Answer:
The efficiency of the engine is 39.53%
Explanation:
data given by the exercise
HV = heating value = 44000 kJ/kg
Vfuel = volume flow rate of fuel = 22 L/h = 0.0061 L/s
W = power output = 55 kW
p = density = 0.8 g/cm³
the mass flow rate of fuel is equal to
the rate of heat supplied is
the efficiency is equal to
%
Answer:
Efficiency of the engine is 39.51 %
Explanation:
Mass of fuel consumed per hour = 22 L/h * 0.08 Kg/L = 17.6 Kg/h
Total Energy Consumed per hour = Mass of fuel consumed per hour * Heating value of fuel
Total Energy Consumed per hour = 17.6 Kg/h * 44,000 KJ/Kg = 774,400 KJ/h
since, 1 KW = 3600 KJ/h
774,400KJ/h = 774,400/3600 kW = 215.1 kW
Efficiency = 85/215.1 = 39.51 %
Answer:
Rate of Entropy =210.14 J/K-s
Explanation:
given data:
power delivered to input = 350 hp
power delivered to output = 250 hp
temperature of surface = 180°F
rate of entropy is given as
T = 180°F = 82°C = 355 K
Rate of heat = (350 - 250) hp = 100 hp = 74600 W
Rate of Entropy
Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
