Air is cooled and dehumidified as it flows over the coils of refrigeration system at 100 kPa from 30 ºC and relative humidityof
84%to 15 ºC and relative humidity of 100%. The mass flow rate of dry air is 0.4 kg/s.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
1 answer:
Answer:
(a) The mass glow rate of water is 0.0032kg/s
(b) The heat removal from air is 14.17KJ/sec
Explanation:
In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.
To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature
Please check attachment for complete solution and step by step explanation
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Answer:
(a) 0.12924
(b) Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Explanation:
(a)
n=200 for fair coin getting head, p= 0.5
Expectation = np =200*0.5=100
Variance = np(1 - p) = 100(1-0.5)=100*0.5=50
Standard deviation,
Z value for 108,
P( x ≥108) = P( z >1.13)= 0.12924
(b)
Taking into consideration significance level of 0.05 yet the value of p is greater than 0.05, it suggests that the coin is fair hence the coin can be used at the beginning of any sport event.
Answer:
In Btu:
Q=0.001390 Btu.
In Joule:
Q=1.467 J
Part B:
Temperature at midpoint=274.866 C
Explanation:
Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)=
Thermal Conductivity is SI units:
Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft
Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft
T_1=500 C=932 F
T_2=50 C= 122 F
Part A:
In Joules (J)
Heat Q is:
In Btu:
Heat Q is:
PArt B:
At midpoint Length=L/2=0.1 m
On rearranging: