Answer:
used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.
Explanation:
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=-
a=
b=
from virial equation

And

a=
b=
Now calculate V1 and V2 at given condition

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

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

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

Now

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

a=
b=
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
Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress =
......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Answer:
A pitot tube is used to measure fluid flow in engineering
Answer:
Δ enthalpy = -23 Btu/Ibm
Explanation:
Given data:
Pressure ( P1 ) = 250 psi
Initial Temperature ( T1 ) = 175°F
Final temperature ( T2 ) = 20°F
<u>Calculate the change in the enthalpy of R-134a </u>
From R-134 table
h1 = 129.85 Btu/Ibm
s1 = 0.23281 Btu/Ibm.R
note : entropy is constant
hence ; s1 = s2
by interpolation ; h2 = 106.95
Δ enthalpy = h2 - h1
= ( 106.95 - 129.85 ) = -23 Btu/Ibm