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ehidna [41]
2 years ago
14

Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select fro

m a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Engineering
1 answer:
Vinvika [58]2 years ago
7 0

Answer:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read  = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

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A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl
bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1.  Steady operating conditions exist.

2.  Kinetic and potential energy changes are negligible.

Properties:  The specific heat of geothermal water ( c_{geo}[) is taken to be 4.18 kJ/kg.ºC.  

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

PP_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The  isentropic  efficiency, n_{T} = \frac{h_{3}-h_{4}  }{h_{3}- h_{4s} }

==\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788

b. Pump

h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.}  w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.}  _{net} = W^{.} _{T, out} - W^{.}  _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\

c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

7 0
3 years ago
Read 2 more answers
What are the Basic requirements of drinking Water ?
iragen [17]

Answer:

1) free of contaminants, 2) alkaline, and 3) micro-clustered

Explanation:

Hope it helps you

4 0
2 years ago
How should you move your board through the planer? (Pick two choices.)
iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
A single-cylinder pump feeds a boiler through a delivery
Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 m^3/hr

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m

We know that area is given by

A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2

We know that discharge is given by Q=AV, here A is area and V is velocity

So Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec

So net discharge in 1 hour = 981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour

8 0
3 years ago
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