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.
Answers
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
Related Questions
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.
Answers
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
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
Answers
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 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?
Answers
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.