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VladimirAG [237]

3 years ago

15

You have removed a very large, thick plate of steel (AISI 1010) from your heat treat oven and have placed it on a large insulate

d surface; the steel is now at a uniform initial temperature of 600.0°C. You now cool the exposed top surface of the steel using high-velocity cooling fluid; which keeps the cooled surface at a constant 30.0°C (assume the other surface is not losing any heat). How long will you need to run the coolant to cool the steel at a depth of 4.20cm to a temperature of 60.0°C?

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

for got sorry

Explanation:

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Q4. (20 points) For a bronze alloy, the stress at which plastic deformation begins is 271 MPa and the modulus of elasticity is 1

babunello [35]

Answer:

a) P = 86720 N

b) L = 131.2983 mm

Explanation:

σ = 271 MPa = 271*10⁶ Pa

E = 119 GPa = 119*10⁹ Pa

A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²

a) P = ?

We can apply the equation

σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N

b) L₀ = 131 mm = 0.131 m

We can get ΔL applying the following formula (Hooke's Law):

ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)

⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm

Finally we obtain

L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm

3 0

3 years ago

Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$

$\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}$

$\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}$

$G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}$

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

$P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))$

The rate of heat:

$Q=G_{m}(h_{3}-h_{2})$

The thermal efficiency:

$\mu_{ther}=\frac{P_{net}}{Q}$

3 0

3 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

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3 years ago

Which option identifies the next step in the following scenario?

Whitepunk [10]

Answer: The engineer will create a detailed sketch that labels all of the visual components.

Explanation:

It should be noted that the reverse engineering is required for the replacement and the modification of an existing product.

With regards to the question, the correct answer is option A "The engineer will create a detailed sketch that labels all of the visual components".

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3 years ago

The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer c

Nesterboy [21]

Answer:

attached below

Explanation:

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