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Karolina [17]
3 years ago
13

A steam turbine operates with 1.6 MPa and 350°C steam at its inlet and saturated vapor at 30°C at its exit. The mass flow rate o

f the steam is 22 kg/s, and the turbine produces 12,350 kW of power. Determine the rate at which heat is lost through the casing of this turbine.
Physics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

638.8kW

Explanation:

The flow rate of the steam m = 22kg/s

The Pressure of the steam at the inlet of the turbine P1 = 1.6MPa

The temperature of the steam at the inlet of the turbine T1 = 350*C

Steam quality at the exit of the turbine x2 = 1.0

The temperature of the steam at the exit of the turbine T2 = 30*C

Power produced = 12,350kW

Assuming the turbine is running on a steady state, hence we neglect the effect of kinetic and potential energy we get:

If you refer to the superheated steam table for the specific enthalpy at a pressure of 1.6MPa and temperature of 350*C, we get

h1 = 3,146kJ/kg

Refer to the steam table for saturated gas at temperature 30*C to get the specific enthalpy value h2 = 2,556.81kJ/kg

The heat that comes out from the turbine can be defined from the balance of energy in the system, and is represented as

Ein - Eout = change in system Energy = 0

Thus Ein = Eout

mh1 = mh2 + Wout + Qout

Qout = m(hi-h2) - Wout

Qout = 22 x (3146-255.6) - 12350

Qout = 638.8kW

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Explanation:

The point is that water is moving smoothly but that the solutes are not.Even though the containers are chemically different (chemical disequilibrium), once all the solutes in one container are contrasted to all the solutes in another container, both have the same total solutes concentrations (this means that they are in osmotic balance).

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2 years ago
Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.
aliya0001 [1]

Answer:

2.2 x 10-19

Explanation:

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3 0
3 years ago
Consider the interactions involved when you use a TV remote control to change the channel. Classify each interaction as long ran
makvit [3.9K]

Explanation:

Following are two interactions that are generally involved when we use a TV remote control to change the channel :

1. Figure touches remote buttons, and its a short range interaction.

2. Now remote sends signal to Television, this is a long range interaction.

7 0
2 years ago
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
2 years ago
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its
Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m  = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \  \times \ \ \ \ 1   \ lbf\ }

m  = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71  \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm

Therefore, the mass of the object is 5.045 lbm.

6 0
2 years ago
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