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3 years ago

For H2O, determine the specified property at the indicated state.

Engineering

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

3.613 bar

1.0290 m3/kg

0.038378 m3/kg

1.363 m3/kg

0.012166 m3/kg

0.002083 m3/kg

0.001177 m3/kg

Explanation:

part a

Table A-3: vf = 1.0435 x 10-3 m 3/kg , vg = 1.673 m 3 /kg. Since vf < v < vg, the state is in the two phase liquid-vapor region, as shown.

From Table A-3, the pressure is the saturation is pressure at 140 C: p = 3.613 bar.

Answer: p = 3.613 bar

part b

The pressure is higher than the critical pressure, as shown on the diagram. Hence, the state is in the compressed liquid region.

From Table A-5: v = 1.0290 m3 /kg.

Answer: v = 1.0290 m3 /kg

part c

Since the temperature is higher than Tsat at 100 bar, the state is super-heated vapor. T sat = 311.1 C

Interpolating in Table A-4, we get

v = 0.038378 m3/kg

Answer: v = 0.038378 m3/kg

part d

vx = vf + x(vg – vf)

v = 1.0291 x 10-3 + (0.4)(3.407 - 1.0291 x 10-3 )

v = 1.363 m3/kg

Answer: v = 1.363 m3/kg

a ) T = 440 C , p = 20 MPa

Tsat(@20 MPa) = 365.75 C

T > Tsat(@20 MPa) hence, super-heated steam

Interpolate the results

v = 0.012166 m3/kg

Answer: v = 0.012166 m3/kg

b) T = 160 C , p= 20 MPa

Tsat(@20 MPa) = 365.75 C

T < Tsat(@20 MPa) hence, compressed liquid region

v = 0.002038 m3/kg

Answer: v = 0.002038 m3/kg

c) T = 40 C , p = 2 MPa

Tsat(@2 MPa) = 212.38 C

T < Tsat(@2 MPa) hence, compressed liquid region

v = 0.001177 m3/kg

Answer: v = 0.001177 m3/kg

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Explanation:

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brackets and head of the screw are made of material with T_fail=120 Mpa

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=0.0024×πm^2

now we have to find shear stress for both head and plate

For head:

T_allow=V/A(head)

48 Mpa=P/0.003×π ..(V=P)

P =48 Mpa×0.003×π

=452.16 KN

For plate:

T_allow=V/A(plate)

48 Mpa=P/0.0024×π ..(V=P)

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4 years ago

A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu

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Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

Area of A is = $\frac{\pi}{4}(0.01)^{2} m^{2}$

= $7.85 *10^{-5}m^{2}$

Velocity of liquid through A = 0.2 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $0.2 *7.85*10^{-5} \frac{m^{3}}{s}$

The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $1039.8 * 0.2 * 7.85 *10^{-5} Kg/s$

= 0.016324 $\frac{Kg}{s}$

From the question the diameter of B = 2 cm = 0.02 m

Area of B = $\frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}$

Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume = $\frac{Volume of Inlet }{time}$ = $Velocity * Area$ i.e is $m^{2} * \frac{m}{s} = \frac{m^{3}}{s}$

= $3.14*10^{-4} *0.01 \frac{m^{3}}{s}$

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

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= 0.00330642 $\frac{Kg}{s}$

From the question The flow rate in term of volume of the outflow at the time of measurement is given as = 0.5 L/s

And also from the question the mass of potassium chloride at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

= $13\frac{g}{L} * 0.5 \frac{L}{s}$

= $\frac{6.5}{1000}\frac{Kg}{s}$ Note (1 Kg = 1000 g)

= 0.0065 kg/s

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Let denote the rate at which liquid flows in terms of mass as as $\frac{dm}{dt}$ i.e change in mass with respect to time hence

Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

(0.016324 + 0.00330642) - 0.0065 = $\frac{dm}{dt}$

$\int\limits {\frac{dm}{dt} } \, dx =\int\limits {0.01313122} \, dx$

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=> t = 1500.2414 sec

t = .4167337 hours (1 hour = 3600 seconds)

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