Answer: a 8143.71 kJ/kg
b 393.15 K
Explanation:
This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists
a since no change in pressure =0 the system does work thus
FOR HELIUM properties in standard thermodynamic chart
cv = 3.1 kJ/kgK
M = Molar mass = 4 kg/kmol
R = Universal gas constant = 8.314 kJ/kg K
cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK
Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol = 20.714 kJ/kgkmol
T = 120 °C to Kelvin = 120 + 273.15k = 393.15 K
W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg
b convert T °C = T K thus 120 + 273.15 K = 393.15 K
P₁/T₁ = P₂/T₂
200 kPa/ 393.15 K = 200 kPa/T₂
T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k
Answer:
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0 it is a incompressible flow.
c) ap = (16/3) i + (32/3) j + (16/3) k
Explanation:
Given
V = xy² i − (1/3) y³ j + xy k
a) The flow has three dimensions (3 coordinates).
b) ∇V = 0
then
∇V = ∂(xy²)/∂x + ∂(− (1/3) y³)/∂y + ∂(xy)/∂z
⇒ ∇V = y² - y² + 0 = 0 it is a incompressible flow.
c) ap = xy²*∂(V)/∂x − (1/3) y³*∂(V)/∂y + xy*∂(V)/∂z
⇒ ap = xy²*(y² i + y k) - (1/3) y³*(2xy i − y² j + x k) + xy*(0)
⇒ ap = (xy⁴ - (2/3) xy⁴) i + (1/3) y⁵ j + (xy³ - (1/3) xy³) k
⇒ ap = (1/3) xy⁴ i + (1/3) y⁵ j + (2/3) xy³ k
At point (1, 2, 3)
⇒ ap = (1/3) (1*2⁴) i + (1/3) (2)⁵ j + (2/3) (1*2³) k
⇒ ap = (16/3) i + (32/3) j + (16/3) k
Answer:
The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.
<u>Explanation:</u>
To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is
, we know ρ=density of fluid=1000
,
g=acceleration due to gravity=9.81
, h=height of the pool=2 m and l=length of the pool=8 m.
hydrostatic force on each wall=
= 313920 N.
<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool.
</em>
At 1 m from the ground this hydrostatic force is acted on each wall.
The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =
=
= 1255680 N. This is 4 times 313920 N.
Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.