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Rina8888 [55]
4 years ago
12

A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and

the surrounding air is at 75°F. Determine (a) the reversible power, (b) the irreversibility, and (c) the second-law efficiency of this freezer.
Engineering
1 answer:
Igoryamba4 years ago
8 0

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL /  COP r  where 75 Btu/min = 1.76856651 hp  where W actual = 0.70 hp

a) W rev =  1.76856651 hp  /  8.73  = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual =  0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

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Mila [183]

Answer:

As many variables as we can coherently communicate in 2 dimensions

Explanation:

Visualization is a descriptive analytical technique that enables people to see trends and dependencies of data with the aid of graphical information tools. Some of the examples of visualization techniques are pie charts, graphs, bar charts, maps, scatter plots, correlation matrices etc.

When we utilize a visualization on paper/screen, that visualization is limited to exploring as many variables as we can coherently communicate in 2-dimensions (2D).

6 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
Write a method printShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, prin
kirill [66]

Answer:

// The method is defined with a void return type

// It takes a parameter of integer called numCycles

// It is declared static so that it can be called from a static method

public static void printShampooInstructions(int numCycles){

// if numCycles is less than 1, it display "Too few"

   if (numCycles < 1){

       System.out.println("Too few.");

   }

// else if numCycles is less than 1, it display "Too many"

    else if (numCycles > 4){

       System.out.println("Too many.");

   }

// else it uses for loop to print the number of times to display

// Lather and rinse

  else {

       for(int i = 1; i <= numCycles; i++){

           System.out.println(i + ": Lather and rinse.");

       }

       System.out.println("Done");

       

   }

}

Explanation:

The code snippet is written in Java. The method is declared static so that it can be called from another static method. It has a return type of void. It takes an integer as parameter.

It display "Too few" if the passed integer is less than 1. Or it display "Too much" if the passed integer is more than 4. Else it uses for loop to display "Lather and rinse" based on the passed integer.

8 0
3 years ago
How do Consumers sometimes interact with a producers?
Vika [28.1K]
I really don’t know good luck
7 0
3 years ago
The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The b
Wewaii [24]

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁  and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0   (+↑)

Nef*Cos 12º -  F₂*Sin 12º = W

⇒   Nef*Cos 12º -  (0.6*Nef)*Sin 12º = 18

⇒   Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒  P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒  P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

7 0
3 years ago
Read 2 more answers
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