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blagie [28]
3 years ago
14

Two engineers are to solve an actual heat transfer problem in a manufacturing facility. Engineer A makes the necessary simplifyi

ng assumptions and solves the problem analytically, while engineer B solves it numerically using a powerful software package. Engineer A claims he solved the problem exactly and, thus, his results are better, while engineer B claims that he used a more realistic model and, thus, his results are better. Will the experiments prove engineer B right
Engineering
1 answer:
deff fn [24]3 years ago
5 0

Answer:

Engineer A results will be more accurate

Explanation:

Analytical method is better than numerical method. Engineer A has used analytical method and therefore his results will be more accurate because he used simplified method. Engineer B has used software to solve the problem related to heat transfer his results will be approximate.

You might be interested in
Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)
IceJOKER [234]

Answer:

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

\mathbf{M_n = 49163.56431  \ g/mol }

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa)            Number- Average Molecular Weight  (g/mol)

82                                                  12,700

156                                                 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

T_S = T_{S \infty} - \dfrac{A}{M_n}

where;

T_S = Tensile Strength

T_{S \infty} = Tensile Strength (Infinity)

M_n = Number- Average Molecular Weight  (g/mol)

SO;

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

From equation (1) ; collecting the like terms; we have :

T_{S \infty} =82+ \dfrac{A}{12700}

From equation (2) ; we have:

T_{S \infty} =156+ \dfrac{A}{28500}

So; T_{S \infty} = T_{S \infty}

Then;

T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}

Solving by L.C.M

\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}

\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}

By cross multiplying ; we have:

({4446000 + A})*  {12700} ={28500} *({1041400 + A})

(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)

Collecting like terms ; we have

(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)

2.67843*10^{10}  = 15800 \ A

Dividing both sides by 15800:

\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}

A = 1695208.861

From equation (1);

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

Replacing A = 1695208.861 in the above equation; we have:

82= T_{S \infty} - \dfrac{1695208.861}{12700}

T_{S \infty}= 82 + \dfrac{1695208.861}{12700}

T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}

T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}

T_{S \infty}= \dfrac{2736608.861 }{12700}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

From equation(2);

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

Replacing A = 1695208.861 in the above equation; we have:

156= T_{S \infty} - \dfrac{1695208.861}{28500}

T_{S \infty}= 156 + \dfrac{1695208.861}{28500}

T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}

T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}

T_{S \infty}= \dfrac{6141208.861}{28500}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

T_S = T_{S \infty} - \dfrac{A}{M_n}

Then:

181 = 215.481- \dfrac{1695208.861 }{M_n}

Collecting like terms ; we have:

\dfrac{1695208.861 }{M_n} = 215.481-  181

\dfrac{1695208.861 }{M_n} =34.481

1695208.861= 34.481 M_n

Dividing both sides by 34.481; we have:

M_n = \dfrac{1695208.861}{34.481}

\mathbf{M_n = 49163.56431  \ g/mol }

5 0
3 years ago
Led test lights are used to test circuits that include controllers and computers. True or false
Marianna [84]

Answer:

True

Explanation:

An LED test light is a piece of electronic test equipment used to determine the presence of electricity in a piece of equipment under test, making this statement true.

3 0
2 years ago
In an ideal gas, specific enthalpy is a function of i. Entropy ii. Temperature iii, Pressure iv. Mass
Mice21 [21]

Answer:

Temperature

Explanation:

In an ideal gas the specific enthalpy  is exclusively a function of Temperature only this can be also written as h = h(T)  

A gas is said be ideal gas if obeys PV= nRT law

And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.

5 0
3 years ago
Water exerts little pressure on a building so it has no implications on foundation design.
777dan777 [17]

Fact

Explanation:

<em>If </em><em>wrong </em><em>Im </em><em>sorry </em><em>Correct </em><em>me </em><em>In </em><em>The</em><em> </em><em>c</em><em>om</em><em>m</em><em>ent </em><em>please </em>

7 0
3 years ago
The typical area of a commercial airplane's passenger window is 80.0 in^2 . At an altitude of 3.00 × 104 ft above the sea level,
nikdorinn [45]

Answer:

The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.

Explanation:

The net force over the window is calculated by multiplying the difference in pressure by the area of the window:

F = Δp*A

The pressure inside the plane is around 1 atm, hence the difference in pressure is:

Δp = 1atm - 0.35 atm = 0.65 atm

Expressing in the EE unit system:

Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2

Replacing in the force:

F = 9.55 lbf/in^2 * 80 in^2  = 764 lbf

For the international unit system, we re-calculate the window's area and the difference in pressure:

A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2

Δp =  0.65 atm * 101325 Pa  = 65861 Pa  = 65861 N/m^2

Replacing in the force:

F = 65861 N/m^2  *0.0516 m^2  = 3398 N

3 0
3 years ago
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