All categories

3 years ago

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters

Engineering

1 answer:

KonstantinChe [14]3 years ago

4 0

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

Background image

I only know this

sorry for errors

You might be interested in

Joey has a car that uses the hand crank to open the windows. Joey is wondering where the energy comes from to open the windows.T

Sedaia [141]

Explanation:

Joey has a car that uses the hand crank to open the windows. Joey is wondering where the energy comes from to open the windows.The sunHuman-powered energy from JoeyThe hand crankThe moving car

5 0

3 years ago

What kind of energy transformation happens when a boy uses energy from a sandwich to run a race

Semmy [17]

A boy eat a energy of a sandwich to run a race because when they eat a sandwich it helps them to help it mid workout and real nutritions of NYC and bring extra fuel and eating the right thing
I hope this help

4 0

3 years ago

Read 2 more answers

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

Yakvenalex [24]

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

7 0

3 years ago

Read 2 more answers

Vẽ thủ tục cho một cuộc gọi thuê bao

shusha [124]

Lo siento, no sé qué estás diciendo.

8 0

3 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

3 0

3 years ago