.543794342927

The probability of at least 1 = 1 - The probability of 0 defective in 20

∴ The probability of at least 1 = 1 - 0.543794342927 = 0.45621

The probability of at least 1 defective ≈ 0.45621 = 45.621%

2) The probability of at least 1 defective in a shipment, p ≈ 0.45621

Therefore, the probability of not exactly 1 defective = q = 1 - p

∴ q ≈ 1 - 0.45621 = 0.54379

The probability of exactly 3 shipment with at least 1 defective, P(Exactly 3 with at least 1) is given as follows;

P(Exactly 3 with at least 1) = ₁₀C₃(0.45621)³(0.54379)⁷ ≈ 0.160212

Therefore, the probability that there will be exactly 3 shipments each containing at least 1 defective device among the 20 devices that are tested from the shipment is 16.0212%

4 0
3 years ago
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