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arlik [135]
3 years ago
11

Biblical studies of john​

Engineering
1 answer:
nevsk [136]3 years ago
5 0

Answer:

<h2>the answer of sols brother is correct</h2><h3>hope it helps you have a good day</h3><h2 />

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Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,
solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

8 0
2 years ago
Please help I need it by today!!!
AfilCa [17]

Answer:

if engineering disappeared for a day i would be at a loss. i wouldnt know what to do with myself considering engineering is my life. one way that engineers improve my life is they help me to understand enything end everything

Explanation:

7 0
3 years ago
¿Cuál era probablemente el activo más valioso de Persia?
laila [671]

Answer:

su suministro de aceite

Explanation:

(it's oil supply)

8 0
3 years ago
Read 2 more answers
QUESTION:
svet-max [94.6K]
OA bloom is smaller than a bar
6 0
2 years ago
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify t
raketka [301]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

5 0
3 years ago
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