What financial arguments could you use to justify your proposed
1 answer:
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Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 × ..............2
P2 = 95 ×
P2 = 3.9 MPa
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:
R = 148.346 N
M₀ = - 237.2792 N-m
Explanation:
Point O is selected as a convenient reference point for the force-couple system which is to represent the given system
We can apply
∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N
∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N
Then
R = √(Rx²+Ry²) ⇒ R = √((66.8556 N)²+(132.4264 N)²)
⇒ R = 148.346 N
Now, we obtain the moment about the origin as follows
M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)
We can see the pic shown in order to understand the question.
