To solve this problem it is necessary to apply the concepts related to thermal transfer given by the thermodynamic definition of heat as a function of mass, specific heat and temperature change.
Mathematically this is equivalent to
Where
m = mass
Specific Heat
Change at temperature
In mass terms (KJ / g) this can be expressed as
Our values are given as
Replacing,
Therefore the heat of combustion per gram on the material is 55.51KJ/g
Answer:
engine B is more efficient.
Explanation:
We know that Carnot cycle is an ideal cycle for all working heat engine.In Carnot cycle there are four processes in which two are constant temperature processes and others two are isentropic process.
We also kn ow that the efficiency of Carnot cycle given as follows
Here temperature should be in Kelvin.
For engine A
For engine B
So from above we can say that engine B is more efficient.
Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
Answer:21.3%
Explanation:
Given
80 % reduction in tool life
According to Taylor's tool life
=c
where V is cutting velocity
T=tool life of tool
80 % tool life reduction i.e. New tool Life is 0.2T
Thus
=1.213V
Thus a change of 21.3 %(increment) is required to reduce tool life by 80%