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Oksanka [162]
3 years ago
5

The component of a fluid system where a fluid is stored, but not under pressure, is called a container.

Engineering
1 answer:
forsale [732]3 years ago
6 0

Answer:

The answer is true

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Explanation:

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A 2-m3insulated rigid tank contains 3.2 kg of carbon dioxide at 120 kPa.Paddle-wheel work is done on the system until the pressu
AleksandrR [38]

Answer:

The change in entropy is found to be 0.85244 KJ/k

Explanation:

In order to solve this question, we first need to find the ration of temperature for both state 1 and state 2. For that, we can use Charles' law. Because the volume of the tank is constant.

P1/T1 = P2/T2

T2/T1 = P2/P1

T2/T1 = 180 KPa/120KPa

T2/T1 = 1.5

Now, the change in entropy is given as:

ΔS = m(s2 - s1)

where,

s2 = Cv ln(T2/T1)

s1 = R ln(V2/V1)

ΔS = change in entropy

m = mass of CO2 = 3.2 kg

Therefore,

ΔS = m[Cv ln(T2/T1) - R ln(V2/V1)]

Since, V1 = V2, therefore,

ΔS = mCv ln(T2/T1)

Cv at 300 k for carbondioxide is 0.657 KJ/Kg.K

Therefore,

ΔS = (3.2 kg)(0.657 KJ/kg.k) ln(1.5)

<u>ΔS = 0.85244 KJ/k</u>

3 0
3 years ago
Which one of the following questions about population growth is the only TRUE statement?A) The size of a population can never ex
tatyana61 [14]

Answer:

Explanation:

5

6 0
3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

5 0
3 years ago
Lab scale tests performed on a cell broth with a viscosity of 5cP gave a specific cake resistance of 1 x1011 cm/g and a negligib
insens350 [35]

Answer:

5.118 m^3/hr

Explanation:

Given data:

viscosity of cell broth = 5cP

cake resistance = 1*1011 cm/g

dry basis per volume of filtrate = 20 g/liter

Diameter = 8m ,  Length = 12m

vacuum pressure = 80 kpa

cake formation time = 20 s

cycle time = 60 s

<u>Determine the filtration rate in volumes/hr  expected fir the rotary vacuum filter</u>

attached below is a detailed solution of the question

Hence The filtration rate in volumes/hr expected for the rotary vacuum filter

V' = ( \frac{60}{20} ) * 1706.0670

   = 5118.201 liters  ≈ 5.118 m^3/hr

4 0
2 years ago
In wet mill ethanol plants, a total energy of 74,488 Btu (British thermal units, a common energy unit) is used to produce 1 gall
V125BC [204]

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield (\% e), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy (E_{g}), measured in Btu, and the energy needed to produce the ethanol (E_{p}), measured in Btu, divided by the energy needed to produce the ethanol. That is:

\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\% (1)

If we know that E_{g} = 76330\,Btu and E_{p} = 74488\,Btu, then the net energy yield of 1 gallon of ethanol:

\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%

\%e = 2.473\,\%

The energy yield for one gallon of ethanol is 2.473 %.

4 0
3 years ago
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