Answer:
work=281.4KJ/kg
Power=4Kw
Explanation:
Hi!
To solve follow the steps below!
1. Find the density of the air at the entrance using the equation for ideal gases
where
P=pressure=120kPa
T=20C=293k
R= 0.287 kJ/(kg*K)=
gas constant ideal for air
2.find the mass flow by finding the product between the flow rate and the density
m=(density)(flow rate)
flow rate=10L/s=0.01m^3/s
m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s
3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow
Work
w=Cp(T1-T2)
Where
Cp= specific heat for air=1.005KJ/kgK
w=work
T1=inlet temperature=20C
T2=outlet temperature=300C
w=1.005(300-20)=281.4KJ/kg
Power
W=mw
W=(0.0143)(281.4KJ/kg)=4Kw
Answer:
<em>The normal stress is 0.7821 MPa</em>
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Explanation:
The external diameter D = 160 mm
The thickness t = 3.8 mm = 3.8 x 10^-3 m
gauge pressure P = 78 kPa = 78 x 10^3 Pa
The maximum shear stress τmax = ?
The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm
The internal radius of the shell r = R - t
==> 80 - 3.8 = 76.2 mm
Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm
==> d = 152.4 x 10^-3 m
The normal stress σ = = = 782052.63 Pa
==> σ = <em>0.7821 MPa</em>
Answer:
a) 14.285
b) 8.956
Explanation:
Given :
The combustion of the ethanol is with air
Air is 21% oxygen
and, 79% nitrogen
thus, for 1 O₂ we have (79/21)N₂
thus,
the stochiometric equation for the combustion is as:
C₂H₅OH + 3[O₂ + (79/21)N₂] ⇒ 2CO₂ + 3H₂O + 3 × (79/21)N₂
Now,
the molecular weight of the fuel (C₂H₅OH) = (2× 12) + (5 × 1) + 16 + 1 = 46 g/mol
Molecular weight of the air = (2 × 16) + ((79/21) × 28) = 137.33 g/mol
a) air/fuel ratio on a molar basis
we have
air-fuel ratio = moles of air / moles of fuel
or
air-fuel ratio = [3 × 1 + 3 × (79/21)] / 1 = 14.285
b) air/fuel ratio on a mass basis = Mass of air / mass of fuel
or
air/fuel ratio on a mass basis = (number of moles of air × molar mass of air) / (number of moles of fuel × molar mass of fuel)
on substituting the values, we have
air/fuel ratio on a mass basis = (3 × 137.33) / (1 × 46) = 8.956
Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test