Answer:
The percent elongation in the length of the specimen is 42%
Explanation:
Given that:
The gage length of the original test specimen = 50 mm
The final gage length = 71 mm
The area = 206 mm²
maximum load = 162,699 N
To determine the percent elongation in %, we use the formula:
The percent elongation in the length of the specimen is 42%
Answer:
(1) 5.74
(2) 5.09
(3) 3.05×10⁻⁵ kg/s
(4) 0.00573 kW
Explanation:
The parameters given are;
Working temperature, = -15°C = 258.15 K
Temperature of the cooling water, = 30°C = 303.15 K
(1) The Carnot coefficient of performance is given as follows;
(2) For ammonia refrigerant, we have;
s₂ = s₁ = 4.9738 kJ/(kg·K)
0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738
∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89
h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg
(3) The circulation rate is given by the mass flow rate, as follows
The refrigeration capacity = 105 kJ/h
The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg
Therefore;
= 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s
(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg
The rating power = Work done per second = W×
∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.
Answer:
hello the figure attached to your question is missing attached below is the missing diagram
answer :
i) 1.347 kW
ii) 1.6192 kW
Explanation:
Attached below is the detailed solution to the problem above
First step : Calculate for Enthalpy
h1 - hf = -3909.9 kJ/kg ( For saturated liquid nitrogen at 600 kPa )
h2- hg = -222.5 kJ/kg ( For saturated vapor nitrogen at 600 kPa )
second step : Calculate the rate of heat transfer in boiler
Q1-2 = m( h2 - h1 ) = 0.008( -222.5 -(-390.9) = 1.347 kW
step 3 : find the enthalpy of superheated Nitrogen at 600 Kpa and 280 K
from the super heated Nitrogen table
h3 = -20.1 kJ/kg
step 4 : calculate the rate of heat transfer in the super heater
Q2-3 = m ( h3 - h2 )
= 0.008 ( -20.1 -(-222.5 ) = 1.6192 kW
