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o-na [289]
4 years ago
6

An engineer designing a hydroelectric facility recalls that the flow rate (in gallons per second) is four times the value of the

head (in feet) but has forgotten both numbers. If the facility is expected to produce 1.5 MW of power with an efficiency of 88%, determine both the flow rate (in gal/s) and the head (in ft).

Engineering
2 answers:
GarryVolchara [31]4 years ago
7 0

Answer:

Flow rate = 86.48 gal/s

Head of water = 21.62 ft

Explanation:

Detailed explanation and calculation is shown in the image below.

bekas [8.4K]4 years ago
5 0

Answer: flowrate=505.228

head=101.307ft

Explanation:P=power,m=flowrate,w=work,h=head,g=acceleration due to gravity

P=mw

W=gh

P=1.5×10^6 ×88%

=1320000Watts

g=9.8×3.281 ft/s

=32.154ft/s

m=4h

1320000=32.154×4h^2

4h^2=1320000/32.154

=41052.44

h^2=41052.44/4

=10263.109

h=√10263.109

=101.307ft

m=4h

m=4×101.307

m=405.228gal/s

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Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conduct
N76 [4]

Answer:

a. \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

<u>Assumption</u>

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

\alpha/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

\frac{d^2T}{dx^2}=0..............1

and \frac{-kdT(0)}{dx} =q_{0}=5000W/m^2

from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

-kC1=qo

C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

50000*(0.006-x)/60+112

833.3(0.006-x)+112

c. inner surface plate temperature is

T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)

117 deg C

6 0
3 years ago
A 50kg block of nickel at 90°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. Determine the fol
Valentin [98]

Answer:

a).Control volume

b). C_{nickel} = 502.416 J/kg-K

c).  C_{water} = 4.187 kJ/kg-K

d). T_{2} = 87.91°C

Explanation:

a). It is a control volume system because mass is varying in the system.

b). Specific heat of nickel is C_{nickel} =   502.416 J/kg-K

c). Specific heat of water is  C_{water} = 4.187 kJ/kg-K

d).We know that

   net  energy transfer = change in internal energy

m_{nickel}\times c_{nickel}(T_{2}-T_{nickel})=m_{water}\times c_{water}(T_{2}-T_{water})

m_{nickel}\times c_{nickel}(T_{2}-T_{nickel})=(volume_{water}\times density_{water})\times c_{water}(T_{2}-T_{water})

50\times 502.416\times (T_{2}-90)=(0.5\times 1000)\times 4.187\times (T_{2}-25)

25120.8\times (T_{2}-90)=2093.5\times  (T_{2}-25)

T_{2} = 87.91°C

6 0
3 years ago
Read 2 more answers
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Tom [10]

Answer:

Market Researcher

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Explanation:

5 0
3 years ago
If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
Brut [27]

Answer:

Part 1: It would be a straight line, current will be directly proportional to the voltage.

Part 2: The current would taper off and will have negligible increase after the voltage  reaches a certain  value. Graph attached.

Explanation:

For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.

V=I*R

where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.

In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.

7 0
4 years ago
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