I need help for these questions D B E U.S. 2% Sun 6:20 PM Soud Alkandari a E Ado

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I need help for these questions

D B E U.S. 2% Sun 6:20 PM Soud Alkandari a E Adobe Reader File Edit View Window Help Assignment 17-3.pdf 1 130% Tools Sign Comment 1. Read through section 5.4 2. A throttling device has water as the fluid. The inlet conditions are 100 psia and 320OF. The outlet pressure is 10 psia. Estimate the outlet temperature. If the inlet velocity of the water is 10 ft/s and the flow area remains constant, what and by how much does this affect the result from before where you neglected Ake? 3. A mixing chamber for air has two inlets and one outlet. Inlet one has a volumetric flow rate of 0.2 m3/ a pressure of 200 kPa, and a temperature of 75OC. The other inlet has a volumetric flow rate of 0.3 m3/s, a pressure of 200 kPa, and a temperature of 20%. The outlet pressure is 180 kPa. Neglecting Ake and Ape, and assuming the mixing chamber is adiabatic, determine the outlet temperature 4. An insulated nozzle with a circular cross section carries superheated steam. The inlet conditions are 1 Mpa, 400OC, with a velocity of 10 m/s. The outlet pressure is 0.6 Mpa. If the inlet diameter is 40 cm and the outlet diameter is 5 cm, what is the outlet temperature?

Explanation / Answer

2)

In throttling, Enthalpy remains constant

At P1 = 100 psia and T = 320 F

h1 = 290.4 Btu/lbm

Now

h2 = h1 = 290.4 Btu/lbm

Now at

P2 = 10 psia

Saturation Properties are

Tsat = 193.2 F

hf = 161.2

hg = 1143

Since

hf < h2 < hg

The state is Liquid-vapor mixture

h2 = (1-x)*hf + x*hg

From this

x = 0.1316

And

T2 = Tsat = 193.2 F

Specific Volume at inlet

v1 = 0.01765 ft^3/lbm

At oulet

vf = 0.01659

vg= 38.42

v2 = (1-x)*vf + x*vg

v2 = 5.07 ft^3/lbm

Mass Flow remains Constant

A*V1/v1 = A*V2/v2

V2 = (v2/v1)*V1 = 2872.5 m/s

Change in KE

dKE = V2^2/2 - V1^2/2 = 4125.64 kJ/kg

3)

suffix 1 and 2 for inlet and 3 for oultet

At P1 = 200 kPa and T1 = 75 deg

From Tables of air

h1 = 348.9 kJ/kg

At P2 = 200 kPa and T2 = 20 deg

h2 = 293.6 kJ/kg

Given

Q1 = 0.2 m^3/s

Q2 = 0.3 m^3/s

From

P*v = R*T

P1*v1 = R*T1

200*v1 = 0.287*(273 + 75)

v1 = 0.49938 m^3/kg

Similarly

200*v2 = 0.287*(273 + 20)

v2 = 0.4205 m^3/kg

Mass Flow rates

m1 = Q1/v1 = 0.2/0.49938

m1 = 0.4 kg/s

m2 = Q2/v2 = 0.3/0.4205

m2 = 0.7 kg/s

Also

m3 = m1 + m2

m3 = 1.1 kg/s

In adiabatic mixture

m1*h1 + m2*h2 = m3*h3

1.1*h3 = 0.4*348.9 + 0.7*293.6

h3 = 313.71 kJ/kg

Now at P3 = 180 kPa and h3 = 313.71 kJ/kg

T3 = 40 deg

4)

Given

P1 = 1 MPa

T1 = 400 deg

V1 = 10 m/s

P2 = 0.6 MPa

D1 = 40 cm

D2 = 5 cm

Mass flow rate is constant (assuming incrompressible flow)

A1*V1 = A2*V2

D1^2*V1 = D2^2*V2

V2 = 640 m/s

At P1 = 1 MPa and T1 = 400 deg (From steam tables)

Enthalpy

h1 = 3264 kJ/kg

Energy Equation

h1 + V1^2/2 = h2 + V2^2/2

3264*1000 + 100/2 = h2 + 204800

h2 = 3059.25 kJ/kg

Now At P2 = 0.6 MPa

hf = 670.5 kJ/kg

hg = 2757 kJ/kg

Since

h2 > hf The state is Superheated

Now

At P = 0.6 MPa and Tx = 250 deg

hx = 2957 kJ/kg

At P = 0.6 MPa and Ty = 300 deg

hy = 3062 kJ/kg

By Linear Interpolating

T2 = Tx + (Ty-Tx)*(h2-hx)/(hy-hx)

T2 = 298.9 Deg

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