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alukav5142 [94]

3 years ago

11

Steam enters a heat exchanger at a pressure of 1.5 bar and a temperature of 400°C with a mass flow rate of 0.05 kg/s and exits a

t a temperature of 120°C and at the same inlet pressure. The steam heat transfer takes place with refrigerant R22 in a separate stream. The refrigerant enters the heat exchanger as liquid at 4.5 bar and leaves at the same pressure and at a temperature of 10°C. The heat exchanger is not well designed and there is stray heat transfer to the surroundings equal to 10% of the heat transferred from the steam.
Determine

(a) The rate of stray heat transfer, in kW.

(b) The rate of heat transfer to the refrigerant, in kW

(c) The mass flow rate of the refrigerant, in kg/s

1 answer:

Alenkinab [10]3 years ago

7 0

Answer:

a. 2.832kw

b. 25.488kw

c. 0.118kg/s

Explanation:

Please look at attachment carefully

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A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

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Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

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$F_T = mw^2 e$

$F_T = mew^2$

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F_T =428.36 N

Transmission ratio $=\frac{300}{428.36} = 0.7$

also

transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

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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:

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

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

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