How to measure the quality of the output signal in ADC?

Subcooled liquid water flows adiabatically in a constant diameter pipe past a throttling valve that is partially open. The liqui

Llana [10]

Answer:

hi-he = 0

pi-pe = positive

ui-ue = negative

ti-te = negative

Explanation:

we know that fir the sub cool liquid water is

dQ = Tds = du + pdv ............1

and Tds = dh - v dP .............2

so now for process of throhling is irreversible when v is constant

then heat transfer is = 0 in irreversible process

so ds > 0

so here by equation 1 we can say

ds > 0

dv = 0 as v is constant

so that Tds = du .................3

and du > 0

ue - ui > 0

and

now by the equation 2 throttling process

here enthalpy is constant

so dh = 0

and Tds = -vdP

so ds > 0

so that -vdP > 0

as here v is constant

so -dP =P1- P2

so P1-P2 > 0

so pressure is decrease here

5 0

2 years ago

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

2 years ago

Select the right answer<br>

Kruka [31]

Answer:

for 1st question the answer is 5th option.

for 2nd question the answer is 2nd option

hope it helps you mate

please mark me as brainliast

5 0

3 years ago

You don't have to notify employees that a lockout/tagout is about to begin?

Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

3 0

2 years ago

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement b, no actuator movement c.

Volgvan

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement.

<h3>What is a pilot-operated check valve (POC)?</h3>

Pilot operated test valves paintings through permitting loose float from the inlet port via the opening port. Supplying a pilot strain to the pilot port permits float withinside the contrary direction. Air strain on pinnacle of the poppet meeting opens the seal permitting air to float freely.

An actuator fault is a form of failure affecting the machine inputs. Due to strange operation or fabric aging, actuator faults might also additionally arise withinside the machine. An actuator may be represented through additive and/or multiplicative fault.

Read more about the pilot-operated check valve:

brainly.com/question/13001928

#SPJ1

7 0

1 year ago