Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Technician A says that a seal can be pried out of a bore using a sharp chisel. Technician B says that smaller metal-backed seals
2 answers:
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To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.
By definition heat exchange in terms of mass flow can be expressed as
Where
Specific heat
= Mass flow rate
= Change in Temperature
Our values are given as
Specific heat of air
From our equation we have that
Rearrange to find
Replacing
Therefore the exit temperature of air is 53.98°C
Answer:
1. = 7.161458
in.⁴
= 36.661458
in.⁴
Iₓ = 28.6458 in.⁴
= 138.6548
in.⁴
2. = 114.
in.⁴
= 37.
in.⁴
Iₓ = 457. in.⁴
= 149.
in.⁴
3. The maximum deflection of the beam is 2.55552 inches
Explanation:
1. The height of the beam having a rectangular cross section is h = 2.5 in.
The breadth of the beam, is = 5.5 in.
The moment of inertia of a rectangular beam through its centroid is given as follows;
= b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458
= 7.161458
in.⁴
= h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458
= 36.661458
in.⁴
The moment of inertia about the base is given as follows;
Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458
Iₓ = 28.6458 in.⁴
= h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548
= 138.6548
in.⁴
2. The height of the beam having a rectangular cross section is h = 7 in.
The breadth of the beam, b = 4 in.
The moment of inertia of a rectangular beam through its centroid is given as follows;
= b·h³/12 = 4 × 7³/12 = 114.
= 114.
in.⁴
= h·b³/12 = 7 × 4³/12 = 37.
= 37.
in.⁴
The moment of inertia about the base is given as follows;
Iₓ = b·h³/3 = 4 × 7³/3 = 457.
Iₓ = 457. in.⁴
= h·b³/3 = 2.5 × 5.5³/3 = 149.
= 149.
in.⁴
3. The deflection, , of a simply supported beam having a point load at the center is given as follows;
The given parameters of the beam are;
The length of the beam, L = 22 ft. = 264 in.
The applied load at the center, W = 750 lbs
The modulus of elasticity for Cedar = 10,000,000 psi
The height of the wood, h = 3 in.
The breadth of the wood, b = 5 in.
The moment of inertia of the wood, = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴
By plugging in the given values, we have;
The maximum deflection of the beam, = 2.55552 inches
Answer:
a) 764.45K
b) 210.48 kJ/kg
c) 30.14%
Explanation:
pressure ratio = 10
minimum temperature = 295 k
maximum temperature = 1240 k
isentropic efficiency for compressor = 83%
Isentropic efficiency for turbine = 87%
<u>a) Air temperature at turbine exit </u>
we can achieve this by interpolating for enthalpy
h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air
T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) = 764.45K
<u>b) The net work output </u>
first we determine the actual work input to compressor
Wc = h2 - h1 ( calculated values )
= 626.57 - 295.17 = 331.4 kJ/kg
next determine the actual work done by Turbine
Wt = h3 - h4 ( calculated values )
= 1324.93 - 783.05 = 541.88 kJ/kg
finally determine the network output of the cycle
Wnet = Wt - Wc
= 541.88 - 331.4 = 210.48 kJ/kg
<u>c) determine thermal efficiency </u>
лth = Wnet / qin ------ ( 1 )
where ; qin = h3 - h2
<em>equation 1 becomes </em>
лth = Wnet / ( h3 - h2 )
= 210.48 / ( 1324.93 - 626.57 )
= 0.3014 = 30.14%
