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3 years ago

Engineering Specificaitons is ...

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5 0

"Engineering specifications" means those records which provide detailed documentation of the construction, wiring, arrangement and related engineering details of the information processing equipment.

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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<em> to</em> the surroundings.

8 0

3 years ago

Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressur

Paha777 [63]

Answer:

Explanation:

Answer:

Explanation:

Answer:

Explanation:

This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.

(a). here we are asked to determine the Temperature and Pressure.

Given that the properties of Air;

ha = 230.02 KJ/Kg

Ta = 230 K

Pra = 0.5477

From the energy balance equation for a diffuser;

ha + Va²/2 = h₁ + V₁²/2

h₁ = ha + Va²/2 (where V₁²/2 = 0)

h₁ = 230.02 + 220²/2 ˣ 1/10³

h₁ = 254.22 KJ/Kg

⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg

from this we have;

Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]

Pr₁ = 0.77759

therefore T₁ = 254.15K

P₁ = (Pr₁/Pra)Pa

= 0.77759/0.5477 ˣ 26

P₁ = 36.91 kPa

now we calculate Pr₂

Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349

⇒ now we obtain properties of air at

Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg

calculating the enthalpy of air at state 2

ηc = h₁ - h₂ / h₁ - h₂

0.85 = 254.22 - 505.387 / 254.22 - h₂

h₂ = 549.71 KJ/Kg

to obtain the properties of air at h₂ = 549.71 KJ/Kg

T₂ = 545.15 K

⇒ to calculate the pressure of air at state 2

P₂/P₁ = 11

P₂ = 11 ˣ 36.913

p₂ = 406.043 kPa

but pressure of air at state 3 is the same,

i.e. P₂ = P₃ = 406.043 kPa

P₃ = 406.043 kPa

To obtain the properties of air at

T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5

for cases of turbojet engine,

we have that work output from turbine = work input to the compressor

Wt = Wr

(h₃ - h₄) = (h₂ - h₁)

h₄ = h₃ - h₂ + h₁

= 1515.42 - 549.71 + 254.22

h₄ = 1219.93 kJ/Kg

properties of air at h₄ = 1219.93 kJ/Kg

T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

T₄ = 1150.58 K

Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]

Pr₄ = 200.5636

Calculating the ideal enthalpy of the air at state 4;

Лr = h₃ - h₄ / h₃ - h₄*

0.9 = 1515.42 - 1219.93 / 1515.42 - h₄

h₄* = 1187.09 kJ/Kg

now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg

P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]

P₄* = 181.316

P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process

P₄ = 181.316/450.5 * 406.043

P₄ = 163.42 kPa

For the 4-5 process;

Pr₅ = (P₅/P₄)Pr₄

Pr₅ = 26/163.42 * 200.56 = 31.9095

to obtain the properties of air at Pr₅ = 31.9095

h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]

h₅ = 734.09 KJ/Kg

T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]

T₅ = 719.32 K

(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;

QH = m(h₃-h₂)

QH = 25(1515.42 - 549.71)

QH = 24142.75 kW

(c). To calculate the velocity at the nozzle exit;

we apply steady energy equation of a flow to nozzle

h₄ + V₄²/2 = h₅ + V₅²/2

h₄ + 0 = h₅₅ + V₅²/2

1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2

therefore, V₅ = 985.74 m/s

cheers i hope this helps

6 0

3 years ago

A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.

Alex73 [517]

Answer: you can watch a video on how to solve this question on you tube

6 0

3 years ago

What kinds of problems or projects would a civil engineer work on?

lisov135 [29]

Answer:

simple projects bovonhztisgx

8 0

3 years ago

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

$A=\frac{\pi d^{2}}{4}$

$A=\frac{\pi\times(10.2)^{2}}{4}$

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

$D=\frac{A_{i}-A_{f}}{A_{i}}\times100$

$D=\frac{81.713-52.7}{81.713}\times100$

D = 35.5%.

Thus, the percentage ductility is 35.5%.

5 0

3 years ago