Answer:
A) 3783.952 kJ/kg
B) 34.5%
C) 2476.67 kJ/kg
Explanation:
A) Determine the rate of heat addition entering the first-stage turbine
Qin ( 1st stage ) = ( h1 - h6 ) + ( h3 - h2 )
= ( 3321.4 - 164.07 ) + ( 3438.566 - 2811.944 )
= 3783.952 kJ/kg
<em>h values are gotten from super heated table </em>
B) Determine the thermal efficiency
n = 34.5%
attached below
C) Rate of heat transfer from working fluid passing through the condenser to the cooling water
Qout = h4 - h5
= 2628.2 - 151.53
= 2476.67 kJ/kg
Answer:
See attached picture for answer.
Explanation:
See attached picture for explanation.
Answer:
axial stress in bar B = 25Mpa.
Deformation of bar A = 0.4mm.
Explanation:
PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.
So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;
The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.
STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,
Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.
STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,
The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.
Answer:
C) Dependent on the mass flows in and out.
Explanation:
Lets take control volume(CV)
Take
=inlet mass flow rate
=exit mass flow rate
If we take unsteady flow process then inlet mass can not be equal to exit mass.Some mass can store if inlet mass flow rate is high and exit mass flow rate is low.
So mass of control volume
.
so above we can say that mass of control volume dependent on inlet and exit mass.
