Please help this is due today!!!!!

8. Air at 25C, 100 kPa and air at 50C, 200 kPa at 1 to 1 volume ratio are mixed inside an adiabatic compressor to 55C, 500 kPa a

quester [9]

Answer:

Workdone, w = 68.935 kJ

Explanation:

m1h1 + m2h2 + Q = m3h3 - Wc

Final mass of mixture,

m3 = m1 + m2 = 5kg

p1V1 = m1R1T1

p2V2 = m2R2T2

Since they are at 1:1,

V1 = V2 and R1 = R2

Comparing equations,

p1/p2 = (m1/m2) x (T1/T2)

m1/m2 = (100/200) x ((50 + 273)/(25+273))

m1/m2 = 0.542

m1 = 0.542m2

m3 = m1 + m2

5 = m2 + 0.542m2

m2 = 3.243kg

m1 = 1.757 kg

Workdone is given as,

Wc = m3h3 - m1h1 - m2h2

h = Cp x T, since air is an ideal gas

Cp = 1.005kJ/kGK

Wc = (5 x 1.005 x (55+273)) - (1.757 x 1.005 x 298) -(3.243 x 1.005 x 323)

Wc = 68.935 kJ

8 0

4 years ago

A 20 mm diameter rod made of ductile material with a yield strength of 350 MN/m2 is subjected to a torque of 100 N.m, and a bend

vodka [1.7K]

Answer:

a) 42.422 KN

b) 44.356 KN

Explanation:

Given data :

Diameter = 20 mm

yield strength = 350 MN/m^2

Torque ( T ) = 100 N.m

Bending moment = 150 N.m

Determine the value of the applied axial tensile force when yielding of rod occurs

first we will calculate the shear stress and normal stress

shear stress ( г ) = Tr / J = [( 100 * 10^3) * 10 ] / $\pi /32$ * ( 20)^4

= 63.662 MPa

Normal stress( Гb + Гa ) = MY/ I + P/A

= [( 150 * 10^3) * 10 ] / $\pi /32$ * ( 20)^4 + 4P / $\pi * 20^2$

= 190.9859 + 4P / $\pi * 20^2$ MPa

a) Using MSS theory

value of axial force = 42.422 KN

solution attached below

b) Using MDE theory

value of axial force = 44.356 KN

solution attached below

5 0

3 years ago

The files provided in the code editor to the right contain syntax and/or logic errors. In each case, determine and fix the probl

Yanka [14]

Question Continuation

public class DebugOne3{

public static void main(String args){

System.out.print1n("Over the river");

System.out.pr1ntln("and through the woods");

system.out.println("to Grandmother's house we go");

}

Answer:

Line 2: Invalid Syntax for the main method. The main method takes the form

public static void main (String [] args) { }

or

public static void main (String args []) { }

Line 3: The syntax to print is wrong.

To print on a new line, make use of System.out.println(".."); not System.out.print1n();

Line 4:

To print on a new line, make use of System.out.println(".."); not System.out.pr1ntln();

Line 5:

The case of "system" is wrong.

The correct case is sentence case, "System.out.println" without the quotes

The correct program goes, this:

public class DebugOne3{

public static void main(String [] args){

System.out.println("Over the river");

System.out.println("and through the woods");

System.out.println("to Grandmother's house we go");

}

Explanation:

3 0

3 years ago

Air at 300 K and 100 kPa steadily flows into a hair dryer having electrical work input of 1500 W. Because of the size of the air

dezoksy [38]

Answer:

0.0280Kg/s

Explanation:

Given:

W = 1500

V2 = 300

V1 = 0

Q = 0 ( adiabatic)

T1 = 300

T2 = 20 m/s = converting to degrees we have 353°c

Let's use the energy equation

$Q - W = m_• *( Cp(dT) + 0.5*(V2^2 - V1^2) Q = 0$

[/tex] m_• = W / (Cp(dT) + 0.5*V2^2) [/tex]

$= 1500 / (1005(353 - 300) + 0.5*21^2)$

$= 1500 W / 53485.5 =0.0280 kg/s$

8 0

4 years ago

Read 2 more answers

Why does teachers grade things that are not due yet

pav-90 [236]

I think because if you’ve already turned it in they might as well grade asap instead of waiting

4 0

3 years ago

Read 2 more answers