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topjm [15]
3 years ago
14

2- A 2-m3 insulated tank containing ammonia at -20 C, 80% quality, is connected by a valve to a line flowing ammonia at 2 MPa, 6

0 C. The valve is opened, allowing ammonia to flow into the tank. At what pressure should the vale be closed if the manufacturer wishes to have 15 kg of ammonia inside at the final state?​
Engineering
1 answer:
alekssr [168]3 years ago
8 0

Answer:

The valve should be closed at 1081 kPa ⇒ 11 bar

Explanation:

Q_{C_V} + m_ih_i = m_2u_2 - m_1u_1 + W_{C_V}

Q_{C_V} = W_{C_V} = 0

m_1 = \frac{V}{v_1}  = \frac{2}{0.49927} = 4.006 kg

m_i = m_2 - m_1 = 15 - 4.006 = 10.994 kg

u_1 = 1057.5, h_i = 1509.9

u_2 =  m_ih_i + m_1u_1

m_2 = \frac{[10.994*1509.9] + [4.006*1057.5]}{15}  = 1389.1 kJ/kg

v_2 = \frac{V}{m_2} = \frac{2}{15} = 0.1333 m^3/kg

Therefore, v₂, u₂ fix state 2.  

​By trial and error, P₂ = 1081 kPa & T₂ = 50.4° C

1081 kPa ⇒ 11 bar

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miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
An incandescent light bulb can be regarded as a resistor. If its power output is 100W, calculate the resistance of the light bul
stira [4]

Answer:121\ \Omega

0.909\ A

Explanation:

Given

Power P=100\ W

Voltage applied V=110\ V

Resistance of the bulb is given by

P=\frac{V^2}{R}

100=\frac{110^2}{R}

R=\frac{12100}{100}

R=121\ \Omega

Current drawn by the Power source is given by

P=V\cdot I

I=\frac{P}{V}

I=\frac{100}{110}

I=0.909\ A

8 0
3 years ago
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r
Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

K = \sigma Y \sqrt{\pi a}

where;

fracture toughness K = 137 MPam^{1/2}

geometry factor Y = 1

applied stress \sigma = ???

crack length a = 2mm = 0.002

∴

137 =\sigma \times 1  \sqrt{ \pi \times 0.002 }

137 =\sigma \times 0.07926

\dfrac{137}{0.07926} =\sigma

\sigma = 1728.489 MPa

Now, the tensile impact obtained is:

\sigma = \dfrac{P}{A}

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

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3 years ago
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siniylev [52]

Answer:

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Explanation:

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2 years ago
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what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .
NISA [10]

Loaded, H_{Loaded}(s) = \frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL) = \frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S) is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

Learn more about Loaded here:

brainly.com/question/20039214

#SPJ4

3 0
11 months ago
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