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3 years ago

British standered institution

Engineering

1 answer:

Scorpion4ik [409]3 years ago

6 0

Answer:

The British Standards Institution, is the national standards body of the United Kingdom. BSI produces technical standards on a wide range of products and services and also supplies certification and standards-related services to businesses

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Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

mixas84 [53]

Answer:

work done = 48.88 × $10^{9}$ J

Explanation:

given data

mass = 100 kN

velocity = 310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J

work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J

work done = 48.88 × $10^{9}$ J

6 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0

3 years ago

On what frequency can you expect to monitor air traffic in and around<br> Lincoln Airport?

Phantasy [73]

Answer:

118.5

Explanation:

Hope this helps!

5 0

3 years ago

A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The

AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = $\frac{60+180}{2}$ = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0

3 years ago

A rich industrialist was found murdered in his house. The police arrived at the scene at 11:00 PM. The temperature of the corpse

d1i1m1o1n [39]

Answer:

The dude was killed around 6:30PM

Explanation:

Newton's law of cooling states:

$T = T_m + (T_0-T_m)e^{kt}$

where,

$T_0$ = initial temp

$T_m$ = temp of room

$T$ = temp after t hours

$k$ = how fast the temp is changing

$t$ = time (hours)

$T_0 = 31$ because the body was initlally 31ºC when the police found it

$T_m = 22$ because that was the room temp

$T = 30$ because the body temp drop to 30ºC after 1 hour

t = 1 because that's the time it took for the body temp to drop to 30ºC

$k=???$ we don't know k so we must solve for this

rearrange the equation to solve for k

$T = T_m + (T_0-T_m)e^{kt}$

$T - T_m= (T_0-T_m)e^{kt}$

$\frac{T - T_m}{(T_0-T_m)}= e^{kt}$

$ln(\frac{T - T_m}{T_0-T_m})=kt$

$\frac{ln(\frac{T - T_m}{T_0-T_m})}{t}=k$

plug in the numbers to solve for k

$k = \frac{ln(\frac{T - T_m}{T_0-T_m})}{t}$

$k = \frac{ln(\frac{30 - 22}{31-22})}{1}$

$k=ln(\frac{8}{9})$

Now that we know the value for k, we can find the moment the murder occur. A crucial information that the question left out is the temperature of a human body when they're still alive. A living human body is about 37ºC. We can use that as out initial temperature to solve this problem because we can assume that the freshly killed body will be around 37ºC.

$T_0 = 37$ because the body was 37ºC right after being killed