5) Two concentric spheres of diameter D1= 70 cm and D2= 120 cm are separated by an air space and have surface temperatures of T1
= 117 C and T2= 27 C. (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres, in W? (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with ε1= 0.5 and ε2= 0.05, in W? (c) What is the net rate of radiation exchange if D2 is increased to 2000 cm, with ε2= 0.05, ε1= 0.5, and D1= 70 cm, in W? (d) What is the net rate of radiation exchange if the larger sphere behaves as a black body (ε2= 1.0) and with ε1= 0.5, D2= 2000 cm, and D1= 70 cm, in W?
1 answer:
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Answer:
Explanation:
From the question we are told that:
Water flow Rate
Initial Temperature
Final Temperature
Let
Specific heat of water
And
Generally the equation for Heat transfer rate of water is mathematically given by
Heat transfer rate to water= mass flow rate* specific heat* change in temperature
Therefore
complete question
A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?
Answer:
3.03 V 0.184 W
2.499 mV 125*10^-9 W
Explanation:
First, apply voltage-divider principle to the input circuit: 1
*5
= 4.545 V
The voltage produced by the voltage-controlled source is:
A_voc*V_i = 4.545 V
We can find voltage across the load, again by using voltage-divider principle:
V_o = A_voc*V_i*(R_o/R_l+R_o)
= 4.545*(100/100+50)
= 3.03 V
Now we can determine delivered power:
P_L = V_o^2/R_L
= 0.184 W
Apply voltage-divider principle to the circuit:
V_o = (R_o/R_o+R_s)*V_s
= 50/50+100*10^3*5
= 2.499 mV
Now we can determine delivered power:
P_l = V_o^2/R_l
= 125*10^-9 W
Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.
