Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
the required documents is 3000
Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation

under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that

hence , actual velocity will be equal to 8.39 m/s
Answer:

Explanation:
From the question we are told that:
Thickness 
Internal Pressure
Shear stress 
Elastic modulus 
Generally the equation for shear stress is mathematically given by

Where
r_i=internal Radius
Therefore


Generally



Generally the equation for outer diameter is mathematically given by


Therefore
Assuming that the thin cylinder is subjected to integral Pressure
Outer Diameter is

Answer:
Hello your question has some missing information below are the missing information
The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump
answer : 2.49
Explanation:
For vapor-compression refrigeration cycle
P1 = P4 ; P1 = 140 kPa
P2( pressure at inlet ) = P3 ( pressure at outlet ) ; P2 = 800 kPa
<u>From pressure table of R 134a refrigerant</u>
h1 ( enthalpy of saturated vapor at 140kPa ) = 239.16 kJ/kg
h2 ( enthalpy of saturated liquid at P2 = 800 kPa and t = 60°C )
= 296.8kJ/kg
h3 ( enthalpy of saturated liquid at P3 = 800 kPa ) = 95.47 kJ/kg
also h4 = 95.47 kJ/kg
To determine the coefficient of performance
Cop = ( h1 - h4 ) / ( h2 - h1 )
∴ Cop = 2.49