Answer: heat loss through wall is 16.58034kW
Temperature of inside wall surface is 47°c
Temperature of outside wall surface is -2.7°c
Explanation:detailed calculation and explanation is shown in the image below.
A city emergency management agency and a construction company have formed a public-private partnership. The construction company
1 answer:
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Answer:
a) 570 kWh of electricity will be saved
b) the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) $1.296 can be earned by selling the SO₂ saved by a single CFL
Explanation:
Given the data in the question;
a) How many kilowatt-hours of electricity would be saved?
first, we determine the total power consumption by the incandescent lamp
= 75 w × 10,000-hr = 750000 wh = 750 kWh
next, we also find the total power consumption by the fluorescent lamp
= 18 × 10000 = 180000 = 180 kWh
So the value of power saved will be;
=
-
= 750 - 180
= 570 kWh
Therefore, 570 kWh of electricity will be saved.
now lets find the heat of electricity saved in Bituminous
heat saved = energy saved per CLF / efficiency of plant
given that; the utility has 36% efficiency
we substitute
heat saved = 570 kWh/CLF / 36%
we know that; 1 kilowatt (kWh) = 3,412 btu per hour (btu/h)
so
heat saved = 570 kWh/CLF / 0.36 × (3412 Btu / kW-hr (
heat saved = 5.4 × 10⁶ Btu/CLF
i.e eat of electricity saved per CLF is 5.4 × 10⁶
b) How many 2,000-lb tons of SO₂ would not be emitted
2000 lb/tons = 5.4 × 10⁶ Btu/CLF
0.6 lb SO₂ / million Btu = x
so
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ / million Btu )] / 2000 lb/tons
x = [( 5.4 × 10⁶ Btu/CLF ) × ( 0.6 lb SO₂ )] / [ ( 10⁶) × ( 2000 lb/ton) ]
x = 3.24 × 10⁶ / 2 × 10⁹
x = 0.00162 ton/CLF
Therefore, the amount of SO₂ not be emitted or heat of electricity saved is 0.00162 ton/CLF
c) If the utility can sell its rights to emit SO2 at $800 per ton, how much money could the utility earn by selling the SO2 saved by a single CFL?
Amount = ( SO₂ saved per CLF ) × ( rate per CFL )
we substitute
Amount = 0.00162 ton/CLF × $800
= $1.296
Therefore; $1.296 can be earned by selling the SO₂ saved by a single CFL.
Answer:
The required size of column is length = 15 ft and diameter = 4.04 inches
Explanation:
Given;
Length of the column, L = 15 ft
Applied load, P = 10 kips = 10 × 10³ Psi
End condition as fixed at the base and free at the top
thus,
Effective length of the column, = 2L = 30 ft = 360 inches
now, for aluminium
Elastic modulus, E = 1.0 × 10⁷ Psi
Now, from the Euler's critical load, we have
where, I is the moment of inertia
on substituting the respective values, we get
or
I = 13.13 in⁴
also for circular cross-section
I =
thus,
13.13 =
or
d = 4.04 inches
The required size of column is length = 15 ft and diameter = 4.04 inches