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Crank
3 years ago
15

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

ir at 6 bar, 295 K. The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of 350 K. Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in kJ
Engineering
1 answer:
bazaltina [42]3 years ago
8 0

Answer:

Q_{cv}=-339.347kJ

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

P_iV_i=m_iRT_i (i could be 1 for initial and 2 for the end)

State1

1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295

m_1=232kg

State2

8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350

m_2=11.946

So, the total mass of the aire entered is

m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

u_1=210.49kJ/kg

For Specific enthalpy

h_1=295.17kJ/kg

For the second state the Specific internal Energy (6bar, 350K)

u_2=250.02kJ/kg

At the end we make a Energy balance, so

U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e

No work done there is here, so clearing the equation for Q

Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)

Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)

Q_{cv}=-339.347kJ

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

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Answer:

Its B).  The efficiency rating of the dishwasher was significantly improved, meaning it would save him money and be more eco-friendly.

Explanation:

On EDG

8 0
2 years ago
Read 2 more answers
A coal-burning power plant generates electrical power at a rate of 650 megawatts (MW), or 6.50 × 108 J/s. The plant has an overa
Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

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          Energy produce in one year =20.49 x 10¹⁶ J/year

7 0
3 years ago
A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f
KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) \frac{dm}{2}     ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = \frac{L}{\pi dm}      ...............2

here lead L = n × p

so tan(α) = \frac{2\times 2}{\pi 19}

α = 3.83°  

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and  now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) \frac{19\times 10^{-3}}{2}  

T = 1.59 Nm

so

power = \frac{2\pi N \ T }{60}     .................3

put here value

power = \frac{2\pi \times 300\times 1.59}{60}

power = 49.95 W

and

as φ > α

so it is self locking screw

 

8 0
3 years ago
Subject: Electronics
Maslowich

Answer:

U just believe in yourself ..........

Explanation:

<em>If </em><em>there </em><em>a</em><em>r</em><em>e </em><em>more </em><em>electrons </em><em>than </em><em>protons </em><em>in </em><em>a </em><em>piece </em><em>of </em><em>matter </em><em>it </em><em>will </em><em>have </em><em>a </em><em>negative</em><em> </em><em>charge </em><em>.</em><em> </em><em>i</em><em>f</em><em> </em><em>there </em><em>are </em><em>fever </em><em>it </em><em>will </em><em>have </em><em>positive</em><em> </em><em>charge </em><em>and </em><em>if </em><em>there </em><em>are </em><em>e</em><em>qual </em><em>numbers </em><em>it </em><em>will </em><em>have </em><em>neutral</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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hope it was helpful to you.....

6 0
2 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
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