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bogdanovich [222]
3 years ago
5

A customer got debited twice, how would you investigate and resolve this conflict on the customer side and on the company side?

Engineering
1 answer:
eimsori [14]3 years ago
8 0

Answer:

First, the question seems to assume that the debit is erroneous. Because, if it's intentional, then there is little need for an investigation.

So if there was an unintentional duplication of transaction, this would totally depend on what kind of debit it is. So to answer the question, we'll explore as many scenarios as possible.

Debit on a Bank Account (Business to Bank)

If it is a debit on the bank account, then the statement of account from the company side and the account activity(ies) leading up to the debits will reveal why there is a double debit. It is possible the transaction was entered twice. This can happen when there is a bad connection that creates a failed transaction message on the customer's end but still carries through with the payment instruction. In this case, the customer will think the debit or payment request didn't go through and proceed to repeat the same transaction.

Banks do execute debits on accounts for charges such as Bank Account Maintenance Charges, Credit card charges etc. Errors could also arise internally that led to the duplication of a debit instruction either due to a bug in the system.

So in this case, we'll need to access the customer's account internally whilst externally, the client will need to provide the message containing the alerts, date and time of debits etc. to help the banks further the investigation.

Business to Business

This could arise out of a "failed POS" transaction. Similar to the condition stated above,  when there is a glitch with the network and the POS fails, it should print out a receipt that reads failed or declined.

This receipt will contain references to the transaction. This will aid the investigation of the debits to the account.

Explanation:

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Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

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3 years ago
Which term refers to the practical applications of scientific knowledge
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Answer:

Technology

Explanation:

Technology is defined as the application of scientific knowledge for practical purposes, especially in industry . Sometimes applying scientific knowledge means using less of what we traditionally think of as technology.

7 0
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Why is the face of the claw on a claw hammer usually a smooth curve? Why isn't it straight or some other shape?
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3 years ago
A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

5 0
3 years ago
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