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Elan Coil [88]
3 years ago
13

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar abs

orber plate. The absorber plate has a surface area of 2 m2 with emissivity and absorptivity of 0.9. The surface temperature of the absorber is 35°C, and solar radiation is incident on the absorber at 500 W/m2 with a surrounding temperature of 0°C. The convection heat transfer coefficient at the absorber surface is 5 W/m2 ∙K, while the ambient temperature is 25°C. The net heat rate absorbed by the solar collector heats the water from an inlet temperature (Tin) to an outlet temperature (Tout). If the water flow rate is 5 g/s with a specific heat of 4.2 KJ/kg *K, determine the temperature rise of the water.

Physics
1 answer:
hjlf3 years ago
6 0

Answer:

The temperature rise is given as 21.35 K

Explanation:

The explanation is shown on the first second and third uploaded image

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12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res
Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster  due to its height is given by;

P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J

P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)

K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\  v_{max} = 24.2 \ m/s

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0
2 years ago
5) When a substance changes its state, all BUT ONE of these statements is true That is A) Mass remains constant B) Only the form
expeople1 [14]
C) kinetic energy changes. Gas>liquid>solid
7 0
3 years ago
SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
2 years ago
What is amperage?
Shtirlitz [24]

question one b

question 2 i think a

3  d

4 c

5 not sure but wanting to say d

6  letter b

7 not sure

8 idk

9 i have no idea

3 0
3 years ago
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