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Usimov [2.4K]
3 years ago
9

John wants to construct a device using quartz crystal, Which device can he construct?

Engineering
2 answers:
kupik [55]3 years ago
6 0

Answer:

the answer inst d stg

Explanation:

tatiyna3 years ago
4 0

Answer: Option D, piezoelectric pressure guage

Explanation: Quartz crystal possess a very useful quality in science as they can generate small charges when pressure is applied to them or when they are hit. This property can be harnessed to construct a piezoelectric pressure gauge which would be used to measure and indicate changes in pressure, the quartz crystal releases little voltage each time there is an applied pressure . This device would be able to sense changes in pressure as there would voltage proportional to the applied pressure.

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Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow
melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

m_{in}h_{in}=W_{out}+m_{out}h_{out}

Whereas the mass flow is just the same, which means we have:

W_{out}=m_{out}(h_{out}-h_{in})

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

s_{out}= 2.2096BTU/lbm-R

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

h_{out}=1659.15BUT/lbm

Then, the isentropic work (negative due to convention) is:

W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU

And the real produced work is:

W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU

Finally, in horsepower:

W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP

Regards!

6 0
3 years ago
A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per
mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0
4 years ago
Read 2 more answers
1. One of these is NOT a type of pneumatic tool. Which one?
Serggg [28]

Answer:Circular

Explanation:

It’s the only thing not list under pneumatic tools‍♂️

5 0
3 years ago
A circular pipe of 25-mm outside diameter is placed in an airstream at 25 °C and 1-atm pressure. The air moves in cross flow ove
Scrat [10]

Answer: a) Fd = 3.24 N/m

b) Q = 520 w/m

Explanation: please find the attached files for the solution

7 0
3 years ago
Just to let you know Christmas is in 10 days&lt;3<br><br> lol
Harrizon [31]

Answer:

yay yay

Explanation:

im so excited i cant wait

7 0
3 years ago
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