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Gnoma [55]
3 years ago
13

A furnace uses preheated air to improve its fuel efficiency. Determine the adiabatic flame temperature when the furnance is oper

ating at a mass air-fuel ratio of 16 for air preheated to 600 K. The fuel enters at 300 K. Assume the following simplified thermodynamic properties:
Tref = 300 K MWFUEL=MWAIR=MWPROD = 29 kg/kmol cp,fuel =cp,air= cp,prod = 1200 J / kg-K
hf,air = hf,prod = 0 hf,fuel = 4x107 J/kg
Engineering
1 answer:
balu736 [363]3 years ago
8 0

Answer:

2543 k

Explanation:

This problem can be resolved by applying the first law of thermodynamics

<u>Determine the adiabatic flame temperature</u> when the furnace is operating at a mass air-fuel ratio of 16 for air preheated to 600 K

attached below is a detailed solution

cp = 1200

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A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is sub
lara [203]

Answer:

hello your question is incomplete attached below is the complete question

<em>answer</em> :

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

Explanation:

a) Determine the surface temperature To and TL based on the known conditions provided in the drawing

To ( inside temperature ) = 598 K

TL ( outside temperature ) = 594 k

attached below is the detailed solution

5 0
3 years ago
The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm
oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s

The exit temperature is:

T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25°C

b) The mean fluid velocity is:

V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s

The hydraulic diameter is:

D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m

The Reynold´s number is:

Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36

Assuming fully developed turbulent flow, the Nusselt number is:

Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69

h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C

The highest component surface temperature is:

T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22°C

6 0
4 years ago
Please help will give brainliest please answer all 3
larisa [96]

Answer: For #1 I'm going to go with A because that has to do with biology

For #2 I'm going to go with B oceans because that has to do with plant life (and life in general).

For #3 I'll say marine/maritime engineer (you can just say marine)

Hope it helps!

8 0
4 years ago
2. The unthreaded part of a bolt or screw is called the
aksik [14]

Answer:

The grip

Explanation:

the head of all headed bolt (except countersunk head bolt)

3 0
3 years ago
Read 2 more answers
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
KiRa [710]

Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

h2=3116KJ/kg

T2=Temperature(Water;P=P2;s=s1)

T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

7 0
3 years ago
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