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Scilla [17]
3 years ago
12

Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA

Engineering
1 answer:
weqwewe [10]3 years ago
5 0
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.


It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
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In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2
sp2606 [1]

Answer:

l=24mm

Explanation:

From the question we are told that:

Plane strain fracture toughness of T=55 MPa-m1/2

Y value Y=1.0

Stress level of\sigma =200 MPa

Generally the equation for length of a surface crack is mathematically given by

l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2

l=\frac{1}{3.142}(\frac{55}{1*200})^2

l=0.024m

Therefore

in mm

l=24mm

6 0
3 years ago
A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest
Citrus2011 [14]

Answer:

Following are the responses to the given question:

Explanation:

7 0
2 years ago
A turbojet aircraft flies with a velocity of 800 ft/s at an altitude where the air is at 10 psia and 20 F. The compressor has a
nika2105 [10]

Answer:

Pressure = 115.6 psia

Explanation:

Given:

v=800ft/s

Air temperature = 10 psia

Air pressure = 20F

Compression pressure ratio = 8

temperature at turbine inlet = 2200F

Conversion:

1 Btu =775.5 ft lbf, g_{c} = 32.2 lbm.ft/lbf.s², 1Btu/lbm=25037ft²/s²

Air standard assumptions:

c_{p}= 0.0240Btu/lbm.°R, R = 53.34ft.lbf/lbm.°R = 1717.5ft²/s².°R 0.0686Btu/lbm.°R

k= 1.4

Energy balance:

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} + \frac{v_{a} ^{2} }{2}\\

As enthalpy exerts more influence than the kinetic energy inside the engine, kinetic energy of the fluid inside the engine is negligible

hence v_{a} ^{2} = 0

h_{1} + \frac{v_{1} ^{2} }{2} = h_{a} \\h_{1} -h_{a} = - \frac{v_{1} ^{2} }{2} \\ c_{p} (T_{1} -T_{a})= - \frac{v_{1} ^{2} }{2} \\(T_{1} -T_{a}) = - \frac{v_{1} ^{2} }{2c_{p} }\\ T_{a}=T_{1} +  \frac{v_{1} ^{2} }{2c_{p} }

T_{1} = 20+460 = 480°R

T_{a}  =480+  \frac{(800)(800}{2(0.240)(25037}= 533.25°R

Pressure at the inlet of compressor at isentropic condition

P_{a } =P_{1}(\frac{T_{a} }{T_{1} }) ^{k/(k-1)}

P_{a} = (10)(\frac{533.25}{480}) ^{1.4/(1.4-1)}= 14.45 psia

P_{2}= 8P_{a} = 8(14.45) = 115.6 psia

4 0
3 years ago
Read 2 more answers
Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per
kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

                                                             $= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$. Then,

Power $ \propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

            $P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

           $P_2 = \text{const.} \times 1000 \ W$

5 0
3 years ago
I'm bored. I want to talk to you humans! weird. STUPID QUARANTINE!!! So. What are yall up to?
kiruha [24]

Answer:

nothing much what class r u in

3 0
3 years ago
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