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laila [671]
2 years ago
8

11. What are restrictions when building or completing a challenge?

Engineering
1 answer:
katrin [286]2 years ago
3 0

Explanation:

The minimum exterior open spaces around buildings that are 55 metres or more, should be 16 metres. On sides where no habitable rooms face, a minimum space of 9 metres shall be left for heights above 27 metres.

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Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
Using Java..
uysha [10]

Answer:

The source code files for this question have been attached to this response.

Please download it and go through each of the class files.

The codes contain explanatory comments explaining important segments of the codes, kindly go through these comments.

Download java
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
7 0
3 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0
3 years ago
An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in
Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, T_{a} = 140^{\circ}C = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, T_{i} = 60^{\circ}C = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)

Thus

T_{e} = 109.71^{\circ}C = 273 + 109.71 = 382.71 K

where

T_{e} = Equilibrium temperature

Now,

To calculate the changer in entropy:

\Delta s = \Delta s_{a} + \Delta s_{i}

Now,

For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K

For Iron:

\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

6 0
3 years ago
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