8. The operation of a TXV is controlled by the

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

Rufina [12.5K]

Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

6 0

3 years ago

Drag the tiles to the boxes to form correct pairs. Identify the designations of the three employees in an automobile company fro

aniked [119]

Answer:

is the fare of our responsibility towards

7 0

3 years ago

Please help me do this with my exam tomorrow.

blagie [28]

Answer:

ddddddddddddddddddddddddddddd

Explanation:

cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

6 0

2 years ago

In several sentences, please discuss:

Bezzdna [24]

Answer:

Everything is explained below in the Explanation section.

Explanation:

a) TMDL stands for = Total Daily Maximum Loads

TMDL is the measure of total maximum amount of pollutant allowed in the water body. TMDL is a important tool to maintain the quality of water and measure the aquatic pollution.

Whereas, the formula to calculate TMDL is as follows:

TMDL = Sum of waste load allocations (point source) + Sum of load allocations (non point sources and background) + Margin of Safety

b) Dead Zones :

Dead zones are zones which are termed as hypoxic which are low in oxygen because of the extreme aquatic pollution caused by human activities which in result deplete the oxygen level above the and below the oceans most required by the marine life. In addition, dead zones are those areas where oxygen is low and life in danger is high comparatively to other areas of oceans.

The main cause of that dead zones is because of eutrophication which means adding dangerous chemical nutrients in the water in exceeding amounts.

c) Why Ammonia-nitrogen is detrimental to waterbodies:

1. Ammonia nitrogen is a very toxic pollutant often found in landfills and sewage and waste products.

2. Ammonia nitrogen reduces the ability of water to disinfect its inhabitants and it reduces the purity. Furthermore, it changes the smell of the water and pollutes it.

3. Last but not the least, if you increase the concentration of ammonia nitrogen, the dissolved oxygen rate will be decreased and it has a inverse relation which is very dangerous to marine life underwater.

d) Streeter-Phelps DO curve:

Please refer to the attachment, I have attached the curve.

This curve is used to evaluate the dissolved oxygen in the water with varying distance.

It can be further studied from the curve that, how much loss sewage and other pollutants can endure upon the marine life.

8 0

3 years ago