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zhuklara [117]
1 year ago
15

consider a simple ideal rankine cycle with fixed boiler and condenser pressures. what is the effect of superheating the steam to

a higher temperature on
Engineering
1 answer:
Aleonysh [2.5K]1 year ago
3 0

The net work output and cycle efficiency will both increase with superheating the steam. Additionally, the steam at the turbine exit has less moisture in it.

The superheating raises the mean temperature of heat addition, which raises the efficiency of the superheat Rankine cycle over that of the simple Rankine cycle. On the T-s diagram to the left, the impact of decreasing condenser pressure on the Rankine cycle efficiency is depicted. Steam leaves the condenser as a saturated mixture at the saturation temperature corresponding to the condenser pressure. Rankine cycle in thermodynamics Roy Mech's phrase "Superheated" This cycle demonstrates the different phases of operation in a turbine plant using superheated steam. F -> G is a symbol for the turbine's enthalpy reduction. The effect of superheated steam on the Rankine cycle's efficiency is that the efficiency of the cycle rises as the superheat of the steam rises.

Learn more about superheating here:

brainly.com/question/14718830

#SPJ4

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Technician A says that a circuit with continuity reads 0 ohms. Technician B says that an open circuit reads 0 ohms. Who is corre
patriot [66]

Answer:

  A

Explanation:

An open circuit has infinite resistance, not 0. A circuit with continuity usually measures a few ohms or less, depending on the length and gauge of the wire and the condition of the connections. Often, on a general-purpose ohmmeter, the reading is very near 0 ohms.

Technician A is correct.

8 0
2 years ago
Question #4
Gemiola [76]

Answer:

The correct option is;

Fluid power systems are much more efficient with regards to energy costs and overall cost savings

Explanation:

The option that will provide a valid reason for the larger proportion of the consumers to change to the fluid power trucks from the diesel powered trucks is the possibility of increased efficiency in the cost of energy, and the cost savings to be made from making the switch to the fluid power trucks, due to the lower input required and the faster rate at which the consumer budget will be rebalanced leading to increased return on investment and improved rate of return.

5 0
3 years ago
Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5
pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

Aa.Va=Ab.Vb=Q

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3

Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3

Using the volume rate:

Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s

Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s

Assuming no losses, the energy equation for fluids can be written as:

Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m

Pb = 70000\ Pa-27303\ Pa - 14715\ Pa

Pb = 27,996\ Pa = 28\ kPa

6 0
3 years ago
For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft
ivanzaharov [21]

Answer:

total time  = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - e^{(-kt^n)}   ...........1    

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}    

solve it we get

k = 0.003786  = 37.86 × 10^{4}

so here

y = 1 - e^{(-kt^n)}

1 - y  =  e^{(-kt^n)}

take ln both side

ln(1-y) = -k × t^n  

so

t = \sqrt[n]{-\frac{ln(1-y)}{k}}    .............2

now we will put the value of y = 87% in equation  with k and find out t

t = \sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}

total time  = 304.21 s

7 0
3 years ago
One of the testing equipments used for inspection is? Test tubes Measuring jar Strain gauges None of the mentioned
Blizzard [7]

Answer:

The answer is the <u>strain gauges. </u>

Explanation:

Inspection systems work or are performed to measure the characteristics of a product, to verify if it meets specified requirements, all using benchmarks and test equipment.

The strain gauges are part of the test equipment used for inspection. These are sensors that measure deformation, pressure and load in resistance tests of materials.

7 0
3 years ago
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