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DerKrebs [107]
3 years ago
14

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

l that transfers energy to the CO at a constant rate of 14 W for 1 h. During the process, the specific internal energy of the carbon monoxide increases by 10 kJ/kg. If no overall changes in kinetic and potential energy occur, determine: (a) the specific volume of the carbon monoxide at the final state, in m3/kg. (b) the energy transfer by work from the carbon monoxide, in kJ. (c) the energy transfer by heat transfer to the carbon monoxide, in kJ.
Engineering
1 answer:
Juli2301 [7.4K]3 years ago
7 0

Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

-The mass of C_o2 = 1 kg  

-The volume of the tank V_tank = 1 m^3  

-The added energy E = 14 W  

-The time of adding energy t = 10 s  

-The increase in specific internal energy Δu = +10 kJ/kg  

-The change in kinetic energy ΔKE = 0 and The change in potential energy  

ΔPE =0  

<u>Required  </u>

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.  

(c)The energy transferred by the heat transfer W in kJ and the direction of  

the heat transfer.  

Assumption  

-Quasi-equilibrium process.  

<u>Solution</u>  

(a) The volume and the mass doesn't change then, the specific volume is constant.

 v= V_tank/m ---> 1/1= 1 m^3/Kg  

(b) The added work is defined by.  

W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  

(c) From the first law of thermodynamics.  

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

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The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

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The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

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also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

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solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

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2 years ago
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A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c
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Answer:

V1=5<u>ft3</u>

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n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5<u>ft3</u>

state 2

V2=m.v2

V2=4lb*0.5ft3/lb=  <u> 2ft3</u>

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

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3 years ago
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Answer:

Technician b is correct

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Therefore, technician b is correct.

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3 years ago
What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical
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