Answer back to question for la ,lot points

Ethylene (C2H4) gas enters a well-insulated reactor and reacts completely with 400% of theoretical air, each at 25°C, 2 atm. The

GuDViN [60]

Answer:see explanation.

Explanation:

(a).The equation of reaction is;

C2H4 + 3(O2 + 3.76N2) -----> 2CO2 + 2H2O + 11.28 N2.

Theoretical air is the amount of air that will allow the complete combustion of the fuel. Air contains 21% of oxygen,79% of Nitrogen. Thus, the molecular mass of air = 29 [Kg/Kmol] . The number of oxygen needed to oxidize the hydrocarbon(C2H4) is 79/21= 3.76 moles of Nitrogen.

(b). N(total) =2+2+11.28 =15.28 kmoles of product

15.28[h(T) - h°] of air

= 15.28(Mass of air) (Cp,1000k)

T(adiabatic) = 802,310 kJ/k. Mole.

When mass of air= 29 Kg/ k.mol

Cp,1000k = 1.142 kJ.k which is the specific heat capacity of air.

T(adiabatic) = 1275k -----> assuming all the products are air.

(C). 2{∆h} co2 + 2{∆h} H20 +11.28 {∆h} N2

After series of calculations and checking of tables:

S(o2) = 320.173 - 8.314 ln 0.12= 337.46 kJ/ K.mol.k

S(H2O)= 273.986- 8.314 ln 0.1406 = 290.30 KJ/ kmol. K

S(N2) = 258.503- 8.314 ln 0.7344= 261.07 kJ/kmol.k

= 2(337.46) + 2(290.30)+11.28(261.07)-360.79-3(218.01)-11.28(193.46)

=1003.3378 kJ/kmol. K

5 0

3 years ago

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is

nexus9112 [7]

Answer:

The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa.

4 0

3 years ago

The ???? − i relationship for an electromagnetic system is given by ???? = 1.2i1/2 g where g is the air-gap length. For current

Artemon [7]

Answer:

a) The mechanical force is -226.2 N

b) Using the coenergy the mechanical force is -226.2 N

Explanation:

a) Energy of the system:

$\lambda =\frac{1.2*i^{1/2} }{g} \\i=(\frac{\lambda g}{1.2} )^{2}$

$\frac{\delta w_{f} }{\delta g} =\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }$

If i = 2A and g = 10 cm

$\lambda =\frac{1.2*i^{1/2} }{g} =\frac{1.2*2^{1/2} }{10x10^{-2} } =16.97$

$f_{m}=-\frac{g^{2}\lambda ^{3} }{3*1.2^{2} }=-\frac{16.97^{3}*2*0.1 }{3*1.2^{2} } =-226.2N$

b) Using the coenergy of the system:

$f_{m}=- \frac{\delta w_{f} }{\delta g} =-\frac{1.2*2*i^{3/2} }{3*g^{2} }=-\frac{1.2*2*2^{3/2} }{3*0.1^{2} } =-226.2N$

8 0

3 years ago

A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500

marysya [2.9K]

It’s is 2000 because it is 20000

7 0

3 years ago

Why dont some vehicles have power steering

daser333 [38]

Answer:

Because older cars didn't really have much power steering, and parking or even turning at low speeds could require a lot of strength to move the steering wheel on a big, heavy car. This moves the rack to the right or left, which turns the vehicle's wheels. Without power assist, you'd need grunt force to turn the pinion and move the rack.

Explanation:

4 0

4 years ago