Answer:
represnt
Explanation:
A utility generates electricity with a 36% efficient coal-fired power plant emitting the legal limit of 0.6 lb of SO2 per million Btus of heat into the plant. Suppose the utility encourages its customers to replace their 75-W incandescents with 18-W compact fluorescent lamps (CFLs) that produce the same amount of light. Over the 10,000-hr lifetime of a single CFL.
Required:
a. How many kilowatt-hours of electricity would be saved?
b. How many 2,000-lb tons of SO2 would not be emitted?
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?
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
[tex]P_{incandescent}[/tex] = 75 w × 10,000-hr = 750000 wh = 750 kWh
next, we also find the total power consumption by the fluorescent lamp
[tex]P_{fluorescent}[/tex] = 18 × 10000 = 180000 = 180 kWh
So the value of power saved will be;
[tex]P_{saved}[/tex] = [tex]P_{incandescent}[/tex] - [tex]P_{fluorescent}[/tex]
[tex]P_{saved}[/tex] = 750 - 180
[tex]P_{saved}[/tex] = 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.
What's the ampacity of a No. 10 type TW copper Wire in a raceway containing six wires and located in an area where the ambient Temperature is 50 C
A. 15.1 A
B. 8.6 A
C. 13.9 A
D. 10.8 A
Determine the convection heat transfer coefficient, thermal resistance for convection, and the convection heat transfer rate that are associated with air at atmospheric pressure in cross flow over a cylinder of diameter D = 100 mm and length L = 2 m. The cylinder temperature is Ts = ° 70 C while the air velocity and temperature are V = 3 m/s and T[infinity] = 20°C, respectively. Plot the convection heat transfer coefficient and the heat transfer rate from the cylinder over the range 0.05 m ≤ D ≤ 0.5 m.
Answer:
attached below
Explanation:
Attached below is a detailed solution to the question above
Step 1 : determine the Reynolds number using the characteristics of Air at 45°c
Step 2 : calculate the Nusselt's number
Step 3 : determine heat transfer coefficient
Step 4 : calculate heat transfer ratio and thermal resistance
Repeat steps 1 - 4 for each value of diameter from 0.05 to 0.5 m
attached below is a detailed solution
Many households in developing countries prepare food over indoor cook stoves with no ducting system to exhaust the combustion products outside. A woman in a household lights a fire in her cookstove at 4 pm. The fire emits benzo(a)pyrene, a carcinogenic compound, at a rate of 0.01 ng/min. The house has a size of 40 m3 and a ventilation rate of 1 m3/h. The ambient indoor air concentration of benzo(a)pyrene was 0.2 ng/m3 when she started the fire, but no significant concentration of benzo(a)pyrene is found in the outdoor air. What is the indoor benzo(a)pyrene concentration in the household at 6 pm?
Answer:
C = 0.22857 ng / m³
Explanation:
Let's solve this problem for part the total time in the kitchen is
t = 2h (60 min / 1h) = 120 min
The concentration (C) quantity of benzol pyrene is the initial quantity plus the quantity generated per area minus the quantity eliminated by the air flow. The amount removed can be calculated assuming that an amount of extra air that must be filled with the pollutant
amount generated
C = co + time_generation rate / (area_house + area_flow)
C = 0.2 + 0.01 120 / (40+ 2)
C = 0.22857 ng / m³
Choose two other elements from the periodic table that you predict should react to form something like table salt
Please pleassssss helppp
I give branlistttttt
A cylindrical specimen of some metal alloy 10 mm in diameter and 150 mm long has a modulus of elasticity of 100 GPa. Does it seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Answer:
N0
Explanation:
It does not seem reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen diameter of 0.08 mm
Given data :
Diameter ( d ) = 10 mm
length ( l ) = 150 mm
elasticity ( ∈ ) = 100 GPa
longitudinal strain ( б ) 200 MPa
Poisson ratio ( μ ) ( assumed ) =0.3
Assumption : deformation totally elastic
attached below is the detailed solution to why it is not reasonable .
The Sd value = 0.08 > the calculated Sd value ( 6*10^-3 ) hence it is not reasonable to expect a tensile stress of 200 MPa to produce a reduction in specimen
A water jet pump involves a jet cross-sectional area of 0.01 m^2, and a jet velocity of 30 m/s. The jet is surrounded by entrained water. The total cross-sectional area associated with the jet and entrained streams is 0.075 m^2. These two fluid streams leave the pump thoroughly mixed with an average velocity of 6 m/s through a cross-sectional area of 0.075 m^2. Determine the pumping rate (i.e., the entrained fluid flowrate) involved in liters/s.
Answer:
the entrained fluid flowrate is 150 liters/s
Explanation:
Given the data in the question;
we determine the flow rate of water though the jet by using the following expression;
Q₂ = A₂ × V₂
where Q₂ is the flow rate of water though the jet, A₂ is the cross sectional area of the jet( 0.01 m² ) and V₂ is the jet velocity( 30 m/s )
so we substitute
Q₂ = 0.01 m² × 30 m/s
Q₂ = 0.3 m³/s
Next we determine the flow rate of water through the pump by using the following expression
Q₃ = A₃ × V₃
where Q₃ is the flow rate of water though the pump, A₃ is the cross sectional area of the pump( 0.075 m² ) and V₃ is the average velocity of mixing( 6 m/s )
so we substitute
Q₃ = 0.075 m² × 6 m/s
Q₃ = 0.45 m³/s
so to calculate the flow pumping rate of water into the water jet pump, we use the expression;
Q₁ + Q₂ = Q₃
we substitute
Q₁ + 0.3 m³/s = 0.45 m³/s
Q₁ = 0.45 m³/s - 0.3 m³/s
Q₁ = 0.15 m³/s
we know that 1 m³/s = 1000 Liter/second
so
Q₁ = 0.15 × 1000 Liter/seconds
Q₁ = 150 liters/s
Therefore, the entrained fluid flowrate is 150 liters/s
The output of a first order transducer is to be connected to a signal conditioner which also has first order dynamic characteristics. The transducer has known time constant of x milliseconds and static sensitivity of xx V/kPa while the signal conditioner has a time constant of y milliseconds and a static sensitivity of yy V/V. Determine the steady state response of this measurement system to an input signal of the form: P(t)=(A+B*sin(w*t) ) kPa.
Answer:
hello your question is poorly written attached below is the correct question
answer : y(t) = 1.5 + 0.025 sin( 600t - 0.259 ) v
Explanation:
Given data:
Time constant = 5msec
static sensitivity = 0.05 v/°c
f = 100 hz
ε = 0.8
T(t) = ( 30 + 2.5 sin600t )°C
attached below is a detailed solution to the question above
(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows the pouring cup and flows into the downsprue. The cross section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 20 cm long, determine the proper diameter at its base so as to main- tain the same volume flow rate.
Answer:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top [tex]d_{top}[/tex] = 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
[tex]V_{base}[/tex] = √2gh
we substitute
[tex]V_{base}[/tex] = √(2 × 9.81 m/s² × 0.2 m )
[tex]V_{base}[/tex] = 1.98091 m/s
[tex]V_{base}[/tex] = 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π[tex]d_{bottom}^2[/tex]/4) × [tex]V_{base}[/tex]
[tex]d_{bottom}[/tex] = √(4Q/π[tex]V_{base}[/tex])
we substitute our values into the equation;
[tex]d_{bottom}[/tex] = √(4(400 cm³/s) / (π×198.091 cm/s))
[tex]d_{bottom}[/tex] = 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm