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kvv77 [185]
2 years ago
12

A single Gates Rubber V belt is to be selected to deliver engine power to the wheel-drive transmission of a riding tractor. A 5-

hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation. How man

Engineering
1 answer:
beks73 [17]2 years ago
7 0

Answer:

Hello there, please check step by step explanations to get answers.

Explanation:

Given that:

5-hp single-cylinder engine is used. At most, the belt transmits 60 percent of this power. The driving sheave has a diameter of 6.2 in. and the driven, 12.0 in. The belt selected should be as close to 92 in pitch length as possible. The engine speed is governor-controlled to a maximum of 3100 rev/min. Select a satisfactory belt, and specify it using the standard designation.

See attached documents for clearity and step by step procedure to answer

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4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and
hodyreva [135]
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
6 0
2 years ago
The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0
2 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
2 years ago
(a) If 15 kW of power from a heat reservoir at 500 K is input into a heat engine with an efficiency of 37%, what is the power ou
julsineya [31]

Answer:

(A) Power output will be 5.55 KW (b) lower temperature will be 315 K

Explanation:

We have given efficiency of heat engine \eta =37% = 0.37

Input power = 15 KW

Temperature of heat reservoir T_H=500K

(A) We know that \eta =\frac{output}{input}

So  [text]0.37=\frac{output}{15}[text]

Output = 5.55 KW

(B) We also know that [text]\eta =1-\frac{T_L}{T_H}0.37=\frac{output}{15}[text], here T_L  is lower temperature and T_H is higher temperature

So 0.37=1-\frac{T_L}{T_H}

0.37=1-\frac{T_L}{500}

T_L=315K

5 0
2 years ago
A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 1.5 ft3/s
svetoff [14.1K]

Answer:

11.52 hp

Explanation:

<u><em>Givens: </em></u>

p_1  = 15 pisa

p_2  = 70 pisa

V_ol=1.5 ft^3/s

<u><em>Solution:  </em></u>

Note: m = p x V_ol (assuming in compressible flow —> p =const)  

The total change in the system mechanical energy can be calculated as follows,  

Δ e= (p_2 - p_1 ) /p

The power needed can be calculated as follows

P = W =mΔ e  = p x  V_ol x(p_2 - p_1 ) /p

  = V_ol x (p_2 - p_1 )

  = 44 pisa. ft^3/s

  = 44 x (1 btu/5.404pisa. ft^3) x (1 hp/0.7068btu/s)

  = 11.52 hp

3 0
2 years ago
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