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2 years ago

Which of the following is an example of a computer simulation?

Engineering

1 answer:

nata0808 [166]2 years ago

5 0

Answer:

b)prototype

Explanation:

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A pump is used to transport water from a reservoir at one elevation to another reservoir at a higher elevation. If the elevation

erastova [34]

Answer:b

Explanation:

We know power delivered by Pump is

$P=\rho \times Q\times g\times \Delta H$

where $P\rho$ =Density of fluid

$Q$ =Flow rate

$g$ =acceleration due to gravity

$\Delta H$ =Change in Elevation

If $\Delta H$ is increased by 4 time then

$P'=\rho \times Q\times g\times (4\Delta H)$

$P'=4 P$

So power increases by four times.

4 0

2 years ago

The flowchart below shows the design steps required to build a working model.

AlekseyPX

Question:

1) test the model and analyze the results of the test

2) building the model and observing it

3) observing the model and reporting results

4) designing the model and drawing conclusions

Answer:

The correct option is;

1) Test the model and analyze the results of the test

Explanation:

Based on the flowchart, a model improvement process involves the implementation of a process or model improvement cycle such as the Plan Do Study Act, PDSA cycle, however feedback to the process will be be gotten from testing the model and analyzing the results of the tests. When grey areas or aspects of the model are found that cause the system to malfunction are determined, steps should then be taken to improve the performance of the model.

4 0

2 years ago

Please help is due tonight

Kipish [7]

Answer:

tHE answer is b

Explanation:

7 0

2 years ago

Read 2 more answers

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.

Oxana [17]

Answer:

COP of heat pump=3.013

COP of cycle=1.124

Explanation

W = Q2 - Q1 ----- equation 1

W = work done

Q2 = final energy

Q1 = initial energy

A) calculate the COP of the heat pump

COP =Q2/W

from equation 1

Q2 = Q1 + W = 15 + 7.45 = 22.45 KW

therefore COP =22.45/7.45 = 3.013

B) COP when cycle is reversed

COP = Q1/W

from equation 1

Q1 + W = Q2 ------ equation 2

Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2

Q1 = 8.375 KW

COP =8.375/7.45 = 1.124

6 0

2 years ago

Read 2 more answers

For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu

bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6 mo1^-2

Now according isothermal work of one mole methyl gas is

W=- $\int\limits^a_b {P} \, dV$

a= $v_2\\$

b= $v_1$

from virial equation

$\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\$

And

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a= $v_2\\$

b= $v_1$

Now calculate V1 and V2 at given condition

$\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}$

Substitute given values $P_1\\$ = 1 x 10^5 , T = 373.15 and given values of coefficients we get

$10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2$

Solve for V1 by iterative or alternative cubic equation solver we get

$v_1=30780 cm^3/mol$

Similarly solve for state 2 at P2 = 50 bar we get

$v_1=241.33 cm^3/mol$

Now

$W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV$

a=241.33

b=30780

After performing integration we get work done on the system is

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get

dV=RT(-1/p^2+0+C')dP

Hence work done on the system is

$W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP$

a= $v_2\\$

b= $v_1$

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series

8 0

2 years ago