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3 years ago

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There are two machines for sale that you are considering purchasing for your sawmill to produce hardwood flooring. You want to f

ind the one that has a higher process capability index, or Cpk. The goal is to produce flooring that is between 46 and 50 millimeters thick. The first machine is more accurate on average, producing to a mean of 48 millimeters…but unfortunately it has more variation with a standard deviation of .7 millimeters. The second machine is not as accurate, with a mean of 47mm, but does deliver a more consistent output, with standard deviation of .3mm.

1 answer:

devlian [24]3 years ago

4 0

Answer:

Machine 2 has a higher process capability index, it would be best considered for purchase.

Explanation:

Process capability index: Cpk= Min [(mean-L spec)/3sd; (U spec-mean)/3sd]

For machine 1, mean= 48mm and L spec= 46 and U spec= 50, Standard deviation sd= 0.7

Cpk= [0.952;0.952]= 0.952

For machine 2, mean= 47 and L spec= 46 and U spec= 50, Standard deviation sd= 0.3

Cpk= [1.111;3.333]= 1.111

It is clearly observed from the calculations above that the Cpk value of machine 2 is higher than that of machine 1.

Since machine 2 has a higher process capability index, it would be best considered for purchase.

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Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

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3 years ago

A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.

Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is 1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

= 15 + 7.45

= 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

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3 years ago

Read 2 more answers

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

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Answer:

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Explanation:

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please rate brainliest if helps and follow

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