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3 years ago

13

Are engineers needed in today’s society ? Why or why not ? I need a short three paragraph essay !!! Please help me !!!

1 answer:

masha68 [24]3 years ago

8 0

Of course they are needed because without them the society wouldn’t be as nice as it is right now and plus there would be no more buildings ! :)

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Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,

aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

System.out.println(birthMonth+"/"+birthYear);

}

}

3 0

3 years ago

Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.

saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 $\mathbf{m^{3}}$

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C ; $v_1 = vg_1$ = 0.0996 $\mathbf{m^{3}}$ / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 $\mathbf{m^{3}}$ / kg

From temperature T₂ = 200⁰ C

$v_f_2 = 0.0016 \ m^3/kg$

$vg_2 = 0.127 \ m^3/kg$

Since $vf_2 < v_2$ , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality $x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}$

$x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}$

$x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}$

$\mathsf{x_2 =0.78}$

At temperature T₂, the specific internal energy $u_f_2 = 850.6 \ kJ/kg$ , also $ug_2 = 2594.3 \ kJ/kg$

Thus,

$u_2 = uf_2 + x_2 (ug_2 -uf_2)$

$u_2 =850.6 +0.78 (2594.3 -850.6)$

$u_2 =850.6 +1360.086$

$u_2 =2210.686 \ kJ/kg$

<u>At state 3:</u>

Temperature $T_3=T_2 = 200 ^0 C ,$

$V_3 = 2V_1 = 0.06 \ m^3$

Specific volume $v_3 = 0.2 \ m^3/kg$

Thus; $vg_3 =vg_2 = 0.127 \ m^3/kg$ ,

SInce $v_3 > vg_3$ , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at $v_3 = 0.2 \ m^3/kg$ and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 $\ m^3/kg$

The specific internal energy $u_3$ at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

$u_2-u_1$

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

$u_3-u_2$

= (2622.3 - 2210.686) kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0

3 years ago

Which of the following is a variable expense for many adults?

sertanlavr [38]

But where are the options?

7 0

2 years ago

Read 2 more answers

“In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set

artcher [175]

Answer:

no i cant answer it sorry

7 0

2 years ago

134a refrigerant enters an adiabatic compressor at 140kPa and -10C, the refrigerant is compressed at 0.5kW up to 700kPa and 60C.

vichka [17]

Answer:

(a) 65.04%

(b) 16.91%

Solution:

As per the question:

At inlet:

Pressure of the compressor, P = 140 kPa

Temperature, T = $- 10^{\circ}C$ = 263 K

Isentropic work, W = 700 kPa

At outlet:

Pressure, P' = 700 kPa

Temperature, T' = $60^{\circ}C$ = 333 K

Now, from the steam table;

At the inlet , at a P = 700 kPa, T = $60^{\circ}C$ :

h = 243.40 kJ/kg, s = 0.9606 kJ/kg.K

At outlet, at P = 140 kPa, T = $- 10^{\circ}C$ :

h' = 296.69 kJ/kg, s' = 1.0182 kJ/kg.K

Also in isentropic process, s = $s'_{s}$ and $h'_{s} = 278.06 kJ/kg.K$ at 700kPa

(a) Isentropic efficiency of the compressor, $\eta_{s} = \frac{Work\ done\ in\ isentropic\ process}{Actual\ work\ done}$

$\eta_{s} = \frac{h'_{s} - h}{h' - h} = frac{278.06 - 243.40}{296.69 - 243.40} = 0.6504 = 65.04%$

(b) The temperature of the environment, $T_{e} = 27^{\circ}C$ = 273 + 27 = 300 K

Availability at state 1, $\Psi = h - T_{e}s = 243.40 - 300\times 0.9606 = - 44.78 kJ/kg$

Similarly for state 2, $\Psi' = h' - T_{e}s' = 296.69 - 300\times 1.0182 = - 8.77 kJ/kg$

Now, the efficiency of the compressor as per the second law;

$\eta' = \frac{\Psi' - \Psi}{h' - h} = \frac{- 8.77 - (- 44.78)}{296.69 - 243.40} = 0.6757$ = 67.57%

4 0

3 years ago