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3 years ago

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An insulated rigid tank is initially evacuated. A valve is opened, and atmospheric air at 95 kPa and 17 ºC enters the tank until

the pressure in the tank reaches 95 kPa, at which point the valve is closed. Determine the final temperature of the air in the tank. Assume constant specific heats.

1 answer:

Darya [45]3 years ago

5 0

Answer:

$T2= 406k$

Explanation:

The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.

Given:

T1=170C

To convert to Kelvin

= 17+273 =290K

T1 = 290K

Pressure (P)= 95KPa

Specific heat ratio = CP/CV= K

WhereK=1.005/0.718

K = 1.4

The final temperature can be calculated using the formula below.

T2 = CP/CV × T1

=. K × T1

T2 = 1.4 × 290

$T2= 406k$

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Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at

a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

8 0

3 years ago

A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac

antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a ) ------ ( 1 )

From the above data given will be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1 becomes

Py = -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

= 3270 N/m^2

6 0

3 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

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2 years ago

A gas stream flowing at 1000 cfm with a particulate loading of 400 gr/ft3 discharges from a certain industrial plant through an

Makovka662 [10]

<u>Solution and Explanation:</u>

Volume of gas stream = 1000 cfm (Cubic Feet per Minute)

Particulate loading = 400 gr/ft3 (Grain/cubic feet)

1 gr/ft3 = 0.00220462 lb/ft3

Total weight of particulate matter = $1000 \mathrm{cfm} \times 400 \mathrm{gr} / \mathrm{tt} 3 \times .000142857 \mathrm{lb} / \mathrm{ft} 3 \times 60=3428.568 \mathrm{lb} / \mathrm{hr}$

Cyclone is to 80 % efficient

So particulate remaining = $0.20 \times 3428.568 \mathrm{lb} / \mathrm{hr}=685.7136$

emissions from this stack be limited to = 10.0 lb/hr

Particles to be remaining after wet scrubber = 10.0 lb/hr

So particles to be removed = 685.7136- 10 = 675.7136

Efficiency = output multiply with 100/input = 98.542 %

4 0

3 years ago

2. A counter flow tube-shell heat exchanger is used to heat a cold water stream from 18 to 78oC at a flow rate of 1 kg/s. Heatin

Anastaziya [24]

Answer:

a) $L = 220\,m$ , b) $U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

Explanation:

a) The counterflow heat exchanger is presented in the attachment. Given that cold water is an uncompressible fluid, specific heat does not vary significantly with changes on temperature. Let assume that cold water has the following specific heat:

$c_{p,c} = 4.186\,\frac{kJ}{kg\cdot ^{\textdegree}C}$

The effectiveness of the counterflow heat exchanger as a function of the capacity ratio and NTU is:

$\epsilon = \frac{1-e^{-NTU\cdot(1-c)}}{1-c\cdot e^{-NTU\cdot (1-c)}}$

The capacity ratio is:

$c = \frac{C_{min}}{C_{max}}$

$c = \frac{(1\,\frac{kg}{s} )\cdot(4.186\,\frac{kW}{kg^{\textdegree}C} )}{(1.8\,\frac{kg}{s} )\cdot(4.30\,\frac{kW}{kg^{\textdegree}C} )}$

$c = 0.541$

Heat exchangers with NTU greater than 3 have enormous heat transfer surfaces and are not justified economically. Let consider that $NTU = 2.5$ . The efectiveness of the heat exchanger is:

$\epsilon = \frac{1-e^{-(2.5)\cdot(1-0.541)}}{1-(2.5)\cdot e^{-(2.5)\cdot (1-0.541)}}$

$\epsilon \approx 0.824$

The real heat transfer rate is:

$\dot Q = \epsilon \cdot \dot Q_{max}$

$\dot Q = \epsilon \cdot C_{min}\cdot (T_{h,in}-T_{c,in})$

$\dot Q = (0.824)\cdot (4.186\,\frac{kW}{^{\textdegree}C} )\cdot (160^{\textdegree}C-18^{\textdegree}C)$

$\dot Q = 489.795\,kW$

The exit temperature of the hot fluid is:

$\dot Q = \dot m_{h}\cdot c_{p,h}\cdot (T_{h,in}-T_{h,out})$

$T_{h,out} = T_{h,in} - \frac{\dot Q}{\dot m_{h}\cdot c_{p,h}}$

$T_{h,out} = 160^{\textdegree}C + \frac{489.795\,kW}{(7.74\,\frac{kW}{^{\textdegree}C} )}$

$T_{h,out} = 96.719^{\textdegree}C$

The log mean temperature difference is determined herein:

$\Delta T_{lm} = \frac{(T_{h,in}-T_{c, out})-(T_{h,out}-T_{c,in})}{\ln\frac{T_{h,in}-T_{c, out}}{T_{h,out}-T_{c,in}} }$

$\Delta T_{lm} = \frac{(160^{\textdegree}C-78^{\textdegree}C)-(96.719^{\textdegree}C-18^{\textdegree}C)}{\ln\frac{160^{\textdegree}C-78^{\textdegree}C}{96.719^{\textdegree}C-18^{\textdegree}C} }$

$\Delta T_{lm} \approx 80.348^{\textdegree}C$

The heat transfer surface area is:

$A_{i} = \frac{\dot Q}{U_{i}\cdot \Delta T_{lm}}$

$A_{i} = \frac{489.795\,kW}{(0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C} )\cdot(80.348^{\textdegree}C) }$

$A_{i} = 9.676\,m^{2}$

Length of a single pass counter flow heat exchanger is:

$L =\frac{A_{i}}{\pi\cdot D_{i}}$

$L = \frac{9.676\,m^{2}}{\pi\cdot (0.014\,m)}$

$L = 220\,m$

b) Given that tube wall is very thin, inner and outer heat transfer areas are similar and, consequently, the cold side heat transfer coefficient is approximately equal to the hot side heat transfer coefficient.

$U_{o} \approx 0.63\,\frac{kW}{m^{2}\cdot ^{\textdegree}C}$

5 0

3 years ago