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kaheart [24]
3 years ago
14

Waste cooking oil is to be stored for processing by pouring it into tank A, which is connected by a manometer to tank B. The man

ometer is completely filled with water. Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa. To what height h can waste oil be poured into tank A? If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height?

Engineering
1 answer:
likoan [24]3 years ago
7 0

KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.

==================================

Answer:

(1). 1.2 metres.

(2). There is going to be the same pressure.

Explanation:

From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;

" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."

=> Also, the density of oil = 930

That is if Pressure, P in B > 18kpa there will surely be a burst.

The height, h the can waste oil be poured into tank A is;

The maximum pressure  = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).

18 × 10^3 = (height, h ×  10 × 930) + 10 × (2 - 1.25) × 1000.

When we make height, h the Subject of the formula then;

Approximately, Height, h = 1.2 metres.

(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.

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Dovator [93]

Answer:

The amount of money saved is $22,075.2

Explanation:

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The percentage of the energy received by the solar panels converted into electricity, η = 15% = 0.15

The cost of electricity, C = $0.12/kWh

The number of days in a year, t = 365 days

Therefore, the maximum area covered by the solar panel = A = 500 m²

The total solar photovoltaic energy received by the solar panels, 'E', in a year is given as follows;

E = S_E × A × t

By plugging in the values of the variables, we get;

E = 6,720 Wh/m²/day × 500 m² × 365 days = 1,226,400,000 Wh

The electrical energy generated, E_e = η × E

∴ E_e = 0.15 × 1,226,400,000 Wh = 183960000 Wh = 183,960 kWh

The amount of money saved = The cost of electricity generated = C × E_e

C × E_e = $0.12/kWh × 183,960 kWh = $22,075.2

∴ The amount of money saved = C × E_e = $22,075.2

3 0
2 years ago
What is the federal E-Rate program?
Annette [7]

Answer:

c. A program that offers discounts to libraries and schools ensuring they have affordable access to modern telecommunications and information services.

Explanation:

Federal E-Rate program refers to the Schools and Libraries Program of the Universal Service Fund managed by the Universal Service Administrative Company (USAC) and being directed by the Federal Communications Commission (FCC).

The program offers telecommunications and internet access to schools and libraries in the United States at discounts of between 20% and 90% in order to make the services affordable to them.

The discounts received by each of he beneficiary schools receive which is between the rage of 20% and 90% is determined by the degree of poverty and the urban/rural status of the population or students being served.

In the program, connectivity and maintenance services are provided by the Schools and Libraries Program, while school that applied to the program has to provide other items like software, hardware (e.g. computers), and among other items that will make then to use the connectivity provided.

I wish you the best.

8 0
3 years ago
For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?
juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= 4\sqrt{2} cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as P = \dfrac{\pi ^2EI}{L_e^2}      

If area moment of inertia(I) is high then buckling load will be high.

We know that  area moment of inertia(I)

For circular tube I = \dfrac{\pi }{64}(D_o^4-D_i^4)

For circular bar I = \dfrac{\pi }{64}D^4  

Now by putting the values

    For circular tube I=62.83 cm^4

  For circular bar I=50.26 cm^4

So we can say that for same cross sectional area the  area moment of inertia(I) is high for tube as compare to bar.So buckling load  will be higher in tube as compare to bar.

3 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

m = 1.06Kg

T_H = 1.2T_L

T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

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= 285.4K

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6 0
3 years ago
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can
Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

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take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0
3 years ago
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