Answer:
The efficiency of the engine is 22.5%.
Explanation:
Efficiency = power output ÷ power input
power output = 55 kW
power input = specific energy×volumetric flow rate×density
specific energy = 44,000 kJ/kg
volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s
density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3
power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW
Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%
Answer:
The heater load =35 KJ/kg
Explanation:
Given that
At initial condition
Temperature= 15°C
RH=80%
At final condition
Temperature= 50°C
We know that in sensible heating process humidity ratio remain constant.
Now from chart
At temperature= 15°C and RH=80%
At temperature= 50°C
The heater load = 73 - 38 KJ/kg
The heater load =35 KJ/kg
Explanation:
I won't answer each of these, but will give you an explaination of how to solve for each.
You'll need to use Ohm's Law and Kirchhoff's Voltage Law.
Remember Ohm's Law as .
Kirchoff's Voltage Law says that the sum of voltages for a given circuit "loop" must equal zero. In the circuit shown, this means that the voltage provided by the battery (E_T) equals the voltage drop across each of the three resisters in the loop.
A couple of other helpful notes:
These three resistors are in <em>series</em> which means that the current flowing through them is equal.
So it is easy to see that...
Solve for the voltage across E_2 and E_3. The sum of the three voltages equals the voltage of the battery (E_T).