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2 years ago

Given in the following v(t) signal.

Engineering

1 answer:

tatuchka [14]2 years ago

3 0

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

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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

bazaltina [42]

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 to the surroundings.

8 0

2 years ago

The drag force, Fd, imposed by the surrounding air on a

Ad libitum [116K]

Answer:

a) 23.551 hp

b) 516.89 hp

Explanation:

given:

$F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2$

required:

the power in hp

solution:

$(F_{d})_{a} =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}$ .............(1)

by substituting in the equation (1)

=353.27 lbf

$(F_{d})_{b} =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}$ ..........(2)

by substituting in the equation (2)

= 2769.29 lbf

power is defined by

P=F.V

$P_{a}=$ 353.27*36.67

=12954.411 lbf.ft/s

=12954.411*.001818

=23.551 hp

$P_{a}=$ 2769.29*102.67

= 284323 lbf.ft/s

= 284323*.001818

= 516.89 hp

3 0

2 years ago

Two rods, with masses MA and MB having a coefficient of restitution, e, move along a common line on a surface, figure 2. a) Find

ahrayia [7]

Answer:

A.) Find the answer in the explanation

B.) Ua = 7.33 m/s , Vb = 7.73 m/s

C.) Impulse = 17.6 Ns

D.) 49%

Explanation:

Let Ua = initial velocity of the rod A

Ub = initial velocity of the rod B

Va = final velocity of the rod A

Vb = final velocity of the rod B

Ma = mass of rod A

Mb = mass of rod B

Given that

Ma = 2kg

Mb = 1kg

Ub = 3 m/s

Va = 0

e = restitution coefficient = 0.65

The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.

Please find the attached files for the solution

6 0

2 years ago

What are the four categories of engineering materials used in manufacturing?

alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They have low weight and are poor conductors of electricity and heat

8 0

2 years ago

A gasoline engine takes in air at 290 K, 90 kPa and then compresses it. The combustion adds 1000 kJ/kg to the air after which th

Inessa [10]

Answer:

attached below

Explanation:

4 0

2 years ago