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jenyasd209 [6]
3 years ago
12

A south-facing collector at latitude 40◦ is tipped up at an angle equal to its latitude. Compute the following insolations for J

anuary 1st at solar noon:
a. The direct beam insolation normal to the sun’s rays.

b. Beam insolation on the collector.

c. Diffuse radiation on the collector.

d. Reflected radiation on the collector with ground reflectivity 0.2.
Engineering
1 answer:
BartSMP [9]3 years ago
3 0

Answer:

Explanation:

(c). looking for the radiation of the collector is given thus

C = 0.095 + 0.04 sin [360/365(n-100)] = 0.095 + 0.04 sin [360/365(1-100)]

C = 0.05535

∴ Diffuse radiation of the collector Idc = C*Ib + (1+cosσ/2)  

Idc = 0.5535 * 908.7 (1+cos40/ 2) = 44.41 W/m²

Idc = 44.41 W/m²

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Answer:

Q1 true

Q2 true

And other I am confuse

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3 years ago
A discrete MOSFET common-source amplifier has RG = 2 MΩ, gm = 5 mA/V, ro = 100 kΩ, RD = 20kΩ, Cgs = 3pF, and Cgd = 0.5pF. The am
Papessa [141]

Answer:

a) -36.36 V/V

b) 15.17 kHz

c) 1.6 GHz

Explanation:

See attached picture.

7 0
3 years ago
A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i
lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

4 0
3 years ago
Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
butalik [34]

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

4 0
4 years ago
Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w
I am Lyosha [343]

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

6 0
3 years ago
Read 2 more answers
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