Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
a) the power consumption of the LEDs is 0.25 watt
b) the LEDs drew 0.0555 Amp current
Explanation:
Given the data in the question;
Three AAA Batteries;
<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------
so V_total = 3 × 1.5 = 4.5V
a) the power consumption of the LEDs
I_battery = 1000 mAh / 18hrs { for 18 hrs}
I_battery = 1/18 Amp { delivery by battery}
so consumption by led = I × V_total
we substitute
⇒ 1/18 × 4.5
P = 0.25 watt
Therefore the power consumption of the LEDs is 0.25 watt
b) How much current do the LEDs draw
I_Draw = I_battery = 1/18 Amp = 0.0555 Amp
Therefore the LEDs drew 0.0555 Amp current
Answer:
The shaft work generated per kilogram is ![-1.3 \frac{MJ}{kg}](https://tex.z-dn.net/?f=-1.3%20%5Cfrac%7BMJ%7D%7Bkg%7D)
Explanation:
Given:
Temperature
K
Initial Pressure
MPa
Final pressure
MPa
From the table superheated,
and
![\frac{K J}{Kg}](https://tex.z-dn.net/?f=%5Cfrac%7BK%20J%7D%7BKg%7D)
Work done by shaft is,
![W = h_{f} - h_{i}](https://tex.z-dn.net/?f=W%20%3D%20h_%7Bf%7D%20-%20h_%7Bi%7D)
![W = 2706.54 - 4635](https://tex.z-dn.net/?f=W%20%3D%202706.54%20-%204635)
![W = -1928.46 \frac{kJ}{kg}](https://tex.z-dn.net/?f=W%20%3D%20-1928.46%20%5Cfrac%7BkJ%7D%7Bkg%7D)
But here efficiency is 0.56,
So work generated per kg is,
Work = ![0.56 \times(- 1928.46)](https://tex.z-dn.net/?f=0.56%20%5Ctimes%28-%201928.46%29)
Work =
![\frac{MJ}{kg}](https://tex.z-dn.net/?f=%5Cfrac%7BMJ%7D%7Bkg%7D)
Therefore, the shaft work generated per kilogram is ![-1.3 \frac{MJ}{kg}](https://tex.z-dn.net/?f=-1.3%20%5Cfrac%7BMJ%7D%7Bkg%7D)