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Zielflug [23.3K]
3 years ago
12

A strip of AISI 304 stainless steel, 2mm thick by 3cm wide, at 550°C, continuously enters a cooling chamber that removes heat at

a rate of 120 kW. How fast must it move to achieve a temperature of 140 °C at the exit?
Engineering
1 answer:
Zanzabum3 years ago
6 0

Answer:

V = 1.23 m/s

Explanation:

Given data:

AISI 304 steel

thickness of steel = 2 mm

width  =  3 cm

Temperature =  550 degree celcius

wer know that heat is given as Q  

Q = m\times Cp \Delta T

\Delta T = 550 - 140 = 410 \degree\ celcius

for AISI 304 steel Cp is 502 J/kg . K

We know that

\dot mass  =\rho A V

\rho = 7.9\times 10^3 kg/m^3

A = 2\times 10^{-3} \times 3\times 10^{-2} m^2

therefore VELOCITY V  will be

120\times 10^{3} = 7.9\times 10^{3} \times 6\times 10^{-5} \times V\times 502\times 410

solving for V we get

V = 1.23 m/s

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Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
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Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

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- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

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