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abruzzese [7]
3 years ago
9

A rigid tank contains 3 kg of water initially at 43.97% quality and at a temperature of 120°C. The water is heated until it reac

hes a pressure of 7 bar. Neglect changes in kinetic and potential energy.
a. At what temperature, in °C, will the water become saturated vapor?
b. Determine the heat transfer, in kJ.
c. The work of the cycle, in kJ

Engineering
1 answer:
makkiz [27]3 years ago
6 0

Explanation: see attachment below

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Define the overall heat transfer coefficient.
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Answer and Explanation:

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It gives the measure of the heat transfer as a result of  convection or conduction. The coefficient of overall heat transfer depends on surface area, resistance of the material, temperature difference, thickness, etc.

It is given by:

Q = UA\Delta T

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U = overall heat transfer coefficient

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The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water i
olganol [36]

Answer:

m_added = 2 kg

Explanation:

From the question, we are told that the cylinder is allowed to fall 800 m in height. Thus, the potential energy will be converted into heat energy which will increase the temperature of water .

Now, let the mass of the falling cylinder be denoted by "m1" and let h be the height of fall.

Thus;

Formula for potential energy = mgh

Thus, as said earlier it's converted to heat generated. So heat generated = m1gh

Now let's calculate the heat absorbed;

heat absorbed = (m2)cΔt

Where;

ΔT is change in temperature

c is specific heat of water .

m2 is mass of water

Heat absorbed = heat generated

Thus;

(m2)cΔt = m1gh

Δt = m1gh/(m2•c)

Now, in both cases of the water and cylinder, m1, g , h and c are constant

Thus, we have;

Δt = (m1gh/m2) × 1/c

Where;

(m1gh/m2) is denoted as a constant k.

Thus;

Δt = K/m

For the first experiment, we have;

m = 2 kg

Δt = 2.4

Thus;

2.4 = K/2

Multiply both sides by 2 to get;

K = 4.8

For the second experiment, we have;

Δt = 1.2

Also,we have seen that K = 4.8

Thus;

Δt = K/m

Thus;

1.2 = 4.8/m

m = 1.2

m = 4 kg

Thus,mass added is;

m_added = 4 - 2

m_added = 2 kg

6 0
3 years ago
The acceleration of a particle as it moves along a straight line is given
BARSIC [14]

Answer:

V_t=6 = 32 m/s

Explanation:

The origin is at point 0 with the positive motion to the right  

The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:  

a=dv/dt

From here we can express dv as:

dv = a dt

Replace a by 2t — 1

dv = (2t — 1) dt

Integrate both sides of equation  

\int\limits^v_a  {2t-1} \, dv

v=t

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putting these value in integral

<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>

We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values

v — 2 = (t^2 — t) — (0^2 — 0)

From here we can write the expression for v as:  

v_t=6=6^2-6+2                             (1)  

So the velocity at t = 6 s is:

v_t=6 = 32 m/s

V_t=6 = 32 m/s

In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero  

To do that, we'll just replace v by 0 in expression (1)

0 = t^2 — t + 2

The roots of the quadratic equation are:

t_1/2=1±  √(1^2-4*2*1)/2

Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right  

Now that we have all the details, we can correctly draw the path of the particle  

We can see from the sketch that the total distance traveled is:  

s^T=Δs_0-1

s^T=| s_1 - s_0 |

Replace s_0 by its value  

s^T=| s_1 - 1 |                                        (2)  

In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time  

Since we have already wrote expression for v as function of time (step 2), we'll use expression  to get the expression for s

v= ds/dt  

Multiply both sides of equation by dt

v dt = ds

Replace v by expression (1)

(t^2 — t + 2) dt = ds

Integrate both sides of equation  

\int\limits^t_b {x} \, dx

t=s

b=(s=0)

x=(t^2 — t + 2)

dx=ds

putting these value in integral

(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0

Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6  

(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1

4 0
3 years ago
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