An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this f
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Answer:
Total saving would be of 36.917 $\yr
Explanation:
Given Data:
Degree F
Degree F
A = 30 ft^2
%
Heat loss before previous final value
= 360 Btu/hr
Heat loss after new value
saving would be
= 616.1782 kw hr/yr
$
= 36.917 $\yr
Answer:
Fa = 57.32 N
Explanation:
given data
mass = 5 kg
acceleration = 4 m/s²
angular velocity ω = 2 rad/s
solution
first we take here moment about point A that is
∑Ma = Iα + ∑Mad ...............1
put here value and we get
so here I = ( ) × m × L² ................2
I = ( ) × 5 × 0.8²
I = 0.267 kg-m²
and
a is = r × α
a = 0.4 α
so now put here value in equation is 1
0 = 0.267 α + m r α (0.4) - m A (0.4)
0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4
so angular acceleration α = 7.5 rad/s²
so here force acting on x axis will be
∑ F(x) = m a(x) ..............3
a(x) = m a - m rα
put here value
a(x) = 5 × 4 - 5 × 0.4 × 7.5
a(x) = 5 N
and
force acting on y axis will be
∑ F(y) = m a(y) .............. 4
a(y) - mg = mrω²
a(y) - 5 × 9.81 = 5 × 0.4 × 2²
a(y) = 57.1 N
so
total force at A will be
Fa = ...............5
Fa =
Fa = 57.32 N
