Ask question

Ask question

All categories

3 years ago

14

2- A 2-m3 insulated tank containing ammonia at -20 C, 80% quality, is connected by a valve to a line flowing ammonia at 2 MPa, 6

0 C. The valve is opened, allowing ammonia to flow into the tank. At what pressure should the vale be closed if the manufacturer wishes to have 15 kg of ammonia inside at the final state?

1 answer:

alekssr [168]3 years ago

8 0

Answer:

The valve should be closed at 1081 kPa ⇒ 11 bar

Explanation:

$Q_{C_V} + m_ih_i = m_2u_2 - m_1u_1 + W_{C_V}$

$Q_{C_V} = W_{C_V} = 0$

$m_1 = \frac{V}{v_1} = \frac{2}{0.49927} = 4.006 kg$

$m_i = m_2 - m_1 = 15 - 4.006 = 10.994 kg$

$u_1 = 1057.5, h_i = 1509.9$

$u_2 = m_ih_i + m_1u_1$

$m_2 = \frac{[10.994*1509.9] + [4.006*1057.5]}{15} = 1389.1 kJ/kg$

$v_2 = \frac{V}{m_2} = \frac{2}{15} = 0.1333 m^3/kg$

Therefore, v₂, u₂ fix state 2.

By trial and error, P₂ = 1081 kPa & T₂ = 50.4° C

1081 kPa ⇒ 11 bar

You might be interested in

Which of the following is not a primary or fundamental dimension? (a)-mass m (b)-length L (c)- timer t (d)-volume V

jolli1 [7]

Answer:

Volume is not the fundamental dimension. So, option d is correct.

Explanation:

Step1

Fundamental dimension is the dimension in which other quantities depend. These are the basic dimensions. Three basic fundaments dimensions are mass, length and time that is represented as MLT respectively.

Step2

Volume is not the fundamental dimension among them as the volume is cube of length dimension. Thus, volume depends upon length.

Thus, Volume is not the fundamental dimension. So, option d is correct.

8 0

3 years ago

As you have learned, not all motor oils are the same. What are two things that make them different?.

saul85 [17]

Answer:

quality ingredients and exceeding industry standards

Explanation:

5 0

2 years ago

Technician A says that the most commonly used combustion chamber types include hemispherical, and wedge. Technician B says that

Inessa05 [86]

Answer:

Technician A and Technician B both are correct.

Explanation:

Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.

Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.

6 0

3 years ago

A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul

Ghella [55]

Answer:

a) 1.16 m/s

b) 1/216000

c) (√15)/6480000

Explanation:

The parameters given are;

Length of boat prototype, lp = 30 m

Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

(vm/vp) = ∝^(1/2) = √∝ = (1/60)^(1/2)

vm = 9 × (1/60)^(1/2) = 1.16 m/s

b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

Fm/Fp = (1/60)³ = 1/216000

c) The ratio of the model to prototype power pm/p $_p$ = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power pm/p $_p$ = √(1/60) × (1/60)³

pm/p $_p$ = √(1/60) × (1/60)³ = (√15)/6480000

6 0

3 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago