Steam enters a long, horizontal pipe with an inlet diameter of D1 = 16 cm at 2 MPa and 300°C with a velocity of 2.5 m/s. Farther downstream, the conditions are 1.8 MPa and 250°C, and the diameter is D2 = 14 cm. Determine

(a) the mass flow rate of the steam and
(b) the rate of heat transfer. For this pipe system, the kinetic energy of the steam is negligible.

m = 0.4005 kg/s

Q_out = 45.1 KJ/s

Explanation:

Given

Pipe inlet diameter D1 = 16 cm

Steam inlet pressure P1 = 2 Mpa

Steam inlet temperature T1 = 300 °C

Pipe outlet diameter D2 = 14 cm

Steam inlet velocity V1 = 2.5 m/s

Steam outlet pressure P2 = 1.8 MPa

Steam outlet temperature T2 = 250 °C  

Required

Determine

(a) The mass flow rate of steam.

(b) The rate of heat transfer.  

Assumptions

Kinetic and potential energy changes are negligible.

This is a steady flow process.

There is no work interaction.  

Solution

Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C

vl = 0.12551 m^3/Kg

h1 = 3024.2 KJ/Kg

The mass flow rate of steam could be defined as the following  

m = 1/v1*A1*V1

m = 0.4005 kg/s

Part b We take the pipe as our system.The energy balance could be defined as the following  

E_in -E_out =ΔE_sys = 0

E_in = E_out

mh_1 = Q_out + mh_2

Q_out = m(h_1-h_2)

From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C  h2= 2911.7 KJ/Kg The heat transfer could  be defined as the following

Q_out = m(h_1-h_2)

Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz