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Morgarella [4.7K]
4 years ago
15

Steam enters an adiabatic nozzle at 2.5 MPa and 450oC with a velocity of 55 m/s and exits at 1 MPa and 390 m/s. If the nozzle ha

s an inlet area of 6 cm2, determine (a) the exit temperature.
Engineering
1 answer:
luda_lava [24]4 years ago
4 0

Answer:

The value of exit temperature from the nozzle = 719.02 K

Explanation:

Temperature at inlet T_{1} = 450°c = 723 K

Velocity at inlet V_{1} = 55 \frac{m}{sec}

velocity at outlet V_{2} = 390 \frac{m}{sec}

Specific heat at constant pressure for steam  C_{p}  = 18723 \frac{J}{kg k}

Apply steady flow energy equation for the nozzle

h_{1} + \frac{V_{1} ^{2} }{2} = h_{2} + \frac{V_{2} ^{2} }{2}

C_{p} T_{1}  + \frac{V_{1} ^{2} }{2} = C_{p} T_{2} + \frac{V_{2} ^{2} }{2}

Put all the values in the above formula we get,

⇒ 18723 × 723 + \frac{55^{2} }{2} = C_{p} T_{2} + \frac{390^{2} }{2}

⇒   T_{2} = 719.02 K

This is the value of exit temperature from the nozzle.

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Explanation:

For true Strain:

step 1:

E true = Ln(1 + 0.5 ) = 0.40

Step 2:

E true = Ln(1 + 0.33 ) = 0.29

By single step process:

E true = Ln(1 + 1 ) = 0.69

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3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
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Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

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4 years ago
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