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3 years ago

Drop the name below the corresponding part. (Look at the picture above to answer)

Engineering

1 answer:

Vlad1618 [11]3 years ago

7 0

Answer:the answer is d

Explanation: because it is the right answer

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Engineers will redesign their products when they find flaws. TRUE O False

nataly862011 [7]

Answer:

true

Explanation:

6 0

3 years ago

Sketch the velocity profile for laminar and turbulent flow.

Margarita [4]

Answer:

The laminar flow is generally given in high viscosity fluids such as honey or oil, it has the characteristic of flowing in an orderly manner, the walls of the tube have a zero speed while in the center it has a maximum speed.

turbulent flow is characterized by fluid velocity vectors presenting themselves in a disorderly manner and in all directions.

I attached the drawings for the velocity profile in laminar and turbulent flow.

4 0

3 years ago

A 10-m-long countercurrent-flow heat exchanger is being used to heat a liquid food from 20 to 808C. The heating medium is oil, w

kolezko [41]

Answer:

mlf=0.5038kg/s

Explanation:

a. Please kindly check attachment for the step by step solution

b. B) in concurrent flow heat exchanger same exit temperatures for both fluids cannot be obtained. Since tho<tco

Case ii) only liquid food is considered tci=20 &tco=80

Cp=3.9kj/kgk

&If Q is same then mlf=0.5038kg/s this is same as case a

5 0

3 years ago

Consider three charged sheets, A, B, and C. The sheets are parallel with A above B above C. On each sheet there is surface charg

babunello [35]

Answer:

Explanation:

To find the electrical force per unit area on each sheet we start defining our variables,

$\sigma_A = -4.10^{-5}C/m^2$

$\sigma_B= -7.10^{-5}C/m^2$

$\sigma_C = -3.1^{-5}C/m^2$

We find the electric field for each one, this formula is given by,

$E= \frac{\sigma_i}{2\epsilon_0}$

Substituting each value from the three charged sheets, we have

$E_A= \frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_B= \frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})$

$E_C= \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

The electric field is

$E_{NET}= E_A+E_B+E_C$

$E_{NET}=\frac{-4.10^{-5}C/m^2}{2\epsilon_0}(\hat{j})+\frac{-7.10^{-5}C/m^2}{2\epsilon_0}(-\hat{j})+ \frac{-3.1^{-5}C/m^2}{2\epsilon_0}(\hat{j})$

$E_{NET} = 0$

Force on each sheet is,

$F=E_{NET}\sigma ds$

$F=0$

The total force is 0

5 0

3 years ago

Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and

jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ = 0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ + 17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0

3 years ago