Complete Question
The complete question is shown on the first uploaded image
Answer:
The velocity of the sandbag as it descends is mm/s
The overall efficiency of the device is %
Explanation:
The explanation is shown on the second uploaded image
Answer:
A) 337.21 kj/h
b) -224.823 kj/h
Explanation:
Steady state temperature = 27⁰C
Pressure = 0.3 MPa
exit temperature = 12⁰c
Refrigerant 134a :
entering pressure = 0.4 MPa
mass flow rate = 35 kg/h
quality = 0.3
inlet temp = 8.93⁰c
A) Determine the rate of heat transfer in kJ/h for Refrigerant 134a stream
calculate the specific enthalpy of refrigerant at the inlet
hm = hf + x ( hg - hf )
at 0.4 MPa : hf = 64 kj/kg , hg = 256 kj/kg
x = 0.3 ( quality )
hence : hm = 64 + 0.3 ( 256 - 64 ) = 64.3 + 57.6 = 121.9 kj/kg
next calculate the specific enthalpy at outlet
Tout = 10⁰c , T sat = 8.93⁰c
since the Tsat < Tout the refrigerant is super heated and from the super heated refrigerant table at P = 0.4 MPa, hout = 257 kj/kg
Heat gained by refrigerant ( Qin)
Qin = mref ( hout - hin )
= 35 ( 257 - 112) = 5075 kj/h
To determine the rate of heat transfer we have to apply the equation below
heat gained by refrigerant = heat gained by air
Qin = Qout
5075 kj/h = mair ( hout - hin ) ------------ 1
considering ideal conditions
at T = 300 k , hout = 300.19 kj/kg ( specific enthalpy )
at T = 285 k , hin = 285.14 kj/kg ( specific enthalpy )
back to equation 1
5075 kj/h = mair ( 300.19 - 285.14 )
mair ( rate of energy transfer ) = 5075 / 15.05 = 337.21 kj/h
B) evaluating the change in flow energy rate in kJ/h
attached below is the detailed solution
Answer:
Please see attachment.
Explanation:
