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3 years ago

11

). A company periodically tests its product for tread wear under simulated conditions. Thirty random samples, each containing 5

items, were taken in a control chart application and it was found that the grand average is

1 answer:

Luden [163]3 years ago

5 0

Answer:

See explanation for step by step answer.

Explanation:

Given that:

A company periodically tests its product for tread wear under simulated conditions. Thirty random samples, each containing 5 items, were taken in a control chart application and it was found that the grand average is....

See attacent for complete answer

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Air enters a cmpressor at 20 deg C and 80 kPa and exits at 800 kPa and 200 deg C. The power input is 400 kW. Find the heat trans

aksik [14]

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, $T_{i} = 20^{\circ}C$ = 273 + 20 = 293 K

Temperature at the outlet, $T_{o} = 200{\circ}C$ = 273 + 200 = 473 K

Pressure at inlet, $P_{i} = 80 kPa = 80\times 10^{3} Pa$

Pressure at outlet, $P_{o} = 800 kPa = 800\times 10^{3} Pa$

Speed at the outlet, $v_{o} = 20 m/s$

Diameter of the tube, $D = 10 cm = 10\times 10^{- 2} m = 0.1 m$

Input power, $P_{i} = 400 kW = 400\times 10^{3} W$

Now,

To calculate the heat transfer, $Q$ , we make use of the steady flow eqn:

$h_{i} + \frac{v_{i}^{2}}{2} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}$

where

$h_{i}$ = specific enthalpy at inlet

$h_{o}$ = specific enthalpy at outlet

$v_{i}$ = air speed at inlet

$p_{s}$ = specific power input

H and H' = Elevation of inlet and outlet

Now, if

$v_{i} = 0$ and H = H'

Then the above eqn reduces to:

$h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}$

$Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}$ (1)

Also,

$p_{s} = \frac{P_{i}}{ mass, m}$

Area of cross-section, A = $\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}$

Specific Volume at outlet, $V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s$

From the eqn:

$P_{o}V_{o} = mRT_{o}$

$m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s$

Now,

$p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg$

Also,

$\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg$

Now, using these values in eqn (1):

$Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW$

Now, rate of heat transfer, q:

q = mQ = $0.925\times 813.33 = 752.33 kW$

4 0

2 years ago

A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok

galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0

3 years ago

Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in

salantis [7]

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

class Program

{

static void Main(string[] args)

{

int[] Number = new int[10]; // Declare array of 10 elements

//Accept Input

int counter = 0;

while(counter<10)

{

Console.WriteLine("Enter Number " + (counter + 1)+": ");

string var = Console.ReadLine();

Number[counter] = Convert.ToInt32(var);

counter++;

}

//Initialize largest to first element of the array

int Largest = Number[0];

//Determine Largest

for(int i=0;i<10;i++)

{

if(Largest < Number[i])

{

Largest = Number[i];

}

}

//Print Largest

Console.WriteLine("The largest input is "+ Largest);

Console.ReadLine();

}

}

}

8 0

3 years ago

What is a material safety data sheet? A. a categorized list of hazardous substances in an industry B. a document with safety inf

horsena [70]

Answer: A.

Explanation:

safety data sheet are documents that list information relating to occupational safety and health for the use of various substances and products.

3 0

3 years ago

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m ? K), and the wire/sheath interface

Svet_ta [14]

Question

A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.

If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

Answer:

a. 4.52W/m

b. 13mm

Explanation:

Given

Diameter of electrical wire = 2mm

Wire Thickness = 2-mm

Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)

Thermal contact resistance = 3E-4m².K/W

Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,

Temperature of the ambient air = 20°C.

Maximum Allowable Sheet Temperature = 50°C.

From the thermal circuit (See attachment), we my write

E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)

= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))

Where r in,i = D/2

= 2mm/2

= 1 mm

= 0.001m

r in,o = r in,i + t = 0.003m

T in, i = Tmax = 50°C

Hence

q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]

= 30/[(Ln3/0.26π) + 1/0.06π)]

= 30/[(1.34) + 5.30)]

= 30/6.64

= 4.52W/m

The critical radius is unaffected by the constant resistance.

Hence

Critical Radius = k/h

= 0.13/10

= 0.013m

= 13mm

5 0

3 years ago