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Alekssandra [29.7K]
3 years ago
9

Air as an ideal gas in a closed system undergoes a reversible process between temperatures of 1000 K and 400 K. The beginning pr

essure is 200 bar. Determinc the highest possible ending pressure for this process.If the ending pressure is 3 bar, determine the heat transfer and work per unit mass, if the boundary of the system is in constant contact with a reservoir at 400 K.

Engineering
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

highest possible ending pressure for this process is 8.0954 bar

Explanation:

We can say that Heat transfer is any or all of several kinds of phenomena, considered as mechanisms, that convey energy and entropy from one location to another. The specific mechanisms are usually referred to as convection, thermal radiation, and conduction.

Please see attachment for the solution.

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A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
Volgvan

Answer:

a) 22.5number

b) 22.22 m length

Explanation:

Given data:

Bridge length = 500 m

width of bridge = 12 m

Maximum temperature = 40 degree C

minimum temperature  = - 35 degree C

Maximum expansion can be determined as

\Delta L = L \alpha (T_{max} - T_{min})

where , \alpha is expansion coefficient = 12\times 10^{-6} degree C

SO, \Delta L = 500\times 12\times 10^{-6}\times ( 40 - (-35))

\Delta L = 0.45 m = 450 mm

number of minimum expansion joints is calculated as

n = \frac{450}{20} = 22.5

b) length of each bridge

Length = \frac{500}{22.5} = 22.22 m

8 0
3 years ago
A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete
Alex

Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

d is grain diameter to be achieved after heating in mm

d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

t₂=?

n=2.2

First find K

K=(d²-d²₀ )/t₁

K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3

K=(0.051²-0.024²) ÷25/2

K=0.000243 mm²/h

Re-arrange equation for K ,to get the equation for d as;

d=√(d₀²+ Kt)  where now t=t₂

d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min

4 0
3 years ago
Which is not required when working in a manufacturing facility?
Artyom0805 [142]
Flip flops are not required
5 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
What is the smallest variable type I can use to represent the number 27?
oksano4ka [1.4K]

Answer:3

Explanation:

Cuz

3 0
3 years ago
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