Answer:
i) 43.55 kg/s
ii) 40 m/s
iii) -199.32 KW
Explanation:
To resolve the above question we have to make some assumptions :
- mass flow through the system is constant
- The only interactions that are between the system and the surrounding are work and heat
- The fluid is uniform
i) first we have to determine the mass flow rate of the air
M = 
=
---------- (1) hence M = 43.55 kg/s
ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s ( inlet velocity )
input this value into equation 1
iii) Next we will determine the power required to run compressor
attached below
power required = -199.32 KW ( this value indicates that there is power supplied )
Answer:
Explanation:
From the given information:
Strain fracture toughness
= 75 MPa
Tensile stress
= 361 MPa
Value of Y = 1.03
Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:
![a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%28%20%5Cdfrac%7Bk_k%7D%7B%5Csigma%20Y%7D%29%5E2%20%5C%5C%20%5C%5C%20%20a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%5CBig%20%5B%20%5Cdfrac%7B75%20%5Ctimes%20%5Csqrt%7B10%5E3%7D%7D%7B361%20%5Ctimes%201.03%20%7D%20%5CBig%5D%5E2%20%5C%5C%20%5C%5C%20%20a_c%20%3D%20%5Cdfrac%7B1%7D%7B%5Cpi%7D%20%5CBig%20%5B%206.378474693%5CBig%5D%5E2%20%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%20a_c%20%3D%2012.95%20%5C%20mm%7D)
Answer:
Δw =7.95 kg/1000m^3
q = 62362.3 kg/1000m^3
Explanation:
To solve this problem we first need to use the psycrometric chart to determine the enthalpy h1, specific volume vi and absolute humidity col by using the given temperature T1 = 32°C and the relative humidity Ф1 = 95%.
h_1 = 106.5 kJ/kg
v_1 = 0.91 m^3/kg
w_1 = 0.02905
We will also need the enthalpy h2 and the absolute humidity w_2 at the exit point. We will again use the pyscrometric chart and the given temperature T_2 = 24°C. From the problem we also know that the exit relative humidity is = 60%.
h2 = 52.6 kJ/kg
w_2 = 0.01119
We need to express the final results in units per 1000 m^3. To do that we will need the mass m of this volume of air V and to calculate that we will use the given pressure p = 1 atm = 101.3 kPa.
m = R_a*T_1/V.p
m = 1000*101.3/0.287*305K
m = 1157 kg
Because it is a closed system, the amount of water removed Δw can be calculated as:
Δw =w_1 - w_2
Δw =0.02905- 0.01119
Δw =0.00687 kg/kg* 1157kg/1000m^3
Δw =7.95 kg/1000m^3
From the energy balance equation we can calculate the specific heat q removed from the air.
q = h_1 - h_2
q = 106.5 kJ/kg - 52.6 kJ/kg
q = 53.9 kJ/kg * 1157kg/1000m^3
q = 62362.3 kg/1000m^3