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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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A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
mash [69]
<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

\\

Explanation:

<h2>Given:</h2>

a_1 \:=\:0.4\:m/s²

t_1 \:=\:60\:s

v_{i1} \:=\:0\:m/s

a_2 \:=\:0\:m/s²

t_2 \:=\:25\:min\:=\:1500\:s

a_3 \:=\:-0.8\:m/s²

v_{f3} \:=\:0\:m/s

\\

<h2>Required:</h2>

Distance from Station A to Station B

\\

<h2>Equation:</h2>

a\:=\:\frac{v_f\:-\:v_i}{t}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v\:=\:\frac{d}{t}

\\

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for v_{f1}

a\:=\:\frac{v_f\:-\:v_i}{t}

0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}

24\:m/s\:=\:v_f\:-\:0\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave1}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_1

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{60\:s}

720\:m\:=\:d

d_1\:=\:720\:m

\\

<h3>Distance when a = 0 m/s²</h3>

v_{f1}\:=\:v_{i2}

v_{i2}\:=\:24\:m/s

\\

Solve for v_{f2}

a\:=\:\frac{v_f\:-\:v_i}{t}

0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}

0\:=\:v_f\:-\:24\:m/s

v_f\:=\:24\:m/s

\\

Solve for v_{ave2}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}

v_{ave}\:=\:24\:m/s

\\

Solve for d_2

v\:=\:\frac{d}{t}

24\:m/s\:=\:\frac{d}{1500\:s}

36,000\:m\:=\:d

d_2\:=\:36,000\:m

\\

<h3>Distance when a = -0.8 m/s²</h3>

v_{f2}\:=\:v_{i3}

v_{i3}\:=\:24\:m/s

\\

Solve for v_{f3}

a\:=\:\frac{v_f\:-\:v_i}{t}

-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}

(t)(-0.8\:m/s²)\:=\:-24\:m/s

t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}

t\:=\:30\:s

\\

Solve for v_{ave3}

v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}

v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}

v_{ave}\:=\:12\:m/s

\\

Solve for d_3

v\:=\:\frac{d}{t}

12\:m/s\:=\:\frac{d}{30\:s}

360\:m\:=\:d

d_3\:=\:360\:m

\\

<h3>Total Distance from Station A to Station B</h3>

d\:= \:d_1\:+\:d_2\:+\:d_3

d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m

d\:= \:37,080\:m

d\:= \:37.08\:km

\\

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>
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This last code segment creates an array using the new keyword. The new keyword is generally used in constructing objects.

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