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Maksim231197 [3]
3 years ago
8

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

at the top of the tank. Determine the pressure at a closed valve attached to the tank 10 ft above its bottom.
Engineering
1 answer:
Yuliya22 [10]3 years ago
5 0

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

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A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh
sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C

to kelvin = (5 + 273)K = 278 K

From  the " thermophysical properties of gases at atmospheric pressure" table; At T_f = 278 K ; by interpolation; we have the following

\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}  → v 13.93 (10⁻⁶) m²/s

\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})} → k = 0.0245 W/m.K

\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})} → ∝ = 19.6(10⁻⁶)m²/s

\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720} → Pr = 0.713

\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}

The Rayleigh number for vertical cavity

Ra_L  = \frac{g \beta (T_1-T_2)L^3}{\alpha v}

= \frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}

= 8.38*10^5

\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}

NU_L= \frac{hL}{k}= 4.878

\frac{hL}{k}= 4.878

\frac{h*0.06}{0.0245}= 4.878

h = \frac{4.878*0.0245}{0.06}

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0
3 years ago
Which type of engineering design uses an already existing design
Snezhnost [94]

Answer:

Reverse engineering

Explanation:

Reverse Engineering is the remaking of already made products following the deconstruction and examination of the product to make known the product design, code and architecture features, gain knowledge of the composition and construction in a scientific research approach

Reverse engineering is also known as back engineering and consists of three main stages

1) Recovery implementation

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3 0
2 years ago
Air enters a compressor operating at steady state at 1 atm with a specific enthalpy of
Savatey [412]

Answer:

option C

option A

Explanation:

Enthalpy gained by air= 1023-290

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Rate of energy gain= mass flow rate × Enthalpy gained by air

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                            = 73.3 kJ/s

rate of heat transfer between compressor and air= 77kW

Heat loss by air to surroundings= 77-73.3

                                                     =3.7kW

Enthalpy lost by steam in turbine= 1407.6-1236.4

                                                     = 171.2 Btu/lb

Rate of energy transfer to turbine= Enthalpy lost by steam× mass flow rate

                                                    = 171.2×5

                                                     = 856 Btu/s

Net rate of energy transfer to turbine=rate of  Energy transfer to turbine- rate of heat transfer to turbine

                          = 856-40

                         = 816 Btu/s

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The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water i
olganol [36]

Answer:

m_added = 2 kg

Explanation:

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Now, let the mass of the falling cylinder be denoted by "m1" and let h be the height of fall.

Thus;

Formula for potential energy = mgh

Thus, as said earlier it's converted to heat generated. So heat generated = m1gh

Now let's calculate the heat absorbed;

heat absorbed = (m2)cΔt

Where;

ΔT is change in temperature

c is specific heat of water .

m2 is mass of water

Heat absorbed = heat generated

Thus;

(m2)cΔt = m1gh

Δt = m1gh/(m2•c)

Now, in both cases of the water and cylinder, m1, g , h and c are constant

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Δt = (m1gh/m2) × 1/c

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(m1gh/m2) is denoted as a constant k.

Thus;

Δt = K/m

For the first experiment, we have;

m = 2 kg

Δt = 2.4

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2.4 = K/2

Multiply both sides by 2 to get;

K = 4.8

For the second experiment, we have;

Δt = 1.2

Also,we have seen that K = 4.8

Thus;

Δt = K/m

Thus;

1.2 = 4.8/m

m = 1.2

m = 4 kg

Thus,mass added is;

m_added = 4 - 2

m_added = 2 kg

6 0
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A fuel oil is burned with air in a furnace. The combustion produces 813 kW of thermal energy, of which 65% is transferred as hea
DENIUS [597]

Answer:

im not sure

Explanation:

8 0
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