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3 years ago

Describe (in narrative) the essential funct tion of a Heat Engine?

Engineering

1 answer:

Vesna [10]3 years ago

7 0

Answer and Explanation:

The heat engine are used for converting the heat energy into mechanical energy .When heat energy is converted into mechanical energy then this mechanical energy are used to do the useful work. Heat engine working based on the three process that are head addition , expansion and heat rejection.

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A set of bus bars also can be called a

Vikki [24]

Answer:

Explanation:

3 0

3 years ago

Read 2 more answers

Consider a NACA 2412 airfoil in a low-speed flow at zero degrees angle of attack and a Reynolds number of 8.9·106 . Calculate th

Fofino [41]

Answer:

a) pressure drag is zero (0)

b) pressure drag is 20%

Explanation:

Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation Ans) Given,

NACA 2412 airfoil

Re = 8.9 x 10^6

We know, for turbulent flow ,drag coefficient, Cdf = 0.074 / Re^0.2

=> Cdf = 0.074 / (8.9 x 10^6)^0.2

=> Cdf = 0.003

For both side of plate, Cd = 2 x 0.003 = 0.006

For zero degree angle of attack for NACA 2412, Cdf = 0.006

Also, Cd = Cdf + Cdp

=> 0.006 = 0.006 + Cdp

=> Cdp = 0

Hence, pressure drag is zero

Now, for zero degree angle of attack for NACA 2412, Cd = 0.0075

Also, Cd = Cdf + Cdp

=> 0.0075 = 0.006 + Cdp

=> Cdp = 0.0075 - 0.006

=> Cdp = 0.0015

Hence, Pressure drag percentage = (Cdp / Cdf) x 100

=> Pressure drag percent = (0.0015/0.0075) x 100 = 20 %

Hence, pressure drag is 20% of pressure drag due to flow seperation

3 0

3 years ago

3. A power cycle operates between hot and cold reservoirs at 500 K and 310 K respectively. At steady state the power output deve

grin007 [14]

Answer:

0.163 MW

Explanation:

To get the minimum rate of energy rehection by heat transfer to cold reservoir, it implies that the power cycle operates in reversible cycle. The efficiency of reversible cycle whose value will be same as efficiency of power cycle will be given by

Efficiency of reversible cycle

n= (Th-Tc)/Th where T represent temperature, n efficiency, subscripts h and c hot and cold respectively

Substituting 500 and 310 for hot and cold temperatures respectively then efficiency

n=(500-310)/500=0.38

Efficiency of power cycle, n= Power output/Qh

Qh= 0.1/0.38= 0.263

Net power output, W= Qh- Qc

Qc=Qh-W= 0.263-0.1=0.163 MW

4 0

4 years ago

Chaplets are used to support a sand core inside a sand mold cavity. The design of the

ziro4ka [17]

Answer:

a) 2 chaplets

b) 9 chaplets

Explanation:

Before we can determine the minimum number of chaplets that should be placed beneath and above the core, we must know the mass of the sand used to make the surface of the mold cavity as well as the mass of the steel metal poured inside the mold.

Density is defined as the ratio of mass of a substance to its density.

Density = Mass/Volume

For the STEEL:

Density of steel = 7.82g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of steel = 7.82×5450

Mass of steel = 42619g

Mass of steel in kg = 42.619kg

For the SAND:

Density of sand = 1.6g/cm³

Volume = volume of the core = 5450cm³

Mass = Density × Volume

Mass of sand = 1.6×5450

Mass of sand = 8720g

Mass of sand in kg = 8.72kg

a) Since the the chaplet support the sand from beneath the core, and each chaplet weighs 45N, we need to know the amount of force possessed by the sand.

Since the mass of the sand is 87.2kg

Weight = mass × acceleration due to gravity

Weight = 8.72×9.81

Weight of sand used to mold the core = 85.54N

Since 1 chaplet weighs 45N

This means that (85.54N/45N) i.e 1.9 which is approximately 2 chaplets must be placed beneath the core to sustain it before the steel metal is poured.

b) Since the metal poured in the core is steel, this means that the chaplet placed above the core must be able to withstand the strength of the steel.

Weight of steel = mass of steel × acceleration due to gravity

Weight of steel = 42.619×9.81

Weight of steel = 418.09N

Since one chaplet weigh 45N, the amount of chaplets that must be placed above the core is;

418.09/45

= 9.3

Therefore the minimum number of chaplets that should be placed above the core after pouring steel metal is 9chaplets.

4 0

3 years ago

An operator decided to develop a subsea drill center of 4 oil wells. The production from the drill center is to be transported t

Vikki [24]

Answer:

The possible options the operator can choose from for the manifold system are;

1) Cluster manifold system

2) Template manifold system

Explanation:

Given that a subsea drill center is to be used for the development of the field that have four wells, we have;

1) Clustered system

In the clustered system design, the wells are situated and drilled around the designated area where the manifold will eventually be installed, such that there is increased flexibility in investment such that the economy of the development can be factored in as the field is being developed favoring the suitability of the cluster field system for a field with a few number of wells such as the one in question.

However for shallow wells and in a situation of high financial uncertainties such that the wells can be drilled at the same time the the manifolds are being constructed saving costs of waiting for the template the cluster manifold system will be more appropriate

Also as there are few wells the cluster manifold system can be more cost effective in terms of scheduling and resource allocation.

2) Template System

In the template manifold system design, the wells are drilled in a prefabricated well template housing which will hold the completion tools of the well, As such the well completion are well arranged and interconnected within the template design

Whereby the aim is for a fast an economic as well as a well built system, then the right choice is the template manifold design where there is direct flow from the wells to the template manifold improving flow assurance, and reducing installation costs as the system does not require jumper installation which is costly

3 0

3 years ago